polygons, polyhedra, patterns & beyond

POLYGONS,POLYHEDRA,PATTERNS&BEYOND

EthanD.Bloch

Spring2015

Contents

TotheStudent v

Part I POLYGONS &POLYHEDRA 1

1 GeometryBasics 31.1 EuclidandNon-Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 LinesandAngles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Polygons 252.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3 GeneralPolygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.5 ThePythagoreanTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3 Polyhedra 733.1 Polyhedra–TheBasics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.2 RegularPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.3 Semi-RegularPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.4 OtherCategoriesofPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . 843.5 EnumerationinPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.6 CurvatureofPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

iv Contents

Part II SYMMETRY &PATTERNS 105

4 Isometries 1074.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1074.2 Isometries–TheBasics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.3 RecognizingIsometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.4 CombiningIsometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1254.5 GlideReflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1304.6 Isometries—TheWholeStory . . . . . . . . . . . . . . . . . . . . . . . . . . 136

5 SymmetryofPlanarObjectsandOrnamentalPatterns 1475.1 BasicIdeas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.2 SymmetryofRegularPolygonsI . . . . . . . . . . . . . . . . . . . . . . . . . 1545.3 SymmetryofRegularPolygonsII . . . . . . . . . . . . . . . . . . . . . . . . 1605.4 RosettePatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1705.5 FriezePatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1785.6 WallpaperPatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1875.7 ThreeDimensionalSymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 204

6 Groups 2156.1 Thebasicidea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.2 Clockarithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2186.3 TheIntegersMod n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2236.4 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2286.5 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2376.6 SymmetryandGroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

SuggestionsforFurtherReading 245

Bibliography 251

TotheStudent

I havewrittentheselecturenotesbecauseI havenotfoundanyexistingtextforMath107thatwasadequateintermsofthechoiceofmaterial, levelofdifficulty, organizationandexposition.MuchoftheapproachinPartII ofthesenotesisinspiredbythetextI previouslyusedforMath107(Farmer, David, “GroupsandSymmetry,” AMS,1996). I hopethatthesenoteswillfittheneedsofthiscourse, andwillhelpyoulearnthismaterial.Ifthereareanyerrorsinthetext, oranythingthatisnotclearlywritten, pleaseacceptmyapolo-

gies. I wouldverymuchappreciateyourfeedbackonthistext, bothintermsoferrorsthatyoufindandsuggestionsforchangesoradditionsthatyoumighthave. Commentscanbeforwardedtomeinperson(afterclass, orinmyoffice, Albee317), orbyemailat [email protected].

Prerequisites

ItisassumedthatanyoneusingthistexthaspassedPartI oftheMathematicsDiagnosticExam.Noparticularbackgroundinmathematicsbeyondthatisrequired. Onsomeoccasionswewillmakeuseofhighschoolalgebra(forexample, thequadraticformula)andhighschoolgeometry(forexample, thePythagoreanTheorem). Forthemostpart, however, wewilltreatmaterialthat,whiletouchinguponsomeverysubstantialideas, doesnotrequiremuchinthewayofalgebraorgeometrybackground. Precalculus(includingtrigonometry, logarithms, andthelike)isnotrequired. Onafewoccasionswewillmentiontrigonometry, butthosebriefreferencescaneasilybeskipped.Whatisneededtoreadthistextisawillingnesstolearnnewideas, tothinkthroughsubtleties,

toworkhard, andtouseyourimagination. Muchofthematerialinthetextisveryvisual, andmakingdrawings, andmentallyimagininggeometricobjects, iscrucial. Someofthearguments

vi TotheStudent

inthetext, thoughnotrequiringmuchinthewayoftechnicalbackground, arenonethelessquitetricky, andrequirecarefulattentiontothedetails.

Exercises

Likemusicandart, mathematicsislearnedbydoing, notjustbyreadingtextsandlisteningtolectures. Inmathematicscoursesweassignexercisesnotbecausewewanttoputthestudentsthroughsomesortofmathematicalbootcamp, butbecausedoingexercisesisthebestwaytoworkwiththematerial, andtoseewhatisunderstoodandwhatneedsfurtherstudy. Doingtheexercisesisthereforeacrucialpartoflearningthematerialinthistext. Exercisesrangefromtheroutinetothedifficult. Whenansweringanexercise, youmayuseanyfactsinthetextuptillthen(includingpreviousexercises).Onefeatureoftheexercisesinthistextisworthmentioning. Inmanyhighschoolmathematics

courses, thetexthasavarietyofworked-outexamples, andthenthehomeworkexercisesareoftenvirtuallyidenticaltotheworked-outexamples, butwithdifferentnumbers. Thestudentsthendotheexercisesbysimplymimickingtheexamplesinthetext. Studentsareoftensatisfiedwiththesetypesofexercises, butfromthepointofviewofintellectualgrowth, suchexercisesaresorelylacking. Thepointoflearningmathematicsisnottolearntoimitatewhattheteacherorthebookdoes, butrathertounderstandnewconceptsandbeabletoapplythem. Hence, inthistext, manyoftheexercisesdonotaskthestudentstomimicwhatisdoneinthetext, butrathertothinkforthemselves. Many(thoughnotall)oftheexercisesinthistextare, purposely,notidenticaltoworkedoutexamplesinthetext, butratheraretobesolvedbythinkingaboutthematerial. Someoftheexercisesinthistextrequirecreativityandimagination, andothersareopenended.

Part IPOLYGONS &POLYHEDRA

1

1GeometryBasics

1.1 EuclidandNon-Euclid

Ancient Greekmathematics was put into its ultimate deductive form by Euclid, who livedroughlyaround300BCE.Inhiswork“TheElements,” Euclidtookanalreadydevelopedlargebodyofgeometry, andgaveitlogicalorderbyisolatingafewbasicdefinitionsandaxioms, andthendeducingeverythingelsefromthesedefinitionsandaxioms. Thestatements thatEucliddeducesfromhisaxiomsanddefinitionsarecalledpropositions(inmoderntextbookstheyareoftenreferredtoastheorems, whichmeansthesamething). WewilllookverybrieflyatsomeaspectsofEuclid’sElements. Thismassiveworkisdividedupinto13“Books;” ourconcernisprimarilywithBookI.Following[Har00](thoughothersemphasizethispointaswell), westressthatEuclid, andthe

ancientGreeksgenerally, viewedgeometryratherdifferentlythanwecurrentlydo. Theonlyquantities thatwereof interestweregeometricones, andof those, geometricquantities thatcouldbeconstructedwithstraightedgeandcompasswereofparticularinterest. Numbersfortheirownsakeseemedlessofinterest(perhapsbecausetheydidnothaveadevelopedunder-standingofnumbers, orperhapsthatiswhytheydidnotdevelopsuchanunderstanding); asolidunderstandingofnumberscamelaterinhistory. Forexample, considerthefamousPythagoreanTheorem, whichisdemonstratedinSection 2.2. Incontemporarylanguage, thistheoremstatesthatifarighttrianglehassidesoflength a and b, andhypotenuseoflength c, then a2+b2 = c2.Sostated, thistheoremtellsussomethingaboutthreenumbers a, b and c: ifthesethreenum-berssatisfyacertaincondition(namelybeingthelengthsofthesidesandhypotenuseofarighttriangle), then theymustalsosatisfy thealgebraicequation a2 + b2 = c2. ToEuclid, suchastatementaboutnumberswouldnothavemadesense. HisversionofthePythagoreanThe-oremis: “Inright-angledtriangles thesquareonthesidesubtendingtherightangleisequaltothesquaresonthesidescontainingtherightangle.” (AllquotesfromEuclidaretakenfrom

4 1. GeometryBasics

[Euc56], whichisthestandardEnglishtranslation.) Euclid’sversionofthetheoremisastatementaboutthreesquares, namelythatonesquare(theoneonthehypotenuse)“isequal”totwoothersquares(thoseonthesides)puttogether. Euclidisinterestedintherelationbetweenthesethreesquares, whicharegeometricobjects, andnotnumbers.ItmightseemfromamodernperspectivethatEuclid’sversionofthePythagoreanisaboutarea,

andthusreallydoesinvolvenumbers, becausetheareaofaplanarfigure(thatis, afigureintheplane)isanumber. Infact, EucliddoesnotmentiontheconceptofareaatallinhisversionofthePythagoreanTheorem. WhenEuclidsaysthattwoplanarfigures(suchassquares)areequal,heisnotmakingastatementaboutthenumericalvaluesoftheirareas(forexampleinsquareinches), butratherissayingthatonefigurecanbecutupintotriangles, andreassembledintotheotherfigure. Thus, Euclid’sversionofthePythagoreanTheoremisstrictlyaboutgeometricobjects. Today, wehavetheconceptofassigningtoeachplanarfigureitsarea(whichisanum-ber), andwerestatevariousgeometricpropertiesintermsofnumericalproperties, butthatisnotthewayEuclid(andtheotherancientGreeks)viewedthings. See[Har00]forathoroughdiscussionofthisissue, andmoregenerallyonwhatEuclidsaid, andhowheunderstoodge-ometry. (Wehighlyrecommend[Har00]asacompaniontoreadingEuclid; thoughmuchofthebookisaimedatamathematicallysophisticatedaudience, somepartsareveryaccessible, andextremelyinsightful.)Without question, “The Elements” is oneof themost important, and influential, works of

mathematicseverwritten—itisarguablyoneofthemostinfluentialintellectualachievementsofhumancivilizationasawhole, notjustofmathematics. Euclid’streatmentofgeometrybe-cametheuniversallyacceptedmethodofdoinggeometryforalmosttwomillenia, uptillthe19thcentury. Moreover, notonlywasEuclideangeometryacceptedasunquestionablytrue, butEuclid’smethodofdeductivereasoningwasconsideredamodeloflogicalargumentation, andanexampleofreasoningthatproducedtheoremsthatwereunquestionablytrue. Itturnsout,however, thatneitheroftheseattributesof“TheElements”istrue. WithoutdiscountingfromtheenormousintellectualandhistoricalimportanceofEuclid’swork, fromamodernvantagepointwecanidentifythreefundamentalflawsinEuclid, theresolutionsofwhichdidnottakeplaceuntilroughlytwomilleniaafterEuclid’stime.Euclidtriedtogiveprecisedefinitionsforgeometricconcepts; hetriedtogiveasetofaxioms

thatdescribeplanar(andspatial)geometry; andhetriedtoproveallotherresultsingeometryinarigorousfashionbasedonlyonhisdefinitionsandaxioms. ItisnowunderstoodthatthereareflawsinEuclid’sdefinitions; hisaxiomsareneithercompletenornecessarilytrue; andsomeofhisproofshavegaps. Fromamodernpointofview, Eucliddidnotreallyachievethelevelof rigor thathas traditionallybeenascribed tohim. Noneof this is todeny thegreatnessofEuclid’sachievement—itisindeedmagnificent—butunderstandingtheproblemsin“TheEle-ments”helpsgiveanaccurateassessmentofEuclid, anditpointsthewaytolaterdevelopmentsingeometry.LetusturntothefirstfourofthedefinitionsfromBookI of“TheElements.” Euclidhasmore

definitionsthanthefourwequote, buttheyaresufficienttoillustratewhatEuclidwasattemptingtoaccomplish.

1.1EuclidandNon-Euclid 5

Definitions

1. A pointisthatwhichhasnopart.

2. A lineisbreadthlesslength.

3. Theextremitiesofalinearepoints.

4. A straightlineisalinethatliesevenlywiththepointsonitself.

Euclidwantstodosomethingverynicewithhisdefinitions, namelydefineallthemathemati-caltermsthatheuses, suchaspointandline. Unfortunately, hedoesnotadequatelyaccomplishhistask. Whenhesaysthat“apointisthatwhichhasnopart,” heneveractuallysayswhatapoint is, onlywhat itdoesnothave, andeven that isunclear. Whenhe says that “a line isbreadthlesslength,” whereheusestheword“line”tomeanwhatwewouldcallacurve, hedoesnottelluswhat“breadth”is, andsowedonotreallyknowwhatalineis. Similarly, hesay“astraightlineisalinethatliesevenlywiththepointsonitself,” butwhatdoesitmeanforsomethingto“lieevenly”withthepointsonitself—otherthantobestraight, butnowwearegoingincircles.Thebottomline is that, bymodernstandards, Euclid’sdefinitionsaremeaningless. In fact,

wenowunderstandthatjustasitisnecessarytostartwithsomeunprovedaxiomsasabasisforalltheothertheoremstobeproved, sotoodoweneedtostartwithsomeundefinedtermsasabasisforallotherdefinitions. Euclid’sdefinitionsaredoomedtofail, becausehetriestodefineeverything. Inthemodernaxiomaticapproachtogeometry, westartwithsomeundefinedterms(forexample, “point”and“line”), thoughwehypothesizevariousaxiomaticpropertiesfortheseundefinedterms(forexample, weassumethateverytwodistinctpointsarecontainedinauniqueline). Whatcountsisnotwhatpointsandlinesare, buthowtheybehave. Iftheybehaveaspointsandlinesoughtto, thenwearesatisfied. AlthoughEuclid’sdefinitionsdonotworkasstated, therearemodernaxiomschemes(suchastheonesbyHilbert in1899andBirkhoffin1932)inwhichthedefinitionsareworkedoutproperly—andtheydojustwhatEuclidwasattemptingtodo. Inotherwords, itispossiblepatchupEuclid’sdefinitions, withthecaveatthatsometermsareleftundefined.WenowturntoEuclid’saxioms. Theaxiomsarebrokenupinto“commonnotions”and“pos-

tulates.” Thecompletelistofcommonnotionsandpostulatesisasfollows.

TheCommonNotions

1. Thingsthatareequaltothesamethingsareequaltooneanother.

2. Ifequalsbeaddedtoequals, thewholesareequal.

3. Ifequalsbesubtractedfromequals, theremaindersareequal.

4. Thingsthatcoincidewithoneanotherareequaltooneanother.

5. Thewholeisgreaterthanthepart.

6 1. GeometryBasics

ThePostulates

1. Todrawastraightlinefromanypointtoanypoint.

2. Toproduceafinitestraightlinecontinuouslyinastraightline.

3. Todescribeacirclewithanycenterandradius.

4. Thatallrightanglesareequaltooneanother.

5. That, ifastraightlinefallingontwostraightlinesmakestheinterioranglesonthesameside less than tworightangles, thestraight lines, ifproduced indefinitely,meetonthatsideonwhicharetheangleslessthanthetworightangles.

Thecommonnotions, whicharenotaboutgeometryperse(thoughEuclidmighthavethoughtofthemgeometrically), seemreasonableenoughasstated. Thepostulates, bycontrast, needsomeexplanationinmodernterminology; Euclid’swordingisdifferentfromwhatwewouldusetoday. Euclid, andingeneraltheancientGreeks, wereveryconcernedwithgeometricconstruc-tionsusingstraightedgeandcompass(wepurposelysay“straightedge”andnot“ruler,” becausetheydidnotallowtheuseofarulertomeasurethings, onlyastraightedgetodrawstraightlinesbetweengivenpoints).Thefirstthreepostulatesinvolvestraightedgeandcompassconstructions. TheFirstPostulate

statesthat, giventwodifferentpoints, wecanconstruct(usingastraightedge)alinesegmentfromonepointtotheother. Inmodernterminology, wherewefocusonpropertiesoflinesandpointsandnotonconstructions, theFirstPostulateisoftenrephrasedas“anytwodistinctpointsarecontainedinone, andonlyone, line.” TheSecondPostulatestatesthatifwearegivenalinesegment(whichisfiniteinlength), wecanextendthelinesegment. TheThirdPostulatestatesthatgivenapoint, andgivenaradius, wecandrawthecirclethathasthegivenpointasitscenter, andhasthegivenradius.TheFourthandFifthPostulatesarenotconcernedwithstraightedgeandcompassconstruc-

tions. TheFourthPostulatesaysthatanytworightangles, nomatterwheretheyarelocatedintheplane, areequaltooneanother. Thisstatementmayseemratherobvious, butthereisinfactsomethingtobehypothesizedhere; wewilldiscussthispostulateinmoredetailinSection 1.2,wherewediscussangles.TheFifthPostulate, bycontrast, requiressomeexplanation. Tounderstandwhatthepostulate

says, supposewearegivena line, say k, and two lines that intersect k, say m and n. SeeFigure 1.1.1. Let α and β betheanglesshowninthefigure. TheFifthPostulatestatesthatifα + β < 180◦, thenthelines m and n willeventuallyintersectonthesamesideof k as αand β. (Notethat 180◦ equals“tworightangles.”) Statedthisway, Euclid’sFifthPostulatedoesmakesenseintuitively. Itwillturnout, asdiscussedlaterinthissection, thattheFifthPostulatehasgreathistoricalsignificance, muchmorethantheotherfourpostulates.

Havingstatedhisdefinitions, commonnotionsandpostulates, Euclidgoesontoprovemanypropositions. Tomakeeverythingcompletelyrigorous, EuclidprovesProposition 1usingonlyhisdefinitions, commonnotionsandpostulates. Proposition 2isprovedusingProposition 1,


k

m

n

α

β

Figure1.1.1

togetherwiththedefinitions, commonnotionsandpostulates. Andsoon. Someofthepropo-sitionsin“TheElements”areveryfamiliartoustoday, forexamplethePythagoreanTheorem,whichisProposition 47inBookI of“theElements,” beingthepenultimatetheoreminthatbook;thelastpropositionisaconversetothePythagoreanTheorem(whichwestateanddemonstrateinourProposition 2.5.2). ThefirsttwopropositionsinBookI areasfollows.

Propositions

1. Onagivenfinitestraightline, toconstructanequilateraltriangle.

2. Toplaceatagivenpoint(asanextremity)astraightlineequaltoagivenstraightline.

Aswithsomeofthepostulates, thesefirsttwopropositionsinvolveconstructionswithstraight-edgeandcompass(thoughnotallthepropositionsinvolveconstructions). Proposition 1ofEu-clidsaysthat, givenalinesegment, wecanconstructanequilateraltrianglethathasthelinesegmentasoneofitsedges. Proposition 2saysthat, givenalinesegment, andapointsome-whereintheplane, wecanconstructanewlinesegmentthatisequalinlengthtothegivenone, andhasthegivenpointasoneofitsendpoints.Inmodernterminology, theproofofEuclid’sProposition 1hastwostages. Wearegivena

linesegmentintheplane. First, Euclidtellsushowtoconstructacertaintrianglethathasthelinesegmentasoneofitsedges; second, heprovesthatthetrianglesoconstructedisindeedequilateral. Whatconcernsusisthefirststageofthisproof. Theideaissimple. Supposewehavealinesegmentwithendpoints A and B, asshowninFigure 1.1.2 (i). First, drawanarc(whichissimplypartofacircle)usingacompasswithcenterat A, andwithradiusthelengthfrom A

to B; thendrawanarcwithcenter B, andthesameradius. SeeFigure 1.1.2 (ii). Let C denotethepointwherethetwoarcsintersect. Usingastraightedge, drawthelinesegmentsfrom A toC, andfrom B to C. SeeFigure 1.1.2 (iii). Wenowhaveatrianglewithvertices A, B and C. Inthesecondstageofhisproof, Euclidshowsthatthistriangleisindeedequilateral.

8 1. GeometryBasics

(i) (ii)

(iii)

A B A B

C

A B

Figure1.1.2

BEFORE YOU READ FURTHER:

Euclid’sproofofProposition 1isintuitivelycompletelycorrect, butfromarigorouspointofview, thereisaflaw. Trytoseeifyoucanfigureoutwhattheproblemwiththisproofis.

TheproblemwithEuclid’sproofofProposition 1ultimatelygoesbacktothefactthathewantshisprooftorelyonlyonhisdefinitions, postulatesandcommonnotions. ThatwecandrawthetwoarcsshowninFigure 1.1.2 (ii) indeedfollowsfromtheThirdPostulate, andthatwecandrawthelinesegmentsfrom A to C andfrom B to C followsfromtheFirstPostulate. WhatisnotexplicitlyguaranteedbyanyofEuclid’spostulatesorcommonnotionsisthatthetwoarcsweconstructedactuallyintersect. Wesimplyassumedthatthetwoarcintersect, andlabeledthepointofintersection C. Howdoweknowthatthetwoarcsreallydointersect? ItcertainlylooksasiftheydoinFigure 1.1.2 (ii), butthatisnotaconvincingargument, becausewemight


havedrawnthefigureincorrectly. Moreover, ifaproofisgenuinelyrigorous, itoughtnottorelyonapicture—thepictureissimplymeanttohelpourintuition.Itiscertainlynottruethatanytwocirclesintheplaneintersect, forexampletwocircleswith

radius 1 incheach, andwithcenters 10 inchesapart. Hence, toguaranteethattwoarcs(whicharepartsofcircles)intersect, wewouldneedtoknowsomethingspecificaboutthemthatinsuresintersection. Forexample, ifweknewsomethingabouttherelationbetweenthecentersofthecirclescontainingthearcsandtheradiiofthecircles, thenwemightbeabletodemonstratethatthearcsintersect; ifthedistancebetweenthecentersofthecirclesistoolargeinrelationtotheradii, thenthearcsmightnotintersect. Itis, infact, possibletogivecriteriaontheradiiandcentersoftwocirclesthatinsurethattwocirclesintersect. Thereis, however, amoresubtleproblemwiththisaspectofEuclid’sproof. Notonlydoweneedtoinsurethattheradiianddistancebetweenthecentersoftwocirclesareappropriateinordertoinsurethatthecirclesintersect, butwealsoneed toknowthatcirclesare“continuous,” that is, that theyhaveno“gaps”inthemrightwheretheintersectionissupposedtotakeplace. Thisissueofgapsisverysubtle, andwedonothavethespacetodiscussithere. Inhispostulatesandcommonnotions,Euclidaddressesneither theissueofappropriateradiiandcenters, nor theissueofnogaps,andthereforehehasnotrigorouslyprovedhisProposition 1. ModernaxiomatictreatmentsofEuclideangeometrysuccessfullyavoidEuclid’sinsufficientaxiomsbygivingmoreaxiomsthanEuclidgave. ItisnotthatwhatEuclidsaidwaswrong; itissimplyinsufficient.The twoproblemswithEuclidmentionedso far, namely thedefinitions thatdonotdefine

anything, andtheinsufficientaxioms, canbothberemedied. TwothousandyearsafterEuclid,mathematicianshaveshowedthatwithregardtothesetwoissues, Euclidwascorrect, justmiss-ingsomedetails. Thereis, however, another, moretricky, problemwithEuclid. TheproblemconcernsEuclid’sFifthPostulate. A lookatthefivepostulatesquicklyrevealsthatthefifthissomehowdifferentfromthefirstfour. Thefirstfouraresimpletostate, andimmediatelybeliev-able. Thefifth, bycontrast, ismuchlongertostate, and, whilecertainlybelievable, doesnotseemasimmediatelyobviousasthefirstfour. MathematiciansthroughoutthecenturiesafterEuclidnoticedthisproblem. Thereisnothinginherentlywrongwithapostulatethatiscompli-cated, asisEuclid’sFifthPostulate, butitisbothersome. OneofthereasonspeoplelikedEuclid’sgeometryisthat(ignoringtheflawsmentionedabove, whichseemtohavebeennoticedonlylateron), itseemedtobeamodelforprovingthatcertainfactsareindisputablytrue. Euclidwastheultimateexampleofhowhumanbeingscouldobtaincertainknowledge. Ifonestartswithindisputablytrueaxioms, andproceedsinanairtightlogicalfashiontodeducethingsfromtheaxioms, thenwhateveronededucesmustalsobeindisputablytrue. However, ifEuclidistobeviewedinthisway, thenitiscrucialthathisaxiomsareindisputablytrue. Thecommonnotionsandthefirstfourpostulatesseemquiteconvincing. TheFifthPostulate, bycontrast, doesnotseemquiteasindisputable, givenitsmorecomplicatednature.WhatcanbedoneabouttheFifthPostulate? Itcannotsimplybedropped—itismostdefinitely

usedinsomeofEuclid’sproofs. WhatanumberofpeopleafterEuclidtriedtodowastodeducetheFifthPostulatefromtheotherfour. IftheFifthPostulatecouldbededucedfromtheotherfour, thenanythingprovableusingallfivepostulatescouldbeprovedusingthefirstfour, which

10 1. GeometryBasics

wouldaddtotheindisputabilityofwhateverwasprovedbyEuclid. Overtheyears, anumberofpeopleclaimedtohavededucedEuclid’sFifthPostulatefromtheotherfour. Wenowknow,however, that theywereallmistaken. Asdiscovered independentlybyKarl FriedrichGauss(1777-1855), JanosBolyai (1802-1860)andNikolai IvanovitchLobatchevsky (1793-1856) intheearly19thcentury, itispossibletoconceiveofperfectlygoodgeometriesthatinvolvethefirst fourpostulates, but somethingother than theFifthPostulate (wewillmentionhow thiscouldhappenverybrieflybelow). Thisdiscoverywasextremelyrevolutionary. Infact, Gauss,arguablythegreatestmathematicianofalltimes, butsomeonewhoseemstohavebeenquiteconcernedabouthisreputation, didnotinitiallypublishhisdiscoveryofwhatisnowcallednon-Euclideangeometry, for fearof thepublic reaction. For the twothousandyearsprior toGauss, “TheElements”hadbeentakenassomethingapproachingasacred, non-challengeable,text. TochallengeEuclid, asGaussdidprivately, andsubsequentlyBolyaiandLobatchevskydidpublicly(andindependently)in1831and1829respectively, wasseenasalmostashereticalasDarwinwaslaterinthe19thcentury.Wecannotgiveherethedetailsofthenon-Euclideanrevolution, excepttosaythatitusheredin

acompletelyneweraofgeometry. Newapproachestogeometry, suchastheuseofisometries(acrucialtoolinourinourtreatmentofsymmetryinChapters 4 and5), andRiemanniangeometry(thatwaslaterusedbyEinsteininhisgeneraltheoryofrelativity)flourishedinthe19thcentury,oncethestrangle-holdofEuclid’sapproachwaslifted. (Ofcourse, haditnotbeenforEuclid,geometrymightnothave reached the19thcentury inasdevelopeda formas itdid, soweshouldnotbelittleEuclid’sfundamentalimportancetogeometry, andallofmathematics; therearesimplyotherapproachestogeometryaswell.) Thediscoveryofnon-Euclideangeometryhadphilosophicalimportancebeyondjustgeometry, orevenmathematics. IfEuclidwasonceheldupasamodelforabsolutetruth, andifwenowknowthatothertypesofgeometryarepossible, thenweneedtorethinkwhat, ifanything, wecanknowwithabsolutecertainty. See[Tru87]or[Gre93]formoredetailsaboutthenon-Euclideanrevolution.TogetabitmoreofafeelforwhatmakesEuclideangeometrydistinctfromnon-Euclidean

geometry, wementionanimportantresultinEuclideangeometryknownasPlayfair’sAxiom. AdemonstrationofPlayfair’sAxiomwillbegiveninSection 1.2.

Proposition 1.1.1 (Playfair’sAxiom). Supposem isaline, andA isapointnotonm. Thenthereisoneandonlyonelinethrough A thatisparallelto m.

Fromourpointofview“Playfair’sAxiom”innotanaxiomatall, butatheoreminEuclideangeometry; however, thenameistraditional, andweuseitwhetherornotitmakesliteralsense.Actually, Playfair’sAxiomisnotonlyatheoreminEuclideangeometry, but, morestrongly, itisequivalenttoEuclid’sFifthPostulate. BythatwemeanthatEuclid’sfivepostulatesimplyPlay-fair’sAxiom, andEuclid’sfirstfourpostulatestogetherwithPlayfair’sAxiomimplyEuclid’sFifthPostulate. (Infact, insomemoderngeometrytexts, thestatementofPlayfair’sAxiomisincor-rectlyreferredtoasEuclid’sFifthPostulate. AlthoughlogicallyitiscorrecttosubstitutePlayfair’sAxiomforEuclid’sFifthPostulate, becausethetwostatementsimplyeachother, historicallyitiscompletelyinaccuratetoreplacetheonestatementwiththeother.)


Non-Euclideangeometryresults fromtakingEuclid’sfirst fourpostulates, butreplacingtheFifthPostulateby somethingelse. It is easier tounderstandwhathappens innon-Euclideangeometry ifweconsideralternatives toPlayfair’sAxiom. Supposethat m isa line, and A isapointnoton m. IfPlayfair’sAxiomwerenottrue, thentherearetwopossiblecases: eitherthereismorethanonelinethrough A thatisparallelto m, orthereisnolinethrough A thatisparallelto m. IftheformerpossibilityistakenasanaxiominsteadofPlayfair’sAxiom, theresultinggeometryiscalledhyperbolicgeometry; ifthelatterpossibilityistakenasanaxiominsteadofPlayfair’sAxiom, theresultinggeometryiscalledsphericalgeometry. InSection 2.2,wewillseethatPlayfair’sAxiomimpliesthatthesumoftheanglesinatriangleis 180◦. Hence,inEuclideangeometrythesumoftheanglesinatriangleis 180◦. Bycontrast, itcanbeprovedthatinhyperbolicgeometry, thesumoftheanglesinatriangleisalwayslessthan 180◦ (theprecisesumcanvaryfromtriangletotriangle); insphericalgeometry, thesumoftheanglesinatriangleisalwaysgreaterthan 180◦ (again, theprecisesumcanvaryfromtriangletotriangle).Itishardtoimaginehowhyperbolicorsphericalgeometrywouldworkifweusethefamiliar

sortofstraightlinesfoundintheplane, butthereisnoneedtorestrictourselvesonlytothemostfamiliarsituation. Forexample, thinkofthesurfaceofasphereasauniverse, andthinkofgreatcirclesas“straightlines”inthisuniverse(greatcirclesarestraightfromthepointofviewofabuglivingonthesphere). Thegeometrythatusesgreatcirclesonthesphereturnsouttobewhatwenowcallsphericalgeometry. Modelsforhyperbolicgeometrycanalsobefound. A detaileddiscussionofnon-Euclideangeometrywould takeus too far afield; see [Mar75]or [Gre93]forfurtherdetails. Onepointworthmentioningisthatitcanbeproved, thoughthisisfarfromobvious, thatEuclideangeometryisnomoreorlessvalidthanhyperbolicorsphericalgeometry.Thatis, ifweacceptEuclideangeometry, weneedtoacceptnon-Euclideangeometryaswell; ifwedonotacceptnon-Euclideangeometry, thenwecannotacceptEuclideangeometryeither.Mathematicallyspeaking, thereismorethanonepossiblevalidgeometry. Whichgeometryourphysicaluniversesatisfiesisanothermatter, onewhichphysicists, notmathematicians, needtodecidebyuseofexperiments. Onadailybasis, ouruniverseiseitherEuclideanorcloseenoughtoitthatitisEuclideanforallpracticalpurposes.InthistextwewillbeworkingwithinwithintheframeworkofEuclideangeometry, though

wewillnotbeapproachingthingsaxiomatically(akasynthetically). However, wewantedtohaveabriefoverviewofwhat theaxiomaticpropertiesof Euclideangeometryare, becausethesefeaturesultimatelyunderlieeverythingthatwedo, evenwhentheyarenotmentionedexplicitly. Forexample, wewillseewhereEuclid’sFifthPostulatecomesintoplayinthestudyofparallellinesinSection 1.2. Ourstudyofsymmetry, inPartII ofthisbook, usesanapproachtogeometrythatonlycameintobeingafterthediscoveryofnon-Euclideangeometry. However,eventhoughourstudyofsymmetrymayseemdifferentfromtheapproachfoundinEuclid, ittooisultimatelyEuclidean.Finally, weend this sectionwith thehope thatour remarksabout theflaws inEuclidwill

notdiscouragethereaderfromtakinganinterestin“TheElements.” Euclidcontainsawealthofsubstantialmathematicalideas, thoughthedryandsometimestediousstyledonotalwaysmaketheseideasapparentuponfirstencounter. Anyonewishingtoread“TheElements”would


dowelltoreadboththeactualtextfoundin[Euc56], togetherwithacompaniontoEuclid, suchastheexcellent[Har00].

1.2 LinesandAngles

The twomost fundamentalnotions in thegeometryof theplaneare point and line. By theword“line”wealwaysmeanastraightline(asisstandardusagetoday, thoughdifferentfromEuclid’susage). Asdiscussed inSection 1.1, Euclidattempted todefine theseconcepts, butdidnotsucceedindoingsoinameaningfulway. Wetakethemodernapproach, andsimplyassumethattherearesuchthingsaslinesandpointsintheplane, andthatthereisarelationbetweenpointsand lines, namely thatgivenapointandgivena line, either thepoint isonthelineoritisnot. Further, werequiretherelationbetweenpointsandlinestosatisfycertainfamiliarproperties, suchasthefactthatanytwodistinctpointsarecontainedinoneandonlyoneline(thisisessentiallyEuclid’sFirstPostulate). Wedonotcaresomuchwhatpointsandlinesare, buthowtheybehave. Formostofthistext(except, forexample, inChapter 3), wewillberestrictingourattentiontopointsandlinesintheplane. Itisalsopossibletodiscusspointsandlines(andothergeometricobjects)inthreedimensionalspace, andhigherdimensionstoo.Whennototherwisenoted, thereadershouldassumewearediscussingtheplane.Westartourdiscussionoflineswithsomenotation. Giventwodistinctpoints A and B inthe

plane, weknowthatthereisauniquelinecontaining A and B. Wedenotethislineby←→AB.

WhenitisnotnecessarytospecifythepointsA and B, wewillalsousesingleletterssuchasmtodenotelines. Intuitively, aline“goesonforever”intwodirections. Giventwodistinctpoints

A and B, wecanalsohavethatpartoftheline←→AB thatstartsat A, and“goesonforever”in

thedirectionof B. Suchanobjectiscalledthe ray from A through B, andisdenotedby−→AB.

Wecall A the startingpoint oftheray−→AB. Wecanalsolookatthatpartoftheline

←→AB that

startsat A andendsat B (orvice-versa). Suchanobjectiscalledthe linesegment from A to B,andisdenotedby AB. SeeFigure 1.2.1 forallthreetypesofobjects. Wecallthepoints A andB the endpoints ofthelinesegment AB.

Oneofthemostbasicquestionaboutlinesintheplaneiswhetherornottwolinesintersect,whichmeansthatthetwolineshaveapointincommon. Ourfirstresultisthefollowingratherobviousfact; evenobviousresultsneedtobeproved, however, becauseourintuitionaboutwhatis“obvious”issometimeswrong(forexample, peopleusedtothinkthattheearthwasflat).

Proposition 1.2.1. Twodistinctlinesintersectinatmostonepoint.

Demonstration. Supposem and n aredistinctlines. Supposefurtherthattheyintersectinmorethanonepoint. Hence, thereareatleasttwodifferentpoints, sayA and B, thatarecontainedinboth m and n. Itfollowsthateachof m and n isalinecontainingthepoints A and B, whichcontradictsthefactstatedabovethatanytwodistinctpointsarecontainedinoneandonlyoneline. Henceitmustbethecasethat m and n donotintersectinmorethanonepoint.

1.2LinesandAngles 13

A B

AB

A B

AB

A B

AB

Figure1.2.1

Becauseoftheaboveproposition, ifwearegiventwodistinctlines, eithertheydonotintersect,ortheyintersectinpreciselyonepoint. Wesaythattwolinesare parallellines iftheydonotintersect; iftwolinesarenotparallel, wesaythattheyare intersectinglines. SeeFigure 1.2.2.Noticethatthedefinitionofparallellinesdoesnotmentionanythingaboutparallellines“goinginthesamedirection,” or“keepingconstantdistancefromeachother.” Bothoftheseideasaretrueaboutparallellinesintheplane, buttheyarenotpartofthedefinition, andneedtobeproved. WewillessentiallyprovethefirstoftheseideasinProposition 1.2.3, andthesecondinProposition 2.2.6. Noticealsothattwoequallinesarenotconsideredparallel, becausetheycertainlydointersect.

parallel lines intersecting lines

Figure1.2.2

Strictlyspeaking, theterm“parallel”appliesonlytolines, andnottolinesegments. However,itmakesintuitivesensetodiscusslinesegmentsbeingparallelornot, andwewillsaythattwolinesegmentsare parallel ifthelinescontainingthelinesegmentsareparallel.Havingbrieflydiscussedlines, wenowturntoanothertypeoffundamentalgeometricobject,

namelyangles. An angle isaregionoftheplanethatisbetweentworaysthatintersectinacommonstartingpoint. Forexample, theshadedregioninFigure 1.2.3 isanangle. Thepointofintersectionofthetworaysiscalledthe vertex oftheangle. Giventworaysthatintersectinacommonstartingpoint, thereareactuallytwoanglesthattheraysdetermine, forexampletheshadedregioninFigure 1.2.3, andtheunshadedregion. Itisthereforenecessarytospecify


whichoftheregionsisbeingreferredto. Thewaytoavoidthissortofambiguityistospecifyananglenotonlybytworaysthatintersectinacommonstartingpoint, butalsotospecifyadirection, eitherclockwiseorcounterclockwise. InFigure 1.2.4 (i)weseeananglespecifiedbytworaysandtheclockwisedirection(indicatedbythecurvedarrow); inFigure 1.2.4 (ii)weseeadifferentanglespecifiedbythesametworaysasinPart (i)ofthefigure, butwiththeoppositedirection.

Figure1.2.3

(i) (ii)

Figure1.2.4

Anangleisageometricobject. Justaswecanmeasurethelengthsoflinesegments, wecanalsomeasureangles, notinunitsoflength(forexample, feetormeters), butinunitsofangularmeasure. Measuringageometricobjectmeansassigningtotheobjectanumber, whichinsomesensetellsusthe“size”oftheobject. Therearetwostandardunitsofangularmeasure, degreesand radians. Degreesare simpler toexplain, andare thereforeused regularly inelementaryandsecondaryschools; radiansareprefered inadvancedmathematics, forvarious technicalreasonswecannotdiscusshere. Weassumethatthereaderisfamiliarwithdegreemeasureofangles, andwewillusedegreesinthistext. (Weshouldmentionthatthenumber 360 usedinthecontextofdegreemeasureiscompletelyarbitrary, andariseshistorically, ratherthanoutofanyrealreason. Itwouldbeequallyvalidtobreakupthetotalanglegoingaroundapointintoanyothernumberof“degrees,” but 360 isfamiliartoeveryone, andwewillstickwithit.)Theonepointaboutmeasuringangles(bydegreesoranyothermethod)thatweneedtostress

involvesourpreviousobservation that thereare two“directions” inwhichananglecanbespecified, namelyclockwiseandcounterclockwise. Bothdirectionsareequallyvalid, butitis


convenienttopickonedirectionashavingpositivemeasure, andonedirectionhavingnegativemeasure. Wewill take theclockwisedirectionaspositive, becausethat is themorefamiliardirectionindailylife. (It iscompletelyarbitrarywhichdirectionis takenaspositive, aslongasweallagreeonthechoice. Inmoreadvancedmathematics texts, it iscustomarytohavecounterclockwisebethepositivedirection.) InFigure 1.2.5 (i)weseeananglethathasmeasure45◦, andinFigure 1.2.5 (ii)weseeananglethathasmeasure −45◦.

-45°45°

(i) (ii)

Figure1.2.5

Degreesareunitsformeasuringangles. Itisimportanttodistinguishbetweenanangleanditsmeasure. Anangleisageometricobject; itsmeasureindegreesisanumberthatweassociatewith theangle. Analogously, theheightofaperson isanumber thatweassociatewith thatperson. Justastwopeoplecanhavethesameheight, butstillbedifferentpeople, similarlytwoanglescanhavethesamemeasure, butstillbedifferentangles. Forexample, inFigure 1.2.6 weseetwodifferentangles, eachofwhichhasmeasure 60◦. Ofcourse, iftwoangleshavedifferentmeasure, theymustbedifferentangles. Iftwoangleshavethesamemeasure, theyneednotbethesameangle, butitisalwaysthecasethattheyarecongruent, whichmeansintuitivelythatoneanglecouldbe“pickedupandplacedpreciselyontopoftheother.” Wewillnotneedamoreformaldefinitionofcongruencehere, andwillsticktotheintuitivenotion. (Theconceptofisometry, whichisdiscussedindetailinChapter 4, providesonewayofdefiningcongruence.)Thebottomlineisthatwhenwesaythat“twoangleareequal”wemeanthattheyarecongruent,andinparticularthattheyhavethesamemeasureindegrees. Forthesakeofbrevity, wewillsometimes“abuse”terminologyandsay, forexample, ofananglethat“is” 90◦, whenweshouldmoreproperlysaythattheangle“hasmeasure” 90◦. Suchabuseofterminologyisverycommon,andshouldcausenoconfusion.

LetusnowreturntoEuclid’sFourthPostulate, whichsays“Thatallrightanglesareequaltooneanother.” Tounderstandthispostulate, weneedtoknowwhatarightangleis. Todaywetendtothinkofarightangleasbeingdefinedtobeanangleof 90◦, butthatisnottheproperwaytounderstandrightangles, becauseitonlytellsusthemeasureofrightangles, notwhattheyaregeometrically. Thegeometricideaofarightangleisbasedonwhathappenswhentwolinesintersectinapoint. AsweseeinFigure 1.2.7 (i), twolinesintersectingdividetheplaneintofourangles. Thefouranglesarenotnecessarilyallequaltoeachother. However, ifitdoeshappenthatallfouranglesareequaltoeachother, thenwecalleachofthefouranglesa rightangle.


Figure1.2.6

SeeFigure 1.2.7 (ii). WhatEuclid’sFourthPostulatesaysisthatanysuchrightangle, formedbyanytwointersectinglines, isequaltoanyotherrightangleformedbytwootherintersectinglines. Asforthedegreemeasureofarightangle, becausethetotalanglegoingaroundapointis 360◦, andbecausetherearefourrightanglesatapoint, itfollowsthatthemeasureofanyrightangleis 360◦/4 = 90◦. Usingthestandardabuseofterminology, wewillfollowcommoncustomandsimplysaythatanyrightangle“is” 90◦. Similarly, theanglealonganystraightlineis 180◦.

(i) (ii)

Figure1.2.7

Wenowmaketwomoregeometricdefinitionsconcerningangles. First, supposewehavetwoanglesthatare“alongaline;” thatis, twoanglesthatareformedwhenaraystartsatapointonaline. Twosuchanglesarecalled supplementaryangles. Theanglesα andβ inFigure 1.2.8 (i)aresupplementaryangles. Next, supposewehavetwointersectinglines. Ofthefouranglesformed,therearetwopairsof“oppositeangles;” thatis, anglesthatintersectonlyinacommonpoint.Suchanglesarecalled verticalangles. Theangles α and β inFigure 1.2.8 (ii)areverticalangles.

Wenowstateaverysimpleresultaboutangles, usingtheconceptsjustdefined.

Proposition 1.2.2.

1. Supplementaryanglesaddupto 180◦.

2. Verticalanglesareequal.


(i) (ii)

αα β β

Figure1.2.8

Demonstration.

(1). Thisisevident, becausetheanglealongastraightlineis 180◦.

(2). InFigure 1.2.9 weseeangles α and β, whichareverticalangles. Noticethattheangleγ isasupplementaryangletoeachof α and β. Hence, byPart (1)ofthisproposition, weknowthat α + γ = 180◦ and γ + β = 180◦. Wededucethat α = 180◦ − γ and β = 180◦ − γ,andtherefore α = β.

α βγ

Figure1.2.9

TheproofofProposition 1.2.2 isverysimple. Moreover, itdoesnotmakeuseofthefullstrengthofEuclid’spostulates, inthattheFifthPostulateisnotused. Ournextresultaboutanglesismoresubstantial, andtheFifthPostulateiscrucialtoitsproof. Weareinterestedinthesituationwheretwolines, saym andn intersectathirdline, say k; seeFigure 1.2.10. Thelinesm and k formfourangles, andthelines n and k formfourmoreangles, allofwhicharelabeledinFigure 1.2.10.Wecallangles x and y interioralternatingangles, andwealsocallangles z and w interioralternatingangles. Similarly, wecallangles α and β exterioralternatingangles, andwealsocallangles δ and ϵ exterioralternatingangles. Finally, wecallangles x and β correspondingangles, andwealsocallangles δ and w, angles z and ϵ, andangles α and y, correspondingangles. Thefollowingpropositionsaysthattheabovesortsofanglesareparticularlynicewhenwestartwithtwoparallellinesthatintersectathirdline.

Proposition 1.2.3. Supposetwoparallellinesintersectathirdline. Thentheinterioralternatinganglesareequal; theexterioralternatinganglesareequal; thecorrespondinganglesareequal.


α

ε β

δ

y w

z x

m

n

k

Figure1.2.10

Demonstration. InFigure 1.2.10 weseeangles x and y, whichareinterioralternatingangles,angles α and β, whichareexterioralternatingangles, andangles x and β, whicharecorre-spondingangles. Wewilldemonstratethepropositionwithregardtothesethreepairsofangles;theotherappropriatepairsofanglesaresimilar, andwewillskipthedetailsforthem.Westartwiththeobservationthatoneofthefollowingthreecasesmustcertainlyhold: either

x+w < 180◦, or x+w = 180◦, or x+w > 180◦. Supposefirstthat x+w < 180◦. RecallEuclid’sFifthPostulate, whichsays“That, ifastraightlinefailingontwostraightlinesmakestheinterioranglesonthesamesidelessthantworightangles, thestraightlines, ifproducedindefinitely, meeton that sideonwhichare theangles less than the two rightangles.” Thispostulateispreciselysuitedtoourcurrentsituation, anditimpliesthat m and n intersectonthesideof x and w. Wehavethereforereachedalogicalimpossibility, because m and n areassumedtobeparallel, andhencecannotintersect. Weconcludethatitcannotbethecasethatx+w < 180◦.Nowsupposethat x+w > 180◦. Clearly y = 180◦ −w and z = 180◦ − x. Therefore

y+ z = (180◦ −w) + (180◦ − x) = 360◦ − (x+w) < 180◦.

ThenbyEuclid’sFifthPostulate, itwouldfollowthat m and n intersectonthesideof y and z,whichagaincannotbe, becausethetwolinesareparallel. Theonlyremainingoptionisthatx + w = 180◦. Therefore x = 180◦ − w. Becauseweknowthat y = 180◦ − w, itfollowsthat x = y. Hence, interioralternatinganglesareequal. Because β and y areverticalangles,weknowthat β = y. Itissimilarlyseenthat α = x. Wededucethat β = x, whichsaysthatcorrespondinganglesareequal, andthat α = β, whichsaysthatexterioralternatinganglesareequal.

TheproofofProposition 1.2.3 verymuchdependsuponEuclid’sFifthPostulate, andthereforeanyotherpropositionthatisdemonstratedusingProposition 1.2.3 mustalsodependuponEu-clid’sFifthPostulate. Forexample, Proposition 2.2.1, whichdiscussesthesumoftheanglesin


atriangle, usesProposition 1.2.3 initsproof. Thus, weseethattheFifthPostulateiscrucialinthestudyofplanargeometry.Amongotherthings, Proposition 1.2.3 showsthatourdefinitionofparallellines, whichwas

simplyintermsofnon-intersection, correspondstoourintuitionthatparallellines“gointhesamedirection.”

Exercise 1.2.1. Findangles x and y asshowninFigure 1.2.11. Thelines m and n areparallel.

y

x

m

n45°

60°

Figure1.2.11

Exercise 1.2.2. Findangles α, β and γ asshowninFigure 1.2.12. Thelines p and q areparallel.

ItturnsoutthattheconversetoProposition 1.2.3 isalsotrue. Thatis, iftwolinesmakeap-propriateangleswithagivenline, thentheyareparallel. Moreprecisely, wehavethefollowingproposition.

Proposition 1.2.4. Supposetwodifferentlinesintersectathirdline. Ifthetwolineshaveequalinterioralternatingangles, orequalexterioralternatingangles, orequalcorrespondingangles,thenthetwolinesareparallel.

Interestingly, eventhoughProposition 1.2.3 reliescruciallyonEuclid’sFifthPostulate, itturnsoutthatProposition 1.2.4 doesnotrelyupontheFifthPostulate(andisthereforetrueinotherge-ometriesforwhichthefirstfourpostulateshold, butthefifthdoesnot). However, aparticularly


γ

β

α

p

q

80° 115°

Figure1.2.12

easydemonstrationofProposition 1.2.4 canbe foundusing theFifthPostulate; thisdemon-stration is left to the readerasExercise 2.2.1, where ituses the fact, proved inSection 2.2,thatthesumoftheanglesinatriangleaddupto 180◦. (SeeTheorem 3.4.1of[WW98]forademonstrationofProposition 1.2.4 thatdoesnotmakeuseofEuclid’sFifthPostulate.)AnimmediateconsequenceofProposition 1.2.4 is thefollowingresult. First, weneedthe

followingstandardbitofterminology. Giventwolinesintheplane, wesaythattheyare per-pendicular iftheymakerightangleswitheachother.

Proposition 1.2.5. Supposetwodifferentlinesarebothperpendiculartoathirdline. Thenthetwolinesareparallel.

WecannowgiveademonstrationofPlayfair’sAxiom (Proposition 1.1.1). Inthisdemonstra-tion, andinothersubsequentplaces, wewillmakeuseofthefactthatifwearegivenalinemandapoint A (eitheronoroff m), wecanconstructalinethrough A thatisperpendiculartom. Wetakethispropertytobeaxiomatic. Also, wenotethatthisconstructionisveryeasytodousingstraightedgeandcompass, thoughwewillnotneedthedetailshere.

DemonstrationofProposition 1.1.1. Supposethatm isaline, andthatA isapointnotonm.Weneedtoshowtwothings: (1)thereisalinethrough A thatisparallelto m; and(2)thereisonlyonesuchline.ToshowPart (1), westartbyconstructingalinethrough A thatisperpendicularto m. Call

thisnewline n. Next, constructalinethrough A thatisperpendicularto n (thefactthat A ison n causesnoproblem). Callthisnewline p. SeeFigure 1.2.13 (i). Byconstruction, weseethatboth m and p areperpendicularto n. ItfollowsfromProposition 1.2.5 that m and p areparallel. Wehavethereforeconstructedalinethrough A, namely p, thatisparallelto m.

ToshowthePart (2), weneedtoshowthattheline p weconstructedaboveistheonlylinethatcontains A andisparallelto m. Let t beanylinethatpassesthrough A otherthan p. SeeFigure 1.2.13 (ii). Giventhat t isnotthesameas p, then, asseeninthefigure, itcannotbethat tisperpendicularto n, because p isperpendicularton. Itnowfollowsthat t andm donotmake

1.3Distance 21

m

p

n

A

m

p

tn

A

(i) (ii)

Figure1.2.13

equalcorrespondingangleswith n, because m makesall 90◦ angleswith n, and t doesnotmake 90◦ angleswithn. Wededucethatm and t arenotparallel, becauseiftheywereparallel,thenbyProposition 1.2.3 itwouldfollowthatm and t wouldhaveequalcorrespondingangleswith n, whichtheydonot. Itfollowsthat p istheonlylinethrough A thatisparalleltom.

Exercise 1.2.3. Westatedabove that ifwearegivena line m andapoint A, wecanconstructalinethrough A thatisperpendicularto m. Showthatthereisonlyonesuchline. Thedemonstrationhastwosubcases, dependinguponwhether A ison m ornot.

1.3 Distance

PlanargeometryasformulatedbyEuclidrests, fundamentally, ontheideathattherearethingscalledpointsandthingscalledlines, andthatthesethingshavesomerelationtoeachother(forexample, anytwodistinctpointsarecontainedinauniqueline). Insomeofthemoremodernapproachestogeometryotherfundamentalnotionshavealsocomeintoplay. Oneoftheideasthathasprovedtobeparticularuseful, forexampleinthestudyofsymmetry(aswewillseeinChapters 4 and5), isthenotionofdistancebetweenpoints. Certainly, theideaoflengthsoflinesegmentsisinEuclid, andhencethedistancebetweentheendpointsofalinesegmentisimplicitinEuclid, butmodernmathematicsdealswiththeconceptofdistancemoreexplicitly. Wewillrestrictourattentiontothedistancebetweenpointsintheplane, thoughitisalsopossibletolookatdistanceinothersituations.Thereare, infact, twodifferentapproachestothenotionofdistancebetweenpoints. First,

wecoulduseCartesiancoordinates, accordingtowhichweassigneverypointintheplaneapairofnumbers (x, y). (Weassumethereaderis familiarwithsuchcoordinates.) Wecouldthendefine thedistancebetween twopoints (x1, y1) and (x2, y2) by thestandarddistanceformula

√(x2 − x1)2 + (y2 − y1)2. (ThisformulafollowsdirectlyfromthePythagoreanThe-

orem.) However, giventhatwearenotgoingtobeusingcoordinatesinothersituations, this


approachtodistancewillnotbetheoneweuse. Alternatively, justasweassumedsomeax-iomaticpropertiesforpointsandlines, wecansimplyhypothesizethatitispossibletoassignauniquedistancebetweenanytwopointsintheplane, andthatthismeasureofdistancesatis-fiescertainproperties. Wewilltakethelatterapproach, thoughwewillnotgiveanaxiomatictreatmentofdistance(whichissurprisinglytricky).SupposeA andB arepointsintheplane.Wethenassigntothesepointsarealnumberdenoted

d(A,B), calledthe distance betweenA and B. Amongthepropertiesthatdistancesatisfiesarethefollowing, where A, B and C arepointsintheplane.

1. d(A,B) ≥ 0;

2. d(A,A) = 0, and d(A,B) = 0 whenever A = B;

3. d(A,B) = d(B,A);

4. d(A,B) ≤ d(A,C) + d(C,B).

Thefirstthreeofthesepropertiesaresimple; thefourth(calledtheTriangleInequality)takesabitmoreofanexplanation. Consideratrianglewithvertices A, B and C, asinFigure 1.3.1.ThenProperty(4)saysthatthelengthoftheside AB ofthetriangleislessthanorequaltothesumofthelengthsoftheothertwosidesofthetriangle(wehaveequalityonlyifthetriangleis“degenerate”). Byrearrangingtheletters, weseethatthesameinequalityholdsfortheothertwosidesofthetriangleaswell. (HadwetakentheapproachofusingCartesiancoordinatestodefinedistancebetweenpoints, itwouldhavebeenpossibletoprovetheabovefourpropertiesusingthedistanceformula.)

A

B

C

Figure1.3.1

Oncewehaveanotionofdistancebetweenpoints, it ispossibletoformulatemanyotherbasicgeometricconceptssuchaslines, raysandlinesegmentsintermsofdistance. Forexample,supposewearegiventwopoints A and B intheplane.Thenthelinesegmentfrom A to B isthecollectionofallpoints X intheplanesuchthatthe

equation d(A,X)+d(X,B) = d(A,B) holds. TherayfromA through B isthecollectionofallpointsX intheplanesuchthatpreciselyoneofthetwoequationsd(A,X)+d(X,B) = d(A,B)or d(A,B)+d(B,X) = d(A,X) holds. ThelinethroughA and B isthecollectionofallpoints

1.3Distance 23

X intheplanesuchthatpreciselyoneofthethreeequations d(A,X) + d(X,B) = d(A,B),or d(A,B)+d(B,X) = d(A,X), or d(B,A)+d(A,X) = d(B,X) holds. (Ifonewantstobecompletelydetailedaxiomatically—asisimportantinmoreadvancedtreatmentsofgeometry—thereareanumberofpossibleapproacheswhen itcomes to the relationof theconceptofdistancetotheconceptsofpointsandlines: onecandefinepointsandlinesaxiomatically, thenconstructadistancefunctionfromtheaxioms, andthenshowthatourformulationoflinesintermsofdistanceisconsistentwiththelinesasgivenbytheaxioms; alternatively, onecantakedistanceas thebasicaxiomaticallydefinedconcept, thenuse theaboveapproach todefinelines, andthenshowthatlinesdefinedinthiswaybehaveaslinesoughtto; or, onecandefinebothlinesanddistanceintermsofcoordinates, andthenshowthattheaboveformulationoflinesintermsofdistanceisvalid. Wewillnotgointosuchdetailsinthistext.)Circlescanalsobedefinedusingthenotionofdistance. Givenapoint A intheplane, and

anon-negativerealnumber r, thenthecircle withcenter A andradius r isthecollectionofallpoints X in theplane such that theequation d(A,X) = r holds. It is evenpossible tocomputeanglesstrictlyintermsofdistance. Considerthetrianglewithvertices A, B and C,showninFigure 1.3.1. Itisthenpossibletocomputetheanglesateachof A, B and C usingonlythelengths d(A,B), d(A,C) and d(B,C). Themethodforsuchcalculationsuses theLawofCosines, studiedintrigonometry. ThislawisstatedinProposition 2.5.3, anditmayalsobefoundinanytextbookontrigonometry. (Ifyouareunfamiliarwiththislaw, simplyignoretheformulaweareabouttostate; wewillnotbeusingthisformulaagain.) UsingtheLawofCosines, theformulafortheangleat A isseentobe

arccos

([d(B,C)]2 − [d(A,B)]2 − [d(A,C)]2

2d(A,B)d(A,C)

).

Wenowturntoanissuethatisphrasedintermsofdistancebetweenpoints, andthatwillbeofusetousinourstudyofisometriesinChapter 4, whichinturnisusedinourstudyofsymmetryinChapter 5. Supposewearegiventwopoints A and B intheplane. Wewouldliketofindallpointsintheplanethatareequidistantto A and B; thatis, wewanttofindallpoints X intheplanesuchthat d(X,A) = d(X,B). Onesuchpointisveryeasytofind, namelythemidpointofthelinesegment AB. Thereare, however, otherpointsintheplanethatareequidistanttoA and B aswell. Thefollowingpropositiontellsusaneasywaytofindallsuchpoints. Inthispropositionweuse the followingterminology. Givena linesegment AB, the perpendicularbisector ofthelinesegmentisthelinethatcontainsthemidpointofthelinesegment, andisperpendiculartothelinesegment. SeeFigure 1.3.2. Asisstandard, weuseasmallsquaretodenotearightangleinthefigure. Wecannowstateourresult.

Proposition 1.3.1. Supposethat A and B aredistinctpointsintheplane. If X isapoint, thend(X,A) = d(X,B) ifandonlyif X isontheperpendicularbisectorof AB.

ThedemonstrationofthispropositionislefttothereaderinExercise 2.2.9 (whichisputofftillSection 2.2, becausesomefactsaboutcongruenttrianglesareneeded).


A

B

Figure1.3.2

2Polygons

2.1 Introduction

A polygon isaregionoftheplanethatisboundedbyafinitenumberoflinesegmentsthataregluedtogether. Wehavethreerequirementsaboutthewayinwhichwegluethelinesegmentstogether.

(1) Linesegmentsaregluedendpoint-to-endpoint.

(2) Everyendpointofalinesegmentisgluedtopreciselyoneotherendpointofalineseg-ment.

(3) Notwolinesegmentsintersectexceptpossiblyattheirendpointswheretheyareglued.

SomepolygonsareshowninFigure 2.1.1. Somenon-polygonsareshowninFigure 2.1.2;theobjectinPart (i)ofthisfigurehasedgesthatarenotgluedendpoint-to-endpoint, theobjectinPart (ii)hasthreeendpointsoflinesegmentsgluedtogether, andtheobjectinPart (iii)haslinesegmentsintersectingnotattheirendpoints. (Itispossibletolookatpolygonswithself-intersections, thatis, inwhichrequirement(3)isdropped, butwewillnotbelookingatsuchpolygonsinthistext.)

Foreachpolygon, the edges ofthepolygonarethelinesegmentsthatboundit; theedgesaresometimescalled“sides,” thoughwewillmostlyavoidthatterm. The vertices (thesingularofwhichis“vertex”) ofapolygonarethepointswhereedgesmeet. Forexample, thepentagonshowninFigure 2.1.1 (i)hasfiveverticesandfiveedges.Itisnothardtofigureoutwhatallpolygonsare. NoticethatallthepolygonsshowninFig-

ure 2.1.1 haveedgesthatforma“circuit.” Thatis, ifyoustartatoneedge, youcanthengoto

26 2. Polygons

(i) (ii) (iii)

Figure2.1.1

(i) (ii) (iii)

Figure2.1.2

oneoftheedgesitmeets, andfromtheretothenextedge, andthenext, andsoonuntilyoucomeallthewayaroundbacktotheedgethatyoustartedwith. Infact, allpolygonsworkthisway, whichwesummarizeinthefollowingproposition, whichwegivewithoutdemonstration.

Proposition 2.1.1. Supposewearegivenapolygon.

1. Theverticesofthepolygoncanbelabeledas A1, A2, A3, . . . , An, wheretheedgesofthepolygonare A1A2, A2A3, A3A4, . . ., An−1An, An A1.

2. Thepolygonhasthesamenumberofedgesasvertices.

SeeFigure 2.1.3 foranillustrationoftheabovepropositioninthecasewhere n = 5. Thereadermightreasonablyask, inlightofProposition 2.1.1 (1), whywedidnotjustdefinepolygonstobethesortoffiguregivenbytheproposition. Theansweristhatwecouldhavedoneso, butwegavethedefinitionofpolygonsthatwedidinordertogiveadefinitionthatismoresimilartothedefinitionofpolyhedrainSection 3.1. Inanycase, wecannowproceedwithanunderstandingofpolygonsasgivenintheaboveproposition.

Sometypesofpolygonsareveryfamiliar, suchastriangles, whicharediscussedinmorede-tailinSection 2.2. Polygonswithfoursidesarereferredtoas quadrilaterals, ofwhichsomeofthemorefamiliartypesaresquares, rhombuses, rectangles, parallelograms, andtrapezoids. Asquare isaquadrilateralinwhichallfouredgesareequalandallfouranglesareequal; a rhom-bus isaquadrilateralinwhichallfouredgesareequal, butthefouranglesarenotnecessarilyequal; a rectangle isaquadrilateralinwhichallfouranglesareequal, butthefouredgesarenotnecessarilyequal; a parallelogram isaquadrilateralinwhichbothpairsofoppositeedgesareparallel; a trapezoid isaquadrilateralinwhichonepairofoppositeedgesisparallel. AnexampleofeachofthesetypesofquadrilateralisshowninFigure 2.1.4 (i)–(v); anexampleofa

2.1Introduction 27

A1

A5

A4

A3A2

Figure2.1.3

quadrilateralthatisnoneofthesetypesisshowninPart(vi)ofthefigure. Ingeneral, polygonsarenamedbythenumbersofedgesthattheyhave. A polygonwithfiveedgesiscalleda pen-tagon, apolygonwithsixedgesiscalleda hexagon, apolygonwitheightedgesiscalledanoctagon, etc. ThenamesofpolygonswithfiveormoreedgesarebasedonGreek, ratherthanLatin, roots. A polygonwith n edges, where n isanintegersuchthat n ≥ 3, iscalledan n-gon.

(i) (ii) (iii)

(iv) (v) (vi)

Figure2.1.4

Exercise 2.1.1. [UsedinSections 2.3, 2.4 and4.5] Supposethataparallelogramhasatleastone 90◦ angle. Showthatallfouranglesmustbe 90◦, andhencetheparallelogramisarectangle. (Useonlyresultsfromthissectionandprevioussections.)

28 2. Polygons

Exercise 2.1.2. Supposethatahexagonhasthepropertythateachpairofoppositeedgesisparallel.

(1) Showthatoppositeanglesinthishexagonareequal.

(2) Mustitbethecasethatoppositeedgeshaveequallengthsinthishexagon? Ifyes,showwhy. Ifnot, giveanexample.

2.2 Triangles

Thestudyoftrianglesgoesbacktotheancientworld. Trianglesplayed, andcontinuetoplay, acentralroleingeometry. Therearemanyimportantresultsabouttriangles, ofwhichwehavethespacetomentiononlyafew. Seemoststandardgeometrytextsformorefactsabouttriangles.A triangle isapolygonwiththreeedges(andhencethreevertices). Ifatrianglehasvertices

A, B and C, wedenotethetriangleby △ABC. Atanyvertexofatriangle, thetwoedgesofthetrianglethatcontainthevertexformananglethatisinsidethetriangle, calledthe interiorangle atthevertex. SeeFigure 2.2.1. Becauseinterioranglesarethemostcommonlyusedtypesofanglesintriangles, ifwesimplyreferto“theangle”atavertex, wewillalwaysmeantheinteriorangle(andsimilarlyforotherpolygons). Ifwemeansomeotherkindofangle, wewillalwayssaysoexplicitly. Observethateachedgeofatriangleislocatedoppositepreciselyoneofthevertices(andhenceoneoftheinteriorangles)ofthetriangle. Weusethenotation |AB|

todenotethelengthoftheedgeAB, andweusethenotation ∡A todenotethemeasureoftheangleat A.

A

B

C

Figure2.2.1

Thereareanumberofspecialtypesoftriangles. An equilateral trianglehasallthreeedgesequal. An isosceles trianglehastwoequaledges. A right trianglehasoneanglearightangle(thatis, a 90◦ angle). A righttrianglecanhaveonlyonerightangle. Itisalsocommontodistinguishbetween acute triangles (thathaveallangleslessthan 90◦), and obtuse triangles (thathaveoneanglegreaterthan 90◦).

2.2Triangles 29

Inadditiontotheinteriorangleateachvertexofatriangle, wecanalsoformanotherangleateachvertex, asfollows. Ateachvertex, extendbeyondthevertexoneof theedgesof thetrianglecontainingthevertex, andformthesupplementaryangle totheinteriorangle; thisangleiscalledthe exteriorangle atthevertex. InFigure 2.2.2 (i)weseeatriangle △ABC, withtheinteriorangleat A denoted α, andtheexteriorangleat A denoted δ. Notethattheinteriorandexterioranglesatavertexaddupto 180◦. Thecarefulreadermightwellaskwhatwouldhappenifwehadextendedtheotherpossibleedgeateachvertex, resultingindifferentexteriorangles. Thereareindeedtwopossibleexterioranglesateachvertex, buttheyareverticalangles.See Figure 2.2.2 (ii), where the twopossible exterior angles at A aredenoted δ and ϵ. ByProposition 1.2.2 (2)itfollowsthatthetwochoicesofexteriorangleareequal, andhence, itdoesnotmatterwhichexterioranglewechoose.

A

B

C

α

δA

B

C

δ

ε

(i) (ii)

Figure2.2.2

Wearenowreadytostateaveryimportantresultabouttheanglesinatriangle.

Proposition 2.2.1.

1. Thesumoftheinterioranglesofatriangleis 180◦.

2. Thesumoftheexterioranglesofatriangleis 360◦.

Demonstration. Supposewehaveatriangle△ABC. AsshowninFigure 2.2.3 (i), let α, β andγ betheinterioranglesofthetriangle, andlet x, y and z betheexterioranglesofthetriangle.

(1). AsshowninFigure 2.2.3 (ii), letn bethelinecontainingtheverticesB andC. ByPlayfair’sAxiom (Proposition 1.1.1), thereisalinethroughvertexA thatisparallelton. Callthisnewlinem. Labelangles δ and ϵ asshowninFigure 2.2.3 (ii). Weseethatβ and δ areinterioralternatingangles, andthat γ and ϵ areinterioralternatingangles. ByProposition 1.2.3 wededucethatβ = δ and γ = ϵ. Because δ, α and ϵ makeupastraightline, wehave δ+ α+ ϵ = 180◦. Itfollowsthat β+ α+ γ = 180◦, whichiswhatwewantedtoprove.

30 2. Polygons

A

BC

α

β γ

x

yz

A

B C

α

β γn

mδ ε

(i) (ii)

Figure2.2.3

(2). Weknowthat α + x = 180◦, that β + y = 180◦, andthat γ + z = 180◦. Hencex = 180◦ − α, and y = 180◦ − β, and z = 180◦ − γ. UsingPart (1)ofthisproposition, wethenseethat x+ y+ z = (180◦ −α) + (180◦ −β) + (180◦ − γ) = 540◦ − (α+β+ γ) =540◦ − 180◦ = 360◦.

TheproofoftheabovepropositionusesProposition 1.2.3 andProposition 1.1.1, bothofwhichrelyuponEuclid’sFifthPostulate. Indeed, Proposition 2.2.1 doesnotholdinsphericalorhyper-bolicgeometry, wheretheFifthPostulateisreplacedwithotheraxioms(seeSection 1.1 forabriefmentionofthesealternativegeometries).

Exercise 2.2.1. [UsedinSection 1.2] UseProposition 2.2.1 (1)todemonstrateProposi-tion 1.2.4.

Exercise 2.2.2. WeknowfromProposition 2.2.1 thattheanglesinatrianglearerelatedtoeachother; inparticular, wecouldnottakethreearbitraryangles, andexpecttheretobeatrianglewiththosethreeangles. Thisexerciseconcernsrelationshipsbetweenthelengthsoftheedgesofatriangle.

(1) Isthereatrianglewithedgesoflengths 2, 3 and 4? Explainyouranswer.

(2) Isthereatrianglewithedgesoflengths 2, 3 and 6? Explainyouranswer.

(3) Whatcanyousayabouttherelationshipbetweenthelengthsoftheedgesofatri-angle. Inparticular, trytocomeupwithcriteriaonthreenumbers a, b and c thatwouldguaranteethatthereexistsatrianglewithedgesoflength a, b and c. Explainyouranswer.

2.2Triangles 31

Whatdoesitmeanfortwotrianglestobe“thesame”? Clearly, if twotriangleshaveedgesofdifferentlengths, oranglesofdifferentmeasures, thenthetwotrianglesarenotthesame.Whatabouttwotriangleswithedgesthathavethesamelengths, andanglesthathavethesamemeasures? Forexample, inFigure 2.2.4 weseetwotrianglesthathavethesamelengthsofedgesandthesamemeasuresofangles(inthiscase 30◦, 60◦ and 90◦). Thesetwotrianglesarenottheexactsame, becausetheyarelocatedindifferentplaces, buttheyare“essentiallythesame.”SimilarlytowhatwesaidaboutanglesinSection 1.2, wesaythattwotrianglesare congruent,if, intuitively, onetrianglecouldbe“pickedupandplacedpreciselyontopoftheother.” Asbefore, wewillnotneedamore formaldefinitionofcongruencehere, andwill stick to theintuitivenotion; theconceptofisometry, whichisdiscussedindetailinChapter 4, providesonerigorouswayofdefiningcongruence, thoughwewillnothavethespacetoprovidethedetails.

A

A’B

CC’

B’

Figure2.2.4

Supposethattriangles△ABC and△A ′B ′C ′ arecongruent, wherevertex A correspondstovertex A ′, wherevertex B correspondstovertex B ′, andwherevertex C correspondstovertexC ′. SeeFigure 2.2.4. Then |AB| = |A ′ B ′|, and |AC| = |A ′ C ′|, and |BC| = |B ′ C ′|, and∡A = ∡A ′, and ∡B = ∡B ′, and ∡C = ∡C ′. Theconverseisalsotrue. Thatis, supposewearegiventwotriangles △ABC and △A ′B ′C ′, andweknowthat |AB| = |A ′ B ′|, that|AC| = |A ′ C ′|, that |BC| = |B ′ C ′|, that ∡A = ∡A ′, that ∡B = ∡B ′, andthat ∡C = ∡C ′.Itwillthenbethecasethatthetwotrianglesarecongruent.Actually, wecandobetter thantheabovestatement. Wejustassertedthat ifweknowsix

thingstobetrueabouttwotriangles(namelytheequalityofthelengthsofthethreeedges, andtheequalityofthemeasuresofthethreeangles), thenthetriangleswillbecongruent. Itturnsout,thoughthisisbynomeansobvious, thatcertainpartialknowledgeaboutthesesixequalitiessufficestoguaranteethattwotrianglesarecongruent. Thefollowingpropositionisonesuchresult.

Proposition 2.2.2 (Side-Side-SideTheorem). Supposethat△ABC and△A ′B ′C ′ aretriangles.Supposethat |AB| = |A ′ B ′|, that |AC| = |A ′ C ′|, andthat |BC| = |B ′C ′|. Then ∡A =∡A ′, and ∡B = ∡B ′, and ∡C = ∡C ′.

Wewillnotdemonstratetheaboveproposition, becauseitwouldtakeustoofarafield. TheSide-Side-SideTheoremsaysthatiftwotriangleshaveedgesthathavethesamelengths, then

32 2. Polygons

thetwotrianglesarecongruent. Anotherwayofthinkingaboutthistheoremisthatitsaysthattrianglesarerigid, inthefollowingsense. Supposeyoutakethreesticks, andjointhemtogetherintoatriangle. Evenifyouweretojointhestickswithhinges, thetrianglecouldnotbede-formed. Iftriangleswerenotrigid, thenknowingthelengthsoftheedgesofatrianglewouldnotuniquelydeterminetheangles, anditwouldthenbepossibletohavetwotriangleswhoseedgeshavethesamelengths, butwithdifferentangles, andthatwouldcontradicttheSide-Side-SideTheorem. Bycontrast, ifyouweretotakefoursticks, andjointhemtogetherwithhingesintoaquadrilateral, thenthefigurecouldbedeformed. SeeFigure 2.2.5. Thefactthattrianglesarerigid, butotherpolygonsarenot, issomethingthatiswellknowninreallife. Asaresult, weoftenseetriangularformsusedinconstructionoftressesandthelike.

Figure2.2.5

TheSide-Side-SideTheoremisnottheonlyresultthatguaranteesthattrianglesarecongruent.Twootherequallyusefulcongruencetheoremsarethefollowing.

Proposition 2.2.3 (Side-Angle-SideTheorem). Supposethat △ABC and △A ′B ′C ′ aretrian-gles. Suppose that |AB| = |A ′ B ′|, that ∡A = ∡A ′, and that |AC| = |A ′ C ′|. Then∡B = ∡B ′, and |BC| = |B ′ C ′|, and ∡C = ∡C ′.

Proposition 2.2.4 (Angle-Side-AngleTheorem). Suppose that △ABC and △A ′B ′C ′ are tri-angles. Suppose that ∡A = ∡A ′, that |AB| = |A ′ B ′|, and that ∡B = ∡B ′. Then|BC| = |B ′ C ′|, and ∡C = ∡C ′, and |AC| = |A ′ C ′|.

Wewillnotdemonstratetheabovetwocongruencetheorems.

Exercise 2.2.3. Findexamplestoshowthatthereisno“Angle-Side-SideTheorem.” Thatis,findtwotriangles△ABC and△A ′B ′C ′ suchthat ∡A = ∡A ′, that |AB| = |A ′ B ′|, andthat |BC| = |B ′ C ′|, andyetthetwotrianglesarenotcongruent. Exactmeasurementsofsuchtrianglesarenotneeded; asketchofthetriangles, togetherwithadescriptionofwhatyoumean, wouldsuffice.

Exercise 2.2.4. Istherean“Angle-Angle-SideTheorem”? Eitherdemonstratewhysuchatheoremistrue, orgiveanexampletoshowthatitisnottrue.

2.2Triangles 33

Congruenceoftriangles, andtheabovethreecongruencetheoremsinparticular, areextremelyusefulinprovingmanyresultsingeometry. Westartwiththefollowingverysimpleresults, butwewill usecongruence inproofsofother, more substantial, results lateron. The followingresults, whichmightseemsoobviousthattheydonotneedproof, infactneedproofjustasanyotherfactingeometrythatisnottakenasanaxiom.Ourfirstresultinvolvesparallelograms. Recall, statedinSection 2.1, thataparallelogramisa

quadrilateralinwhichbothpairsofoppositeedgesareparallel.

Proposition 2.2.5.

1. Oppositeedgesinaparallelogramhaveequallengths.

2. Oppositeanglesinaparallelogramareequal.

Demonstration. Supposethatwehaveparallelogram ABCD, asseeninFigure 2.2.6 (i). By

definition, weknowthatthelines←→AB and

←→DC areparallel, andthatthelines

←→AD and

←→BC are

parallel. Drawthelinesegment AC. SeeFigure 2.2.6 (ii). Wethenhavethetriangles △ABC

and △CDA. Letangles α, β, γ and δ beasshowninFigure 2.2.6 (ii). Observethat α and

γ are interioralternatingangles (because lines←→AB and

←→DC areparallel), and thereforeby

Proposition 1.2.3 weknowthat α = γ. Similarly, wededucethat β = δ, usingthefactthat←→AD and

←→BC areparallel.

(i) (ii)

A

D C

B A

D C

B

α

βγ

δ

Figure2.2.6

Wenowclaimthat △ABC and △CDA arecongruent, wherevertex A in △ABC corre-spondstovertex C in△CDA, wherevertex B in△ABC correspondstovertexD in△CDA,andwherevertex C in △ABC correspondstovertex A in △CDA. ThatthesetwotrianglesarecongruentfollowsfromtheAngle-Side-AngleTheorem(Proposition 2.2.4), andthefactthatα = γ, that β = δ, andthat AC isthesameinbothtriangles. Becausethetwotrianglesarecongruent, wededucethatcorrespondingedgeshavethesamelengths. Thatis, weconcludethat |AB| = |DC|, andthat |AD| = |BC|m, whichisPart (1)oftheproposition. Part (2)alsofollowsfromthecongruenceofthetwotriangles; detailsarelefttothereader.

34 2. Polygons

Exercise 2.2.5. Supposethatinaquadrilateral, bothpairsofoppositeedgeshaveequallengths. Showthatthequadrilateralisaparallelogram. (Becarefulwithnotconfusingathe-oremanditsconverse—thisfactcannotbeprovedbysimplyquotingProposition 2.2.5 (1).)

Exercise 2.2.6. Supposethatinaquadrilateral, apairofoppositeedgesareparallelandhaveequallengths. Showthatthequadrilateralisaparallelogram.

Exercise 2.2.7. Showthatthetwodiagonalsinaparallelogrambisecteachother.

AsaconsequenceofProposition 2.2.5 (1), wecandeducethefollowingresultaboutparallellines. RecallfromSection 1.2 thattwolinesaredefinedtobeparalleliftheydonotintersect.Ourintuitivepictureofparallellines, however, involvesmorethanjustthattwolinesdonotintersect, butalsothatthey“keepconstantdistancefromeachother.” Wecannowshowinthefollowingpropositionthatthisintuitivenotionisinfactcorrectforparallellinesintheplane.WewerenotabletogivethispropositioninSection 1.2, wherewefirstdiscussedparallellines,becauseweneedcongruenttrianglestoproveit. Whenyoureadthefollowingproposition, ithelpstoconsiderFigure 2.2.7.

m

n

A

C

B

D

Figure2.2.7

Proposition 2.2.6. Supposem and n areparallellines. LetA and B bepointsonm. Drawlinesthrough A and B respectivelythatareperpendicularto n; let C and D bethepointswheretheseperpendicularlinesintersect n. Then |AC| = |BD|.

Demonstration. First, observethatbecausethelines←→AC and

←→BD arebothperpendicularto

theline n, thentheyareparalleltoeachother(thisisProposition 1.2.5). Giventhat m and n

2.2Triangles 35

areparallel, itfollowsthatthequadrilateral ABCD isaparallelogram. WenowapplyProposi-tion 2.2.5 (1)tothisparallelogram, todeducethat |AC| = |BD|.

Ournextsimpleresultinvolvesisoscelestriangles.

Proposition 2.2.7 (PonsAsinorum). Supposethat △ABC isatriangle. If |AB| = |AC|, then∡B = ∡C.

Demonstration. Wearegiventhetriangle △ABC. Wenowclaimthat △ABC iscongruentwithitself, wherevertex B in △ABC correspondstovertex C in △ABC, wherevertex C in△ABC correspondstovertex B in △ABC, andwherevertex A in △ABC correspondstovertex A in △ABC. That these two trianglesarecongruent follows fromtheSide-Side-SideTheorem(Proposition 2.2.2), andthefactthat |AB| = |AC|, andthat |BC| = |BC|. Itfollowsthat △ABC hasequalangleswithitself, wheretheangleat B correspondstotheangleat C,andvice-versa(theangleatA correspondstoitself, thoughthatisnotofanyuse). Wethereforededucethattheanglesopposite AB and AC areequal, whichiswhatwearesupposedtoshow. (ItmayseemstrangethatweareapplyingtheSide-Side-SideTheoremtoatriangleanditself, ratherthantwodistincttriangles, butnothinginthestatementofthistheoremsaysthatthetwotrianglesunderconsiderationhavetobedistinct, thoughtheymostoftenare. However,eventhoughwearecomparingatrianglewithitself, wereallyareprovingsomething, becausewearehavingdifferentverticescorrespondwitheachotherinthecongruence.)

Thename“PonsAsinorum”means“Ass’Bridge” inLatin. Therearevariousexplanationsforthisname, somereferringtotheappearanceofthetriangleunderdiscussion, otherstothestateofmindofthosetryingtounderstandtheresult.AnimmediateconsequenceofPonsAsinorumisthefirstpartofthefollowingproposition,

whichagainisveryfamiliar; thesecondpartofthepropositionfollowsfromthefirstpart, to-getherwithProposition 2.2.1 (1).

Proposition 2.2.8.

1. Allthreeanglesinanequilateraltriangleareequal.

2. Eachangleinanequilateraltriangleis 60◦.

Exercise 2.2.8. Supposewearegivenatriangle △ABC. Showthatif ∡B = ∡C, then|AB| = |AC| (sothatthetriangleisisosceles).

Exercise 2.2.9. [Used inSection 1.3] Usecongruent triangles todemonstrateProposi-tion 1.3.1.

36 2. Polygons

Exercise 2.2.10. Supposewehaveacircle, andsupposethatA, B and C arepointsonthecirclesuchthatthelinesegmentAB isadiameterofthecircle. Formthetriangle△ABC.SeeFigure 2.2.8. Showthattheangleat C is 90◦.

A B

C

Figure2.2.8

Itisimportantnottogetoverlyconfidentaboutcongruencetheorems. Justknowingthattwotriangleshavethreethings(edgesorangles)equaldoesnotalwaysguaranteethatthetrianglesarecongruent. Forexample, simplyknowingthattwotriangleshavethesameanglesdefinitelydoesnotguaranteethatthetrianglesarecongruent. InFigure 2.2.9 weseetwotriangles, labeled△ABC and △A ′B ′C ′, thathavethesameangles, butdifferentlengthsoftheiredges.

A A’

B

B’

C

C’

Figure2.2.9

Wejustobserved thatknowingonly theangles ina triangledoesnotdeterminewhat thelengthsofitsedgesare. However, eventhoughtwotriangleswiththesameanglesmighthavedifferentsizes, asinFigure 2.2.9, theyhavethesame“shape.” Thefollowingpropositionmakesthisnotionofthesameshapemoreprecise, bylookingatratiosoflengthsofedges. First, wemakethefollowingdefinition. Twotriangles△ABC and△A ′B ′C ′ arecalled similar if ∡A =∡A ′, , and ∡B = ∡B ′, and ∡C = ∡C ′.Thefollowingproposition, whichmakesprecisethenotionthatsimilar triangles“havethe

sameshape,” iswhatmakestheconceptofsimilarityoftrianglessouseful.

2.2Triangles 37

Proposition 2.2.9. Supposethattriangles △ABC and △A ′B ′C ′ aresimilar. Then

|AB|

|A ′ B ′|=

|AC|

|A ′ C ′|=

|BC|

|B ′ C ′|;

equivalently, wehave

|AB|

|AC|=

|A ′ B ′|

|A ′ C ′|,

|AB|

|BC|=

|A ′ B ′|

|B ′ C ′|,

|AC|

|BC|=

|A ′C ′|

|B ′ C ′|.

ThedemonstrationofProposition 2.2.9 willbedelayeduntilSection 2.4, bywhichpointwewillhavediscussedtheareaoftriangles, whichweuseinthedemonstration.Asanexampleof theuseofProposition 2.2.9, supposethatwehavetwosimilar triangles

△ABC and △A ′B ′C ′. Suppose further thatwe are given that the edges of △ABC havelengths |AB| = 5, and |AC| = 6, and |BC| = 8; in △A ′B ′C ′ wearegivenonlythelength|A ′ C ′| = 20. Canwefindthelengthsoftheothertwoedgesof △A ′B ′C ′? Wecan, usingProposition 2.2.9. Bythatproposition, weknowthat

|AB|

|AC|=

|A ′ B ′|

|A ′ C ′|.

Hence, wededucethat5

6=

|A ′ B ′|

20.

Solvingthisequation, weseethat |A ′ B ′| =50

3. Thereadercanusethesametypeofcalculation

toseethat |B ′ C ′| = 15.

Exercise 2.2.11. A treecastsashadowthatis 20 ft.long. Atthesametimeofday, a 3 ft.stickcastsa 5 ft.shadow. Howtallisthetree?

Havingmentionedsimilartriangles, wecannotavoidmentioningoneofthemostimportantusesofsimilartriangles: trigonometry. Noteveryonewhousestrigonometryrecognizesthefun-damental roleplayedbysimilar triangles in trigonometry, andit is that role thatwewant todiscuss. Thestudyoftrigonometrytreatsthesixstandardtrigonometricfunctions, namelysine,cosine, tangent, secant, cosecant andcotangent.Letuslookatthesinefunction(theotherfivefunctionswouldworkcompletelysimilarly).

Todefinethesinefunction, weneedthefollowingterminology. Recallthatarighttriangle isatriangleinwhichoneoftheanglesisarightangle. Inarighttriangle, thetwoedgesthatformtherightanglearecalledthe sides ofthetriangle, andtheedgethatisoppositetherightangleiscalledthe hypotenuse ofthetriangle. (Thedistinctionbetweensidesandhypotenuseholds

38 2. Polygons

onlyinrighttriangles, notinothertriangles.) Inintroductorytreatmentsoftrigonometry, itistypicaltodefinethesinefunctionasfollows. Let α beananglebetween 0◦ and 90◦. Wewanttocomputethesineof α, whichisgoingtobeanumberdenoted sinα. Wecompute sinα byplacing α inarighttriangle, asinFigure 2.2.10, andthenletting sinα betheratioofthelengthofthesideopposite α tothelengthofthehypotenuse. Thatis, welet

sinα =|opposite|

|hypotenuse|.

α

opposite

hypotenuse

Figure2.2.10

Forexample, supposewewantedtocompute sin 45◦. Wecanplace 45◦ inanisoscelesrighttriangle. Onesuchrighttrianglehassidesoflength 1. UsingthePythagoreanTheorem(assumingthereaderisfamiliarwiththisresult, anditsstatementcanbefoundinSection 2.5 forthosewhoarenot), wecomputethatthehypotenuseofthetrianglehaslength

√2. SeeFigure 2.2.11.

Itfollowsthat sin 45◦ = 1√2.

1

12

45°

Figure2.2.11

Sofarsogood. Thereis, however, onepotentiallytroublingaspecttotheaboveapproachtodefiningsine. Givenanangle, wedefinedsineoftheanglebyplacingtheangleinarighttriangle,andtakingtheratioofthelengthofthesideoppositetheangletothelengthofthehypotenuse.Thereare, ofcourse, differentpossibletrianglesthatcouldbeused. Whatwouldhappeniftheratioofthelengthofthesideoppositetheangletothelengthofthehypotenuseinonerighttrianglecontainingtheanglewerenotequaltothecorrespondingratioinanotherrighttrianglecontainingtheangle? Ifthatweretohappen, thedefinitionofsineoftheanglewouldbeinvalid,becauseitoughtonlytodependupontheangleitself, andnothingelse(suchasachoiceof

2.3GeneralPolygons 39

righttriangle). Well, itturnsoutthatthereisnoprobleminthechoiceoftriangle, andhereisthereasonwhy. Supposewehadtworighttrianglescontaininganangleα. Bothtrianglesalsohavea90◦ angle. Giventhatthesumoftheanglesinanytriangleis 180◦ (byProposition 2.2.1 (1)), thenweknowthatthethirdangleineachofthetwotrianglesis 90◦−α. Therefore, thetwotriangleshavethesameangles; hencetheyaresimilartriangles. ItnowfollowsfromProposition 2.2.9thattheratioofthelengthofthesideopposite α tothelengthofthehypotenuseisthesameinbothtriangles, andthatmeansthatthereisnoprobleminourdefinitionofsine. Thesametypeofargumentholdsfortheothertrigonometricfunctions.Wenotethatalthoughthetrigonometricfunctionsaredefinedintermsoftriangles, theyhave

importantusesinmanyareabeyondthestudyoftriangles. Forexample, thetrigonometricfunc-tionsareusedtodescribeoscillatorymotion(forexample, springsandpendulums), andwavephenomena(forexample, soundandlight).

2.3 GeneralPolygons

HavingdiscussedtrianglesinSection 2.2, wenowturntopolygonswithmorethanthreeedges.Somepropertiesoftriangleshaveanalogsforallpolygons, andothersdonot. AswementionedinSection 2.2, onepropertyoftrianglesthatdoesnotholdforallpolygonsisrigidity. Anotherway to state the same fact is to say that theanalogof theSide-Side-SideTheorem (Proposi-tion 2.2.2)doesnotholdforpolygonsotherthantriangles. Forexample, asquarewithedgesoflength 1 andarhombuswithedgesoflength 1 havealledgesofequallength, andyettheydonothaveequalangles. SeeFigure 2.2.5.Ontheotherhand, therearefeaturesofpolygonswithmorethanthreesidesthatdonotappear

inthecaseoftriangles. Wenowturntoonesuchissue. ConsiderthepolygonsinFigure 2.3.1.There isa fundamentaldifferencebetween them: thepolygon inPart (i)has, intuitively, “noindentations,” whereasthepolygoninPart (ii)doeshaveanindentation. Itisnotveryconvenienttechnicallytotrytodefinethenotionof indentationsdirectly, sowecapturetheideaofnothaving indentationsas follows. A polygoniscalled convex ifany twopoints in thepolygonarejoinedbyalinesegmententirelycontainedinthepolygon. WeseeinFigure 2.3.2 howtwopointsinthepolygoninFigure 2.3.1 (ii)arejoinedbyalinesegmentthatisnotentirelycontainedinthepolygon, andthereforethepolygonisnotconvex; thefactthatsomeotherpairsofpointsinthepolygonarejoinedbylinesegmentsentirelycontainedinthepolygondoesnotmakethepolygonsatisfythedefinitionofconvexity. Bycontrast, thepolygoninFigure 2.3.1 (i)isconvex. Wenotethatalltrianglesareconvex, sothedistinctionbetweenconvexvs.non-convexdidnotariseinourdiscussionoftriangles, butitwillbeimportantinourdiscussionofpolygonsandpolyhedra.

Aswasthecasefortriangles, atanyvertexofapolygon, thetwoedgesofthepolygonthatcontainthevertexformananglethatisinsidethepolygon, calledthe interiorangle atthevertex.InFigure 2.3.3 weseeapolygon, withitsinterioranglesdenoted α1, α2, α3, α4 and α5. Wewillshortlydefineexterioranglesforpolygons.

40 2. Polygons

(i) (ii)

convex not convex

Figure2.3.1

Figure2.3.2

α1

α2α3

α4α5

Figure2.3.3

Oneofthethingsinterioranglesareusefulforisthattheygiveaneasywayofdeterminingwhetherornotapolygonisconvex. IfyoulookattheconvexpolygonpicturedinFigure 2.3.1 (i),youwillnoticethateachinterioranglesislessthan 180◦; bycontrast, inthenon-convexpolygonpicturedinFigure 2.3.1 (ii), oneoftheinterioranglesisgreaterthan 180◦. Moregenerally, wehavethefollowingresult, thedemonstrationofwhichweomit.

Proposition 2.3.1. A polygonisconvexifandonlyifallitsinterioranglesarelessthanorequalto 180◦.

Wenow turn to thequestionof angle sums in polygons. It ismuch simpler to dealwiththisquestion forconvexpolygons, sowestartwith thatcase. Weknowthat thesumof theinterioranglesofatriangleis 180◦, andthesumoftheexterioranglesofatriangleis 360◦ (seeProposition 2.2.1). Isthereananalogoftheseresultsforgeneralconvexpolygons? Toanswerthis


question, wefirstneedtodefineexterioranglesforconvexpolygons. Itturnsoutthatthesamedefinitionthatworkedfortrianglesworksforconvexpolygonsingeneral. Morespecifically, ateachvertexofaconvexpolygon, wecanextendbeyondthevertexoneof theedgesof thepolygoncontainingthevertex, andformthesupplementaryangle totheinteriorangle, whichwewillcallthe exteriorangle atthevertex. InFigure 2.3.4 weseeaconvexpolygon, withoneofitsverticeslabeled A, withtheinteriorangleat A denoted α, andtheexteriorangleat Adenoted β. Aswasthecasefortriangles, theinteriorandexterioranglesatavertexaddupto180◦. (Alsojustasfortriangles, itwouldmakenodifferencehadweextendedtheotherpossibleedgeateachvertex.)

αβ

A

Figure2.3.4


TrytofigureoutforyourselfhowtogeneralizeProposition 2.2.1 foranarbitraryconvexn-gon. Thatis, givenaconvex n-gon, whatisthesumofitsinteriorangles, andwhatisthesumofitsexteriorangles? (Theanswersmightinvolvethenumber n.)

ToseehowwemightgeneralizeProposition 2.2.1 tootherconvexpolygons, letuslook, forexample, attheanglesumsfortheoctagonshowninFigure 2.3.5. Itiseasytoseethateachinteriorangleis 135◦, andeachexteriorangleis 45◦. Thesumoftheinterioranglesistherefore8 · 135◦ = 1080◦, andthesumoftheexterioranglesis 8 · 45◦ = 360◦. Interestingly, thesumoftheexterioranglesfortheoctagonisthesameasforatriangle, butthesumoftheinterioranglesfortheoctagonismorethanforatriangle. Asseeninthefollowingproposition, itturnsoutthatthesumoftheinterioranglesofaconvexpolygondependsuponthenumberofedgesofthepolygon, whereasthesumoftheexterioranglesdoesnotdependuponthenumberofedges. Thedemonstrationofthepropositionclarifieswhythisresultholds.

Proposition 2.3.2.

1. Thesumoftheinterioranglesofaconvex n-gonis (n− 2)180◦.

2. Thesumoftheexterioranglesofaconvex n-gonis 360◦.

Demonstration. Suppose that the convex n-gon P has interior angles α1, α2, α3, . . ., αn,andexteriorangles β1, β2, β3, . . ., βn. Forexample, forthepolygonoriginallyshowninFig-ure 2.3.3 (i), weseetheexteriorangleslabeledinFigure 2.3.6 (i).

42 2. Polygons

Figure2.3.5

α1

β1

α2

β2α3 β3

α4

β4α5β5

(i) (ii)

Figure2.3.6

(1). Itispossibletobreakupthepolygon P into n − 2 triangles, wheretheverticesofthetrianglesareallverticesoftheoriginalpolygon. SeeFigure 2.3.6 (ii)foronewayofdoingthis.(Thereismorethanonewaytobreakupthepolygonintotriangles, butitwillnotmatterwhichwayischoosen.) UsingProposition 2.2.1 (1)weknowthatthesumoftheanglesineachofthesetrianglesis 180◦. Becausethereare n− 2 triangles, thesumofalltheanglesinallthetrianglesis (n − 2)180◦. However, asseeninFigure 2.3.6 (ii), puttingalltheanglesinallthetrianglestogetheristhesameasputtingalltheanglesintheoriginalpolygontogether. Hencethesumofalltheanglesintheoriginalpolygonis (n− 2)180◦.

(2). Weknowthat αi + βi = 180◦ foreach i = 1, 2, 3, . . . , n. Therefore βi = 180◦ − αi

foreach i. WethenusePart (1)ofthispropositiontoseethat

β1 + β2 + β3 + · · ·+ βn = (180◦ − α1) + (180◦ − α2) + (180◦ − α3) + · · ·+ (180◦ − αn)

= (180◦ + 180◦ + 180◦ + · · ·+ 180◦︸︷︷︸n times

)− (α1 + α2 + α3 + · · ·+ αn)

= n180◦ − (n− 2)180◦ = 2 · 180◦ = 360◦.


Wenowturn toanglesumsinnon-convexpolygons. Consider thepolygonshowninFig-ure 2.3.7 (i). Theinterioranglesatvertices A, B, C, D, E and F are 135◦ each; theinterioranglesatvertices G and I are 90◦ each; andtheinteriorangleatvertex H is 270◦. Thesumoftheinterioranglesistherefore 6 · 135◦ + 2 · 90◦ + 270◦ = 1260◦. Noticethatthepoly-goninthefigurehas 9 edges. Ifweuse n = 9 intheformulagiveninProposition 2.3.2 (1),wewouldobtain (9 − 2)180◦ = 1260◦, whichispreciselythesumoftheinterioranglesinthepolygoninFigure 2.3.7 (i). Now, theformulainProposition 2.3.2 (1)wasonlyprovedforconvexpolygons, butperhapsitholdsfornon-convexpolygonsaswell. Theproofweusedforconvexpolygonsdoesnotworkfornon-convexpolygons. Forexample, thepolygonshowninFigure 2.3.7 (ii)cannotbedividedupintotrianglesinthesamewaythatwedidintheproofofProposition 2.3.2 (1). Itturnsoutthatanotherproofcanbeused. Thekeyisexteriorangles.

(i) (ii)

A

H I

BC

D

E

F G

Figure2.3.7

Howdowedealwithexterioranglesofnon-convexpolygons? Simplydefiningthemaswedidforconvexpolygonsdoesnotquitework. Recallthatwedefinedtheexteriorangleatthevertexofaconvexpolygonbyextendingbeyondthevertexoneoftheedgesofthepolygoncontainingthevertex, and forming the supplementaryangle to the interiorangle, whichwecalled theexteriorangleatthevertex. SeeFigure 2.3.4. However, ifwelookatthevertexlabeled H inFigure 2.3.7 (i), weseethatextendingeitheroftheedgescontainingthevertexwillgointotheinteriorofthepolygon, asituationthatdoesnotseemquiteright(atleastuponfirstglance).Onewayaroundthisproblemistorecallthattheinteriorandexterioranglesatavertexofaconvexpolygoncanberelatednotonlygeometrically, asinthedefinitionofexteriorangles,butintermsoftheirmeasure. Moreprecisely, themeasuresoftheinteriorandexterioranglesatavertexofaconvexpolygonaddupto 180◦. Therefore, themeasureofanexteriorangleisjust180◦ minusthemeasureoftheinteriorangle. (Westresstheword“measure”heretoemphasizethatitisdistinctfromtheactualangle, but, havingemphasizedithere, wewillreverttostandardterminologyandnotmentionitfurther.) Inthenon-convexcase, wecansimplytakethisrelation

44 2. Polygons

betweeninteriorandexterioranglesasadefinition. Thatis, we define the exteriorangle atthevertexofanypolygon(convexornot)as 180◦ minustheinteriorangleatthevertex.LetuslookatthepolygoninFigure 2.3.7 (i). TheexterioranglesatverticesA, B, C,D, E and

F are 180◦ − 135◦ = 45◦ each; theexterioranglesatvertices G and I are 180◦ − 90◦ = 90◦

each; andtheexteriorangleatvertex H is 180◦ − 270◦ = −90◦. Itmayseemstrangetoobtainanegativeexteriorangleatvertex H, butlookhownicelyitworksout. ThesumoftheexterioranglesofthepolygoninFigure 2.3.7 (i)is 6 · 45◦ +2 · 90◦ − 90◦ = 360◦. Thissumispreciselythesumofexterioranglesforconvexpolygons, asgiveninProposition 2.3.2 (2). Itturnsout, aswewillseeinProposition 2.3.3 below, thatthesameformulasforsumsofinteriorandexterioranglesthatworkforconvexpolygonsalsoworkfornon-convexpolygons—aslongaswedefineexterioranglesfornon-convexpolygonsaswedid, andacceptthefactthatexterioranglescouldbenegativeaswellaspositive.BeforeweturntoProposition 2.3.3, letuslookmorecloselyattheissueofnegativevs.positive

exteriorangles. Thekeyistorecallthat, asdiscussedinSection 1.2, wedistinguishedbetweenclockwisevs.counterclockwiseangles, withtheformerhavingpositivedegreemeasure, andthelatterhavingnegativedegreemeasure. LookatthepolygonshowninFigure 2.3.8 (i). Thinkoftheedgesasgoingaroundthepolygonintheclockwisedirection, asindicatedbythearrows.Wewillonceagainobtainexterioranglesbyextendingedges, aswedidforconvexpolygons,but this timewewillalwaysextend theedge thatcomesbeforeavertex (where“before” iswithrespecttogoingaroundthepolygonintheclockwisedirection). WestartbylookingatthevertexlabeledA. Theinteriorangleatthisvertexislessthan 180◦. InFigure 2.3.8 (ii)weextendtheedgecontaining A thatcomesbefore A. Wecanthenlookattheanglefromtheextendededgetotheedgethatcomesafter A. Thisangleisclockwise, andisthereforeapositiveangle.Thisangleispreciselyequalto 180◦ minustheinteriorangleat A. Next, welookatthevertexlabeled B. Theinteriorangleatthisvertexisgreaterthan 180◦. InFigure 2.3.8 (ii)weextendtheedgecontaining B thatcomesbefore B. Wecanthenlookattheanglefromthisextendededgetotheedgethatcomesafter B. Thisangleiscounterclockwise, andisthereforeanegativeangle. Thisangleisagainpreciselyequalto 180◦ minustheinteriorangleat B. Inbothcases,weseethatitispossibletogiveageometricmeaningtotheexteriorangle, aslongaswetakeintoaccountthedifferencebetweenclockwisevs.counterclockwiseangles.

Wearenowreadyforthefollowingproposition, whichgeneralizesProposition 2.3.2.

Proposition 2.3.3.

1. Thesumoftheinterioranglesofan n-gonis (n− 2)180◦.

2. Thesumoftheexterioranglesofan n-gonis 360◦.

Demonstration. Supposewehavean n-gon P asshowninFigure 2.3.9 (i), withvertices A1,A2, A3, . . ., An, andwithinteriorangles α1, α2, α3, . . ., αn. Welet β1, β2, β3, . . ., βn betheexteriorangles. WewillfirstshowthatPart (2)holds, andthenshowthatPart (1)holdsbyusingPart (2).


(i) (ii)

BB

AA

Figure2.3.8

(i) (ii)

A7

A4

A5

A6

A3A2

A1

A7

A4

A5

A6

A3

A2

A1

α1

α2

α3

α4

α5

α6

α7

β1

β2

β3

β4β5

β6

β7

Figure2.3.9

(2). AsshowninFigure 2.3.9 (ii), weextendeveryedgeof P intheclockwisedirection, andlabeltheexteriorangles. Observethattheanglefromtheextendededgeat A1 totheextendededgeat A2 isequaltotheexteriorangleat A1, whichisdenoted β1. Similarly, theanglefromtheextendededgeatA2 totheextendededgeatA3 isequaltotheexteriorangleatA2, whichisdenoted β2. Hence, theanglefromtheextendededgeatA1 totheextendededgeatA3 isequalto β1 + β2. Bythesameargument, theanglefromtheextendededgeat A1 totheextendededgeat A4 isequalto β1 +β2 +β3. Ifwekeepgoingallthewayaroundthepolygon, weseethattheanglefromtheextendededgeatA1 toitself, goingallthewayaround 360◦, isequaltoβ1 +β2 +β3 + · · ·+βn. Hence β1 +β2 +β3 + · · ·+βn = 360◦. (Notethatthisargumentworksonlybecausetheedgesofthepolygondonotintersecteachother; iftheywereallowedtointersecteachother, theanglefrom A1 toitselfaftergoingallthewayaroundthepolygonmightbetwice 360◦, orthreetimes 360◦, etc.)

(1). Thispartof thedemonstration isabackwardsversionof thedemonstrationofPropo-sition 2.3.2 (2). Weknow, bydefinition, that βi = 180◦ − αi foreach i = 1, 2, 3, . . . , n.

46 2. Polygons

Therefore αi = 180◦ − βi forall i. WethenusePart (2)ofthispropositiontoseethat

α1 + α2 + α3 + · · ·+ αn = (180◦ − β1) + (180◦ − β2) + (180◦ − β3) + · · ·+ (180◦ − βn)

= (180◦ + 180◦ + 180◦ + · · ·+ 180◦︸︷︷︸n times

)− (β1 + β2 + β3 + · · ·+ βn)

= n180◦ − 360◦ = n180◦ − 2 · 180◦ = (n− 2)180◦.

Exercise 2.3.1. Showthataquadrilateralhasatmostoneinterioranglethatisgreaterthan180◦.

Exercise 2.3.2. Recall, statedinSection 2.1, thatarectangle isdefinedtobeaquadrilateralinwhichallfouranglesareequal.

(1) Showthatallfouranglesinarectangleare 90◦.

(2) Showthateveryrectangleisaparallelogram.

(3) Showthatoppositeedgesinarectanglehaveequallengths.

Exercise 2.3.3. Supposethataquadrilateralhastwopairsofequaladjacentangles. Showthatthequadrilateralisatrapezoid.

Exercise 2.3.4. Showthatanyparallelogramisconvex.


Exercise 2.3.5. [UsedinSection 4.5] Supposethataquadrilateralhastwooppositeedgesthathaveequal lengths, and that these twoedgesarebothperpendicular tooneof theedgesthatisinbetweenthem. Thegoalofthisexerciseistoshowthatthequadrilateralisarectangle. Weoutlinetwodemonstrations, oneusingcongruenttriangles(insteps(1)–(3)below), andtheotherusingsimilartriangles(insteps(4)–(8)below); thereaderisaskedtofillinthedetailsofeachstep.Supposethat ABCD isaquadrilateral. Supposethat |AD| = |BC|, andthatboth AD

and BC areperpendicularto AB. SeeinFigure 2.3.10. Wenowproceedasfollows.

(1) Showthattriangles△ABC and△ABD arecongruent. Deducethat |AC| = |BD|.

(2) Showthattriangles △ACD and △BCD arecongruent. Deducethat ∡D = ∡C.(3) ByProposition 2.3.3 (1)we know that the sumof the angle in ABCD is 360◦.

Deducethat ∡C = 90◦ and ∡D = 90◦. Itfollowsthatallfouranglesinthequadri-lateralareequal, andhencethequadrilateralisarectangle.

(4) Wenowgiveanotherdemonstrationofthefactthatthequadrilateralisarectangle.Asafirststep, wewanttoshowthatthequadrilateralisaparallelogram. ItfollowsfromProposition 1.2.5 that AD and BC areparallel. Nowsupposethat AB andCD arenotparallel. Thenextendthemuntiltheymeet, sayinpoint P. Wewillarriveatalogicalcontradiction.

(5) Showthatthetriangles △ADP and △BCP aresimilar.

(6) Deducethat|AD|

|BC|=

|AP|

|BP|.

(7) Use the fact that |AP| > |BP| todeduce that |AD| > |BC|. Explainwhy thisisalogicalimpossibility, giventhehypothesesofthisexercise. Deducethat AB isparallelto CD, andhencethequadrilateralisaparallelogram.

(8) NowuseExercise 2.1.1 toshowthatthequadrilateralisarectangle.

Polygonscanbeveryirregularlooking, forexample, thepolygonshowninFigure 2.3.7 (ii).Wenowwanttoturnourattentiontopolygonsinwhich, asmuchaspossible, onepointlookslikeanyotherpoint. A polygonisa regularpolygon ifthefollowingtwoconditionshold: (1)alledgeshavethesamelength; (2)allinterioranglesareequal. Forexample, asquareandanequilateraltrianglearebothregularpolygons. Thatthefirstpartoftheabovedefinitiondoesnotalonesufficecanbeseenbyconsideringarhombus, inwhichalledgeshavethesamelength,butnotallinterioranglesareequal; arhombusisnotsomethingwewishtocallregular.A squareisaregularpolygonthathasfouredges. Istherearegular n-gonforeachpossible

n? Theanswerisyes. InFigure 2.3.11 weseearegulartriangle(alsoknownasanequilateral

48 2. Polygons

A B

D C

Figure2.3.10

triangle), aregularquadrilateral(alsoknownasasquare), aregularpentagon, aregularhexagon,aregularheptagon(whichhassevenedges)andaregularoctagon. Ingeneral, foragivenpositiveintegern (whichisgreaterthanorequaltothree), wecanmakearegularn-gonasfollows. First,drawacircle. (Theradiusofthecircledoesnotmatter; differentchoicesofradiuswillyielddifferentlysizedpolygons, butthesizeofpolygonsdoesnotmattertoushere.) Next, calculate360◦/n. Thendraw n raysfromthecenterofthecircle, wheretheraysformanglesof 360◦/nbetweenthem. SeeFigure 2.3.12 (i). The n raysintersectthecirclein n points. Wetakethesepointstobetheverticesofapolygon, whichwethenconstructbyjoiningtheseverticeswithedges. Thepolygonthusconstructedisaregular n-gon. SeeFigure 2.3.12 (ii). Wementionthatthisconstructionofaregular n-gonisnotaclassicalstraightedgeandcompassconstructionofthesortusedinancientGreekmathematics, butthereisnoneedforustorestrictourselvestosuchconstructions. (It turnsout that it isnotpossible toconstructall regularpolygons intheancientGreekmanner.) Observe that for the regularpolygonsconstructedby theabovemethod, theverticesalllieonacircle. Itturnsoutthatanyregularpolygon, nomatterhowitwasconstructed, hasallitsverticesonacircle.

Asweseeinthefollowingproposition, theanglesinaregularpolygonaredeterminedbythenumberofedgesinthepolygon. ThisresultwillbeusefultousinSection 3.2.

Proposition 2.3.4.

1. Eachinteriorangleofaregular n-gonis(n− 2)180◦

n.

2. Eachexteriorangleofaregular n-gonis360◦

n.

Demonstration. BothpartsofthispropositionfolloweasilyfromProposition 2.3.3. Allinterioranglesinaregularpolygonareequal, andsoeachinteriorangleequalsthesumoftheinteriorangles, whichis (n−2)180◦, dividedbyn. A similarconsiderationholdsforexteriorangles.

Thefollowingpropositionnowfollowseasilyfromwhatwehavejustseen.

2.4Area 49

Figure2.3.11

(i) (ii)

Figure2.3.12

Proposition 2.3.5. Everyregularpolygonisconvex.

Demonstration. It followsfromProposition 2.3.4 (1) that theinterioranglesineveryregularpolygonarelessthan 180◦. WenowuseProposition 2.3.1 todeducethateveryregularpolygonisconvex.

InTable 2.3.1 wegivethevaluesoftheinterioranglesofsomeoftheregularpolygons, ob-tainedbypluggingintheappropriatevaluesof n intotheformulagiveninProposition 2.3.4 (1).

2.4 Area

Westartthissectionwithadiscussionoftheareasofpolygons. Ourdiscussionofareaultimatelyreliesuponthreebasicideasconcerningarea, whichweassumewithoutproof: (1)arectanglethathaswidth x andheight y hasarea xy; (2)congruentshapeshaveequalareas; and(3)ifa

50 2. Polygons

RegularPolygon NumberofEdges(n) InteriorAngleequilateraltriangle 3 60◦

square 4 90◦

regularpentagon 5 108◦

regularhexagon 6 120◦

regularheptagon 7 128.57◦

regularoctagon 8 135◦

Table2.3.1

shapeisbrokenupintoafinitenumberofpiecesthattogetherexactlyfilluptheoriginalshape,thentheareaoftheoriginalshapeisthesumoftheareasofthepieces.Inthefollowingproposition, wegiveformulasfortheareasoftriangles, trapezoidsandparal-

lelograms(thelasttwoofwhich, ifthereaderneedsreminding, aredefinedinSection 2.1). Inordertostatetheseformulas, weneedtodefinethenotionofanaltitudeineachofthesetypesoffigures. Inallthesecase, thecommonideaisasfollows. Supposewehavealine m andapoint P thatisnoton m. The altitude from P to m isthelinesegmentthatisperpendiculartom, andhasoneendpointin m, andtheotherendpointat P. SeeFigure 2.4.1.

m

P

altitude

Figure2.4.1

Supposewehaveatriangle, andwechooseanedgeofthetriangle. The altitude perpendiculartothatedgeisthealtitudefromthevertexoppositetheedgetothelinecontainingtheedge.Inthetriangle △ABC showninFigure 2.4.2 (i), weseethealtitude, labeled h, perpendiculartoedge AC. InFigure 2.4.2 (ii)weseethatthealtitudeperpendiculartotheedgeofatriangleneednotbeinsidethetriangle. A trianglehasthreedistinctaltitudes, oneperpendiculartoeachedge, andthesethreealtitudeswillingeneralnothavethesamelengths. Aninterestingfactisthatinanytriangle(evenanobtuseone), thelinescontainingthethreealtitudesmeetinasinglepoint, calledthe orthocenter ofthetriangle; see[WW98, Section4.6]formoredetails.

Next, supposewehavea trapezoid. The altitude of the trapezoid isconstructedby takinganypointononeofthetwoparalleledges, andconstructingthealtitudefromthatpointtotheotherparalleledge. SeeFigure 2.4.3 (i). A trapezoidhasmanyaltitudes, but it follows fromProposition 2.2.6 thattheyallhavethesamelength. Inaparallelogram, wecanconstructthe

2.4Area 51

A

B

C A

B

h h

C

(i) (ii)

Figure2.4.2

altitudesperpendiculartoeachpairofedges, justasweconstructedthealtitudeofatrapezoid(whichhasonlyonepairofparallel edges). A parallelogramhas twodistinct altitudes, oneperpendiculartoeachpairofparalleledges, andthesetwoaltitudeswillingeneralnothavethesamelengths. SeeFigure 2.4.3 (ii)foroneofthealtitudesofaparallelogram.

h h

(i) (ii)

Figure2.4.3

Wecannowstateourformulasconcerningareas.

Proposition 2.4.1.

1. Supposethataparallelogramhasanedgeoflength b, andthealtitudeperpendiculartothatedgeoflength h. Thentheareaoftheparallelogramis bh.

2. Supposethatatrapezoidhasparalleledgesoflengthm1 andm2, andaltitudeoflength

h. Thentheareaofthetrapezoidism1 +m2

2h.

3. Supposethatatrianglehasanedgeoflength b, andaltitudeperpendiculartothatedge

oflength h. Thentheareaofthetriangleis1

2bh.

Demonstration.

(1). Thebasicideaistocutupourparallelogram, andrearrangeitintoarectangle. Morepre-cisely, supposethatwehaveaparallelogramwithverticesA, B,C andD, suchthat |CD| = b,andthatthealtitudeperpendicularto CD haslength h. SeeFigure 2.4.4 (i). Drawthelines

52 2. Polygons

through A and B respectively that areperpendicular to the linecontaining CD; let E andF respectivelybe thepointsof intersectionof these lineswith the linecontaining CD. Fig-ure 2.4.4 (ii).

h

AB

C Db

h h

AB

C F D E

(i) (ii)

Figure2.4.4

Observethat BF and AE areparallel, becauseofProposition 1.2.5. Also, weknowthat AB

and F E areparallelbyhypothesis. Hencethat ABFE isaparallelogram. Moreover, becauseBF and AE arebothperpendicularto F E, weseebyExercise 2.1.1 that ABFE isarectangle.Next, byusingProposition 2.2.5 (1)weknowthat |AB| = |CD|, andthat |AD| = |BC|. It

followsfromProposition 2.2.6 that |BF| = |AE|. Moreover, because BC andAD areparallel,and BF and AE areparallelasnotedabove, itcanbeseenthat ∡CBF = ∡DAE (adetaileddemonstrationofthisequalityusesProposition 1.2.3; weleavethesedetailstothereader). ItthenfollowsfromtheSide-Angle-SideTheorem(Proposition 2.2.3)thattriangles△CBF and△DAE

arecongruent. Hence, thesetwotriangleshavethesamearea.WededucethattheparallelogramABCD hasthesameareaastherectangleABFE. WecanthinkoftheedgeAB asthebaseoftherectangleABFE, andAE asthealtitudeofthisrectangle. Usingtheobservationsatthestartofthisparagraph, weseethat |AB| = b, anditfollowsfromProposition 1.2.5 that |AE| = h.Hencetheareaoftherectangle ABFE is bh. Itfollowsthattheparallelogram ABCD hasareabh.

(2). ThereaderislefttoprovidethisdemonstrationinExercise 2.4.1.

(3). Supposethatwehaveatriangle△ABC, that |AC| = b, andthealtitudeperpendiculartoAC haslength h. SeeFigure 2.4.5 (i). WenowformaparallelogrambydrawingthelinethroughB thatisparalleltoAC, anddrawingthelinethroughC thatisparalleltoAB (Playfair’sAxiom(Proposition 1.1.1)guaranteesthatwecandrawtheselines). Let D bethepointofintersectionofthetwonewlines, sothat ABCD isaparallelogram. Figure 2.4.5 (ii).

ItisevidentthattheparallelogramABCD hasedgeAC asitsbase, andhasaltitudeoflengthh. ByPart (1)ofthisproposition, weknowthattheareaoftheparallelogramis bh. ByusingProposition 2.2.5 (1)weknowthat |AB| = |CD|, andthat |AC| = |BD|. It then followsfrom the Side-Side-SideTheorem (Proposition 2.2.2) that triangles △ABC and △BCD are

2.4Area 53

h

A

B

Cb

h

A

B D

Cb

(i) (ii)

Figure2.4.5

congruent. Hence, thesetwotriangleshavethesamearea, andsoeachhashalftheareaofthe

parallelogram ABCD. Wededucethattheareaoftriangle △ABC is1

2bh.

Observethatifwethinkofatriangleasadegeneratetrapezoidinwhichoneoftheparalleledgeshaslengthzero, thentheformulafortheareaofatriangle(Part (3)oftheabovepropo-sition)followsimmediatelyfromtheformulafortheareaofatrapezoid(Part (2)oftheaboveproposition).AnimmediateconsequenceofProposition 2.4.1 isthefollowingresult. SeeFigure 2.4.6 for

anillustrationofthisfact.

Proposition 2.4.2.

1. Supposethattwoparallelogramshaveanedgeofthesamelength, andhavethealtitudesperpendiculartothoseedgeshavingthesamelength. Thentheparallelogramshavethesamearea.

2. Suppose that two triangleshaveanedgeof the same length, andhave thealtitudesperpendiculartothoseedgesofthesamelength. Thenthetriangleshavethesamearea.

h

b b

h

Figure2.4.6

Exercise 2.4.1. [UsedinThisSection] DemonstrateProposition 2.4.2 (2).

54 2. Polygons

Exercise 2.4.2. Supposethattriangles △ABC and △A ′B ′C ′ aresimilar. Let Q and Q ′

respectivelydenotetheareasof △ABC and △A ′B ′C ′. Showthat

Q

Q ′ =|AB|

2

|A ′ B ′|2 .

Wenowhaveformulasfortheareasofrectangles, trianglesparallelogramsandtrapezoids.Whataboutmorecomplicatedpolygons? Itisnotpossibletohaveasimpleformulatocovereachpossibletypeofpolygon. However, itisalwayspossibletofindtheareaofanypolygonbychoppingitupintosimpleshapes(forexample, rectanglesandtriangles), figuringouttheareaofeachofthepieces, andthenaddingtheareasofthepiecesup. SeeFigure 2.4.7 foranexampleofacomplicatedpolygonchoppedupintorectanglesandtriangles.

Figure2.4.7

Exercise 2.4.3. FindtheareasofthepolygonsshowninFigure 2.4.8.

Weremindthereaderthattheperimeter ofapolygonisthesumofthelengthsofitsedges.

Exercise 2.4.4. Showthatiftworectangleshavethesameareaandthesameperimeter,thentheyhavethesamedimensions. (Thisoneusessomealgebra.)

2.4Area 55

(i) (ii) (iii)

23

2

2

1

1

2 2

2

2 2

Figure2.4.8

Exercise 2.4.5.

(1) Findtwopolygonsallofwhoseedgeshaveintegerlengths, whichhavethesameareasandsameperimeters, butthatarenotcongruent.

(2) IfthetwoconvexpolygonsyoufoundinPart (1)werenotconvex, findtwoconvexpolygonsallofwhoseedgeshaveintegerlengths, whichhavethesameareasandsameperimeters, butthatarenotcongruent.

Exercise 2.4.6. A kite isaquadrilateralthathastwopairsofadjacentedgeswithequallengths. Wecallthediagonalthathasonepairofadjacentedgeswithequallengthsononesideandtheotherpairontheothersidethe crossdiagonal; theotherdiagonaliscalledthe maindiagonal. (Thenomenclatureinthisexerciseisnotstandardized.) Forthesakeofthisexercise, assumethatallkitesunderdiscussionareconvex(thatobviatestheneedconsideringdifferentsubcases), thoughtherearealsonon-convexkites.

(1) Showthatthemaindiagonalinakitebreaksthekiteintotwocongruenttriangles.

(2) Showthatthemaindiagonalinakitebisectseachoftheanglesatitsendpoints.

(3) Showthatthemaindiagonalinakitebisectsthecrossdiagonal.

(4) Showthatthecrossdiagonalinakiteisperpendiculartothemaindiagonal.

(5) Showthattheareaofakiteistheonehalftheproductofthelengthsofthediagonals.

Wecanapplytheconceptofareatoregularpolygons, startingwiththefollowingexercise.

56 2. Polygons

Exercise 2.4.7. Findtheareaforeachofthefollowingregularpolygons.

(1) Anequilateraltrianglewithedgesoflength 1.

(2) A regularhexagonwithedgesoflength 1.

(3) A regularoctagonwithedgesoflength 1. (Hint: Donottrytosolvethisproblembydividingtheoctagonintoeightcongruenttriangleswithacommonvertexinthecenteroftheoctagon—thatmethodisquitetricky.)

Oneverynicepropertyofregularpolygonsisthattheyminimizeperimeteramongallpolygonswithagivennumberofedgesandagivenarea. Moreprecisely, supposeweareinterestedinpolygonswith n edges, where n issomepositiveintegergreaterthanorequaltothree. Supposefurtherthatwearegivenanarea A, where A issomepositivenumber. Amongallthe n-gonsthathavearea A, whichonehasthesmallestperimeter? Theansweristheregular n-gonwitharea A. Thoughthisresultseemsreasonableintuitively, thedemonstrationisbeyondthescopeofthistext. Alternatively, ifwearegivenaperimeter P, wecanaskwhich n-gonwithperimeterP hasthelargestarea. Again, theansweristheregular n-gonwithperimeter P.

Exercise 2.4.8.

(1) Supposewehaveacircle, andsupposethatA and B arepointsonthecirclethatarenotdiametricallyoppositeeachother. SupposefurtherthatC isanotherpointonthecirclethatisbetween A and B. Considertheareaofthetriangle △ABC. ExplainwhyofallpossiblechoicesofpointsC, theonewhere△ABC hasthemaximalareaiswhere C ismidwaybetween A and B.

(2) Giveaninformalexplanationthat, amongallpolygonswith n vertices, andwithitsverticesallonagivencircle, theregular n-gonwillhavethemaximalarea.

Wecanalsouseregularpolygonstohelpgiveusanintuitive(thoughnotrigorous)explanationregardingsomeaspectsofthenumber π. Ofcourse, thenumber π relatestocircles. However,wecanuseregularpolygonstohelpusunderstandcirclesbecause, aswementionedabove, ifwestartwithacircle, wecanformaregular n-gonwithverticesonthecircle, foranypositiveinteger n greaterthanorequaltothree. Ifwechoosethenumber n tobeverylarge, thenthen-gonapproximatesthecircleveryclosely. Thelargerthe n, thebettertheapproximation. Nopolygoneverequalsthecircle, thoughforverylarge n itmightbeimpossibletodistinguisharegular n-gonfromacirclewiththenakedeye.Todiscussthenumber π, recallthatitisusuallydefinedastheratioofthecircumferenceto

thediameterofacircle. Thatiscorrect, butthereisanaspecttothisdefinitionthatisusually

2.4Area 57

glossedover—howdoweknowthatinallcircles, thereisthesameratioofthecircumferencetothediameter. Ifthisratioweredifferentindifferentcircles, thenthedefinitionof π wouldnotmakeanysense. Infact, thisratioisthesameinallcircles, ascanbeproved. A rigorousproofcanbefoundusingcalculus, butwecanusetrianglestogiveusanintuitiveideawhythisratioisthesameinallcircles. Beforewestart, wenotethatinsteadoftheratioofthecircumferencetothediameter, wecanjustaswelllookattheratioofthecircumferencetotheradiusofthecircle,whichshouldyield 2π ratherthan π. Ifwecanshowthatthetheratioofthecircumferencetotheradiusisthesameforanytwodifferentcircles, thatwillsuffice.Supposewehavearegiven twocirclesofdifferent sizes. Ineachcircle, constructa regu-

lar n-gonwithitsverticesonthecircle. Onesuchcircleandregular n-gonisshowninFig-ure 2.4.9 (i); inthefigurewehavea 10-gon, thoughitwouldbepossibletohaveanynumberofedges. Ineachofthepolygons, drawlinesegmentsfromthecenterofthecircletotheverticesofthepolygon, thusbreakingthepolygonupinto n isoscelestriangles. InFigure 2.4.9 (ii)weseethesmallerofourtwocircles; thelengthofeachedgeofthepolygonis a, andtheradiusofthecircleis r. InFigure 2.4.9 (iii)weseethelargerofourtwocircles; thelengthofeachedgeofthepolygonis c, andtheradiusofthecircleis s.

(i) (ii) (iii)

ac

r s

Figure2.4.9

Now, comparethetwopolygonsinthedifferentsizedcircles. Theyarebothregular n-gons,eventhoughtheyareofdifferentsizes. Hence, byProposition 2.3.4 (1), bothofthese n-gonshavethesameinteriorangles. Thebaseanglesintheisoscelestrianglesintowhichthepolygonsarebrokenuparethereforealsothesameinbothpolygons. Itfollowsthattheisoscelestrianglesinthesmallerpolygonareallsimilartotheisoscelestrianglesinthelargerpolygon. Inparticular,byusingProposition 2.2.9, weseethat

a

r=

c

s.

Multiplyingeachsideby n yieldsna

r=

nc

s.

58 2. Polygons

Weobservethattheperimeterofthesmallerpolygonis na, andtheperimeterofthelargerpolygonis nc. Hence

Perimeterofthesmallerpolygon

Radiusofthesmallercircle=

Perimeterofthelargerpolygon

Radiusofthelargercircle.

If n isverylarge, thentheperimeterofthepolygonsisveryclosetothecircumferenceofthecircle.Byletting n gotoinfinity, wededucethat

Circumferenceofthesmallercircle

Radiusofthesmallercircle=

Circumferenceofthelargercircle

Radiusofthelargercircle,

whichiswhatweweretryingtoshow. Thisargumentisnotcompletelyrigorousasstated, butitdoesgiveaplausibleargument.Wecantaketheabovesortofreasoningonestepfurther. Supposethatacirclehasradius r.

Aswejustdiscussed, wehave C = 2πr, where C isthecircumferenceofthecircle. Thereis, ofcourse, anotherequallyusefulformulainvolving π, namelyA = πr2, whereA istheareaofthecircle. Wecannottakethisareaformulaasthedefinitionof π, becausewehavealreadydefinedπ in termsof thecircumference. Rather, wecangiveanintuitiveargument for this formula,giventhatwehavealreadyseenwhy C = 2πr oughttobetrue. (Again, ourargumentwillnotbecompletelyrigorous, thougharigorousargumentcanbefoundusingcalculus.)Supposewehaveacircleofradius r. Again, formaregular n-gonwithverticesonthecircle.

Thistime, makesurethat n isanevennumber. Weformisoscelestrianglesasbefore; thistime,wecoloreveryothertriangle. SeeFigure 2.4.10 (i)forthecasewhere n isten. Observethattheareaofthe n-gonisveryclosetotheareaofthecircle; if n isverylarge, thentheapproximationisquiteclose.

(i) (ii)

Figure2.4.10

Now, wetakethe n-gon, andwerearrangeitstrianglesasshowninFigure 2.4.10 (ii). Theshapethattheserearrangedtrianglesformisaparallelogram. However, if n isverylarge, thetriangleswillbeverythin, andtheparallelogramwillbealmostarectangle. Whatisthelength

2.4Area 59

ofthealtitudeoftheparallelogram? If n isverylarge, thenthetriangleswillbeextremelythin,andthealtitudeofeachtrianglewillbeapproximatelythesameasthelengthofitssides, whichisjusttheradiusofthecircle, namely r. Whatisthelengthofthebaseoftheparallelogram? Itishalfthecircumferenceofthecircle, namely πr. Hence, usingProposition 2.4.1 (1), weseethattheareaoftheparallelogramisapproximately πr · r = πr2. If n getslargerandlarger, theapproximationgetsbetterandbetter. Note, however, thattheareaoftheparallelogramisveryclosetotheareaofthecircle, becauseitisthesameastheareaofthepolygoninsidethecircle.Onceagain, as n goestoinfinity, wededucethatareaofthecircleisactuallyequaltotheareaoftheparallelogram, andhencetheareaofthecircleis πr2. Onceagain, thisargumentisnotcompletelyrigorousasstated, butgivesanintuitivepictureofwhatishappening.Weend thissectionbyusing theconceptofarea todemonstratea result thaton the face

ofithaslittletodowitharea, namelyProposition 2.2.9, whichconcernssimilartriangles. Todemonstratethisproposition, weneedtwopreliminaryresults, towhichwenowturn; itisinthedemonstrationofthefirstoftheseresultsthatweencountertheuseofarea. Ourapproachhere follows [Mey99, Section2.4]. To read the statementofourfirstpreliminary result, seeFigure 2.4.11.

A

B

D E

C

Figure2.4.11

Proposition 2.4.3. Supposethatatriangle △ABC hasapoint D onedge AB, andapoint Eonedge AC, placedsothat DE isparallelto BC. Then

|AB|

|AD|=

|AC|

|AE|.

Demonstration. Asafirststep, weaddthelinesegmentBE, asshowninFigure 2.4.11 (i). Next,drawanaltitudefrom E to AB; supposethisaltitudehaslength h, asindicatedinthefigure.

60 2. Polygons

A

B

D E

C

A

B

D E

C

h

(i) (ii)

Figure2.4.12

Considerthetriangle △ABE. UsingProposition 2.4.1 (3), weknowthattheareaofthistri-

angleis1

2|AB|h. Similarly, theareaofthetriangle △ADE is

1

2|AD|h. Itthenfollowsthat

areaof △ABE

areaof △ADE=

12|AB|h

12|AD|h

=|AB|

|AD|.

Next, returntotheoriginalsituationinFigure 2.4.11, andaddthelinesegmentDC, asshowninFigure 2.4.11 (ii). Thesamesortofreasoningasbeforeshowsthat

areaof △ACD

areaof △ADE=

|AC|

|AE|;

weleavethedetailstothereader.Asournextstep, wecomparethetwotriangles △DEB and △DEC showninthetwoparts

ofFigure 2.4.11. Wecanthinkof DE asthebaseofbothtriangles. Moreover, because DE isparallelto BC (byhypothesis), itfollowsthatthealtitudeof △DEB perpendicularto DE hasthesamelengthasthealtitudeof△DEC perpendiculartoDE (thisfactusesProposition 2.2.6).ItthenfollowsbyProposition 2.4.2 (2)that △DEB and △DEC havethesamearea.Finally, giventhattriangles △DEB and △DEC havethesameareas, weseethattriangles

△ABE and △ACD havethesameareas. Pluggingthisobservationsintothetwoformulasfortheratiosofareasthatwepreviouslysaw, wesee

|AB|

|AD|=

areaof △ABE

areaof △ADE=

areaof △ACD

areaof △ADE=

|AC|

|AE|.

Thuswehaveshownthedesiredequation.

2.4Area 61

Our secondpreliminary result is theconverse to thepreviousproposition; again, seeFig-ure 2.4.11.

Proposition 2.4.4. Supposethatatriangle △ABC hasapoint D onedge AB, andapoint Eonedge AC, placedsothat

|AB|

|AD|=

|AC|

|AE|.

Then DE isparallelto BC.

Demonstration. WeseethegivensituationinFigure 2.4.13 (i); wedonotknowyetwhetherDE isparallelto BC ornot(becausethatiswhatwearetryingtoshow), andwehavedrawnthecasewherethetwolinesegmentsarenotparallel.

A

B

DE

C

A

B

D FE

C

(i) (ii)

Figure2.4.13

ByusingPlayfair’sAxiom (Proposition 1.1.1), wecandrawalinecontainingD thatisparalleltothelinecontaining BC. ThislinethroughD intersects AC inapoint, whichwewillcall F.Then DF isparallelto BC. SeeFigure 2.4.13 (ii).WecannowapplyProposition 2.4.3 tothetriangle △ABC withthepoints D and F. We

deducethat|AB|

|AD|=

|AC|

|AF|.

Ontheotherhand, weknowbyourhypothesesthat

|AB|

|AD|=

|AC|

|AE|.

62 2. Polygons

Wecombinethesetwoequationstoderive

|AC|

|AF|=

|AC|

|AE|,

andbycancellingandrearrangingitfollowsthat |AF| = |AE|. Giventhat E and F arebothpointsin AC, weseethat E = F. ByconstructionweknowthatDF isparallelto BC, andwededucethat DE isparallelto BC.

WenowhavealltheingredientsneededforthepromiseddemonstrationofProposition 2.2.9.

DemonstrationofProposition 2.2.9. Supposethattriangles△ABC and△A ′B ′C ′ aresimilar.Wewillfirstshowthat

|AB|

|AB ′|=

|AC|

|AC ′|.

First, wenote that either |AB| = |AB ′| or |AB| = |AB ′|. If ithappens tobe thecasethat |AB| = |AB ′|, thenwecandeducethat △ABC and △A ′B ′C ′ arecongruent, usingtheAngle-Side-AngleTheorem(Proposition 2.2.4). Inthatcase, itwouldfollowthat |AC| = |AC ′|,andthenwewouldseethat

|AB|

|AB ′|= 1 =

|AC|

|AC ′|,

whichiswhatwearetryingtoshow.Nowassumethat |AB| = |AB ′|, because that is theremainingcase. Therearenowtwo

possibilities, namely |AB| > |AB ′| or |AB| < |AB ′|; wewilldiscussonlythefirstofthesetwocases, theothercasebeingvirtuallyidentical. So, assumethat |AB| > |AB ′|.Because |AB| > |AB ′|, wecanfindapoint D on AB sothat |AD| = |AB ′|. SeeFig-

ure 2.4.14. Nowfindapoint E on AC sothat

|AB|

|AD|=

|AC|

|AE|;

Suchapointcanalwaysbefound. Again, seeFigure 2.4.14.

WecannowapplyProposition 2.4.4 tothetriangle△ABC withpointsD and E. Thepropo-sitionimpliesthat DE isparallel to BC. WecannowuseProposition 1.2.3 todeducethattheangle α equalstheangleat B. Byhypothesisthetriangles △ABC and △A ′B ′C ′ aresim-ilar, andhencetheangleat B equals theangleat B ′. It followsthat theangle α equals theangleat B ′. Wealsoknowthattheangleat A equalstheangleat A ′. Giventhatwealsohave|AD| = |AB ′| (whichistruebyvirtueofourchoiceof D), weseethattriangles △ADE and△A ′B ′C ′ arecongruentbytheAngle-Side-AngleTheorem(Proposition 2.2.4). Wederivethat|AE| = |AC ′|. Finally, weknowbyconstructionthat

|AB|

|AD|=

|AC|

|AE|.

2.5ThePythagoreanTheorem 63

A

B

D E

C

A'

B' C'α

Figure2.4.14

Itfollowsthat|AB|

|AB ′|=

|AC|

|AC ′|.

Thislastequationistheoneweweresupposedtodemonstrate.Wenotethatacompletelysimilarargumentcouldbeusedtoshowthat

|AC|

|AC ′|=

|BC|

|B ′C ′|;

wewillskipthedetails. Wehavethereforeshownthat

|AB|

|AB ′|=

|AC|

|AC ′|=

|BC|

|B ′ C ′|,

whichisthefirstdisplayedequationinthestatementofProposition 2.2.9. TheseconddisplayedequationinthestatementofProposition 2.2.9 followsstraightforwardlyfromthefirstdisplayedequation, andwewillleavethattothereader. Henceourdemonstrationiscomplete.

2.5 ThePythagoreanTheorem

This section treats what is probably themost famous theorem about triangles, namely thePythagoreanTheorem. (This theorem seems tohavebeenknownempirically inbothBaby-lonandChinabeforethetimeofPythagoras, thoughthereisnoevidencethatthetheoremwasprovedpriortoPythagoras.) TherearemanyotherequallyimportanttheoremsingeometryotherthanthePythagoreanTheorem, butwefocusonitnowbecauseitissofamiliar, andbecauseitbringstogetheranumberofideaswehaveencounteredsofarabouttriangleandpolygons.

64 2. Polygons

ItisimportanttostatethePythagoreanTheoremcorrectly. WheneverI askstudentsinaclasstostatethePythagoreanTheorem, theresponseI invariablyreceiveis: “a2+b2 = c2.” However,justtostatethisequationisabsolutelynotcorrect. Itisnottruethatthisequationholdsfor anynumbers a, b and c. Theequationholdsonlyforparticularvaluesof a, b and c, namelythosethatcorrespondtothelengthsofthesidesandhypotenuseofarighttriangle. (Recallthatinarighttriangle, thetwoedgesthatformtherightanglearecalledthe sides ofthetriangle, andtheedgethatisoppositetherightangleiscalledthe hypotenuse ofthetriangle.)WearenowreadytostateanddemonstratethePythagoreanTheorem. Therearemanyproofs

ofthistheorem, goingbacktotheancientworld. ForEuclid’sproofseeProposition 47ofBookI of[Euc56], thoughyouhavetolookatsomeofEuclid’spreviouspropositionstofigureoutall thedetailsofhisproofof thePythagoreanTheorem. Wegive twodifferentproofsof thetheorem(bothquitedifferentfromEuclid’sproof), toshowhowthesameresultcanbeprovedbydifferentapproaches. Ourfirstproof(averywidelyusedone)isbasedonareaandcongruenceof triangles; thesecondisbasedonsimilarityof triangles. NeitherofourproofswouldhavemadesensetotheancientGreeks(whodidnothaveouralgebra). See[Loo40]foravarietyofproofsofthePythagoreanTheorem. A curiousfactoidisthatthereisaproofofthePythagoreanTheoremattributedtoPresidentJamesGarfield—perhapstheonlyknownmathematicalproofattributedtoapresidentoftheUnitedStates.

Proposition 2.5.1 (PythagoreanTheorem). Supposethatarighttrianglehassidesoflength a

and b, andhypotenuseoflength c. Then a2 + b2 = c2.

Demonstration. FirstProof: InFigure 2.5.1 (i)weseethetriangleunderconsideration. Giventhatthesumofthethreeanglesinthetriangleis 180◦ (aswesawinProposition 2.2.1 (1)), weknowthat α+ β = 90◦.Wenowconstructa squarewithsidesof length a + b, andbreak itupasshown inFig-

ure 2.5.1 (ii). Weseefourcopiesoftheoriginalrighttriangleinsidethelargersquare. Thatthesefourtrianglesarereallycongruenttotheoriginaltriangleisintuitivelyclear, andformallyfollowsfromtheSide-Angle-SideTheorem(Proposition 2.2.3).

Nowconsidertheangle γ, asindicatedinthefigure. Giventhat α, β and γ togethermakeupastraightline, weknowthat α + β + γ = 180◦. Because α + β = 90◦, aspreviouslymentioned, wededucethat γ = 90◦. A similarargumentshowsthat theother threeanglesbetweenthesidesoflength c arealso 90◦. Itfollowsthatthefigurethathasfouredgesoflengthc isinfactasquare.Weknowthattheentiresquarewithsidesoflength a + b hasarea (a + b)2. Ontheother

hand, wecanalsocompute theareaof theentiresquarebyaddinguptheareasof thefivepieces(onesquareandfourtriangles)intowhichwehavebrokenitup. Thesquarewithsidesoflength c hasarea c2. Eachofthefourcopiesoftheoriginaltrianglehasarea 1

2ab. Thetotal

areaistherefore c2 + 4 · 12ab = c2 + 2ab. Equatingthetwowaysofcomputingthetotalarea,

obtain

(a+ b)2 = c2 + 2ab.


αα

α

α

α

ββ

β

β

β

γ

a

aa

a

a

bb

b

b

b

c

cc

c

c

(i) (ii)

Figure2.5.1

Recallfromalgebratheformula (a+ b)2 = a2 + 2ab+ b2. Wethenobtain

a2 + 2ab+ b2 = c2 + 2ab.

Canceling 2ab fromeachsideofthislastequationyields a2 + b2 = c2, whichiswhatwewantedtoprove.

SecondProof: InFigure 2.5.2 (i)weseethetriangleunderconsideration, wheretheanglebe-tween the sidesof lengths a and b is a right angle. In Figure 2.5.2 (ii)wehavedrawn thealtitudeperpendiculartoedge AB. Let D bethepointwherethealtitudeintersects AB. Intriangle△ABC, let a, b and c respectivelydenotethelengthsoftheedgesoppositeangles A,B and C. Let x denotethelengthof AD, andthusthelengthof BD is c− x.

Considerthetwotriangles△ABC and△DBC. Eachtrianglehasarightangle, andeachhastheangleat B, sotheyhavetwoequalangles. Becausethesumoftheanglesineachtriangleis 180◦, the two triangles in facthaveall threeangles the same. Thus the two trianglesaresimilar, wherevertices A, B and C respectivelyin △ABC correspondtovertices C, B and D

in △DBC. ItnowfollowsfromProposition 2.2.9 that

c

a=

a

c− x.

Itcanalsobeseenthatthetwotriangles △ABC and △ADC aresimilar, wherevertices A, Band C respectivelyin △ABC correspondtovertices A, C and D in △ADC. Itfollowsfrom

66 2. Polygons

(i) (ii)

x c-xDB

b

C

A

a

cB

b

C

A

a

Figure2.5.2

Proposition 2.2.9 thatc

b=

b

x.

Ifwecrossmultiplytheabovetwoequations, weobtain

c(c− x) = a2 and cx = b2.

Multiplyingoutthefirstequation, weobtain

c2 − cx = a2 and cx = b2.

Substitutingthesecondequationintothefirst, weobtain

c2 − b2 = a2.

Moving b2 totheothersideyields a2 + b2 = c2, whichiswhatwewantedtoprove.

Exercise 2.5.1. Thetwosidesofarighttriangleare 6 and 11 inchesrespectively. Howlongisthehypotenuse?

Exercise 2.5.2. A 40 ft.wireisstretchedfromthetopofapoletotheground. Thewirereachestheground 25 ft.fromthebaseofthepole. Howhighisthepole?

Exercise 2.5.3. ProvethePythagoreanTheoremusingFigure 2.5.3 insteadofFigure 2.5.1.


a

ab

aa

bb

b

c

cc

c

Figure2.5.3

ThePythagoreanTheoremstatesthatifarighttrianglehassidesoflength a and b, andhy-potenuseoflength c, then a2+b2 = c2. Couldithappenthatinanon-righttrianglewithedgesoflength a, b and c, theformula a2 + b2 = c2 alsoholds? Thefollowingpropositionsaysthatitcouldnothappen; inotherwords, theformula a2+b2 = c2 istheexclusiveprovinceofrighttriangles. ItisinterestingtonotethatwewillusethePythagoreanTheoremtodemonstratethefactthetheoremdoesnotholdinnon-righttriangles.

Proposition 2.5.2 (Converse to the PythagoreanTheorem). Supposewearegivena triangle△ABC. Let a, b and c respectivelydenotethelengthsoftheedgesoppositeangles A, B andC. Supposethat a2 + b2 = c2. Then C isarightangle.

Demonstration. Wefollow[Bar01, p.10]. Theangle C couldbeeitheracute(lessthan 90◦),obtuse(greaterthan 90◦)orarightangle(equalto 90◦). Wewillprovethatthefirsttwocasescannothappen; itwillthenfollowthat C isarightangle, whichiswhatwearetryingtoprove.Supposefirstthat C islessthan 90◦. SeeFigure 2.5.4 (i). InFigure 2.5.4 (ii)wehavedrawnthe

altitudeperpendiculartoedge BC. LetD bethepointwherethealtitudeintersects BC. Let hdenotethelengthof AD, andlet x denotethelenthof CD. Weseethatthelengthof BD isa − x. Observethatbecause C islessthan 90◦, itfollowsthat x > 0. (If C werearightangle,then D wouldbethesameas C, and x wouldbe 0.)

Thetwotriangles△ADC and△ABD arerighttriangles. ApplyingthePythagoreanTheoremtoeachone, weobtainthetwoequations

x2 + h2 = b2 and (a− x)2 + h2 = c2.

Isolating h2 ineachoftheseequationsyields

h2 = b2 − x2 and h2 = c2 − (a− x)2.

68 2. Polygons

(i) (ii)

x a-xDB

hb

A

C

c

aB

b

A

C

c

Figure2.5.4

Equatingthesetwoexpressionsfor h2 givesus

b2 − x2 = c2 − (a− x)2.

Recallfromalgebratheformula (a− x)2 = a2 − 2ax+ x2. Wethereforeobtain

b2 − x2 = c2 − (a2 − 2ax+ x2).

Distributingthenegativesigninyields

b2 − x2 = c2 − a2 + 2ax− x2.

Cancelling x2 frombothsidesgiveus

b2 = c2 − a2 + 2ax.

Finally, bringthe a2 totheotherside, andwededucethat

a2 + b2 = c2 + 2ax.

Wenowhavealogicalimpossibility. Ontheonehand, wehaveassumedthat a2 + b2 = c2.Ontheotherhand, wejustdeducedthat a2 + b2 = c2 + 2ax. Giventhatneither a nor x is 0,thenneitheris 2ax, andsowehaveanimpossiblesituation. Theonlywayoutofthisproblemistoadmitthatourhypothesisthat C islessthan 90◦ isfalse.A similarargumentshowsthatthehypothesisthatC isgreaterthan 90◦ isalsofalse. (Weleave

ittothereadertosupplythedetails; thedifferenceisthatFigure 2.5.4 needstobemodifiedsothatC isgreaterthan 90◦, whichmakesthealtitudeperpendicularto BC beoutsidethetriangle△ABC.) Asaresult, theonlyremainingpossibilityisthat C isarightangle, whichiswhatwewantedtoshow.

ThePythagoreanTheoremdefinitelydoesnotholdfortrianglesthatarenotrighttriangles.WenowturntotwogeneralizationsofthePythagoreanTheoremthatdoholdforalltriangles.Thefirstofthesegeneralizations, themorewellknownandusefulofthetwo, iscalledtheLaw


ofCosines, anditisessentiallythePythagoreanTheoremwithacorrectionfactorthatmakesitworkforalltriangles. TheLawofCosinesisimportantintrigonometry, andshowsupinotherbranchesofmathematics, andapplicationsofmathematics. TostatetheLawofCosines, weneedtousethetrigonometricfunctioncosine. Forthosenotfamiliarwithcosine, youcansimplyskipthestatementoftheLawofCosinesgivenbelow; wewillnotbeusingthislawatanypointinthistext. However, wementionit, toshowoneofthewaysinwhichthePythagoreanTheoremcanbegeneralizedtonon-righttriangles.Justtoremindthosefamiliarwiththetrigonometricfunctions, cosineisafunctionthatassigns

toeveryangle x anumberdenoted cos x. Forexample, wehave cos 0◦ = 1, and cos 60◦ = 1/2,and cos 90◦ = 0. TheLawofCosinesisasfollows.

Proposition 2.5.3 (LawofCosines). Supposewearegivenatriangle △ABC. Let a, b and c

respectivelydenotethelengthsoftheedgesoppositeangles A, B and C. Then

c2 = a2 + b2 − 2ab cosC.

ForaproofoftheLawofCosines, seemostbooksontrigonometryfordetails. WechosetohighlighttheangleC intheabovestatementoftheLawofCosines, thoughinanon-righttrianglenooneangleisspecial, andtheLawofCosinesalsostatesthat a2 = b2 + c2 − 2bc cosA andb2 = a2 + c2 − 2ac cosB. Next, supposethat △ABC isinfactarighttriangle, with C therightangle. Then C = 90◦, and cosC = 0. Inthatcase, theLawofCosinesjustreducestothePythagoreanTheorem.Wenowturntooursecond, lesswellknown, generalizationof thePythagoreanTheorem.

ThisgeneralizationisknownasPappus’VariationonthePythagoreanTheorem IntheLawofCosines, wemaintainthe a2, b2 and c2 thatareinthestatementofthePythagoreanTheorem,butputinanextracorrectionterm(involvingtrigonometry)toaccountfornon-righttriangles.Recallthatgeometrically, theterms a2, b2 and c2 correspondtotheareasofcertainsquares.InPappus’VariationonthePythagoreanTheorem, statedbelow, wereplacesquaresbycertainparallelograms, andbysodoingwewillbeabletoallowfornon-right triangleswithout theuseofacorrectionterm(andthuswithoutanytrigonometry). WhenreadingthestatementofPappus’VariationonthePythagoreanTheorem, itwillhelptolookatFigure 2.5.5.

Proposition 2.5.4 (Pappus’VariationonthePythagoreanTheorem). Supposewearegivenatri-angle△ABC. FormparallelogramsACDE andBCFG ontheedgesAC andBC respectively.Extendthelinesegments DE and FG untiltheyintersectinthepoint H. Formtheparallelo-gram ABIJ withedgesAJ and B I thatareparalleltoandhaveequallengthasHC. Thentheareaoftheparallelogram ABIJ equalsthesumoftheareasoftheparallelograms ACDE andBCFG.

Demonstration. First, weextend HC untilitcutsthroughtheparallelogram ABIJ, breakingitupintotwosmallerparallelograms AJKL and BIKL. SeeFigure 2.5.6. WewillnowshowthattheareaofparallelogramACDE equalstheareaoftheparallelogramAJKL. A completelyidenticalargument (thedetailsofwhichwewill skip)canbeused to show that theareaof

70 2. Polygons

B

G

FH

J I

C

A

E

D

Figure2.5.5

parallelogram BCFG equalstheareaoftheparallelogram BIKL. Itwillthenfollowthatthesumoftheareasoftheparallelograms ACDE and BCFG equalsthesumoftheareasoftheparallelogramsAJKL and BIKL, whichinturnequalstheareaoftheparallelogramABIJ, thuscompletingtheargument.

B

G

FH

JK

I

C

A L

E

D

Figure2.5.6

Toshowthattheareaofparallelogram ACDE equalstheareaoftheparallelogram AJKL,weneedtomakeonemoreparallelogram. ExtendAJ untilitintersectsthelinecontainingDE


inpointM. WeseethatMA isparalleltoHC. SeeFigure 2.5.7. HenceACHM isaparallel-ogram. Comparetheparallelograms ACDE and ACHM. Theybothhavetheedge AC, andthenbothhavethesamealtitudeperpendiculartothisedge. ItfollowsfromProposition 2.4.2 (1)thatthesetwoparallelogramshavethesamearea.

B

G

FH

JK

I

C

A L

E

D

M

Figure2.5.7

Next, comparetheparallelogramsACHM andAJKL. ObservethattheedgeHC ofthepar-allelogram ACHM isequalinlengthandparalleltotheedge AJ oftheparallelogram AJKL.Moreover, thealtitudesperpendicular to these twoedgesareequal. It follows fromProposi-tion 2.4.2 (1)thatthesetwoparallelogramshavethesamearea. Hence, becausetheparallelo-grams ACDE and ACHM havethesameareas, andtheparallelograms ACHM and AJKL

havethesameareas, itfollowsthattheparallelograms ACDE and AJKL, whichiswhatweneededtoshow.

When we apply Pappus’Variation on the PythagoreanTheorem to a right triangle, andwe choose parallelograms that happen to be squares, then we simply obtain the standardPythagoreanTheorem.

72 2. Polygons

3Polyhedra

3.1 Polyhedra–TheBasics

A polyhedron isasolidregionofspacethatisboundedbyafinitenumberofpolygonsthataregluedtogether. Wehavethreerequirementsaboutthewayinwhichwegluethepolygonstogether.

(1) Polygonsaregluededge-to-edge(thatis, entireedgesaregluedtoentireedges), orvertex-to-vertex.

(2) Everyedgeofapolygonisgluedtopreciselyoneotheredge.

(3) Notwopolygonsintersectexceptpossiblyalongtheiredgeswheretheyareglued.

SomepolyhedraareshowninFigure 3.1.1. Somenon-polyhedraareshowninFigure 3.1.2;theobjectinPart (i)haspolygonsthatarenotgluededge-to-edge, andtheobjectinPart (ii)hasthreeedgesofpolygonsgluedtogether. Notethatthepluralof“polyhedron”is“polyhedra.”Wewillrestrictourattentiontopolyhedramadeupoutofconvexpolygons, thoughitisalsopossibletoconsiderpolyhedrawithnon-convexfaces. (Itisalsopossibletolookatpolyhedrawithself-intersections, thatis, inwhichrequirement(3)isdropped; wewillnotbelookingatsuchpolyhedrainthistext, withonebriefexceptionattheendofSection 3.6.)

Foreachpolyhedron, the faces ofthepolyhedronarethepolygonsthatboundit; the edgesarethelinesegmentswherethefacesmeet; the vertices arethepointswhereedgesmeet. Forexample, thecube showninFigure 3.1.1 hassixsquarefaces, twelveedgesandeightvertices.Thefollowingsimplefactsaboutfaces, edgesandverticesofpolyhedra, whichwillbeofuse

lateron, canbederivedfromourrequirementsonhowpolygonsaregluedtogethertomakepolyhedra.

74 3. Polyhedra

(i) (ii) (iii)

Figure3.1.1

(i) (ii)

Figure3.1.2

Proposition 3.1.1.

1. Everyedgeinapolyhedroniscontainedinprecisely 2 faces.

2. Everyvertexinapolyhedroniscontainedinatleast 3 edges.

3. Everyfaceinapolyhedroncontainsatleast 3 edges.

Just aswe had both convex and non-convex polygons, so toowe can have convex andnon-convexpolyhedra. Theideaofconvexityiscompletelythesameforpolyhedraasforpoly-gons. Intuitively, apolyhedronisconvexifithasno“indentations.” Moreformally, apolyhedronisconvexifanytwopointsinthepolyhedronarejoinedbyalinesegmentcontainedentirelyinthepolyhedron. ThepolyhedrainFigure 3.1.1 (i)and(ii)areconvex, whereasthepolyhedroninPart (iii)ofthefigureisnot. Wewillmostly, thoughnotexclusively, dealwithconvexpolyhedrainthistext. Oneusefulfactaboutconvexpolyhedron, whichwestatewithoutdemonstration,isthefollowing.

Proposition 3.1.2. Atanyvertexofaconvexpolyhedron, thesumoftheanglesatthevertexadduptolessthan 360◦.

3.1Polyhedra–TheBasics 75

Thoughsomepolyhedraarequiteirregular, therearesomenicecategoriesofpolyhedrathatareconvenienttoworkwith. A pyramid isobtainedbytakingapolygonintheplane, takingapoint“above” thepolygon, and joining thispoint to theverticesof thepolygon. Thenewvertexisoftencalledthe conepoint ofthepyramid(itisalsoknownasthe apex ofthepyramid);thepolygonthatwestartedwithiscalledthe base ofthepyramid, andweoftencallsuchapyramidasa“pyramidoverthepolygon.” ThefamouspyramidsinGiza, Egypt, areexamplesofpyramidswithsquarebases(oftencalled“squarepyramids,” or“pyramidsoversquares”);observethatthemathematicaluseofthetermpyramidallowsforpyramidswithanybase, notjustasquarebase. SeeFigure 3.1.3 (i)forapyramidoverapentagon. A bipyramid isobtainedbytakingapolygonintheplane, takingonepoint“above”thepolygonandonepoint“below”it, andjoiningbothpointtotheverticesofthepolygon; thenewverticesarebothcalled conepoints ofthebipyramid(oneisthe“top”conepoint, andoneisthe“bottom”conepoint). SeeFigure 3.1.3 (ii)forabipyramidoverapentagon. A prism isobtainedbytakingapolygonintheplane, placinganidenticalcopyofthepolygondirectlyabovethefirst, andjoiningpairsofcorrespondingverticesofthetwocopiesofthepolygon. SeeFigure 3.1.3 (iii)foraprismoverapentagon. An antiprism isobtainedbytakingaregularpolygonintheplane, placingacopyoftheregularpolygonabovethefirst, butrotatedsothateachvertexoftheupperpolygonisabovethemiddleofanedgeofthelowerpolygon, andthenjoiningeachuppervertextothetwolowerverticesclosesttoit. SeeFigure 3.1.3 (iv)foranantiprismoverapentagon.

(i) (ii)

(iii) (iv)

Figure3.1.3

Theabovecategoriesofpolyhedracanoverlap. Forexample, as thereadercanverify, thebipyramidoverasquare, calledanoctahedron, isinfactalsoanantiprismoveratriangle.

76 3. Polyhedra

Exercise 3.1.1. Foreachofthefollowingquestions, iftheanswerisyes, giveanexample,andiftheanswerisno, explainwhynot. (Toexplainwhyapolyhedroncannotbeintwodifferentcategories, itdoesnotsufficesimplytostatethatthetwocategoriesareconstructeddifferently, becausesometimestwodifferentconstructionscanyieldthesameresult, forexampletheoctahedron, whichisbothabipyramidandanantiprism.)

(1) Canapolyhedronbebothabipyramidandaprism?

(2) Canapolyhedronbebothaprismandanantiprism?

(3) Canapolyhedronbebothapyramidandeitheraprismoranantiprism?

(4) Canapolyhedronbebothapyramidandabipyramid? (Ifyouthinkthattheanswerisno, it isnot sufficient simply to say thatapyramidhasoneconepoint, andabipyramidhastwoconepoints. Perhapsifyoulookatacertainpolyhedroninonewaythereisoneconepoint, andviewedanotherwaytherearetwoconepoints;perhapsnot.)

Exercise 3.1.2. Findallpyramidsthathaveallregularfaces.

Justasthereareformulasfortheareasofsimpletypesofpolygons, therearealsoformulasforthevolumesofsimpletypesofpolyhedra. However, thedemonstrationsofthesevolumeformu-lasaremorecomplicatedthanforareaformulas, andsowewillstatethefollowingpropositionwithoutdemonstration. (See[Har00, Sections26-27]foratechnicaldiscussionofvolumes.) Justaswediscussedthenotionofthealtitudeofatriangle, parallelogramortrapezoid, wecansimi-larlydefinethenotionofanaltitudeforpolyhedrasuchaspyramidsandprisms; weomitfurtherdetails. A bipyramidismadeupoftwopyramidsgluedtogether, andeachofthesepyramidshasanaltitude.

Proposition 3.1.3.

1. Supposethataprismhasabaseofarea b, andaltitudeoflength h. Thenthevolumeoftheprismis bh.

2. Supposethatapyramidhasabaseofarea b, andaltitudeoflength h. Thenthevolumeofthepyramidis 1

3bh.

3. Supposethatabipyramidhasabaseofarea b, andaltitudesoflength h1 and h2 foreachofthetwopyramidsinthebipyramid. Thenthevolumeofthebipyramidis 1

3b(h1+h2).

Itisinterestingtocomparethevolumeformulaforpyramidswiththeareaformulafortriangles,andtocomparethevolumeformulaforprismswiththeareaformulaforparallelograms.

3.1Polyhedra–TheBasics 77

Givenaconvexpolyhedron, wecanformanewpolyhedron, calledits dualpolyhedron, asfollows. First, foreachfaceoftheoriginalpolyhedron, chooseapointinitsinterior(forexample,choosethecenterofgravityoftheface). Thesechosenpointswillbetheverticesofthedualpolyhedron, calleddualvertices. Next, consideranedgeintheoriginalpolyhedron. Thisedgeiscontainedinpreciselytwofacesoftheoriginalpolyhedron. Wethenputanedgeinthedualpolyhedronjoiningthetwodualverticesthatarecontainedinthesetwofacesoftheoriginalpolyhedron. Wethusobtaintheedgesofthedualpolyhedron. Finally, consideravertexintheoriginalpolyhedron. Thisvertexiscontainedinsomefacesoftheoriginalpolyhedron. Wethenputafaceinthedualpolyhedronthathasasitsverticesthedualverticesthatarecontainedinthesefacesoftheoriginalpolyhedron. Wethusobtainthefacesofthedualpolyhedron.Forexample, supposeouroriginalpolyhedronisacube, asshowninFigure 3.1.4 (i). Thedual

ofthecubeisshowninsidethecubeinFigure 3.1.4 (ii). Thisdualpolyhedronhassixvertices,twelveedgesandeightfaces(itiscalledanoctahedron, andwewillencounteritagaininthenextsection).

(i) (ii)

Figure3.1.4

Exercise 3.1.3.

(1) Whatisthedualtoabipyramidoveran n-gon?

(2) Whatisthedualtoapyramidoveran n-gon?

Exercise 3.1.4. Suppose P isaconvexpolyhedron. Whatistherelationbetween P andthedualofthedualof P?

78 3. Polyhedra

3.2 RegularPolyhedra

Justasregularpolygonswerethemost“uniform”polygonspossible, wewanttofindpolyhedrathatareas“uniform”aspossible. Forapolygontoberegular, itneedstosatisfytworequirement,namelythatalltheedgeshavethesamelengths, andthatalltheanglesareequal(requiringonlyedgesofequallengthdoesnotsuffice). Wewillneedsimilarrequirementstoinsurethatapolyhedroniscompletelyuniform. A convexpolyhedronisa regularpolyhedron ifthefollowingthreeconditionshold: (1)everyfaceisaregularpolygon; (2)allfacesareidentical; and(3)allverticesareidentical, whichmeansthatallverticesarecontainedinthesamenumberoffaces.Itisnothardtoseethatinaregularpolyhedron, alltheedgesmusthavethesamelength.Thatthefirsttwopartsoftheabovedefinitiondonotsufficecanbeseenbyconsideringa

bipyramidoveranequilateraltriangle, asshowninFigure 3.2.1. Ifalltheedgesofthispolyhe-dronhaveequallength, thenconditions(1)and(2)oftheabovedefinitionwillbesatisfied, butwestillwouldnotwanttocallthispolyhedronregular, becausetheverticesdonotall“lookthesame.” Moreprecisely, twooftheverticesarecontainedinthreetriangleseach, whereasthreeoftheverticesarecontainedinfourtriangleseach. Henceweneedcondition(3)ofthedefintionofregularpolyhedra.

Figure3.2.1

In thecaseofpolygons, wesawthat therewere infinitelymanydistinct regularpolygons,one foreachpossiblenumberof edges. That is, there is a regular 3-gon (alsoknownasanequilateraltriangle), aregular 4-gon(alsoknownasasquare), aregular 5-gon, aregular 6-gon,etc. (Ofcourse, eachoneofthesepolygonscouldbeconstructedindifferentsizes, butweareonlyinterestedinshapesthataregenuinelydifferent.) Thefollowingresultshows, somewhatsurprisingly, thatwhatholdsforregularpolygonsdoesnotholdforregularpolyhedra.

Proposition 3.2.1. Every regular polyhedron is one of the five polyhedra described inTa-ble 3.2.1.

Demonstration. Recall thatall regularpolyhedraareassumedtobeconvex. Thekeyto thisdemonstrationisProposition 3.1.2, whichsaysthatatanyvertexofaconvexpolyhedron, thesumoftheanglesatthevertexmustadduptolessthan 360◦.

3.2RegularPolyhedra 79

Name Faces FacesperVertextetrahedron 4 triangles 3 trianglespervertexcube(akahexahedron) 6 squares 3 squarespervertexoctahedron 8 triangles 4 trianglespervertexdodecahedron 12 pentagons 3 pentagonspervertexicosahedron 20 triangles 5 trianglespervertex

Table3.2.1

Allthefacesinaregularpolyhedronarethesame(andareregularpolygons), andallverticesarecontainedinthesamenumberoffaces. Letusstartbyexaminingthesituationwhenallthefacesofaregularpolyhedronareequilateraltriangles. Weknowthattheanglesinanequilateraltriangleareall 60◦. Howmanyequilateraltrianglescancontaineachvertexofaregularpoly-hedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Weobservethat 3 · 60◦ = 180◦, that 4 · 60◦ = 240◦, that 5 · 60◦ = 300◦ andthat 6 · 60◦ = 360◦. Hence,weseethatitmightbepossibletohavearegularpolyhedronwithfacesthatareequilateraltri-angles, andwitheither 3, 4 or 5 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareequilateraltriangles, andwith 6 ormorefacescontainingeachvertex. Thereindeedexistregularpolyhedrawithfacesthatareequilateraltriangles, andwitheither 3, 4 or 5 facescontainingeachvertex, namelythetetrahedron, theoctahedronandtheicosahedron. Thereexistnootherpolyhedrasatisfyingthesamecriteriaforthetypesoffacesandvertices. Wehavethereforefoundalltheregularpolyhedrawithfacesthatareequilateraltriangles.Next, letusexaminethesituationwhenallthefacesofaregularpolyhedronaresquares. We

knowthattheanglesinasquareareall 90◦. Howmanysquarescancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan360◦? Weobservethat 3 · 90◦ = 270◦ andthat 4 · 90◦ = 360◦. Hence, weseethatitmightbepossibletohavearegularpolyhedronwithfacesthataresquares, andwith 3 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthataresquares,andwith 4 ormorefacescontainingeachvertex.Weseethatacubeisaregularpolyhedronwithfacesthataresquares, andwith 3 facescontainingeachvertex; thecubeisuniqueinsatisfyingthisproperty. Wehavethereforefoundalltheregularpolyhedrawithfacesthataresquares.Now, let us examine the situationwhenall the facesof a regular polyhedronare regular

pentagons. WeknowfromTable 2.3.1 thattheanglesinaregularpentagonareall 108◦. Howmanyregularpentagonscancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Weobservethat 3 · 108◦ = 324◦ andthat 4 · 108◦ = 432◦. Hence, weseethatitmightbepossibletohavearegularpolyhedronwithfacesthatareregularpentagons, andwith 3 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareregularpentagons, andwith 4 ormore facescontainingeachvertex. A dodecahedron isa regularpolyhedronwith faces thatareregularpentagons, andwith 3 facescontainingeachvertex; thedodecahedronisuniquein

80 3. Polyhedra

satisfyingthisproperty. Wehavethereforefoundalltheregularpolyhedrawithfacesthatareregularpentagons.Finally, letusexaminethesituationwhenallthefacesofaregularpolyhedronareregular

polygonswith 6 ormoreedges. WeknowfromTable 2.3.1 thattheanglesinaregularhexagonareall 120◦, andthattheanglesinaregularpolygonwithmorethan 6 edgesarelargerthan120◦. Howmanyregularpolygonswith 6 ormoresidescancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Theanswerisnone, giventhat 3 · 120◦ = 360◦, and 3 timesananglelargerthan 120◦ wouldbemorethan 360◦. Giventhateveryvertexinapolyhedronmustbecontainedinatleastthreefaces, weseethat itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareregularpolygonswith 6 ormoreedges.Puttingalltheabovetogether, weseethattherearepreciselyfiveregularpolyhedra, aslisted

inTable 3.2.1.

Thefiveregularpolyhedra, whicharelistedinTable 3.2.1, areshowninFigure 3.2.2. Itcanbeshownthat theverticesofeachregularpolyhedronlieonasphere(seeTheorem44.4in[Har00]fordetails). Noticethattheregularpolyhedraarenamedbythenumberoffaceseachonehas. Thefiveregularpolyhedraarealsoknownasthe platonicsolids, inhonorofthean-cientGreekphilosopherPlato. (Forthosefansofchildren’sliterature, youmightknowoftheDodecahedronin“ThePhantomTollbooth;” ifyouarenotfamiliarwiththisbook([Jus61]), itishighlyrecommended.)

Webrieflymentionedthenotionofadualpolyhedron intheprevioussection. Letuslookatthedualsofeachofthefiveregularpolyhedra. Wealreadysawintheprevioussectionthatthedualof thecube is theoctahedron. The reader canverify, by sketching theappropriatepicture, thatthedualoftheoctahedronisthecube. Similarly, itcanbeseenthatthedualofthedodecahedronistheicosahedron, andthedualoftheicosahedronisthedodecahedron.Thedualof the tetrahedron is simply itself. We therefore see that the regularpolyhedraareself-containedinaverynicearrangementwhenitcomestoduality.

Exercise 3.2.1.

(1) Whichoftheregularpolyhedraarepyramids?

(2) Whichoftheregularpolyhedraarebipyramids?

(3) Whichoftheregularpolyhedraareprisms?

(4) Whichoftheregularpolyhedraareantiprisms?

Exercise 3.2.2. Findallconvexpolyhedrathatarebothbipyramidsandantiprisms.

3.3Semi-RegularPolyhedra 81

Tetrahedron Cube Octahedron

Dodecahedron Icosahedron

Figure3.2.2

Exercise 3.2.3. Supposethatyouhaveacubemadeoutofclay; supposefurtherthattheclayisred, buttheoutsideofthecubeispaintedblue. Youthenslicethecubeinastraightlinewithaknife, causingthecubetobreakintotwopieces. Eachpiecehasanexposedredpolygon, wherethecubewassliced. Dependinguponhowyouslicethecube, youmightgetdifferentexposedpolygons; alltheexposedpolygonswillbeconvex. Forexample, ifyousliceparalleltooneofthefacesofthecube, yourexposedpolygonwillbeasquare;ifyousliceoffacornerofthecuberightnexttoavertex, yourexposedpolygonwillbeatriangle. Whatareallthepossibleexposedpolygonsthatcouldbeobtainedbyslicingthecube?

3.3 Semi-RegularPolyhedra

Regularpolyhedra, asdiscussedintheprevioussection, arethemostuniformpolyhedra. Wenowturntoaslightlybroadercategoryofpolyhedra, namelythe semi-regularpolyhedra, whichareconvexpolyhedrathatsatisfythefollowingtwocondition: (1)everyfaceisaregularpoly-gon; (2)allverticesare identical, whichmeans thatallverticesarecontainedinsametypes

82 3. Polyhedra

ofpolygonsarrangedinthesameorder. Observethatthefacesarenotallrequiredtobethesametypeofpolygon. Itisthecasethatalledgesinasemi-regularpolyhedronhavethesamelength. Certainlyeveryregularpolyhedronisalsosemi-regular, buttherearesemi-regularpoly-hedrathatarenotregular. Forexample, weseeinFigure 3.3.1 asemi-regularpolyhedroncalledtherhombicuboctahedron, thefacesofwhichareallequilateraltrianglesandsquares, andtheverticesofwhichareeachcontainedinthreesquaresandonetriangle.

Figure3.3.1

Supposewearegivenavertexinapolyhedron. Wedefinethe vertexconfiguration ofthisvertex tobea listofnumbersof the form (a1, a2, . . . , an), where thenumbers a1 throughan arethenumbersofedgesofthepolygonscontainingthevertex, listedinorderaswegoaroundthevertex(itdoesnotmatterwhichpolygonwestartwith). Forexample, inthepolyhe-dronshowninFigure 3.3.1, everyvertexhasvertexconfiguration (3, 4, 4, 4). Thedefinitionofsemi-regularpolyhedraimpliesthatinanysemi-regularpolyhedron, allverticeshavethesamevertexconfiguration.Thefollowingpropositiontellsusallthepossiblesemi-regularpolyhedra.

Proposition 3.3.1. Everysemi-regularpolyhedronisoneofthefollowing:

(A) A regularpolyhedron.

(B) A prismoveraregularpolygon, withthesidesmadeupofsquares.

(C) Anantiprismoveraregularpolygon, withthesidesmadeupofequilateraltriangles.

(D) Oneofthe 14 polyhedradescribedinTable 3.3.1.

Wewillnotdemonstratetheaboveproposition. Thedemonstrationissimilarto, thoughmorecomplexthan, thedemonstrationshowingthatthereareonlyfiveregularpolyhedra(Proposi-tion 3.2.1); formoredetails, see, forexample, theproofofTheorem46.1in[Har00]. Picturesofthe 14 polyhedralistedinTable 3.3.1 aregiveninFigure 3.3.2. ObserveinTable 3.3.1 thattherhombicuboctahedronandpseudorhombicuboctahedronhavethesamevertexconfigurations,andthesamenumbersoffacesofeachtype; however, thesepolyhedraarenotidentical. TherhombicuboctahedronisshowninthelowerleftcornerofFigure 3.3.2, andthepseudorhom-bicuboctahedronisshowninthelowerrightcornerofthefigure. Thepseudorhombicubocta-hedroncanbeobtainedfromtherhombicuboctahedronbyrotatingthetop“cap”by 45◦.

3.3Semi-RegularPolyhedra 83

Name VertexConfig. TypesofFacestruncatedtetrahedron (3, 6, 6) 4 trianglesand 4 hexagonstruncatedoctahedron (4, 6, 6) 6 squaresand 6 hexagonstruncatedicosahedron (5, 6, 6) 12 pentagonsand 20 hexagonstruncatedcube (3, 8, 8) 8 trianglesand 6 octagonstruncateddodecahedron (3, 10, 10) 20 trianglesand 12 10-gonstruncatedcuboctahedron (4, 6, 8) 12 squares, 8 hexagonsand 6 octagonstruncatedicosidodecahedron (4, 6, 10) 30 squares, 20 hexagonsand 12 10-gonscuboctahedron (3, 4, 3, 4) 8 trianglesand 6 squaresicosidodecahedron (3, 5, 3, 5) 20 trianglesand 12 pentagonsrhombicosidodecahedron (3, 4, 5, 4) 20 triangles, 30 squaresand 12 pentagonssnubcube (3, 3, 3, 3, 4) 32 trianglesand 6 squaressnubdodecahedron (3, 3, 3, 3, 5) 80 trianglesand 12 pentagonsrhombicuboctahedron (3, 4, 4, 4) 8 trianglesand 18 squarespseudorhombicuboctahedron (3, 4, 4, 4) 8 trianglesand 18 squares

Table3.3.1

Figure3.3.2

The14polyhedralistedinTable 3.3.1 areoftencalledthe Archimedeansolids. DonotgetcaughtupindecipheringthenamesoftheArchimedeansolids; itisnotimportant. Moreover,notallauthorsusethesamenamesforthe 14 polyhedralistedinTable 3.3.1. Theonewordfromthenamesofthesemi-regularpolyhedrathatisworthmentioningis“truncated.” Totruncateapolyhedron, wesimplychopoffasmallpiecearoundeachvertex. Forexample, atruncatedcubewillhaveeightsmalltriangles(oneforeachoriginalvertexofthecube), andsixoctagons(oneforeachoriginalsquarefaceofthecube). ThereaderisencouragedtolocatethetruncatedcubeinFigure 3.3.2.

84 3. Polyhedra

The terms “semi-regularpolyhedron”and “Archimedean solid”areusedwith somevaria-tionintheliterature. Sometextsaddapropertycalled“vertextransitivity” tothedefinitionofsemi-regular (thoughwedonot). Wedonotyethave the tools toexplainvertex transitivity,thoughitwillbeexplainedinSection 5.7. ItturnsoutthatallthepolyhedralistedinProposi-tion 3.3.1 exceptforone, thepseudorhombicuboctahedron, satisfyvertextransitivity. Hence,thosetextsthatrequirevertextransitivitylistonly13Archimedeansolids.Observethatthereareinfinitelymanydifferentsemi-regularpolyhedra, becausetherearein-

finitelymanydifferentsemi-regularprismsandantiprisms(therebeingoneofeachforeachpos-sibleregularpolygon). However, becausetheregularpolyhedra, theprismsandtheantiprismsareallotherwiseknownpolyhedra, sometextsfocusontheArchimedeansolidswhentheydis-cusssemi-regularpolyhedra. Wenotethat, aswasthecaseforregularpolyhedra, theverticesofeachsemi-regularpolyhedronlieonasphere(seeCorollary46.2in[Har00]fordetails).

Exercise 3.3.1. WhatcanbesaidaboutthefacesofthedualofanArchimedeansolid?

Exercise 3.3.2. Weknowthatthedualofeachregularpolyhedronisitselfaregularpoly-hedron. Canithappenthatthedualofanon-regularsemi-regularpolyhedronisitselfasemi-regularpolyhedron? Iftheanswerisyes, giveanexample, andiftheanswerisno,explainwhynot.

3.4 OtherCategoriesofPolyhedra

Intheprevioustwosectionswediscussedregularpolyhedraandsemi-regularpolyhedra, whicharetypesofconvexpolyhedrathatsatisfycertainniceproperties. Inthissectionwediscusstwofurthertypesofrelativelynicepolyhedra. Westartwith deltahedra, whichareconvexpolyhedramadeupentirelyofequilateraltriangles. Wehavealreadyencounteredthreesuchpolyhedra,namelythetetrahedron, theoctahedronandtheicosahedron. Therearesomeotherdeltahedra,besidesthesethree, thoughtheyareneitherregularnorsemi-regular(thatis, notallverticesarecontainedinsametypesofpolygonsarrangedinthesameorder).


Therearefivedeltahedrathatareneitherregularnorsemi-regular. Trytofindasmanyoftheseasyoucan.

Thefollowingpropositionlistsallthedeltahedra.

Proposition 3.4.1. Everyconvexdeltahedroniseitheraregularpolyhedron(atetrahedron, anoctahedronoranicosahedron), orisoneofthe 5 polyhedralistedinTable 3.4.1.

3.4OtherCategoriesofPolyhedra 85

Name Facestriangularbipyramid 6 trianglespentagonalbipyramid 10 trianglessnubdisphenoid 12 trianglestricappedtriangularprism 14 trianglesbicappedsquareantiprism 16 triangles

Table3.4.1

Wenotethatnotallauthorsusethesamenamesforthefivenon-regularconvexdeltahedrathatwehaveusedintheabovetable, thoughthereisnodisagreementovertheactualpolyhe-dra, regardlessoftheirnames. Picturesofthe 5 polyhedra listed inTable 3.4.1 areshowninFigure 3.4.1.

triangular bipyramid

tricapped triangular prism

bicapped square antiprismsnub disphenoid

pentagonal bipyramid

Figure3.4.1

Wenextturntoanevenmorebroadcategoryofpolyhedra, namelythe face-regularpolyhe-dra, whicharepolyhedrathathaveallregularfaces, thoughnotallfacesarenecessarilythesame, andnotallverticesnecessarilyhavethesameconfigurationsoffacescontainingthem.(Wefollow[Har00]inusingtheterm“face-regular.”) Thiscategoryincludesallregularpolyhe-dra, allsemi-regularpolyhedraandalldeltahedra, butthereareothersaswell. Forexample,placingapyramidwithsquarebaseandequilateralsidesontopofacubeyieldstheface-regularpolyhedronshowninFigure 3.4.2. Aswasthecaseforregularandsemi-regularpolyhedra, alltheedgesinaface-regularpolyhedronmusthavethesamelength.

86 3. Polyhedra

Figure3.4.2

Itturnsoutthattherearealimitednumberofconvexface-regularpolyhedra, thoughtheproofislengthy, andisbeyondthescopeofthisbook. Fortherecord, however, westatethefollowingProposition.

Proposition 3.4.2. Everyconvexface-regularpolyhedroniseitherasemi-regularpolyhedron,oroneof 91 otherpolyhedrathatarenotsemi-regular.

Ofthe 91 convexnon-semi-regularface-regularpolyhedramentionedintheabovePropo-sition, fivearethenon-regulardeltahedralistedinProposition 3.4.1. Weshouldmentionthatsometextslist 92 convexnon-semi-regularface-regularpolyhedra, becausetheydonotcon-siderthepseudorhombicuboctahedron tobesemi-regular(asmentionedinSection 3.3). These91 (or, insometexts, 92)polyhedraaresometimesreferredtoasthe Johnsonsolids, namedafterthepersonwhofirstpublishedthecompletelistofthesepolyhedra(see[Joh66], whichisverytechnical).HavingshownoneoftheJohnsonsolidsinFigure 3.4.2, weleaveittothereadertofindsome

othersinthefirsttwoofthefollowingExercises. Wenotethatallface-regularpyramidswerefoundinExercise 3.1.2.

Exercise 3.4.1. Findallconvexface-regularpolyhedrathathaveidenticalfaces, otherthanthedeltahedra.

Exercise 3.4.2. FindatleasttwoJohnsonsolidsthatarenotpyramids, bipyramidsordelta-hedra, andaredifferentfromtheoneshowninFigure 3.4.2.

3.5EnumerationinPolyhedra 87

Exercise 3.4.3. A face-regularpolyhedroniscalled elementary ifitcannotbebrokenupintotwoormoreface-regularpolyhedrathatarejoinedalongacommonface. Forexample,theoctahedron isnotelementary, because itcanbebrokenup into twopyramidswithsquarebases. Findatleastoneothernon-elementaryface-regularpolyhedron, andatleastoneelementaryface-regularpolyhedron.

Exercise 3.4.4. Supposewehavetwoface-regularpolyhedra, andoneofthefacesinthefirstpolyhedronisidenticaltooneofthefacesinthesecondpolyhedron. Wecanthengluethetwopolyhedraalongtheiridenticalfaces, yieldingonelargerpolyhedra. Forexample,startingwithacubeandapyramidwithasquarebaseandequilateralsides, andgluingthetwoalongasquarefaceineach, resultsinthepolyhedronshowninFigure 3.4.2.

(1) Is thepolyhedron that results from thegluing two face-regularpolyhedraby thisprocessnecessarilyface-regular? Explainyouranswer.

(2) Supposetheoriginaltwoface-regularpolyhedrawerebothconvex. Isthepolyhe-dronthatresultsfromthegluingnecessarilyconvex? If theanswerisyes, explainwhy, andiftheanswerisno, giveanexampletoshowwhynot.

Exercise 3.4.5. Showthatthereareinfinitelymanynon-convexface-regularpolyhedra.

3.5 EnumerationinPolyhedra

Oneofthenicefeaturesofpolyhedra(asopposedto“smooth”objects, suchasspheres)isthattheyoffersomethingstobecounted, namelythenumberofvertices, thenumberofedgesandthenumberoffaces. Givenapolyhedron, welet V , E and F, respectively, denotethenumberofvertices, edgesandfacesofthepolyhedron. Forexample, foracubewehave V = 8, andE = 12, and F = 6. Forconvenience, wewritethesethreenumbersas (V, E, F), calledthefacevector ofthepolyhedron. Hence, thefacevectorofthecubeis (8, 12, 6).

88 3. Polyhedra

Exercise 3.5.1.

(1) Whatisthefacevectorofeachoftheregularpolyhedra?

(2) Whatisthefacevectorofeachofthesemi-regularpolyhedra?

Exercise 3.5.2. Findaconvexpolyhedronsuchthatneither E nor F isdivisibleby 3.

Exercise 3.5.3.

(1) Whatisthefacevectorofapyramidoveran n-gon?

(2) Whatisthefacevectorofabipyramidoveran n-gon?

(3) Whatisthefacevectorofaprismoveran n-gon?

(Note: Youranswerforeachpartofthisexercisewillinvolve“n.”)

Exercise 3.5.4. Findabipyramidthathasthesamefacevectorastheicosahedron.

Exercise 3.5.5. Supposethataconvexpolyhedron P hasfacevector (V, E, F). Whatisthefacevectorofthedualof P?

Foreachpolyhedron, wecandetermine its facevector. It canhappen, however, that twodifferentpolyhedrahavethesamefacevector(justastwodifferentpeoplecanhavethesameheightandweight). Forexample, thetwopolyhedrashowninFigure 3.5.1 havethesamefacevectors. Ofcourse, iftwopolyhedrahavedifferentfacevectors, theycannotbethesame.

Whatcanbesaidaboutthenumbers V , E and F thatarisefrompolyhedra? Westartwiththefollowingverysimpleresult, whichfollowsfromthefactthatapolyhedronisasolidobjectinthreedimensionalspace.


(i) (ii)

Figure3.5.1

Proposition 3.5.1. Supposethat P isapolyhedron.

1. V ≥ 4.

2. E ≥ 6.

3. F ≥ 4.

Inordertosaymoreabout V , E and F, weneedthefollowingnotion. Ratherthansimplylookingatthenumberoffaces, namely F, wewanttolookmorefinely, andcountthenumberoffaceswith 3 edgeseach, denoted F3, thenumberoffaceswith 4 edgeseach, denoted F4, etc.Ingeneral, foreachpositiveinteger n (where n ≥ 3), let Fn denotethenumberoffaceswith nedgeseach. Similarly, foreachpositiveinteger n (where n ≥ 3), let Vn denotethenumberofverticescontainedin n edgeseach. Forexample, forthepolyhedronshowninFigure 3.4.2, wehave F3 = 4, F4 = 5, F5 = 0, F6 = 0, etc.; wealsohave V3 = 4, V4 = 5, V5 = 0, V6 = 0,etc. Thefollowingresultwillbeusefulinprovingvariouspropositionsofinterest.


1. F = F3 + F4 + F5 + F6 + · · · .

2. V = V3 + V4 + V5 + V6 + · · · .

3. 2E = 3F3 + 4F4 + 5F5 + 6F6 + · · · .

4. 2E = 3V3 + 4V4 + 5V5 + 6V6 + · · · .

Demonstration.

(1). Thisequationisevidentlytrue, becauseeveryfacehasatleastthreeedges(byProposi-tion 3.1.1 (3)), andisthuscountedpreciselyoneamong F3, F4, F5, etc.

(2). Thisequationisevidentlytrue, becauseeveryvertexiscontainedinatleastthreeedges(byProposition 3.1.1 (2)), andisthuscountedpreciselyoneamong V3, V4, V5, etc.

90 3. Polyhedra

(3). Thesum 3F3 + 4F4 + 5F5 + 6F6 + · · · countsalltheedgesthatarecontainedinallthefacesofthepolyhedron. However, eachedgeofthepolyhedroniscontainedinpreciselytwofaces, usingProposition 3.1.1 (1). Therefore, thesum 3F3+4F4+5F5+6F6+ · · · countseachedgetwice, andsoitequalstwicethenumberofedges. Inotherwords, wehave 3F3 + 4F4 +5F5 + 6F6 + · · · = 2E.

(4). ThisargumentisverymuchliketheoneforPart (3)ofthisproposition, usingthefactthateachedgeofthepolyhedroncontainstwovertices.

Part (3)oftheabovepropositionboilsdowntoamuchsimplerstatementinthecasewhereallthefaceshavethesamenumberofedges. Moreprecisely, if P isapolyhedronsuchthatallitsfacesare n-gons, then nF = 2E. ThereaderisaskedtodemonstratethisfactinExercise 3.5.6.Intheparticularcasewhere P hasalltriangularfaces, then 3F = 2E.

Exercise 3.5.6. [UsedinThisSection] Supposethat P isapolyhedron.

(1) Supposethatallthefacesof P are n-gons. Showthat nF = 2E.

(2) Supposethatallthefacesof P aretriangles. Showthat F isdivisibleby 2, and E isdivisibleby 3.

(3) Supposethateveryvertexof P iscontainedin q edges. Showthat qV = 2E.


Arethereanyinterestingrelationsbetweenthethreenumbers V , E and F thatholdforallpolyhedra, or, atleast, allconvexpolyhedra? Trytolookforsuchrelationshipsyourself.Lookat thevaluesof V , E and F forvariousexamplesofpolyhedra, for instance theregularandsemi-regularpolyhedra; canyoufindanypatterns? Canyoufindanyrelationsbetween V , E and F thatholdsinallexamplesyouhaveexamined?

Thereareindeedrelationsbetweenthenumbers V , E and F thatholdforallpolyhedra. Oneexampleofsucharelationisgiveninthefollowingproposition.


1. E ≥ 3

2V .

2. E ≥ 3

2F.

Demonstration.


(1). UsingProposition 3.5.2 (4)(2), inthatorder, weseethat

2E = 3V3 + 4V4 + 5V5 + 6V6 + · · ·≥ 3V3 + 3V4 + 3V5 + 3V6 + · · · = 3(V3 + V4 + V5 + V6 + · · · ) = 3V.

(2). ThiscaseisverysimilartoPart (1)ofthisproposition; thedetailsarelefttothereader.

Itfollowsfromtheabovepropositionthatforanypolyhedron, thenumberofedgesisalwaysgreaterthanboththenumberofverticesandthenumberoffaces.A moresubstantial(andmoredifficulttodemonstrate)relationbetweenthenumbers V , E and

F thatholdsforanyconvexpolyhedronisthefollowingproposition, knownasEuler’sFormula;itisduetothegreatmathematicianLeonhardEuler (1707-1783).

Proposition 3.5.4 (Euler’sFormula). Foranyconvexpolyhedron, wehave

V − E+ F = 2.

Demonstration. Supposewearegivenaconvexpolyhedron P. Wewant tofigureoutwhatV − E+ F equals. Thefirststepinthisdemonstrationistoformtheprojectionof P, whichwenowdescribe.Consider, forexample, acube. Imaginethatthecubeismadenotofsquaresgluedtogether,

butisjustawireframe. Wecouldthenputalightrightabovethewireframecube, andapieceofpaperbelowthecube, asshowninFigure 3.5.2 (i). Thelightcastsashadowonthepaper; theshadowispicturedinFigure 3.5.2 (ii). Wecallthisshadowthe projection ofthecube. Noticethattheprojectionismadeupofedgesandvertices, andthattheseedgesandverticesdivideuptheplaneintoanumberofregions. Weobservefurtherthatthenumberofverticesintheprojectionisthesameasthenumberofverticesintheoriginalcube, namely 8, andthatthenumberofedges in theprojection is thesameas thenumberofedges in theoriginalcube,namely 12. Further, wenoticethattheprojectiondividesuptheplaneinto 6 regions(ofwhich5 are“bounded”and 1 is“unbounded”), andthatthisnumberofregionsequalsthenumberoffacesofthecube.

Wecouldsimilarlyformtheprojectionofanyconvexpolyhedron P. InFigure 3.5.2 (iii)weseetheprojectionoftheregulartetrahedron. Thesameobservationaboutnumbersofvertices,edgesandregionsthatheldfortheprojectionofthecubeholdsfortheprojectionofanyconvexpolygon. Thatis, thenumberofverticesoftheprojectionequals V , thenumberofedgesoftheprojectionequals E, andthenumberofregionsintheplaneformedbytheprojectionequals F.Therefore, tofigureoutV−E+F for P, wecanjustaswellfigureoutV−E+F fortheprojection(where V and E nowmeanthenumberofverticesandedgesrespectivelyoftheprojection, andF nowmeansthenumberofregions). Itturnsoutthattheprojectionismucheasiertoworkwiththantheoriginalpolyhedron.Wenowproceedbymodifyingtheprojectiononestepatatime. Wewillverifythateach

modificationdoesnotchangethesum V − E+ F, thoughitmightchangeindividualvaluesofV , E and F. First, weaskwhethertheedgesontheprojectionhaveanycomplete“loops.” For

92 3. Polyhedra

(i) (ii) (iii)

Figure3.5.2

example, supposetheprojectionisasshowninFigure 3.5.3 (i); thereareanumberofcompleteloopsinthisprojection, forexample, theedgeslabeled a, b, c and d formaloop. Weproceedasfollows. Supposewehavealoopinourprojection. Thenchooseoneoftheedgesthatmakeuptheloop(itdoesnotmatterwhich), andwewillremovethatedge. InFigure 3.5.3 (ii), wehaveremovedtheedgelabeled b. Theresultisanewconfigurationofvertices, edgesandregions.Whathappensto V , E and F asaresultofremovingoneedgefromaloop? Weobservethat Visunchanged, that E decreasesby 1, andthat F decreasesby 1. Hence, thenewvalueofthesumwearecalculatingis

V − (E− 1) + (F− 1) = V − E+ F.

Inotherwords, thesumV−E+F isthesameinthenewconfigurationasintheoldconfiguration.Nowconsiderthenewconfigurationofvertices, edgesandregions. Ithasfewerloopsthan

theoriginalprojection. Iftherearestillloops, thenremoveanotheredgefromoneoftheloops.Keepremovingoneedgeatatimeuntiltherearenomoreloopsleft. Afterallthenecessaryremovals, thevalueof V − E + F isstillunchanged, butwenowhaveasimplersituation, inthattherearenoloops. Forexample, weseeinFigure 3.5.3 (iii)onepossibleresultofremovingedgesfromalltheloopsinFigure 3.5.3 (i). (Therearedifferentchoicesforremovingedgesfromloopsateachstage, sotheresultingconfigurationcanvary, butitwillmakenodifference.)Wenowperformadifferent typeofmodification. Becauseour newconfigurationhas no

loops, itmusthave“free”vertices; thatis, verticesthataretheendpointsofonlyoneedge. Forexample, thevertexlabeled A inFigure 3.5.3 (iii)isafreevertex. Wethenchooseafreevertexinourconfigurationwithoutloops, andthenremoveboththefreevertexandthesingleedgeofwhichthevertexisanendpoint; weleaveinplacetheotherendpointoftheremovededge. InFigure 3.5.3 (iv), wehaveremovedthevertex A andtheedgethathas A asanendpoint. Whathappensto V , E and F asaresultofthisnewtypeofprocedure? Weobservethat V isdecreasesby 1, that E decreasesby 1, andthat F isunchanged. Hence, thenewvalueofthesumweare


calculatingis(V − 1)− (E− 1) + F = V − E+ F.

Inotherwords, thesum V − E + F isonceagainthesameinthenewconfigurationasintheoldconfiguration.Nowconsiderthenewconfigurationofvertices, edgesandregions. Ithasfeweredgesthanthe

originalprojection. Iftherearestilledges, thentheremuststillbefreevertices. Keepremovingoneedgeatatimeuntilthereisonlyoneedgeleft. Then, nomatterwhattheoriginalprojectionwas, thefinalconfigurationwill lookliketheoneshowninFigure 3.5.3 (v). Ateachstepofourprocedure, thevalueof V − E+ F neverchanged. Hence, ifwecancomputethevalueofV − E+ F forthefinalconfiguration, thatwillbethesamevalueasfortheoriginalprojection,andhencefortheoriginalpolyhedron. InFigure 3.5.3 (v)weseethat V = 2, that E = 1 andthat F = 1. Hence V − E + F = 2 − 1 + 1 = 2. Therefore V − E + F = 2 fortheoriginalpolyhedron, whichiswhatwewantedtoshow.

(i) (ii) (iii)

(iv) (v)

a

b

c

d

a c

A

Figure3.5.3

Euler’sFormulahasmanyuses. Forexample, wecanuseittodeducefurtherrelationsbetweenthenumbers V , E and F forconvexpolyhedra. Onesuchresultisthefollowingproposition.

Proposition 3.5.5. Supposethat P isaconvexpolyhedron.

94 3. Polyhedra

1. 4 ≤ F ≤ 2V − 4.

2. 4 ≤ V ≤ 2F− 4.


Demonstration.

(1). IfwemultiplyEuler’sFormulaby 2 wehave

2V − 2E+ 2F = 4.

Next, weknowfromProposition 3.5.3 (2)that 2E ≥ 3F. Ifwesubtracttwonumbersfromthesamenumber, thensubtractingthesmallernumbergivesabiggerresult. So, weseethat

2V − 3F+ 2F ≥ 2V − 2E+ 2F = 4.

Itfollowsthat2V − F ≥ 4.

Rearranging, weobtain2V − 4 ≥ F.

CombiningthislastresultwithProposition 3.5.1 (3), wederivethat

4 ≤ F ≤ 2V − 4.

(2). ThisargumentisverymuchliketheoneforPart (1)ofthisproposition, withtheroleoffacesandverticesinterchanged.

Exercise 3.5.7. ThisexerciseusesExercise 3.5.6.Supposethat P isaconvexpolyhedron.

(1) Supposethatallthefacesof P aretriangles. Findaformulaforeachof E and F intermsof V .

(2) Supposethatallthefacesof P arequadrilaterals. Findaformulaforeachof E andF intermsof V .

(3) Supposethatallthefacesof P arepentagons. Findaformulaforeachof E and F intermsof V .

Exercise 3.5.8. Suppose that P isaconvexpolyhedron, andthatall the facesof P aretriangles. Assumefurtherthat P hasatleastfivevertices. Istherealwaysabipyramidoversome n-gonthathasthesamefacevectoras P? Ifthereis, find n intermsofthenumberofverticesof P.

96 3. Polyhedra

Exercise 3.5.9. Supposethataconvexpolyhedron P isself-dual(thatis, thepolyhedronisit’sowndual). Findaformulaforeachof E and F intermsof V .

Exercise 3.5.10. Find all possible convex polyhedra P that are self-dual (as in Exer-cise 3.5.9), andallthefacesofwhicharetriangles.

Exercise 3.5.11. Supposethat P isaconvexpolyhedron. Showthat E ≤ 3F− 6.

Exercise 3.5.12. Supposethat P isaconvexpolyhedron. Showthat P mustcontainatleastone face thathaseither 3, 4 or 5 edges. (Inotherwords, thisexerciseshowsthat therecannotbeaconvexpolyhedronwithallfaceshaving 6 ormoreedges.)

Wearenow in aposition todiscuss a very interestingquestion regarding face vectors ofconvexpolyhedra. Weknowthateveryconvexpolyhedronhasafacevector (V, E, F). Canwegobackwardsinthisprocess? Thatis, supposewearegiventhreepositiveintegers (x, y, z);dothesethreenumbersnecessarilyformthefacevectorofaconvexpolyhedron? Theanswerisno. Forexample, supposeweweregiventhenumbers (5, 7, 1). Therecannotbeaconvexpolyhedronwiththesenumbersasitsfacevector, because F ≥ 4 foranyanyconvexpolygon(asinProposition 3.5.1 (3)). Howabout (5, 9, 7)? Wedonothaveaproblemwith V , E andF notbeinglargeenough, buttherestillcannotbeaconvexpolyhedronwiththesenumbersasitsfacevector, because 5 − 9 + 7 = 3, andsoEuler’sFormulaisnotsatisfied. Howabout(8, 11, 5)? Euler’sFormulaissatisfiedthistime, buttherestillcannotbeaconvexpolyhedrawiththesenumbersasitsfacevector, because 2 · 5 − 4 = 6, andyet 8 ≰ 6, sothesenumberdonotsatisfyProposition 3.5.5 (2).Theaboveexamplesshowussomeofthereasonswhythreepositiveintegers (x, y, z) might

notbethefacevectorofaconvexpolyhedron. ThefollowingPropositionsaysthatthesearetheonlypossiblethingsthatcangowrong.

Proposition 3.5.6. Supposewearegiventhreepositiveintegers (x, y, z). Thenthesenumbersarethefacevectorofaconvexpolyhedronifandonlyifthefollowingthreecriteriahold:

1. x− y+ z = 2.

2. 4 ≤ z ≤ 2x− 4.

3. 4 ≤ x ≤ 2z− 4.


Inotherwords, if (x, y, z) satisfytheabovethreecriteria, thentheyarethefacevectorofsomeconvexpolyhedron(possiblymorethanone); iftheydonotsatisfyallthreecriteria, thentheyarenotthefacevectorofaconvexpolyhedron. Thedemonstrationoftheabovepropositionisbeyondthescopeofthisbook.Forexample, supposewearegiventhenumbers (5, 9, 6). Itcanbeverifiedthatthesenumbers

satisfyallthreecriteriainProposition 3.5.6, andsotheymustbethefacevectorofsomepolyhe-dron. Thereaderisaskedtofindsuchapolyhedron(hint: tryconstructingpyramids, bipyramids,andthelikewith 5 vertices).

Exercise 3.5.13. Foreachofthesetsofthreenumbersgivenbelow, statewhetherornotitisthefacevectorofaconvexpolyhedron. Ifitisthefacevectorofaconvexpolyhedron,findsuchapolyhedron; ifnot, explainwhynot.

(1) (5, 10, 6).

(2) (12, 18, 8).

(3) (23, 33, 12).

(4) (10, 20, 12).

DoesEuler’sFormulaholdforallpolyhedra? Theanswerisdefinitelyno. Forexample, considerthepolyhedral“torus” showninFigure 3.5.4 (“torus”isthemathematicalnameforanythingshapedlikethesurfaceofabagel). Itisseenfromthefigurethatthefacevectorofthispolyhedronis (16, 32, 16), andhence V − E + F = 16 − 32 + 16 = 0. ThereforeEuler’sFormuladoesnotholdinthiscase. ItturnsoutthatEuler’sFormulaholdsforthosepolyhedrathatdonothave“holes”throughthem, thoughitisbeyondthescopeofthisbooktogivethedetailsofwhythisistrue.

Figure3.5.4

98 3. Polyhedra

EventhoughEuler’sformuladoesnotholdforallpolyhedra, itturnsoutthattheconceptofV − E + F isnonethelessusefulforallpolyhedra. Foranypolyhedron P, wedefinethe Eulercharacteristic of P, denoted χ(P), tobethenumber

χ(P) = V − E+ F.

Foranyconvexpolyhedra P, weknowfromEuler’sFormulathat χ(P) = 2. IfwedenotethepolyhedronshowninFigure 3.5.4 by T , thenwesawthat χ(T) = 0. Thoughwecannotgointofurtherdetailshere, weremarkthattheEulercharacteristic, anditsgeneralizations, isaveryimportantconceptinanumberofbranchesofmodernmathematics.Finally, weuseEuler’s formula togiveanotherproofofProposition 3.2.1, whichdescribes

the five platonic solids. What is interesting about this second proof is that it is entirelycombinatorial—thatis, itisbaseduponwholenumbers—andmakesnouseofgeometry.

DemonstrationofProposition 3.2.1. Suppose P isaregularpolyhedron. Thenbydefinition Pisconvex, allfacesof P areidentical; andallverticesof P arecontainedinthesamenumberoffaces(andhencethesamenumberofedges).Supposethateverypolygonof P has n edges, andthateveryvertexof P iscontainedin q

edges. ThenbyExercise 3.5.6 weknowthat nF = 2E and qV = 2E. Hence F = 2nE and

V = 2qE.

Wenowsubstitutetheaboveformulasfor F and V intoEuler’sformula, obtaining

2

qE− E+

2

nE = 2.

Dividingeverytermintheaboveequationby 2E andcancelingyields

1

q−

1

2+

1

n=

1

E.

Because E isapositivenumber, itfollowsthat

1

q−

1

2+

1

n> 0,

andtherefore1

q+

1

n>

1

2. (3.5.1)

Whatvaluesof n and q couldsatisfyEquation (3.5.1)? Thenumbers n and q arebothwholenumber. Moreover, weknowthat n ≥ 3, becauseeverypolygonhasat least 3 edges, andq ≥ 3, becauseinapolyhedroneveryvertexiscontainedinatleast 3 edges.Coulditbethat n ≥ 6? Supposethatistrue. Then 1

n≤ 1

6. ThenEquation (3.5.1)impliesthat

1

q+

1

6≥ 1

q+

1

n>

1

2,

3.6CurvatureofPolyhedra 99

whichimpliesthat1

q>

1

2−

1

6=

1

3.

Itwouldfollowthat q < 3, whichisimpossible. Hence n ≤ 5. Therefore n isoneof 3, 4 and5.A similarargumentshowsthat q isoneof 3, 4 and 5. Therearethenninepossiblecasesfor

n and q.

1. n = 3 and q = 3. Weverifythat 13+ 1

3= 2

3> 1

2. Thispolyhedronisatetrahedron.


4= 7

12> 1

2. Thispolyhedronisanoctahedron.


5= 8

15> 1

2. Thispolyhedronisatetrahedron.


3= 7

12> 1

2. Thispolyhedronisacube.

5. n = 4 and q = 4. Weobservethat 14+ 1

4= 1

2, andsoEquation (3.5.1)isnotsatisfied.

Hence, thereisnosuchregularpolyhedron.


5= 9




3= 8

15> 1

2. Thispolyhedronisaicosahedron.


4= 9




5= 2



WehavethereforeverifiedthattheregularpolyhedraarepreciselythefivepolyhedralistedinTable 3.2.1.

3.6 CurvatureofPolyhedra

Ifwethinkofthe“surfaceofapolyhedron,” wenoticethatatsomeverticesthesurfaceappearstobe“curving”morerapidly, andinotherplacesitappearstobecurvingless. (Theword“curved”mightseemstrangewhenappliedtosomethingthatismadeupofflatpolygons, butthesametermisalsoappliedtosmoothsurfaces(suchasasphere), anditisquitestandard.) Forexample,considerthesurfaceofthepolyhedronshowninFigure 3.6.1. Intuitively, thesurfaceismoresharplycurvedatthevertexlabeledA thanatthevertexlabeled B. Itwouldbenicetoquantifycurvature byassigningtoeachvertexanumberthattellsushowcurvedthesurfaceisatthevertex. A verynicemethodforsodoing, whichwenowdescribe, goesbacktoDescartes. (See[Fed82]foratranslationandexpositionofDescartes’workonpolyhedra.)

100 3. Polyhedra

A

B

Figure3.6.1

Theideaofcurvatureatavertexofapolyhedronistoseehowfartheneighborhoodofthevertexisfrombeingflat. Iftheneighborhoodisflat, thecurvatureshouldbe 0. Giventhataflatplanehasangle 360◦ aroundanypoint, weuse 360◦ asourbasisofcomparisonformeasuringcurvature. Foranypolyhedron, andforanyvertexofthepolyhedron, wedefinethe angledefectatthevertextobe 360◦ minusthesumofalltheangles(inthevariousfacesofthepolyhedron)thatcontainthevertex. Theangledefectisthecommonmeasureofcurvatureforverticesofpolyhedra. Forexample, if v isavertexinaregularoctahedron, thenthevertexiscontainedinfour 60◦ angles, andthereforetheangledefectatthevertexis

360◦ − (60◦ + 60◦ + 60◦ + 60◦) = 120◦.

Bycomparison, ifw isavertexinaregulardodecahedron, thenthevertexiscontainedinthree108◦ angles, andthereforetheangledefectatthisvertexis

360◦ − (108◦ + 108◦ + 108◦) = 36◦.

Thefactthattheangledefectatthevertexoftheregularoctahedronislargerthantheangledefectatthevertexoftheregulardodecahedroncorrespondstothefactthattheoctahedronisintuitivelymore“sharplypointed”atitsverticesthanthedodecahedron, ascanbeseenbylookingatpicturesofeach(or, evenbetter, lookingatmodelsofthem).

Exercise 3.6.1. Findtheangledefectateachoftheverticesofthefollowingpolyhedra.

(1) A cube.

(2) A regularicosahedron.

(3) ThepolyhedronshowninFigure 3.4.2 (assumingthatallthetrianglesareequilateral).


Itcanbeseen, usingProposition 3.1.2, thatinaconvexpolyhedron, theangledefectatanyvertexispositive. However, innon-convexpolyhedra, itispossibletohaveanegativeangledefectatavertex. Thereadershouldtrytofindanexampleofapolyhedronwithavertexthathasnegativeangledefect.Simplycalculatingangledefects isof interest, but there ismore to the story than that. In

particular, Descartesdiscoveredaverysubtlefactaboutangledefects, whichwenowstate.


Descarteslookedatthesumofalltheangledefectsinaconvexpolyhedron. Examineafewexamplesofpolyhedra, bothregularandnon-regular, and, ineachexample, calculatethesumoftheangledefectsatallthevertices. Doyounoticeapattern?

Ifyoudidtheabovecalculationscorrectly, youshouldhavenoticedthat foreveryconvexpolyhedronthatyoutried, thesumoftheangledefectsatall theverticesis 720◦. Descartesshowedthatthisremarkableresultindeedholdsforallconvexpolyhedra. GiventhatweknowabouttheEulercharacteristic ofpolyhedra(definedinSection 3.5), wecangeneralizeDescartes’resultasfollows. (Descartes, wholivedwellbeforeEuler, wasmostlikelyunawareoftheEulercharacteristic, thoughthereissomedebateaboutthatintheliterature.)

Proposition 3.6.1 (GeneralizedDescartes’Theorem). Supposethat P isapolyhedron. Thenthesumoftheangledefectsatalltheverticesof P equals 360◦ · χ(P).

Demonstration. Suppose a1, a2, . . . , aV aretheverticesof K, andsupposethat t1, t2, . . . , tFarethefacesof K. Notethatthereare V verticesand F faces. Supposethatface tk hasCk edges.Weobservethat

C1 + C2 + · · ·+ CF = 2E;

thisequationcanbeshownverysimilarlytothedemonstrationofProposition 3.5.2 (3); weleavethedetailstothereader.Foravertex ak, welet dk denotetheangledefectat ak; thatis, wehave

dk = 360◦ − (sumoftheanglesatvertex ak).

Thesumoftheangledefects, whichiswhatwearetryingtoevaluate, istherefore d1 + d2 +· · ·+ dV . WenowuseProposition 2.3.3 andtheaboveformulafor C1 +C2 + · · ·+CF toseethat

d1+d2 + · · ·+ dV =

= [360◦ − (sumofanglesat a1)] + · · ·+ [360◦ − (sumofanglesat aV)]

= [360◦ + · · ·+ 360◦︸︷︷︸V times

]− [(sumofanglesat a1) + · · ·+ (sumofanglesat aV)]

= 360◦ · V − (sumofallanglesof P)

102 3. Polyhedra

= 360◦ · V − [(sumofanglesin t1) + · · ·+ (sumofanglesin tF)]

= 360◦ · V − [(C1 − 2)180◦ + · · ·+ (CF − 2)180◦]

= 360◦ · V − [(C1 + · · ·+ CF)180◦ − (2 · 180◦ + · · ·+ 2 · 180◦︸︷︷︸

F times

)]

= 360◦ · V − [2E · 180◦ − 360◦ · F]= 360◦ · V − 360◦ · E+ 360◦ · F= 360◦ · χ(P).

WenotethattheGeneralizedDescartes’TheoremultimatelyreliesuponEuclid’sFifthPostu-late, becauseintheproofofthetheoremweusetheformulainProposition 2.3.3 forthesumoftheinterioranglesinapolygon(whichinturnmakesuseofthefactthatthesumoftheinterioranglesinatriangleis 180◦, whichisprovedusingEuclid’sFifthPostulate).Finally, wecanuse theGeneralizedDescartes’Theorem togiveanotherdemonstrationof

Euler’sFormula(Proposition 3.5.4). Forthisnewdemonstration, weneedthefollowingobser-vationconcerningtheGeneralizedDescartes’Theorem. Recallthatourdefinitionofpolyhedra,asstatedinSection 3.1, involvesthreecriteriaonthefacesofeachpolyhedra, namelythat(1)facesaregluededge-to-edge; (2)everyedgeofafaceisgluedtotheedgeofpreciselyoneotherface; and(3)notwofaces intersectexceptpossiblyalongtheiredgeswhere theyareglued.Thethirdconditionimpliesthatapolyhedrondoesnothaveanyself-intersections. Actually, ifwelookcarefullyatthedemonstrationoftheGeneralizedDescartes’Theorem, itisseenthatwhereascriteria(1)and(2)arecrucial(inshowingthat C1 + C2 + · · · + CF = 2E), weneveractuallyusecriterion(3). Hence, theconclusionofGeneralizedDescartes’Theoremholdsevenforpolyhedrathatdonotsatisfycriterion(3); thatis, forpolyhedrainwhichfacesmightoverlapeachother, thoughwestillonlythinkofthefacesasbeinggluedtoeachotheralongtheiredges.

SecondDemonstrationofEuler’sFormula(Proposition 3.5.4). Suppose that P is a convexpolyhedron. Ourgoalistoshowthat χ(P) = 2.Choosea faceof P; call this face C. We thenstartbyexpanding C in suchaway that it

becomeswiderthantherestof P. SeeFigure 3.6.2 (i)and(ii)foranexampleofsuchstretching.(Notethatthissortofstretchingispossiblepreciselybecause P isconvex.)

Next, wecollapseallof P ontothefaceC, making P completelyflat. Bytheoriginalconvexityof P, weseethatinthecollapsedversionof P therearetwolayersoffaces: wehave C onthebottom, andthentherestof P ontopinasinglelayer. Thecollapsedversionof P isnolongerapolyhedronaswehavediscussedup tillnow, but itdoessatisfycriteria (1)and (2) in thedefinitionofpolyhedra. SeeFigure 3.6.2 (iii) for theresultofcollapsing theexampleshowninFigure 3.6.2 (ii). WhatweseeinFigure 3.6.2 (iii)looksverymuchliketheprojectionswesaw inFigure 3.5.2 (ii)and (iii), althoughweare thinkingof ithere inaverydifferentway.InFigure 3.5.2 (ii)and(iii)wethoughtofthepolyhedronasawireframewithnofaces, andthendrewtheshadowoftheframe; bycontrast, inFigure 3.6.2 (iii)wewanttothinkofthepolyhedronashavingitsfaces, andsimplycollapsingthepolyhedron, facesandall.


(i) (ii) (iii)

C

Figure3.6.2

Observethattheabovestretchingandcollapsingprocesschangesthegeometryof P, butitdoesnotchange V , E or F. Hence, ifwewanttoshowthat χ(P) = 2, itsufficestoworkwiththecollapsedversionof P, insteadoftheoriginal. So, fromnowon, whenwesay“P,” wewillrefertothecollapsedversion.Letusnowcalculatethesumoftheangledefectsin P. Foreachvertexof P thatisintheinterior

of C, weseethat P isflatnear P, andthereforetheangledefectiszero. Hence, theonlyangledefectsthatarenotzeroareontheboundaryofC. Let a1, a2, . . . , an denotetheverticesoftheboundaryof C. Let α1, α2, . . . , αn denotetheinterioranglesat a1, a2, . . . , an respectively,andlet β1, β2, . . . , βn denotethecorrespondingexteriorangles. (SeeFigure 2.3.6 (i) foranexampleoftheseangles.) Considervertex a1. Giventhewaythat P iscollapsedonto C, weseethatthesumoftheanglesatvertex a1 isprecisely 2α1. Hencetheangledefectat a1 is360◦− 2α1. However, wecansimplifythisexpressionas 360◦− 2α1 = 2(180◦−α1) = 2β1,usingwhatweknowaboutexterioranglesfromSection 2.3. Thesameargumentshowsthattheangledefectat a2 is 2β2, andsimilarlyfortherestoftheverticesof C. Finally, weseethatthesumofalltheangledefectsof P equals

2β1 + 2β2 + · · ·+ 2βn = 2(β1 + β2 + · · ·+ βn).

However, wenotethatProposition 2.3.3 (2)tellsusthat β1+β2+ · · ·+βn = 360◦. Itfollowsthatthesumofalltheangledefectsof P equals 720◦.On theother hand, by theGeneralizedDescartes’Theorem (Proposition 3.6.1), which as

mentionedpriortothisdemonstrationcanbeappliedto P evenwhencollapsed, weknowthatthesumoftheangledefectsof P is 360◦ · χ(P). Comparingourtwocalculationsofthesumoftheangledefects, weseethat 360◦ · χ(P) = 720◦. Itfollowsthat χ(P) = 2, whichisthesameas V − E+ F = 2.

104 3. Polyhedra

Part IISYMMETRY &PATTERNS

105

4Isometries

4.1 Introduction

Anexcellentplacetostartthestudyofsymmetry isthebook[Wey52], byHermanWeyl, oneof thegreatmathematiciansof the20thcentury. I recommendacareful readingof thefirstchapter(BilateralSymmetry)only; afterthattheauthorlapsesintosometechnicalitiesthatarebest skippedover, though inbetween the technicalparts therearephilosophical ideas (andpictures)thatarewellworthreading. WebeginwithalengthyquotefromWeyl(pp.3–6); theitalicsareintheoriginal.

“IfI amnotmistakentheword symmetry isusedinoureverydaylanguageintwomeanings. Intheonesensesymmetricmeanssomethinglikewell-proportioned,well-balanced, andsymmetrydenotesthatsortofconcordanceofseveralpartsbywhichtheyintegrateintoawhole. Beauty isboundupwithsymmetry. . . Inthissensetheideaisbynomeansrestrictedtospatialobjects; thesynonym“harmony”pointsmoretowardsitsacousticalandmusicalthanitsgeometricapplications . . .

“Theimageofthebalanceprovidesanaturallinktothesecondsenseinwhichthewordsymmetryisusedinmoderntimes: bilateralsymmetry , thesymmetryofleftandright, whichissoconspicuousinthestructureofthehigheranimals, especiallythehumanbody. Nowthisbilateralsymmetryisastrictlygeometricnotionand, incontrasttothevaguenotionofsymmetrydiscussedbefore, anabsolutelypreciseconcept . . .

“ . . . Symmetry, aswideorasnarrowasyoumaydefineitsmeaning, isoneideabywhichman[sic]throughtheageshastriedtocomprehendandcreateorder,beauty, andperfection.

108 4. Isometries

“. . . FirstI willdiscussbilateralsymmetryinsomedetail . . . Thenweshallgener-alizethisconceptgradually . . . firststayingwithintheconfinesofgeometry, butthengoingbeyondtheselimits throughtheprocessofmathematicalabstractionalongaroadthatwillfinallyleadustoamathematical ideaofgreatgenerality,thePlatonicideaasitwerebehindallthespecialappearancesandapplicationsofsymmetry. Toacertaindegreethisschemeistypicalforalltheoreticknowledge:Webeginwithsomegeneralbutvagueprinciple(symmetryinthefirstsense), thenfindanimportantcasewherewecangivethatnotionaconcreteprecisemean-ing(bilateralsymmetry), andfromthatcasewegraduallyriseagaintogenerality,guidedmorebymathematicalconstructionandabstractionthanbythemiragesofphilosophy; andifweareluckyweendupwithanideanolessuniversalthantheonefromwhichwestarted. Gonemaybemuchofitsemotionalappeal, butithasthesameorevengreaterunifyingpowerintherealmofthoughtandisexactinsteadofvague.”

Thatnicelysumsupwhereweareheading, thoughwewillnotquitemakeittothefullestpossiblelevelofabstraction, thatrequiringagreaterdegreeofmathematicalbackgroundthanweareassuminginthistext. Wewill, nonetheless, getquiteclosetoWeyl’svision.Whentheword“symmetry”isusedcolloquially, itismostofteninreferencetobilateralsym-

metry, alsoknownas left-rightsymmetry. Theroleofbilateralsymmetry inartandnature iswithoutquestionquitelarge, evenifsymmetryisnotverymuchinfavorinthecontemporaryartworld. Indeed, somanyancientculturesusedbilateralsymmetryintheirartandornamen-tationthatitishardnottowonderwhy. Isitbecausethehumanbodyisessentiallybilaterallysymmetric(atleastexternally—theinternalorgansarenotsymmetricallyplaced)? Isitbecausesymmetryhassomearchetypalsymbolism?Biologically, whyisthehumanbodyexternallybilaterallysymmetric(andwhyisitnotsym-

metricinternally)? A relatedissueisthatofleftvs.right. Isthereanyinherentdifferencebetweenthetwo(nottomentionsuperiorityofoneovertheother—recalltheoriginoftheword“sinis-ter”)? Orareleftandrightdistinguishableonlyinthattheyareoppositesofoneanother? Suchphilosophicalquestionsarefascinating, butadiscussionofthemwouldtakeusabitfarafield.Readthefirstchapterof[Wey52]forextremelythoughtfulremarksontheseissues.Theconceptofbilateralsymmetryappliestoplanarobjects, thatis, two-dimensionalobjects

(forexample, adrawing)aswellas tospatialobjects, that is, threedimensionalobjects (forexample, asculpture). Forthesakeofrelativesimplicity, wewillrestrictourattentiontoplanarobjects(exceptinSection 5.7). Notetheword“relative”intheprevioussentence. Aswewillsee, therearemoresubtletiestothestudyofsymmetry—evenofplanarobjects—thanmeetstheeye; planarobjectsareeasiertoworkwiththanspatialones, buteventheyarenottrivial.Thoughtheconceptofbilateralsymmetryisaveryfamiliarone, andmostofuswouldhaveno

troubleidentifyingwhetheranygivenobjectisbilaterallysymmetricornot, aprecisedefinitionofbilateralsymmetrytakessomethought. Imaginethatanintelligentalienlandedinyourbackyard, and, becauseitjusthappenstospeakalanguagethatyouknow, youstartexplainingtoitallsortsofthingsaboutourculture; atsomepointyouusetheword“symmetric”(referring

4.1Introduction 109

tosomethingthatisbilaterallysymmetric), andthealienasksyouformoredetails. Howwouldyouexplain it? Thereareanumberofwaysyoumightexplain theword“symmetric” to thealien(thoughonlyoneofthemwillturnouttobeusefulwhenweconsidersymmetryatitsmostgeneral).


Thinkof variousways inwhichyoucouldexplainwhat itmeans for anobject tobebilaterallysymmetric.

ConsiderthethreeobjectsinFigure 4.1.1; twoofthemarebilaterallysymmetric, andoneisnot(thoughithasothersymmetry). SupposeyouwishtoexplaintoouralienthattheheartinFigure 4.1.1 (i)hasbilateralsymmetry. First, youcouldsaythatitlooksthesamewhenreflectedinamirror (unlike thewritingon thispage, forexample, whichwouldbebackwardswhenreflectedinamirror—unlessyouhappentobeLeonardodaVinci). Ofcourse, thisexplanationwouldnothelpyouralienifithadneverseenamirror. Second, youcouldtakethepieceofpaperwiththeheartonitandfolditinhalfalongtheverticallinethroughthecenteroftheheart,noticingthatthetwohalvesarethusseentobethesame. Thisapproachseemssatisfactory, andwouldprobablymakethingsquitecleartothealien.

(i) (ii) (iii)

Figure4.1.1

A thirdexplanationforthesymmetryoftheheartisthatifyoudrewitonapieceofverythinglass, andthenflippedtheglassoverabouttheverticallinethroughthecenteroftheheart, thedrawingoftheheartwouldlookthesameaftertheflipasbeforeit. Consideredanotherway,supposeyouplayedthefollowinggamewithsomeone. Youdrawafigureonapieceofverythinglass, andyouasktheotherpersontoclosehisorhereyes. Youtheneitherfliptheglassoverabouttheverticallineonthepieceofglassoryoudonot. Next, youaskthesecondpersontoopenhisorhereyes, andtellyouwhetherornotyouflippedtheglass. Ifyouhaddrawnanon-symmetricfigureontheglass, thesecondpersonwouldonlyhavetonotewhetherthefigurelookeddifferenttoseewhetheryouhadflippedtheglassornot. Ontheotherhand, hadyoudrawnaheart(oranyotherbilaterallysymmetricfigure, drawnsothatitslineofsymmetryistheverticallineinwhichyouflipped), thepersoncouldnottellwhetheryouhadflippedthe

110 4. Isometries

glassornot, becausetheappearanceoftheheartdoesnotchangeasaresultofaflipabouttheverticallinethroughitscenter.Thisthirdmethodofexplainingthebilateralsymmetryoftheheartmayseemthemostcom-

plicated, butit isthemostusefulforourpurposes. Wesayingeneralthataplanarobjectisbilaterallysymmetricifthereisalinesothatiftheplaneisflippedaboutthisline, theobjectappearsunchanged. (Startingtheinnextsection, wewillcallsuchaflipbythemorestandardmathematicalterm“reflection.”) InFigure 4.1.2 (i)thelineusedtodetectbilateralsymmetryfortheheartisvertical. AsseeninFigure 4.1.2 (ii), thelineusedtodetectbilateralsymmetryofanobjectneednotbevertical. Anobjectmayalsobebilaterallysymmetricwithrespecttomorethanoneline, asinFigure 4.1.2 (iii), oreveninfinitelymaysuchlines, asisthecaseofthecircle(Figure 4.1.2 (iv)).

(i) (ii) (iii) (iv)

Figure4.1.2

Thereareothertypesofsymmetrythanjustbilateralsymmetry. Withhindsight, mathemati-cianscametounderstandthatthecommonfeatureofalltypesofsymmetryisthattheycanbedetectedthroughcertaintypesof“motionsoftheplane,” ofwhichflippingisonespecialcase.Anotherexampleofthetypeof“motionsoftheplane”usefultothestudyofsymmetryisrota-tion. Notall“motionoftheplane”areuseful, however. Forexample, stretchingortearingtheplane, whileinterestinginothercontexts, isnotofuseinthestudyofsymmetry. InChapter 5wewillhaveadetaileddiscussionofthesymmetryofvariouscategoriesofplanarobjects. InthischapterwelaythegroundworkforChapter 5 bygivingadetaileddiscussionoftherelevanttypesof“motionsoftheplane.”

4.2 Isometries–TheBasics

Intheprevioussectionwesawthatonewaytodetectthebilateralsymmetryofaplanarobjectisbyflippingtheobjectinaline, andseeingwhethertheobjectappearsunchanged. Flippingtheplaneinalineisoneexampleofacertaintypeof“motionoftheplane”thatiscrucialinthestudyofsymmetry. Twootherexamplesof“motionsoftheplane”wouldberotatingtheplane90◦ clockwiseaboutsomepoint, andshiftingtheentireplane 5 inchestotheright. Moremessy

4.2Isometries–TheBasics 111

transformationssuchassquashingtheplanedowntoasinglepoint, stretchingtheplane, etc.,canalsobeimagined.Itisveryimportanttonote, however, thatweareinterestedonlyintheneteffectofa“motion,”

not“howitgetsthere.” Forexample, rotationby 90◦ clockwiseaboutsomepointhasthesameneteffectasrotationby 270◦ counterclockwiseaboutthesamepoint, andweconsiderthesetworotationstobethesame“motion.” Similarly, shiftingtheplane 10 inchestotherighthasthesameneteffectasfirstshiftingtheplane 15 inchestotheright, andthenshiftingit 5 inchestotheleft.Theword“motionof theplane” isactuallyanunfortunate term, in that itsusemightgive

thefalseimpressionthatweareinterestedinhowwedothe“motion,” ratherthanjusttheneteffect. Hence, wewillnotusetheinformalterm“motionoftheplane”anymore, andinsteadwilladheretothemorestandardmathematicalterm transformation oftheplane. Byatransformationoftheplane, wemeanaruleofassignmentthattakeseachpointintheplane, andassignsitsomepointwhereitwillendup. Forexample, shiftingtheplane 10 inchestotherighttakeseachpointintheplaneandassignsitanewlocation, namely 10 inchestotherightofitsinitialposition; firstshiftingtheplane 15 inchestotheright, andthenshiftingit 5 inchestotheleft,againtakeseachpointintheplaneandassignsitanewlocationthatis 10 inchestotherightofitsinitialposition. Althoughweashumanbeingsmightthinkofshiftingtheplane 10 inchestotherightasadifferent process thanfirstshiftingtheplane 15 inchestotherightandthenshiftingit 5 inchestotheleft, fromthepointofviewoftransformations, theprocessisirrelevant, andonlytheassignmentofpointstotheirfinallocationsisofinterestinthestudyofsymmetry. Wedenotetransformationswiththesametypeofnotationasusedforfunctions. Thatis, supposeT isatransformationoftheplane. Then, foreachpoint A intheplane, welet T(A) denotetheresultofapplyingthetransformationto A. Forexample, suppose T isthetransformationobtainedbyshiftingtheplane 10 inchestotheright. IfA isapointintheplane, then T(A) willbethepointthatis 10 inchestotherightof A. Notethat T moveseverypointintheplane 10inchestotheright, notjustsomeofthepoints.Tohelpavoidfurtherconfusionovertheterm“transformationoftheplane,” wesummarize

twoimportantpointsasfollows:

1. A transformationoftheplanetakeseachpointintheplane, andassignsittoapointwhereitendsup. Whatcountsistheneteffectofthetransformation, thatis, whereeachpointendsupinrelationtoitsinitialposition. Wemightthinkgeometricallyoftransformationsasprocesses, butthatprocessisjustforourpersonalintuitivebenefit, andhasnomathe-maticalsignificance. Twotransformationsarethesameiftheyhavethesameneteffects,eveniftheyseemdifferentasprocesses.

2. A transformationofplanetransformsthewholeplane, notjustsomepartoftheplane. Evenwhenwewillbelookingatsymmetriesofspecificobjects, andapplyingtransformationsoftheplanetothem, wealwaysneedtothinkofeachtransformationasbeingappliedtothewholeplane.

112 4. Isometries

Forgettingtheabovetwopointsisacommonmistakeamongstudentsfirstlearningaboutsym-metry, andoftenleadstoalotofconfusion. So, pleasekeepthesetwopointsinmind.Wearenotinterestedinalltransformationsoftheplane, butonlythoserelevanttothestudyof

symmetry. Ifwethinkofflippingtheplaneaboutaline, orrotatingtheplaneaboutsomepoint,weseethatbothofthesetransformationshavethenicepropertythattheydonotstretch, shrinkordistort anything. Bothof these transformationspreserve lengths, angles, sizesand shapesprecisely. Itisthispropertyofnon-distortionthatiscrucialtothestudyofsymmetry. RecallfromSection 1.3 thattheconceptofdistancebetweenpointscanbeusedtodescribeobjectssuchaslinesandcircles, andcanbeusedtodetermineangles. Hence, itshouldnotbesurprisingthatpreservingdistancesbetweenpointsisthekeytodescribingthosetransformationsoftheplanethatdonotstretch, shrinkordistort. Moreprecisely, wesaythatatransformationoftheplaneisan isometry if, foranytwopointsA and B intheplane, theirdistancebeforethetransformationequalstheirdistanceafterthetransformation. UsingournotationfordistancebetweenpointsfromSection 1.3, wesay thata transformation T of theplane isan isometry if, forany twopoints A and B intheplane, wehave d(T(A), T(B)) = d(A,B); equivalently, wecansaythat |T(A) T(B)| = |AB|. (Sometextsusetheterm“rigidmotion” tomeanwhatwecallanisometry, thoughwewillnotusethatterm.)Itisalsopossibletodefinethenotionofisometryforthreedimensional(andhigher)space,

thoughwewillbestickingtoisometriesoftheplane(exceptinSection 5.7); theword“isom-etry”willthereforealwaysrefertoanisometryoftheplane, exceptwhereotherwisenoted. Acompletelythoroughandrigoroustreatmentofisometrieswouldbeverylengthy. Inthischapterwewilldiscusssomeofthebasicideasinvolvingisometries, asmuchasisneededforourstudyofsymmetry. Isometriesare, withoutquestion, thefundamental—andunifying—conceptinthemathematicalstudyofsymmetry, andourtimelookingatisometrieswillbewellspent.Therearemanythingstobesaidaboutisometries, butthemostbasicquestionsis: canwe

figureoutallthetypesoftransformationsoftheplanethatareisometries? ThecompleteanswertothisquestionwillbegiveninSection 4.6. Inthemeantime, wecandescribeindetailthreefamiliartypesofisometries.Thesimplesttypeofisometryiscalled translation. A translationistheresultof“sliding”the

planerigidlyinagivendirection, andbyagivendistance. InFigure 4.2.1 weseetheeffectoftranslatingtheplane 3 inchestotheright. Inthisfigure, asinmanyotherfigurestocome, inordertoseetheeffectoftheisometry, wedrawsomethingontheplane, tobeabletocompareitsinitialpositionwithitsfinalposition. Theplaneitselfisblank, andifwedonotdrawanythingonit, wecannotseeanydifferenceofhowtheplanelooksbeforeandafteranisometry. Imaginetheplaneasaninfinite, verythin, perfectlysmooth, sheetofglass—ifthesheetofglassismoved,itdoesnotlookanydifferent. Instead, wetaketwosheetsofglass, oneontopoftheother. Then,wedrawthesame non-symmetric objectonbothsheetsofglass, directlyontopofeachother.Wewilltypicallydrawtheletter F, becauseitissimple, thoughanynon-symmetricobjectwoulddo. (Weusetheterm“non-symmetric”intuitivelyrightnow; wewilldiscusstheterminmoredetailinSection 5.1.) Wethenperformtheisometryononeofthesheetsofglass. Comparingtheunmovedcopyoftheobject(calledthe initialobject) withthemovedone(calledthe terminal


object), wecanobtainapictureoftheeffectoftheisometry. InFigure 4.2.1 weseelabeledtheinitial F andtheterminal F, illustratingtranslationby 3 inchestotheright. Westress, however,thatitisthewholeplanethatisbeingtranslated, notjusttheletter F.

Finitial

Fterminal

Figure4.2.1

A translationoftheplanecanbeinanydirection, andbyanyamount. Oneusefulwaytodescribeatranslationisasfollows. InFigure 4.2.2 weseetheresultoftranslatingtheplane. Inthisfigure, ratherthandrawinganinitialobjectsuchastheletter F, wesimplydrewonepoint,labeled A; thepoint A wastakentoanewpoint, labeled A ′, bythetranslation. Toseehowtheplanewastranslated, wedrewanarrowfrom A to A ′. Thearrowislabeled v. Thisarrowcompletelycharacterizesthetranslation, inthefollowingsense. Supposewehadstartedwithsomeotherpoint intheplane, say B, insteadof A; let B ′ denotethepoint towhich B wasmovedbythetranslation. Itwouldturnoutthatdrawinganarrowfrom B to B ′ wouldyieldanarrowthatisparalleltothearrowfrom A to A ′. Forourpurposeshere, twoparallelarrowsofthesamelengthareconsideredidentical. Sucharrows, whereweconsiderparallelarrowsofthesamelengthtobeidentical, arecalled vectors. Anytranslationcorrespondstoavector, calledits translationvector obtainedasjustdescribed, andanyvectordeterminesatranslation. Givenavector v, wedenoteby Tv thetranslationcorrespondingtothevector v; thatis, thetranslationobtainedbytakingeverypointintheplane, andmovingitbytheamountanddirectionofthevector v. If A isanypointintheplane, then Tv(A) istheresultoftaking A, andmovingitinthelengthanddirectionof v.

A

A’v

Figure4.2.2

A particularlynoteworthytranslationistranslationbyzerolength(itdoesnotmatterwhichdirection), calledthe trivialtranslation. Thetrivialtranslationdoesnot“moveanything,” thoughitisstillatransformationoftheplane, and, inparticular, anisometry. Recallthatatransformationoftheplaneisaruleofassignmentthattakeseachpointintheplane, andassignsitsomepointwhereitwillendup. Inthecaseofthetrivialtranslation, thetransformationtakeseachpoint

114 4. Isometries

intheplane, andassignsitthefinallocationexactlywhereitstarted. Thisisometrythat“doesnotdoanything”isextremelyimportant(justasthenumberzeroisimportantinthestudyofnumbers). Anothernameforthetrivialtranslationisthe identityisometry, anditisdenoted I.If A isanypointintheplane, then I(A) = A. A translationthatisnotthetrivialtranslationiscalleda non-trivial translation.Anotherfamiliartypeofisometryiscalled rotation. InFigure 4.2.3 weseetheeffectofrotating

theplane 90◦ clockwiseaboutthepoint A; asusual, inordertoseetheeffectofthisisometry,wedrewaletter F ontheplane, andwethenseewherethisletter F endsup. Westress, onceagain, thatitisthewholeplanethatisbeingrotated, notjusttheletter F.

Finitial Fterminal

A

Figure4.2.3

Anyrotationischaracterizedbyknowingtwothings, namelythepointaboutwhichwerotate,calledthe centerofrotation, andtheanglebywhichwerotate. Whenwerotatetheplaneaboutapoint, wecanthinkofourrotationsaseitherclockwiseorcounterclockwise. Becauseweareonly interested in thenet effectof a rotation, not theprocessof rotation, it reallydoesnotmatterwhetherweuseclockwiseorcounterclockwiserotations. Forexample, rotationby 90◦

clockwiseaboutsomepointhasthesameneteffectasrotationby 270◦ counterclockwiseaboutthesamepoint, andsoweconsider these tworotations tobe thesame isometry. Moreover,bothoftheserotationshavethesameneteffectasrotationby 450◦ clockwise. Inordertoavoidredundancy, wewillusuallymakeuseof clockwise rotations; all rotationswillbeassumedclockwise, unlessotherwiseindicated. (Sometextsusecounterclockwiserotations, andsoinanytextyouread, itisimportanttomakesurewhichdirectionofrotationisbeingused.)Thenotationfortherotationbyangle α clockwisewithcenterofrotation A isdenoted RA

α ;ifitisnotimportanttodenotethecenterofrotation, wesometimeswrite Rα. Forexample, therotationshowninFigure 4.2.3 canbewrittenas RA

90◦ . Ifwewanttospecifyacounterclockwiserotation, wewillusenegativeangles. Forexample, therotationshowninFigure 4.2.3 couldalsobewrittenas RA

−270◦ . Insteadofusingdegreestodescribeangles, weoftendescriberotationintermsoffractionsofawhole 360◦ rotation. Forexample, arotationby 90◦ isthesameas 1/4ofawhole 360◦ rotation. Hence, wecanalsowritetherotationshowninFigure 4.2.3 as RA

1/4.


Ingeneral, weusethenotation RA1/n, where n isawholenumber, tomeanaclockwiserotation

by 360◦/n, that is, a rotationby 1/n ofawhole 360◦ turn. Wecommonlycall rotationby180◦ a halfturnrotation (orjust halfturn). Rotationbyangle 0◦, orequivalentlybyanywholenumbermultipleof 360◦, isreferredtoasthe trivialrotation, andit is thesameisometryastheidentityisometry I mentionedpreviously. Notethat 0◦ isamultipleof 360◦, because 0◦ =0·360◦, sowecansimplystatethatatrivialrotationisonewheretheangleisamultipleof 360◦,where“multiple”inthiscontextwillalwaysmeanbyawholenumber. (Itmayseemstrange,buttranslationbyzeroisindeedthesameisometryasrotationbyzero.) A rotationthatisnotthetrivialrotationiscalleda non-trivial rotation.

Exercise 4.2.1. Drawtheeffectontheletter R showninFigure 4.2.4 astheresultofrotatingtheplaneby 60◦ clockwiseabouteachofthepointsshown. (Therewillbefouranswers,oneforeachpoint.)

BD

C

A R

Figure4.2.4

Thereisaveryinterestingdifferencebetweentranslationsandrotations. Whenwetranslatetheplanebyanyamountotherthanzero, nopointendsupwhereitstarted. Bycontrast, whenwerotatebyanyangleotherthanamultipleof 360◦, thereisalwaysone(andonlyone)pointthatdoesendupwhereitstarted, namelythecenterofrotation. A pointthatendsupwhereitstartedafterwedoanisometryoftheplaneiscalleda fixedpoint oftheisometry. If R isanisometry, andif X isapointintheplane, then X isafixedpointof R preciselyif R(X) = X.Usingthisterminology, weseethatanon-trivialtranslationhasnofixedpoints; thatanon-trivialrotationhasonefixedpoint; andthattheidentityisometryhaseverypointasafixedpoint.Althoughanon-trivial translationhasnofixedpoints, observethatanylinethat isparallel

tothedirectionoftranslationistakentoitselfbythetranslation. Forexample, iftheplaneistranslated 5 inchestotheright, thenanyhorizontallineistakenontoitselfbythetranslation;eachnon-horizontallineisnottakenontoitselfbythistranslation. A linethatistakenontoitselfbyanisometryoftheplaneiscalleda fixedline oftheisometry. A fixedlineisalinethatistakenontoitself; theindividualpointsofthefixedlineneednoteachbetakenontothemselves.

116 4. Isometries

Theidentityisometryhaseverylineintheplaneasafixedline. Donon-trivialrotationshavefixedlines? Theanswerdependsupontheangleofrotation. A rotationby 180◦ (oranymultipleof 180◦)takeseverylinecontainingthecenterofrotationontoitself, soalltheselinesarefixedlinesoftherotation; linesthatdonotcontainthecenterofrotationarenotfixed. Ontheotherhand, arotationthatisnotbyamultipleof 180◦ hasnofixedlines.Thethirdtypeofisometrywewishtoexamineiscalled reflection. Reflectionisthemathemat-

icaltermforflippingtheplaneinaline. InFigure 4.2.5 weseetheeffectofflippingtheplaneinthelinelabeled n; asusual, wedrewaletter F ontheplaneasaninitialobject, andwethenseewherethis F endsupasaresultofthereflection. Westress, asalways, thatitisthewholeplanethatisbeingreflected, notjusttheletter F.

initial

terminalF

Fn

Figure4.2.5

A reflectionischaracterizedbythelineinwhichtheplaneisflipped, calledthe lineofreflec-tion (alsoknownasthe lineofsymmetry or mirrorline). Thenotationforareflectioninline nisMn; ifwehaveanumberoflines, forexample L1, . . . , Ls, thenwewillwritethereflectionsintheselinesas M1, . . . ,Ms whenthemeaningisclear.Althoughwethinkofreflectioninalineasflippingtheplaneaboutthatline, thereisanother

wayofthinkingaboutreflectionthatcapturestheideabetter(especiallyifwewanttocomparereflectionoftheplaneinalinewithreflectionofthreedimensionalspaceinaplane—whichisreflectioninamirror). Recallthatwhatcountsinanisometryisonlyitsneteffect, thatis,whereeachpointoftheplaneendsupinrelationtoitsinitialposition, andnotthegeometricprocessusedtovisualizetheisometry(forexample, flippingtheplaneinthecaseofreflection).LetuslookatFigure 4.2.5. Picksomepointintheinitial F, andthenfinditscorrespondingpointintheterminal F; forexample, wewillpickthepointattheverybottomrightoftheinitial F.Whatistherelationofthispointintheinitial F andthecorrespondingpointintheterminal F?InFigure 4.2.6 welabelthechosenpointontheinitial F by A, anditscorrespondingpointontheterminal F by A ′. Wecanthendrawthelinesegment AA ′, asshowninthefigure. Thecrucialobservationisthat AA ′ isperpendiculartothelineofreflectionm, andthatthepointsA and A ′ areeachthesamedistancefromtheline m, thoughonoppositesidesofit. Thatis,ifwelabelthepointofintersectionof AA ′ and m by O, thenthelengthsof AO and A ′ O


areequal. Whatwehavesaidaboutthepoint A alsoholdsforanyotherpointintheinitialF. Therefore, giventheline m andtheinitial F, insteadofobtainingtheterminal F byflippingtheplaneabout m, wecouldhavefoundtheterminal F bytakingeachpointintheinitial F,drawingaperpendicularlineto m fromthepoint, andthenlocatingthecorrespondingpointontheterminal F bycontinuingtheperpendicularpastm thesamedistancewewentfromthestartingpointto m. Bydoingthisprocesstosufficientlymanypointsontheinitial F (oranyotherinitialobject), wecouldconstructtheterminalobject. Thismethodalsoholdsinthreedimensionalspace. Whenyoulookatyourselfinthemirror, yourimageisasfarbehindthemirrorasyouareinfrontofit.

initial

terminalF

Fm

A

A’

O

Figure4.2.6

Exercise 4.2.2. Drawtheeffectontheletter R showninFigure 4.2.7 astheresultofre-flectingtheplaneineachofthelinesshown. (Therewillbethreeanswers, oneforeachline.)

Tocomparereflectionswithtranslationsandrotations, recallthenotionoffixedpoints. Inareflection, everypointonthelineofreflectionisfixed, butnootherpoint isfixed. This factcontrastswithnon-trivial translations, whichhavenofixedpoints, andnon-trivial rotations,whichhaveonefixedpointeach. Theidentityisometryhasallpointsfixed. Observethatunliketranslationsandrotations, whichcanbetrivialornot(thatis, therearetranslationsandrotationsthatequaltheidentityisometry), thereisnotrivialreflection; thatis, noreflectioncanequaltheidentityisometry.

Exercise 4.2.3. Suppose m isalineintheplane. Describeallfixedlines ofthereflectionMm.

118 4. Isometries

Rm

p

n

Figure4.2.7

Exercise 4.2.4. [UsedinSections 4.5, 4.3 and4.6, andAppendixA] Supposethat A andB aredistinctpointsintheplane. Let m betheperpendicularbisectorof AB.

(1) Showthat Mm takes A to B andtakes B to A, andthat Mm istheonlyreflectionoftheplanetodoso.

(2) Showthat Mm fixesanypointthatisequidistantto A and B.

Thereisanotherimportantdistinctionbetweenthetranslationsandrotationsontheonehand,andreflectionsontheotherhand. InFigure 4.2.1 weseetheeffectofatranslationontheletterF; inFigure 4.2.3 weseetheeffectofarotationontheletter F. Certainly, inFigure 4.2.1 theterminal F stilllooksjustlikealetter F. InFigure 4.2.3 theterminal F doesnotlookpreciselyliketheletter F usuallydoes, butifyouturnyourhead(orthepage)justtherightamount, theterminal F doeslooklikeastandardletter F. Bycontrast, inFigure 4.2.5, whichshowstheeffectofareflectionontheletter F, nomatterhowyouturnyourhead, theterminal F doesnotlookright. Thereflection“reverses”theletter F (andanyotherinitialobject), andthusitdoesnotlooklikeastandardletter F. Theterminal F looks like themirror imageof the initial F. Theformalterminologyweuseisthattranslationsandrotationsare orientationpreserving, whereasreflectionsare orientationreversing.Wehavesofardiscussedthreeparticulartypesofisometries: translations, rotationsandre-

flections. Arethereanyothertypesofisometries, ordothesethreetypesincludeallisometries?Intuitively, itishardtoimagineanyothertypeofisometry, butthatisnotarigorousargumentthatwoulddemonstratethatthethreetypesofisometriesaretheonlytypesthatexist. Toobtainabetterfeelforthisquestion, weneedtolookatisometriesfromaslightlydifferentpointofview, asdiscussedinSection 4.3.

4.3RecognizingIsometries 119

4.3 RecognizingIsometries

InSection 4.2, wediscussedisometriesastransformationsoftheplane. Toseewhateffectaparticularisometryhas, wewoulddrawanobjectintheplane, suchastheletter F, andseewhathappenedtotheobjectasaresultoftheisometrybycomparingtheinitialobjectwiththeterminalobject. Wenowwanttotakea“backwards”lookatisometries. Supposetwopeopleweretoplaythefollowinggame. Onepersondrawsaletter F onablackboard, tobeusedasaninitialobject. Thesecondpersonthencloseshereyes. Thefirstpersonchoosessomeparticularisometry, performstheisometry, anddrawstheterminal F ontheboardthatresultsfromapplyingtheisometry. Thefirstpersontheneraseseverythingontheboardotherthantheinitial F andtheterminal F. Thesecondpersonnowopenshereyes, andlooksatthetwoletters F ontheboard. Canthesecondpersonfigureoutwhatisometrythefirstpersonused? Inotherwords,insteadoftakinganisometryandseeingwhatitseffectis, thesecondpersonseestheeffectoftheisometry, andtriestofigureoutwhattheisometryis.Itwillturnoutthatthesecondpersoncanalwaysfigureoutwhatisometrythefirstperson

used; howthesecondpersondoessoisthesubjectofthepresentsection, thoughthecompleteanswerwillbegivenonlyinSection 4.5. Thereis, however, onecaveat. Supposethefirstpersondoesthefollowing. First, shetranslatestheplane 15 inchestotheright. Then, beforethesecondpersonopenshereyes, shetranslatestheplane 5 inchestotheleft, anddrawstheterminal Fafterdoingbothtranslations. Theneteffectwillbethattheterminal F liesexactly 10 inchestotherightoftheinitial F. Thefirstpersontheneraseseverythingontheblackboardotherthantheinitialandterminalletters F. Whenthesecondpersonopenshereyesandtriestofigureoutwhatisometrywasused, shewouldmostlikelythinkthattheisometryusedwastranslationoftheplaneby 10 inchestotheright. Thereisnowaythatthesecondpersoncouldguessthatthefirstpersonstartedbytranslatingtheplane 15 inchestotheright, andthentranslatingtheplane5 inchestotheleft. Theonlythingthatthesecondpersoncanfigureoutistheneteffectofwhatthefirstpersondid, nottheparticularprocess. However, asmentionedinSection 4.2, itisonlyneteffectthatisofinterestinourdiscussionofisometries.Letusnow rephraseourquestion. Supposewehave two identical letters F on theplane,

onelabeledastheinitialobject, andonelabeledastheterminalobject. Canwefindasingleisometryoftheplanethatwouldhavetheeffectoftakingtheinitial F totheterminal F? (Itmightwellbeaskedwhetherwecanfindmorethanoneanswer, butitwillturnoutthatthereisnevermorethanone; thisfactisprovedrigorouslyinAppendix A,butwewillnotgointothedetailshere.)Letusstartbylookingatsomeparticularcases. First, considertheinitialandterminalletters

F showninFigure 4.3.1.

Here it seems fairlyclear thatwecanfindasingle isometryof theplane thatwouldhavetheeffectof taking the initial F to the terminal F, namelya translation. Tofigureoutwhichtranslation, labelsomepointontheinitial F bytheletter A; thenlabeltheexactcorrespondingpointontheterminal F bytheletterA ′. InFigure 4.3.2 wehavetakentheleft-mostpointonthebottomofeachletter F asthepointbeinglabeled. Wecanthentake v tobethevectorfrom A

120 4. Isometries

Finitial F

terminal

Figure4.3.1

to A ′. Thetranslation Tv istheisometrywearelookingfor; thatis, ithastheeffectoftakingtheinitial F totheterminal F. Observethatitwouldnothavemadeanydifferencehadwechosenadifferentpoint A intheinitial F, aslongas A ′ isthepointintheterminal F thatcorrespondstoourchoiceof A.

Finitial

A’

A

vFterminal

Figure4.3.2

Next, considertheinitialandterminalletters F showninFigure 4.3.3. Isthereasingleisometrythatwouldhavetheeffectoftakingtheinitial F totheterminal F?

F Finitial terminal

Figure4.3.3

Letusconsiderthethreetypesofsymmetrieswithwhichwearefamiliar, namelytranslations,rotationsandreflections. A translationcouldnotpossiblytaketheinitial F to theterminal FinFigure 4.3.3, becausethelatterisatananglewiththeformer. A reflectionalsocouldnotpossiblytaketheinitial F totheterminal F, becauseareflectionisorientationreversing, and


yettheterminal F isnotreversed. Hence, theonlytypeofisometrywithwhichwearefamiliarthatcouldpossiblyworkisarotation. Itturnsoutthatarotationdoesindeedwork. Tofindtherotationthatworks, weneedtofinditscenterofrotationandangleofrotation.Westartbyfindingthecenterofrotation. Aswasthecaseinthepreviousexample, weproceed

bylabelingsomepointontheinitial F bytheletter A, andlabelingthecorrespondingpointontheterminal F bytheletterA ′. Whereverthecenterofrotationweareseekingislocated, itmustbeequidistanttothetwopoints A and A ′; thatis, ifapoint X intheplaneisthesoughtaftercenterofrotation, then d(X,A) = d(X,A ′). WenowmakeuseofProposition 1.3.1, whichsaysthatapoint X intheplanehas d(X,A) = d(X,A ′) ifandonlyifitisontheperpendicularbisectorof AA ′. Hence, wecannarrowoursearchforthecenterofrotationtothosepointsontheperpendicularbisectorofAA ′. SeeFigure 4.3.4. Howdoweknowwhichpointonthislineisthecenterofrotation? Thetrickistorepeatourprocedurewithanotherpairofcorrespondingpoints, say B and B ′. Thecenterofrotationisalsoontheperpendicularbisectorof BB ′. ThesetwoperpendicularbisectorsareshowninFigure 4.3.5. Noticethatthesetwolinesintersectinpreciselyonepoint. Becausethecenterofrotationmustbeonthetwoperpendicularbisectors,itmustbepreciselythepointwherethetwoperpendicularbisectorsintersect. Wehavethereforefoundthecenterofrotation. ItturnsoutthatitdoesnotmatterwhichpointsA and B wechooseon the initial F; wewillalwaysobtain thesamecenterof rotation, labeled O in thefigure.Finally, tofindtheangleofrotation, justmeasuretheanglebetweenfromthelinesegmentOA

tothelinesegment OA ′; itwouldworkjustaswelltouse OB and OB ′. Alltold, wehavenowdeterminedtherotationthattakestheinitial F totheterminal F.

FFinitial terminal

AA’

Figure4.3.4

Asourthirdexample, considertheinitialandterminalletters F showninFigure 4.3.6. Onceagainweaskwhetherthereisasingleisometrythatwouldhavetheeffectoftakingtheinitial Ftotheterminal F.

122 4. Isometries

FFinitial terminal

AA’

BB’

O

Figure4.3.5

F

Finitial

terminal

Figure4.3.6

Inthiscase, weseethattheterminal F hasreversedorientationwhencomparedtotheinitialF. Hence, theonlytypeofisometrywithwhichwearefamiliarthatcouldpossiblyworkisareflection. Itturnsoutthatareflectiondoesindeedwork. Howdowefindthelineofreflection?Asbefore, westartby labelingsomepointon the initial F by the letter A, and labeling thecorrespondingpointontheterminal F bytheletter A ′. UsingExercise 4.2.4 (1), weknowthatthelineofreflection, ifthereisone, mustbetheperpendicularbisectorofAA ′. SeeFigure 4.3.7.


Ifanyotherpairofcorrespondingpointsontheinitial F andterminal F ischosen, say B and B ′,thentheperpendicularbisectorof BB ′ isseen(inthecaseofFigure 4.3.6)tobethesamelineastheperpendicularbisectorof AA ′. Wehavethereforefoundthedesiredlineofreflection.

F

Finitial

terminal

A

A’

B

B’

Figure4.3.7

Exercise 4.3.1. IneachofthethreepartsofFigure 4.3.8 areshowninitialandterminalletters F, obtainedbyusingan isometry. Foreachof the threecases, statewhat typeofisometrywasused. Moreover, iftheisometryisarotation, indicateitscenterofrotation;if the isometry isa translation, indicate the translationbyanarrow; if the isometry isareflection, indicatethelineofreflection.

Asourfinalexample, considertheinitialandterminalletters F showninFigure 4.3.9. Isthereasingleisometrythatwouldhavetheeffectoftakingtheinitial F totheterminal F?

Asinthepreviousexample, weseethattheterminal F hasreversedorientationwhencom-paredtotheinitial F. Hence, notranslationorrotationcouldwork. A reflectionmightseemlikeagoodbet, butthattoodoesnotwork. InFigure 4.3.10 weseetwopairsofcorrespondingpointsontheinitial F andterminal F, labeledA andA ′, and B and B ′. Weseeinthiscasethattheperpendicularbisectorof AA ′ isnotthesamelineastheperpendicularbisectorof BB ′. Ifareflectiontooktheinitial F totheterminal F, thenitwouldneedtohavebothperpendicularbisectorsasitslineofreflection, whichmakesnosense. Hence, thereisnoreflectionthattakestheinitial F totheterminal F.

Whatcanwesayasaresultofthislastexample? Wewouldhavetodrawoneoftwopos-sibleconclusions: eitherthattwoidenticalcopiesoftheletter F canbedrawnintheplanein

124 4. Isometries

(iii)

(i) (ii)

FterminalF

initial

FFinitial

Finitial

Fterminal

terminal

Figure4.3.8

Finitial

Fterminal

Figure4.3.9

suchaway thatno isometryof theplane takesone to theother, or, alternatively, that thereareisometriesotherthanthethreetypeswehavediscussedsofar(translations, rotationsandreflections). Whichofthesepossibilitiesisthecorrectone? Wewillseetheanswertothisques-tioninSection 4.5. First, however, wewillneedaveryimportanttool, whichisdescribedinSection 4.4.

4.4CombiningIsometries 125

FF

initial

terminalA

A’

B

B’

Figure4.3.10

4.4 CombiningIsometries

Justasnumbersbecometrulyusefulonlywhenwecanadd, subtract, multiplyanddividethem,thetransformationsoftheplanediscussedabovebecomemuchmoreinterestingwhenweseehowtocombinethem. Ifwethinkofatransformationoftheplaneasawayofmovingthepointsintheplane, thenifwearegiventwotransformations, wecancombinethembyfirstdoingoneandthendoingthesecond. Suppose P and Q aretransformationsoftheplane; welet Q ◦ P

denotethecombinedtransformationobtainedbyfirstdoing P andthendoing Q. WerefertoQ ◦ P asthe composition of P and Q. (Thenotation Q ◦ P mayseem“backwards”atfirstglance, inthatitmeansdoing P firstratherthanQ first, butthisnotationisverystandardinthemathematicalliterature, andisalsoquiteconvenientinsomesituations, sowewillstickwithit.) Thephrases“P followedby Q”and“Q following P”meanthesamethingas Q ◦ P. Onewayofobtainingafeelforthenotation Q ◦ P isbylookingatwhat Q ◦ P doestoapointA intheplane. Theresultofapplyingtheisometry Q ◦ P tothepoint A wouldbedenoted(Q ◦ P)(A). However, weknowthatthepoint (Q ◦ P)(A) isobtainedbyfirstdoing P to A,resultinginthepoint P(A), andthendoing Q tothat, resultinginthepoint Q(P(A)). Hence,weseethat (Q ◦ P)(A) = Q(P(A)).Beforewelookatexamplesofcompositionsofisometries, weneedtoaskthefollowingques-

tion: thecompositionoftwoisometriesisatransformationoftheplane; isitalsoanisometry?Fortunately, theanswerisyes. Intuitively, thisassertionseemsreasonable, becausedoingeachoftwoisometriesindividuallydoesnotchangedistancesbetweenpointsintheplane, sodoingthemconsecutivelyshouldnotchangedistancesbetweenpoints. UsingournotationfordistancebetweenpointsfromSection 1.3, wecanwriteoutourdemonstrationformallyasfollows.

Proposition 4.4.1. Supposethat P and Q areisometries. Then Q ◦ P isanisometry.

Demonstration. Supposethat A and B arepointsintheplane. Weneedtoshowthattheirdis-tancebeforethetransformationQ ◦ P equalstheirdistanceafterthetransformation. Because Pisanisometry, weknowthat d(P(A), P(B)) = d(A,B). Because Q isanisometry, weknow

126 4. Isometries

that d(Q(P(A)), Q(P(B))) = d(P(A), P(B)). Itthenfollowsthat d(Q(P(A)), Q(P(B))) =d(A,B). However, we know that Q(P(A)) can be rewritten as (Q ◦ P)(A), and thatQ(P(B)) canberewrittenas (Q ◦ P)(B). Puttingourobservationstogether, wededucethatd((Q ◦ P)(A), (Q ◦ P)(B)) = d(A,B). Wehavethereforeshownpreciselywhatisneededtoseethat Q ◦ P isanisometry.

Letuslookatsomeexamplesofcompositionofisometries. InFigure 4.4.1 weseefourlineslabeledm, n, k and l, withthelinesallintersectinginapointA. Letusstartwithaverysimplecomposition, namely RA

1/4 ◦ RA1/2. Thiscompositionmeansfirstrotatingtheplaneclockwiseby

1/2 ofawholerotationcenteredatA, andthenrotatingtheplaneclockwiseby 1/4 ofawholerotationcenteredatA. Theneteffectofthiscompositionisclearlyrotatingtheplaneclockwiseby 3/4 ofawholerotationcenteredat A. Insymbols, wehave RA

1/4 ◦ RA1/2 = RA

3/4.

A

n

m

k

l

Figure4.4.1

Next, letustryaslightlymorecomplicatedcomposition, namely Mm ◦ RA1/2. Hereitwould

behardtoguesstheanswer, aswedidinthepreviouscomposition. Rather, wewillcalculatethecompositionbydrawinganobjectontheplane, forexampletheletter F, andthenseeingwhathappenstotheletter F asaresultofdoingeachofthetwoisometriesinthegivenorder.Intheleft-mostpartofFigure 4.4.2, weseethataletter F hasbeendrawn. (Wecouldjustaswellhavedrawntheletter F anywhereelseintheplane.) InthemiddlepartofFigure 4.4.2,weseetheresultofdoing RA

1/2 totheplane. Itisimportanttoobservethatwhiletheletter F

hasbeenmovedasaresultofdoing RA1/2, thelinesofreflectionhavenotmoved; theyarefixed

linesofreference, andarenotpartoftheplanethatistransformedwhenwedoanisometry.Hence, whenwereferto Mm, forexample, wewillalwaysbereferringtothesameline, nomatterwhatotherisometrieswemighthavedonepreviously. (Ifthemeaningofsymbolssuchas“Mm”weretodependuponwhatcamebeforeit, thenthesamesymbolswouldmeandifferentthingsindifferentsituations, whichwouldleadtomuchconfusion.) Intheright-mostpartofFigure 4.4.2, weseetheresultofdoing Mm totheletter F inthemiddlepartofthefigure. Thecomposition Mm ◦ RA

1/2, whichwearetryingtocompute, thereforetakestheletter F shownintheleft-mostpartofthefigure, andmovesittowhereitisshownintheright-mostpartof


thefigure. Now, byProposition 4.4.1, weknowthatthecomposition Mm ◦ RA1/2 isitselfan

isometry. So, weneedtofinda single isometrythatwouldtaketheletter F intheleft-mostpartofthefiguretotheletter F intheright-mostpartofthefigure. TheanswerisseentobeMk. Wethereforeseethat Mm ◦ RA

1/2 = Mk.

l l

k

Am

n

l

k

F

Am

n

Fk

Am

nF

R1/2

AMm

R1/2

AMm Mk° =

Figure4.4.2

Inpractice, it ispossible todo theabovecompositionabitquickerbydoing itall inonedrawing, butlabelingeachstep, asshowninFigure 4.4.3. Ifyoudocompositionsbythisquickermethod, makesureyoulabeleachstep; otherwise, itwillbeimpossibleforyoutogobackandfigureoutwhatyoudid(orforanyoneelsetounderstandwhatyoudid).

Weareallfamiliarwithsomeofthebasicpropertiesofadditionandmultiplicationofnumbers,forexamplethattheorderofadditionormultiplicationdoesnotmatter; thatis, wealwayshavea+b = b+a foranynumbers a and b, andsimilarlyformultiplication. WewilldiscusssuchpropertiesinmoredetailinChapter 6. Doescompositionofisometriessatisfyallthesamebasicpropertiesasadditionandmultiplicationofnumbers? Unfortunately, theanswerisno.ReferringonceagaintoFigure 4.4.1, letusnowcomparethetwoexpressions RA

1/4 ◦ Mm

and Mm ◦ RA1/4. We leave it to the reader to verify (usingdrawings similar to that shown

inFigure 4.4.2) that RA1/4 ◦ Mm = Ml andthat Mm ◦ RA

1/4 = Mn. Hence, theorderofcompositionofisometriesdoesmatter, incontrasttoadditionofnumbers. Itisnotthecasethatordermatterswiththecompositionofeverytwopossibleisometries, forexampleMm ◦ Mk =RA1/2 and Mk ◦ Mm = RA

1/2, butitisthecasethatordersometimesmatters. Hencewedonot

128 4. Isometries

Am

n

l

k

F F

F

1

2

3

Figure4.4.3

haveageneralruleforcompositionofisometriesanalogoustothegeneralrule a+ b = b+ a

fornumbers.Ontheotherhand, somepropertiesofadditionandmultiplicationofnumbersdoholdfor

compositionofisometries. Forexample, weknowthat a + (b + c) = (a + b) + c foranynumbers a, b and c. Letus lookatanexampleofa similarcalculation forcompositionofisometries. StillreferringtoFigure 4.4.1, considerthetwoexpressions Mk ◦ (RA

1/4 ◦ Mn) and

(Mk ◦ RA1/4) ◦ Mn. Computingeachoftheseexpression, wesee

Mk ◦ (RA1/4 ◦ Mn) = Mk ◦ Mm = RA

1/2

and(Mk ◦ RA

1/4) ◦ Mn = Ml ◦ Mn = RA1/2.

(Weleaveittothereadertoverifyeachstepofthesecalculations.) ThegeneralversionofthispropertyisgivenasPart (2)ofthefollowingproposition.

Proposition 4.4.2. Supposethat P, Q and R areisometries.

1. P ◦ I = P and I ◦ P = P.

2. (P ◦ Q) ◦ R = P ◦ (Q ◦ R).

Demonstration.

(1). Wewillshowthat P ◦ I = P; thefactthat I ◦ P = P canbeshownsimilarly, thedetailsbeinglefttothereader. Let A beapointintheplane. Wewillshowthat (P ◦ I)(A) = P(A).BecauseA waschosenarbitrarily, itwillthenfollowthattheisometry P ◦ I equalstheisometryP, becausetheydothesamethingtoeverypointintheplane. Letusexamine (P ◦ I)(A). Aswehaveseenpreviously, wemayrewritethisexpressionas P(I(A)). Wealsoknowthat I(A) = A.Puttingtheselasttwoobservationstogether, weseethat (P ◦ I)(A) = P(I(A)) = P(A), whichiswhatwewantedtoshow.


(2). Let A beapointintheplane. Wewillshowthat ((P ◦ Q) ◦ R)(A) = (P ◦ (Q ◦R))(A). Because A waschosenarbitrarily, itwillthenfollowthattheisometry (P ◦ Q) ◦ R

equalstheisometry P ◦ (Q ◦ R), becausetheydothesamethingtoeverypointintheplane.Startingwiththelefthandsideoftheequationwewishtoprove, wecompute

((P ◦ Q) ◦ R)(A) = (P ◦ Q)(R(A)) = P(Q(R(A)))

= P((Q ◦ R)(A)) = (P ◦ (Q ◦ R))(A).

Proposition 4.4.2 mightnotseemveryimpressive, butitwillbeused(sometimesexplicitly,andmoreoftenimplicitly)throughoutourdiscussionofisometriesandsymmetry. OneuseofPart (2)ofthepropositionthatwementionnowisthatitallowsustowriteexpressionssuchasP ◦ Q ◦ R unambiguously. Thatis, supposewehavethreeisometries, namely P, Q and R,andtwopeopleareaskedtocomposethem, inthegivenorder(whichiswhattheexpressionP ◦ Q ◦ R wouldmean). Giventhatwecanonlycomposetwoisometriesatatime, onepersonmightcompute P ◦ Q ◦ R bydoing (P ◦ Q) ◦ R, and theotherpersonmightcomputeP ◦ Q ◦ R bydoing P ◦ (Q ◦ R). BecauseofPart (2)oftheproposition, weareassuredthatbothpeoplewillobtainthesameanswer, andthereforeitisnotambiguoustowrite P ◦ Q ◦ R

Weend this sectionwith a comment. This section is ratherbrief, andcontainsone fairlysimple idea, namely forming the compositionof two isometriesbyfirst doingone isometryandthendoingtheother. Donotletthesimplicityofthisideafoolyou. Theideaofformingthecompositionof isometries is thecrucialstepthatallowsforamathematical treatmentofsymmetry, and forvarious substantial resultsabout the symmetryofornamentalpatterns, asdiscussedinChapter 5. Indeed, itmightbesaidthatwechoosetostudysymmetryintermsofisometriespreciselybecauseisometriescanbecomposed. Bycomposingisometries, wewillbeabletoviewthevarioussymmetriesofanobjectnotasisolatedthings, butasthingsthatinteractwitheachother, anditispreciselythisinteractionthatwillleadtointerestingresults.

Exercise 4.4.1. ReferringtoFigure 4.4.1, computethefollowingcompositions; thatis, foreachofthefollowingexpressions, findasingleisometrythatisequaltoit.

(1) RA1/4 ◦ RA

1/3.

(2) RA1/4 ◦ Mn.

(3) Mn ◦ Mm.

(4) Ml ◦ RA3/4 ◦ Mn.

(5) RA1/4 ◦ Mm ◦ RA

1/2.

For futureuse, wegive the followingproposition, whichsummarizes theeffectofcompo-sitionofisometriesonthepreservationorreversaloforientation. Thisproposition, Part (3)of

130 4. Isometries

whichintuitivelysays that twonegativesmakeapositive, certainlyseemsplausible, andweomitadetaileddemonstration(whichwouldrequireamoretechnicallysophisticateddefinitionoforientationpreservingandreversingthanwehavegiven).

Proposition 4.4.3. Supposethat P and Q areisometriesoftheplane.

1. If P and Q arebothorientationpreserving, then P ◦ Q isorientationpreserving.

2. If P isorientationpreservingandQ isorientationreversing, orvice-versa, then P ◦ Q isorientationreversing.

3. If P and Q arebothorientationreversing, then P ◦ Q isorientationpreserving.

4.5 GlideReflections

Section 4.3 endedwithaquery: theexampleshowninFigure 4.3.9 demonstratedthateitherwehadtoconcludethattwoidenticalcopiesoftheletter F canbedrawnintheplaneinsuchawaythatnoisometryoftheplanetakesonetotheother, orthatthereareisometriesotherthanthethreetypeswehavediscussedsofar, namelytranslations, rotationsandreflections. Wenowusethenotionofcompositionofisometries, asdiscussedinSection 4.4, toresolvethisquestion.Aspreviouslystated, itisnotpossibletotaketheinitial F totheterminal F inFigure 4.3.9

byatranslation, arotationorareflection. However, drawtheline m halfwaybetweenthetwoletters F, asshowninFigure 4.5.1. Itcanthenbeseenthatfirstreflectingtheplanein m, andthentranslatingtheplanedownwardbytheappropriatedistance, willtaketheinitial F totheterminal F. Itwouldalsoworktotranslatetheplanedownward, andthenreflectinm. Therefore,althoughnosinglereflectionortranslationoftheplanewilltaketheinitial F totheterminal F,itisthecasethatthecompositionofareflectionandatranslationdoestaketheinitial F totheterminal F. Weknowthatreflectionandtranslationareisometries, andsobyProposition 4.4.1,wealsoknowthatthecompositionofareflectionandatranslationisanisometry. Hence, thereisanisometryoftheplanethattakestheinitial F totheterminal F inFigure 4.3.9. Letuscallthisisometry G. Thecrucialpointisthis: eventhough G wasmadeupoutofareflectionandatranslation, itisasingleisometryinitsownright. Recallthatwhatcountsinanisometryisonlyitsneteffect, thatis, whereeachpointoftheplaneendsupinrelationtoitsinitialposition,andnotthegeometricprocessusedtovisualizetheisometry(inthepresentcasefirstreflectingandthentranslating). Giventhatwecannottaketheinitial F totheterminal F inFigure 4.3.9byasingletranslation, rotationorreflection, wededucethattheisometryG isnotatranslation,rotationorreflection. Hence, thereareisometriesthatarenotofthethreefamiliartypes.

Theisometry G discussedaboveisanexampleofnewtypeofisometry, calleda glidereflec-tion. A glidereflectionisanisometrythatisobtainedbyfirstreflectingtheplaneinaline, andthentranslatinginadirectionparalleltothelineofreflection. (Pleasenotetheword“parallel”here.) Thelineofreflectionusedinformingaglidereflectioniscalledthe lineofglidereflection.Ifthetranslationusedinaglidereflectionhaszerolength, theglidereflectioniscalleda trivialglidereflection. Thatis, atrivialglidereflectionisaglidereflectionthatisjustareflection. A

4.5GlideReflections 131

FF

initial

terminal

m

Figure4.5.1

glidereflectionthatisnotjustareflectioniscalleda non-trivial glidereflection. Wenotethatanon-trivialglidereflectionhasnofixedpoints, butthelineofglidereflectionisafixedline.Also, wenotethataglidereflectionreversesorientation.Thenotationforaglidereflection, wherethereflectionisinline m, andthetranslationisby

vector v (whichisparallelto m), is Gm,v. Insymbolswecanwritethat Gm,v = Tv ◦ Mm. Wereiteratetheimportantpointthataglidereflectionisasingleisometry, whichtakeseachpointoftheplane, andassignsitsomepointwhereitwillendup. Thefactthatweconstructaglidereflectionasthecompositionoftwootherisometriesdoesnotdetractfromthefactthattheglidereflectioncanbethoughtofasanisometryinitsownright.Animmediatequestionthatcomestomindconcerningglidereflectionsiswhetheritmatters

ifwefirstreflectandthentranslate, orvice-versa. Althoughingeneraltheorderdoesmatterwhenwecomposetwoisometries, inthecaseofconstructingglidereflections, itturnsoutnottomatter.

Proposition 4.5.1. Supposem isalineintheplane, and v isavectorthatisparalleltom. ThenMm ◦ Tv = Tv ◦ Mm.

Demonstration. Supposem and v arerespectivelyalineandvectorasshowninFigure 4.5.2 (i).Let A beanypointintheplane. Weneedtoshowthatboth Mm ◦ Tv and Tv ◦ Mm take A tothesamepointintheplane. Itwillfollowthat Mm ◦ Tv = Tv ◦ Mm. Ifthepoint A isontheline m, thenitiseasytoseethat Mm ◦ Tv and Tv ◦ Mm take A tothesamepoint; weleavethedetailstothereader. Nowsupposethat A isnotontheline m.IfwefirstdoMm, thepointA istakentothepoint B, asshowninFigure 4.5.2 (ii). Ifwethen

do Tv, thepoint B istakentopoint C, asshowninthefigure. Ontheotherhand, ifwefirstdo Tv, thepoint A istakentothepoint D, asshowninFigure 4.5.2 (ii). Ifwecouldshowthatdoing Mm takes D to C, thenitwouldfollowthatboth Mm ◦ Tv and Tv ◦ Mm take A to D,whichimpliesthat Mm ◦ Tv and Tv ◦ Mm bothtake A tothesamepointintheplane, whichiswhatweneededtoshow.

132 4. Isometries

Considerthequadrilateral ABCD. Asafirststep, wewillshowthatthisquadrilateralisarectangle. First, weuseExercise 4.2.4 (1)toseethat m istheperpendicularbisectorof AB.Next, becausethevector v isparallelto m, itfollowsthateachof AD and BC isparalleltom. Because m isperpendicularto AB, itfollowsbyProposition 1.2.3 that AD and BC arebothperpendicularto AB. Moreover, because Tv takes A to D, and B to C, weseethat AD

hasthesamelengthas BC. WecannowapplyExercise 2.3.5 tothequadrilateral ABCD, andsothisquadrilateralisinfactarectangle.Let P thethepointwherethelinem intersectsthelinesegmentAB; weknowfromprevious

commentsthat P isthemidpointofAB. LetQ bethepointwherethelinem intersectsthelinesegment CD. SeeFigure 4.5.2 (iii). Weknowthat ABCD isarectangle. ByExercise 2.3.2, wededucethat ABCD isaparallelogram. Inparticular, itfollowsthat CD isparallelto AB. Be-causeAB isperpendiculartom, itfollowsfromProposition 1.2.3 that CD isperpendiculartom. ItissimpletoseethatthequadrilateralAPQD isarectangle(itiscertainlyaparallelogram,anduseExercise 2.1.1). Hence AP hasthesamelengthas DQ. Because ABCD isaparal-lelogram, itfollowsfromProposition 2.2.5 (1)that CD hasthesamelengthas AB. BecauseAP hashalfthelengthof AB, andbecause AB and CD havethesamelengths, itfollowsthat DQ hashalfthelengthof CD. Fromalltheabovereasoning, wededucethat m istheperpendicularbisectorof CD. ItfollowsfromExercise 4.2.4 (1)that Mm takes C to D, andthatiswhatweneededtoshow.

D C

v v

(i) (ii) (iii)

m m

D

BA BA P

Q C

m

Figure4.5.2


Exercise 4.5.1. Drawtheeffectontheletter R showninFigure 4.5.3 astheresultofglidereflectingtheplaneusingeachofthepairsoflineofreflectionandtranslationvectorshown.

Rm

p

n

v

u

w

Figure4.5.3

Exercise 4.5.2. Supposethat G isaglidereflection. Whatkindofisometryis G ◦ G?

Exercise 4.5.3. Suppose m isalineintheplane, and v isavectorthatisnotparalleltom. Isitalwaysthecasethat Mm ◦ Tv = Tv ◦ Mm? Explainyouranswer.

Anotherimportantquestionconcerningglidereflectionisthefollowing. Weconstructaglidereflectionbyfirstreflectinginaline, andthentranslatinginadirectionparalleltothelineofreflection. Whatwouldhappen ifwewere to reflect theplane ina line, and then translateinadirectionnotparalleltothelineofreflection. Wouldweobtainyetanothernewtypeofisometry? Itturnsoutthatthecompositionofanyreflectionandanytranslationequalseitherareflectionoraglidereflection(dependingupontherelationshipbetweenthelineofreflectionandthetranslationvector), thoughwewillomitthedetails. Thereforewedonotobtainanynewtypeofisometrybyfirstreflectingandthentranslatinginadirectionnotparalleltothelineofreflection.RecallFigure 4.3.9. Whenwefirstencountered thatexample, wesaw thatno translation,

rotationorreflectionalonecouldtaketheinitial F totheterminal F. Atthetimewewerenotfamiliarwithanyothertypesofisometries, butnowweknowaboutglidereflections, andwe

134 4. Isometries

haveseeninthecurrentsectionthattheinitial F inFigure 4.3.9 canbetakentotheterminal Fbyaglidereflection. Howwasthisglidereflectionfound?Tofindtheisometrythattakestheinitial F totheterminal F inFigure 4.3.9, westartjustas

wedidforreflections, namelybylabelingtwopointsontheinitial F bytheletters A and B,andlabelingthecorrespondingpointontheterminal F bytheletters A ′ and B ′. Wecanthenfindthemidpointsofthetwolinesegments AA ′ and BB ′, anddrawalinethroughthesetwomidpoints. Callthisline m. SeeFigure 4.5.4.

FF

initial

terminal

A

A’

B

B’

m

Figure4.5.4

It is seen that simply reflecting theplane in the line m doesnot take the initial F to theterminal F. However, ifwefirstreflecttheplaneintheline m, andthenfollowthisreflectionbyatranslationparalleltom bytheverticaldistancebetweentheinitial F andterminal F, thenthecompositionofthereflectionandtranslationwillindeedtaketheinitial F totheterminal F.Hence, thereisaglidereflectionthattakestheinitial F totheterminal F.Wecannowstate thecompleteanswer to theproblemof recognizing isometriesby their

effects, whichweleftunfinishedinSection 4.3. ThatouransweriscompletereliesuponPropo-sition 4.6.1 inthenextsection, aswellassomemathematicaldetailsweskipovertoavoidadigression, butwecanstatethecompleteanswerrightnow. Supposewearegivetwoidenti-calletters F intheplane. Wecanthenfindwhichsingleisometrytakesone F totheother, asfollows. (Moreover, thisisometryturnsouttobeunique.) Therearetwocases.

Case1: thetwoletters F havethesameorientation. Connecttwopairsofcorrespondingpointswithlinesegments, andformtheperpendicularbisectors. Ifthetwoperpendicularbisectorsin-tersect, thentheisometryisarotation, withthepointofintersectionbeingthecenterofrotation;theanglecanbefoundbydrawinglinesfromthecenterofrotationtocorrespondingpointsonthetwoletters F, andmeasuringtheanglebetweenthesetwolines. Ifthetwoperpendicularbisectorsareparallel, theisometryisatranslation; simplydrawanarrowfromapointonone Ftoitscorrespondingpointonanother F.


Case2: thetwo Fshaveoppositeorientations. Connecttwopairsofcorrespondingpointswithlinesegments, andfindthemidpointsofthesetwolines. Drawalinethroughthetwomidpoints.If the line through themidpoints isperpendicular to theconnecting linesegments, then theisometryisareflectioninthelinethroughthemidpoints; ifthelinethroughthemidpointsisnotperpendiculartotheconnectinglinesegments, thentheisometryisaglidereflection, obtainedbyfirstreflectinginthelinethroughthemidpoints, andthentranslatingasnecessary.

Exercise 4.5.4. IneachofthethreepartsofFigure 4.5.5 areshowninitialandterminalletters F, obtainedbyusingan isometry. Foreachof the threecases, statewhat typeofisometrywasused. Moreover, iftheisometryisarotation, indicateitscenterofrotation;if the isometry isa translation, indicate the translationbyanarrow; if the isometry isareflection, indicatethelineofreflection; iftheisometryisaglidereflection, indicatethelineofreflectionusedintheglidereflection.

(iii)

(i) (ii)

FterminalFinitial

Finitial

F Finitial

Fterminal

terminal

Figure4.5.5

136 4. Isometries

4.6 Isometries—TheWholeStory

HavinglearnedaboutglidereflectionsinSection 4.5, wearenowfamiliarwithfourtypesofisometries: translations, rotations, reflectionsandglidereflections. Are thereanyother typesofisometries? Asstatedinthefollowingproposition, theanswerisno. Thispropositionisthemostfundamentalfactaboutisometriesoftheplane, anditwillformthebasisformuchofoursubsequentdiscussionofisometriesandofsymmetry(giveninthischapterandthenext). Thedemonstrationofthispropositionissomewhatlengthy, andisgiveninAppendix A.

Proposition 4.6.1. Anyisometryoftheplaneiseitheratranslation, arotation, areflectionoraglidereflection.

WediscussedinSection 4.2 thenotionsoffixedpoints, fixedlines, andorientationpreservingandreversing. Nowthatweknowallthetypesofisometries, wesummarizethesepropertiesforalltypesofisometriesinTable 4.6.1.

IsometryType FixedPoints FixedLines OrientationIdentityisometry I allpoints alllines preservingNon-triv.trans. Tv nopoints linesparallelto v preservingNon-triv.rot. RA

α thepoint A none, oralllinesthrough A preservingRefl. Mm pointson m m, andalllinesperp.to m reversingNon-triv.gl.refl. Gm,v nopoints theline m reversing

.

Table4.6.1

Dowereallyneedallfourtypesofisometries? Theanswerisbothyesandno. Wecouldobtainallisometriesbyusingjusttranslations, rotationsandreflections, ifweallowforcombinationsofthem(becauseglidereflectionsareobtainedbycomposingreflectionsandtranslations). Ac-tually, ifwewanttobethemost“economical”intermsofusingthefewesttypesofisometriestoobtainallothers, wecoulduseonlyreflections. Itturnsout, thoughthisfactisnotatallobvious,thatwecangetalltheotherthreetypesofisometries(andthereforeallisometries)bycombininguptothreereflectionsatatime. Indeed, thebulkoftheproofofProposition 4.6.1, asgiveninAppendix A,involvesshowingthattheneteffectofanyisometryoftheplanecanbeobtainedbyeithertheidentity, orthecompositionofone, twoorthreereflections. If, however, wewanteachisometry(meaningeachpossibleneteffect)tobedescribedbyasingletransformationoftheplane, ratherthanacompositionofothertransformations, thenweneedtranslations, rotations,reflectionsandglidereflections. Ourgoalbeingnotefficiencybutobtaininganunderstandingofisometriesandsymmetry, wewilluseallfourtypesofisometries.Ifwecombinetwoisometries, howdoweknowwhattheresultis? Oneapproachwouldbeto

drawaninitialobjectintheplane, performthetwoisometriesoneaftertheother, andthenlookattheneteffecttakingtheinitialobjecttotheterminalobject. WecouldthenapplythemethodofrecognizingisometriesdiscussedinSections 4.3 and4.5 tofigureoutwhattheresultingsingleisometrywas. However, itwillbemoreusefultobeabletoknowtheresultofcombiningtwo

4.6Isometries—TheWholeStory 137

isometriessimplyfromknowingthetwoisometriesthatarecombined. (Asaroughanalogy, itissimilarlybettertofigureout 547× 23 withpencilandpaper, ratherthanmaking 23 rowsof547 stoneseach, andthencountingthetotalnumberofstones.)Recallingthateveryisometryisoneofourfourtypes, wecanseehowtocombineanytwo

isometriesbybreakingupourdiscussionintovariouscases, dependinguponwhichtwotypesofisometriesarebeingcombined. Insomecaseswecanobtainveryspecificresultsabouttheresultofcomposingtwoisometriesofagiventype; inothercases, itisverydifficulttowriteaformulafortheresultingisometry(unlessweusesomemoreadvancedmathematics), butwecanatleaststateitstype. Insomeofthecases, wewillrestrictourattentiontonon-trivialtranslations,rotationsandglidereflections, toavoidvariousspecialcases. Wewillstateherethethreemostimportantresults, namelythoseconcerningcompositionsof twotranslations, tworeflectionsandtworotations. Itisalsopossibletodiscussthecompositionsoftwoglidereflections, andcompositionsoftwodifferenttypesofisometries; toavoidalengthydigression, wewilldiscusstheseothercasesinAppendix B.Inorder todiscuss the compositionof two translations, weneed todiscuss thenotionof

additionofvectorsintheplane. Supposewehavetwovectors v andw intheplane, representedbyarrows, asshowninFigure 4.6.1 (i). Wecanaddthesetwovectorsasfollows. First, positionthetwoarrowssothattheyhaveacommonstartingpoint, asshownFigure 4.6.1 (ii); thereisnoproblemmovingarrowsthatrepresentvectors, aslongasthearrowsarenotstretchedorshrunk, orrotated. Next, wecanformtheparallelogramwiththetwoarrowsassides, asshownFigure 4.6.1 (iii). Finally, wetakethediagonalintheparallelogramtobeanarrowforthesumof v and w, denoted v +w; seeFigure 4.6.1 (iv). Thistypeofvectoradditionisveryusefulinbothmathematicsandthesciences(forexample, theadditionofforcesinphysics).

(i) (ii)

(iii) (iv)

v

w

vv

w

w

vv

w

w

v

w

v+w

Figure4.6.1

138 4. Isometries

Wecannowuseadditionofvectorstounderstandthecompositionoftranslations.

Proposition 4.6.2. Supposethat Tv and Tw aretranslationsoftheplane. Then Tw ◦ Tv = Tv+w.

Demonstration. Chooseanypoint A intheplane. Thentheresultofdoing Tw ◦ Tv to A isthesameasfirstdoing Tv to A, yielding Tv(A), andthendoing Tw tothat, yielding Tw(Tv(A)).SeeFigure 4.6.2. However, thetriangleshowninthisfigureisthesameasthelowerhalfoftheparallelogramshowninFigure 4.6.1 (iv). Hence Tw(Tv(A)) isthesameas Tv+w(A). Becausethisreasoningholdsforanypoint A intheplane, wededucethattheisometry Tw ◦ Tv hasthesameneteffectastheisometry Tv+w. Itfollowsthat Tw ◦ Tv = Tv+w.

w

v

v+w

Tv(A)A

Tw(Tv(A))

Figure4.6.2

Itisseenthatforanytwovectors v andw intheplane, wehave v+w = w+ v. Combiningthisobservationwiththeaboveproposition, wededucethatforanytwovectors v and w, wehave Tw ◦ Tv = Tv ◦ Tw. Inotherwords, orderdoesnotmatterwhentwotranslationsarecomposed. Bycontrast, orderdoesmatterwhenmostotherisometriesarecomposed.Wenowturntothecompositionoftworeflections; therearethreesubcases, dependingupon

whetherthetwolinesofreflectionareequal, areparallel, orareneitherparallelnorequal.

Proposition 4.6.3. Supposethat Mm and Mn arereflectionsoftheplane.

1. If m = n, then Mn ◦ Mm = I.

2. If m and n areparallel, then Mn ◦ Mm = Tv, where v is thevector that is in thedirectionperpendicularto m and n, andthathaslengthtwicethedistancefrom m ton.

3. If m and n areneitherparallelnorequal, then Mn ◦ Mm = RAα , where A isthepoint

ofintersectionof m and n, andwhere α istwicetheanglefrom m to n.

Demonstration. WeknowthateachofMm andMn areorientationreversing, andtherefore, byProposition 4.4.3 (3)weknowthatMn ◦ Mm isorientationpreserving. GiventhatMn ◦ Mm isanisometrybyProposition 4.4.1, thenitiseitheratranslation, arotation, areflectionoraglidereflectionbyProposition 4.6.1. UsingTable 4.6.1, it followsthat Mn ◦ Mm mustbeeitheratranslationorarotation(ortheidentityisometry, whichisbothatranslationorarotation),becausereflectionsandglidereflectionsareorientationreversing.


(1). Thisisclear, becauseMm istheresultofflippingtheplaneaboutthelinem, andflippingtheplaneaboutthesamelinetwiceleaveseverypointintheplanewhereitstarted.

(2). Weclaimthat Mn ◦ Mm doesnotfixanypointintheplane. Suppose, tothecontrary,that Mn ◦ Mm didfixsomepoint, say B. Thatwouldmeanthat Mn(Mm(B)) = B. WethendeducethatMn(Mn(Mm(B))) = Mn(B). ByapplyingPart (1)ofthispropositiontoMn, wethereforeseethat Mm(B) = Mn(B).Howcould thispossiblyhappen? Thereare threecases, dependinguponwhether B ison

m, ison n orisnotoneitherline. If B ison m, then Mm(B) = B, but Mn(B) = B, butthiscannotpossiblybethecase, giventhatMm(B) = Mn(B). So, weconcludethat B isnotonm.A similarargumentshowsthat B isnoton n. Nowsupposethat B isnotoneitherm or n. ThenMm(B) = B and Mn(B) = B. Inthatcase, weuseExercise 4.2.4 (1)todeducethat m istheperpendicularbisectorofthelinesegmentfrom B to Mm(B), andthat n istheperpendicularbisectorofthelinesegmentfrom B toMn(B). However, giventhatMm(B) = Mn(B), weseethat m mustbethesamelineas n, whichcannotbe, giventhat m and n areparallel, whichmeansthattheyaredistinctlines. Thefinalconclusionofthisargumentisthat Mn ◦ Mm hasnofixedpoints, becauseourassumptiontothecontraryledtoalogicalimpossibility.WeknowthatMn ◦ Mm iseithertheidentityisometry, atranslationorarotation. Theidentity

isometryfixesallpoints, andanon-trivialrotationfixespreciselyonepoint. Hence, weseethatMn ◦ Mm mustbeatranslation. Tofindoutwhichtranslation, wecansimplyseewhathappenstoonepointintheplane. Forconvenience, supposethatboththelines m and n arevertical,asinFigure 4.6.3 (ifnot, simplychangeyourvantagepoint). Consideranypoint X onthelinem. Then Mm fixes X, and Mn takes X toapoint Y thatisdirectlytotherightof X, andatadistancefrom X thatistwicethedistancefromm to n. ItfollowsthatMn ◦ Mm = Tv, wherev isthevectorthatisinthedirectionperpendicularto m and n, andthathaslengthtwicethedistancefrom m to n.

(3). ObservethatthepointA, whichistheintersectionofthelinesm and n, isfixedbybothMm andMn. HencethepointA isfixedbyMn ◦ Mm. SupposeZ isapointonthelinem thatisdifferentfromthepointA. SeeFigure 4.6.4. ThenMm fixesZ, andMn doesnot(becauseZ isnoton n). Hencethepoint Z isnotfixedbyMn ◦ Mm. WethereforeseethatMn ◦ Mm fixessomepointsanddoesnotfixothers. WeknowthatMn ◦ Mm iseithertheidentityisometry, atranslationorarotation. Theidentityisometryfixesallpoints, andanon-trivialtranslationfixesnopoints. ItfollowsthatMn ◦ Mm isanon-trivialrotation. Sucharotationfixespreciselyonepoint, namelyitscenterofrotation. ItthereforemustbethecasethatthepointA isthecenterofrotation. Tofindtheangleofrotation, wecansimplyseewhathappenstoonepointotherthanA. Takethepoint Z chosenabove. SupposethatMn ◦ Mm moves Z tothepointlabeledW in

Figure 4.6.4. Usingthedefinitionofreflection, itcanbeseenthattheanglefromtheline←→AW

totheline n isequaltotheanglefromtheline m (whichisthesameastheline←→AZ)tothe

line n. (A rigorousdemonstrationoftheequalityofthesetwoanglescanbeobtainedbyusingcongruenttriangles; thereaderisaskedtoprovidedetailsinExercise 4.6.1.) Itfollowsthatthe

140 4. Isometries

rotationwithcenterofrotation A thattakes Z to W mustberotationbytheangle α, whichistwicetheanglefrom m to n. Wethereforeseethat Mn ◦ Mm = RA

α .

nm

X Y

Figure4.6.3

Z

W

A

n m

Figure4.6.4


Exercise 4.6.1. [UsedinThisSection] InthedemonstrationofProposition 4.6.3 (3), we

assertedthattheanglefromtheline←→AY totheline n isequaltotheanglefromthelinem

totheline n; seeFigure 4.6.4. Usecongruenttrianglestodemonstratethisclaim.

Weturnnexttothecompositionoftworotations. Onceagainwehavetwomaincases, thistimedependinguponwhetherthetwocentersofrotationarethesameornot. However, inthecasewherethetwocentersofrotationarenot thesame, therearetwosubcases, dependinguponwhether the twoanglesadduptoamultipleof 360◦ ornot. Moreover, whenthe twoanglesdonotadduptoamultipleof 360◦, thenwewillnotbeabletogiveasimpledescriptionoftheresultingisometry, whichisdefinitelyarotation, butforwhichitistrickytodescribethecenterofrotation. A pictorialapproachtofindingthedesiredcenterofrotationisfoundinthedemonstrationofProposition 4.6.4 (thispictorialapproachwillbeusefulinAppendix E).

Proposition 4.6.4. Supposethat RAα and RB

β arerotationsoftheplane.

1. If A = B, then RBβ ◦ RA

α = RAα+β.

2. If A = B, andif α + β isnotamultipleof 360◦, then RBβ ◦ RA

α = RCα+β, where C isa

pointintheplaneuniquelydeterminedbyA, B, α and β (andwhichisdescribedmorepreciselyinthedemonstration).

3. If A = B, andif α+ β isamultipleof 360◦, then RBβ ◦ RA

α = Tv, where v isthevectorfrom A to RB

β(A).

Demonstration. Weknowthateachof RAα and RB

β areorientationpreserving, andtherefore,byProposition 4.4.3 (1)weknowthat RB

β ◦ RAα isorientationpreserving. Giventhat RB

β ◦ RAα

isanisometrybyProposition 4.4.1, thenitiseitheratranslation, arotation, areflectionoraglidereflectionbyProposition 4.6.1. UsingTable 4.6.1, itfollowsthat RB

β ◦ RAα mustbeeither

atranslationorarotation(ortheidentityisometry, whichisbothatranslationorarotation).Considerthefixedpointsof RB

β ◦ RAα , ifthereareany. If R

Bβ ◦ RA

α turnsouttohavenofixedpoints, thenitmustbeanon-trivialtranslation; ifithasatleastonefixedpointandatleastonepointthatisnotfixed, thenitmustbeanon-trivialrotation, andthefixedpointmustbethecenterofrotation; ifithasmorethanonefixedpoint, thenitmustbetheidentityisometry.

(1). Supposethat A = B. Then RBβ ◦ RA

α = RAβ ◦ RA

α . Becauseboth RAβ and RA

α fixthepointA, then RA

β ◦ RAα fixesA aswell. Hence RA

β ◦ RAα hasafixedpoint, andcannotbeanon-trivial

translation; itmustthereforebearotationwithcenterofrotation A, ortheidentityisometry(whichcanalsobethoughtofasarotationwithcenterofrotation A).Nowthatweknowthat RA

β ◦ RAα isarotationwithcenterofrotation A, thequestionisby

whatangle. Drawaninitialobjectintheplane, forexampletheletter F. Ifweperform RAα , then

theresultingimageoftheletter Fwillmakeangle αwiththeoriginalletter F. IfwethenperformRAβ , theresultingimageoftheletter F willnowmakeangle α + β withtheoriginalletter F.

Hence, theonlypossiblerotationthatequals RAβ ◦ RA

α is RAα+β.

142 4. Isometries

(2). Supposethat A = B, andthat α+ β isnotamultipleof 360◦. Byaddingorsubtractingmultiplesof 360◦ toeachofα andβ asnecessary, wecanensurethatα andβ arebothbetween0◦ and 360◦; addingandsubtractingmultiplesof 360◦ doesnotchangetheeffectof RB

β or RAα .

Wenowmakethefollowingconstruction. First, wedrawalinesegmentfrom A to B. Next, wedrawtheangle α atthepointA, andtheangle β atthepoint B, sothatbothanglesarebisectedby AB. SeeFigure 4.6.5. Thetwoanglesintersectinpoints C and D asshown. Inthefigure,forconvenience, wehavepositionedthepoints A and B sothat AB ishorizontal; ifthislinesegmentisnothorizontal, thentheconstructionwouldlookjustlikewhatwehaveshown, butrotatedappropriately. Thereare, infact, anumberofdifferentcasesthatwouldlooksomewhatdifferentfromwhatisshowninthefigure, dependinguponwhethereachoftheangles α andβ islessormorethan 180◦; wehavedrawnthecasewherebothanglesarelessthan 180◦. Thekeypointtonoteisthat, because α + β isnotamultipleof 360◦, thetwoangleswillindeedintersectintwopoints.

A α β B

C

D

Figure4.6.5

Wenowmakethefollowingobservation. Ifweapplytheisometry RAα totheplane, thepoint

labeled C willbetakentothepoint D. Ifwethenapply RBβ, thepoint D willbetakenbackto

C. Hence, thecomposition RBβ ◦ RA

α fixesthepoint C. Ontheotherhand, itcanbeseenthatRBβ ◦ RA

α doesnotfixthepoint D. Therefore, theisometry RBβ ◦ RA

α fixesatleastonepoint,butdoesnotfixallpoints. Wealreadysawthat RB

β ◦ RAα iseitherarotation, atranslationor

theidentityisometry. Because RBβ ◦ RA

α cannotbeanon-trivialtranslation(whichwouldfixnopoints), northeidentityisometry(whichfixesallpoints), itmustbearotation. A rotationfixespreciselyonepoint, namelyitscenterofrotation. Wededucethat RB

β ◦ RAα isarotationwith

centerofrotation C.Nowthatweknowthat RB

β ◦ RAα isarotation, thequestionisbywhatangle. Usingthesame

ideaasinthedemonstrationofPart (1)ofthisproposition, itisseenthattheanglemustbeα+β.Hence, theonlypossiblerotationthatequals RB

β ◦ RAα is RC

α+β.

(3). Supposethat A = B, andthat α+ β isamultipleof 360◦. Because α+ β isamultipleof 360◦, theneither α and β arebothmultiplesof 360◦, orneitheraremultiplesof 360◦. Wehavetolookateachofthesetwocasesseparately.


First, supposethat α and β arebothmultiplesof 360◦. Then RBβ and RA

α areboththeidentityisometry. Itfollowsthat RB

β ◦ RAα isalsotheidentityisometry. Becausetheidentityisometryisa

trivialtranslation, then RBβ ◦ RA

α = Tv, where v haslengthzero.Next, suppose thatneither α nor β is amultipleof 360◦. Thenneither RB

β nor RAα is the

identityisometry. Wewanttoshowthat RBβ ◦ RA

α isatranslation. Wewillshowthisresultbyeliminatingtheotherpossibilities. First, notethat RA

α fixesthepoint A, but RBβ doesnotfixthe

point A, because RBβ isanon-trivialrotation. Itfollowsthat RB

β ◦ RAα doesnotfixthepoint A,

andtherefore RBβ ◦ RA

α isnottheidentityisometry.Now, supposethat RB

β ◦ RAα werearotation. Thenitwouldhavesomecenterofrotation,

sayapoint D. If RBβ ◦ RA

α werearotation, bywhatanglewouldtherotationbe? UsingthesameideaasinthedemonstrationofPart (1)ofthisproposition, itisseenthattheanglemustbeα+ β. Hence, theonlypossiblerotationthatcouldequal RB

β ◦ RAα wouldbe RD

α+β. However,weareassumingthat α+β isamultipleof 360◦. Itwouldthenfollowthat RD

α+β istheidentityisometry, andhencethat RB

β ◦ RAα istheidentityisometry. However, wehaveseenthat RB

β ◦ RAα

doesnotfixthepoint A, soitcouldnotpossiblybetheidentityisometry. Wethereforehavealogicalcontradiction. Theonlyresolutionofthisdilemmaisthat RB

β ◦ RAα cannotbearotation.

Wededucethat RBβ ◦ RA

α mustbeatranslation.Because RB

β ◦ RAα is a translation, it equals Tv, where v is somevector in theplane. To

determinethisvector, wecantakeanypointintheplane, seewhereitendsafterperformingRBβ ◦ RA

α , andthentakethearrowfromthepoint’soriginalpositiontoitsfinalposition. Forconvenience, wechoosethepoint A. Because RA

α fixesthepoint A, thenweseethattheresultofapplying RB

β ◦ RAα to thepoint A results in thepoint RB

β(A). Therefore, wededuce thatRBβ ◦ RA

α = Tv, where v isthevectorfrom A to RBβ(A).

Havingjustgivenpropositionsdescribingthedetailsofsomeofthepossiblecompositionsofisometries, weleavethedetailsoftheothertypesofcompositionstoAppendix B.Whatwestaterightnowisasummaryofallthewaysofcombiningthefourdifferenttypesofisometriesinachart, asseeninTable 4.6.2. Noticethatthistablecanbebrokenintofoursub-boxes, basedonorientationpreservingororientationreversing. Inthistable, weincludetrivialtranslations,rotationsandglidereflections, toavoidspecialcases.

◦ translation rotation reflection glidereflectiontranslation trans. trans.orrot. refl.orgl.refl. refl.orgl.refl.rotation trans.orrot. trans.orrot. refl.orgl.refl. refl.orgl.refl.reflection refl.orgl.refl. refl.orgl.refl. trans.orrot. trans.orrot.

glidereflection refl.orgl.refl. refl.orgl.refl. trans.orrot. trans.orrot.

.

Table4.6.2

144 4. Isometries

Exercise 4.6.2. Supposearotationisfollowedbyaglidereflection, whichisthenfollowedbyareflection. Whattypeofisometry(orisometries)couldbeobtainedasaresultofthiscomposition?

Exercise 4.6.3. Describetheisometrythatresultsfromahalfturnfollowedbyanotherhalf-turn? (Theresultdependsuponwhetherthetwohalfturnshavethesamecenterofrotationornot.)

Exercise 4.6.4. [UsedinSection 5.5] A halfturnisfollowedbyareflection. Supposethatthecenterofrotationof thehalfturnisonthelineofreflection. Showthat theresultingisometryisareflectioninthelinethroughthecenterofrotationandperpendiculartotheoriginallineofreflection.

Exercise 4.6.5. [UsedinSection 5.5] A halfturnisfollowedbyareflection. Supposethatthecenterofrotationofthehalfturnisnotonthelineofreflection. Showthattheresult-ingisometryisaglidereflection, whichhaslineofglidereflectionthroughthecenterofrotation, andperpendiculartotheoriginallineofreflection.

Exercise 4.6.6. [UsedinSection 5.5] A halfturnisfollowedbyaglidereflection. Supposethatthecenterofrotationofthehalfturnisonthelineofglidereflection. Showthattheresulting isometry is a reflection, whichhas lineof reflectionperpendicular to the lineofglidereflection, andatadistancefromthecenterofrotationhalf thedistanceof thetranslationintheglidereflection.

Exercise 4.6.7. [UsedinSection 5.5] A reflectionisfollowedbyaglidereflection. Supposethat the lineof reflection isperpendicular to the lineofglide reflection. Show that theresultingisometryisahalfturn, withthecenterofrotationonthelineofglidereflection.

Finally, we turn to thequestionof inversesof isometries. Bywayofanalogy, consider theintegersandtheoperationofaddition. Giventhenumber 5, isthereanumberthat“cancelsit


out”withrespecttoaddition? Theanswerisyes, namelythenumber−5, because 5+(−5) = 0,and (−5) + 5 = 0. WewilldiscussthisconceptatitsmostgeneralinSection 6.4, butfornow,wewanttoknowwhetherthereisananalogof“cancelingout”intherealmofisometriesandcomposition. Supposewearegivenanisometry P. Canwefindsomeotherisometry Q that“cancels” P out? Thatis, canwefindanisometryQ suchthat P ◦ Q = I andQ ◦ P = I? (Weneedtospecifyboththeseequations, becauseingeneraltheorderofcompositionofisometriesdoesmatter, andwecannotautomaticallydeduceoneoftheseequationsfromtheother.) IfwecanfindsuchanisometryQ, isiscalledan inverseisometry of P; weoftensimplysay inverse ofP forshort. Forexample, considertheisometry P thatistranslationtotherightby 4 inches. IfweletQ betranslationtotheleftby 4 inches, thenclearlyQ cancelsout P, andvice-versa, inthatdoingone, andthentheother, leavesuswiththeidentityisometry. Inotherwords, translationtotheleftby 4 inchesistheinverseoftranslationby 4 inchestotheright.Doeseveryisometryhaveaninverse? Ifso, istheinverseunique? Cantheinverse, ifitexists,

befoundeasily? AsseeninProposition 4.6.5, theanswertoallthesequestionsisyes. Weusethefollowingcommonnotation. Suppose P isanisometry. Theinverseisometry, of P isdenotedP−1. Wethereforehave P ◦ P−1 = I and P−1 ◦ P = I. Theselasttwoequationsmeanthatforanypoint X intheplane, wehave P(P−1(X)) = X and P−1(P(X)) = X.Inorder todiscuss inversesof translations, wewill need the followingnotionconcerning

vectorsintheplane. Supposewehaveavector v intheplane, representedbyanarrow. Wedefinethe inversevector of v tobethevector, denoted−v, representedbythearrowthathasthesamelengthasthearrowrepresenting v, buthavingtheoppositedirection. Clearly v+(−v) = 0and (−v) + v = 0, where 0 heremeansavectorwithlengthzero.

Proposition 4.6.5. Supposethat P isanisometry. Then P hasauniqueinverse. Further, wecanfindtheinverseof P asfollows.

1. I−1 = I.

2. If Tv isatranslation, then (Tv)−1 = T−v.

3. If RAα isarotation, then

(RAα

)−1= RA

−α.

4. If Mn isareflection, then (Mn)−1 = Mn.

5. If Gn,v isaglidereflection, then (Gn,v)−1 = Gn,−v.

Demonstration. WeknowfromProposition 4.6.1 thateveryisometryoftheplaneiseitheratranslation, arotationareflectionoraglidereflection. Hence, oncewedemonstrateParts (2)–(5)ofthisproposition, aswillbedoneshortly, thenitwillbeverifiedthateveryisometryhasaninverse.Toshowthattheinverseofeachisometryisunique, supposetothecontrarythatthereisan

isometry P thathastwodistinctinverses Q and R. Then, bythedefinitionofwhatitmeanstobeaninverse, weknowthat P ◦ Q = I and Q ◦ P = I, andthat P ◦ R = I and R ◦ P = I.Wethenseethat

Q = Q ◦ I = Q ◦ (P ◦ R) = (Q ◦ P) ◦ R = I ◦ R = R,

146 4. Isometries

wherewemakerepeateduseofProposition 4.4.2. Wehavenowderivedthat Q = R, whichisalogicalimpossibility, becauseweassumedthatQ and R weredistinct. Theonlywayoutofthisproblemistodeducethat P cannothavetwodistinctinverses, whichmeansthattheinverseof P isunique.WenowdemonstrateParts (1)–(5)oftheproposition. Thesedemonstrationsareallbasedon

thesameidea, whichisthatinordertoshowthattwoisometriesareinverses, weshowthatthey“canceleachotherout.”

(1). ByusingProposition 4.4.2 (1), weseethat I ◦ I = I. Itfollowsthat I−1 = I.

(2). Supposethat Tv isatranslation. ThenbyProposition 4.6.2 itfollowsthat

T−v ◦ Tv = Tv+(−v) = T0 = I.

A similarargumentshowsthat Tv ◦ T−v = I. Wededucethat (Tv)−1 = T−v.

(3). Supposethat RAα isarotation. ThenbyProposition 4.6.4 (1)itfollowsthat

RA−α ◦ RA

α = RAα+(−α) = RA

0 = I.

A similarargumentshowsthat RAα ◦ RA

−α = I. Wededucethat(RAα

)−1= RA

−α.

(4). SupposethatMn isareflection. ThenbyProposition 4.6.3 (1)itfollowsthatMn ◦ Mn =I. Wededucethat (Mn)

−1 = Mn.

(5). Suppose that Gn,v isaglidereflection. WeknowfromProposition 4.5.1 that Gn,v =Mn ◦ Tv and Gn,v = Tv ◦ Mn, andsimilarlyfor Gn,−v. Wenowcomputethat

Gn,−v ◦ Gn,v = (Mn ◦ T−v) ◦ (Tv ◦ Mn) = Mn ◦ (T−v ◦ Tv) ◦ Mn

= Mn ◦ I ◦ Mn = Mn ◦ Mn = I,

where the thirdequalityholdsbyPart (2)of thisproposition, and thefifthequalityholdsbyPart (4). A similarargumentshowsthat Gn,v ◦ Gn,−v = I. Wededucethat (Gn,v)

−1 = Gn,−v.

5SymmetryofPlanarObjectsandOrnamentalPatterns

5.1 BasicIdeas

Ourgoalinthischapteristoapplythegeneralconceptofmathematicalsymmetrytothestudyofornamentalpatterns. Suchpatternscanbe 1-dimensional, 2-dimensionalor 3-dimensional. Anexampleofa 1-dimensionalornamentalpatternisastringofbeads; anexampleofa 2-dimen-sionalpatternisapieceofwallpaper; anexampleofa 3-dimensionalpatternisapileofcannonballs. Wewillconcentrateon 2-dimensionalornamentalpatterns, that is, planarornamentalpatterns. Suchpatternsarequitecommoninart, crafts, designandarchitecture, andarefoundinmanyculturesaroundtheworld. Forexample, wallpaperisquitecommoninEuropeancul-ture; theMuslimtraditionusescomplicatedgeometricdesignsintheirbuildings; variousAfricanpeoplesuserepeatingpatternsintheirart. (Symmetriesofthreedimensionalobjectshavealsobeenstudied, butaremuchmorecomplicatedthanintheplanarcase, andwewillonlydiscusssuchsymmetryverybrieflyinSection 5.7.) A completestudyofplanarornamentalpatternsin-volvessomefairlyadvancedmathematics, namelygrouptheory(whichwewilldiscussbrieflyinChapter 6, thoughwewillnotbeabletodiscussthefullsetoftechnicalitiesneededforacompletelyrigoroustreatmentofornamentalpatterns). Evenwithoutallthetechnicaltoolsofgrouptheory, however, wecanexamineandanalyzevarioustypesofornamentalsymmetry.Ourmaintoolinthestudyofsymmetryistheconceptofisometry. InChapter 4 wediscussed

someofthefundamentalpropertiesofisometriesoftheplane. AsweviewedthingsinChapter 4,theplaneitselfwasblank, thoughwesometimesdrewafigure(suchastheletter F)onit, inordertoseewhathappenedwhenweperformedanisometry. Now, bycontrast, wewanttostartwithanobjectdrawnontheplane, andthenanalyzethesymmetryofthisobjectbyusingisometries.Bytheterm“object” drawnontheplanewesimplymeanacollectionofpointsintheplane.Thesepointscouldbeisolated(asinFigure 5.1.1 (i)), orcouldformageometricfigure(asinFigure 5.1.1 (ii)), orcouldformapicture(asinFigure 5.1.1 (iii)), orcouldformanythingelse.

148 5. SymmetryofPlanarObjectsandOrnamentalPatterns

Wewilloftenrefertoanobjectdrawnintheplaneasa“planarobject” ora“planarpattern;”wewilloftendroptheadjective“planar,” becauseitwillalwaysbeassumed(exceptforafewcases, wherewewillexplicitlysaythatwearelookingatnon-planarobjects). Wewillignoreissuesofcolor inourdiscussionofornamentalpatterns, andwillconsiderallpatterns tobe“black”pointsonatransparentplane.

(i) (ii) (iii)

Figure5.1.1

RecallourinformaldiscussionofsymmetryinSection 4.1, inwhichwerelatedsymmetrytothenotionoftransformationsoftheplane. Givenaplanarobject, a symmetry oftheobjectisanyisometryoftheplanethatcarriestheobjectontoitself. Thatis, afterperformingtheisometry,theobjectlooksjustasitdidbeforetheisometrywasperformed. Insymbols, if K isanobjectand P isanisometryoftheplane, then P isasymmetryof K preciselyif P(K) = K; wedonotrequirethat P fixeseverypointof K, butonlythat P takesallof K ontoitself. (Althoughweareusingisometriestofindthesymmetryofagivenobject, westressthattheisometriesare, asalways, transformationsofthewholeplane, notjusttheobject.) Notethatwiththisdefinitionof symmetry, weuse theword“symmetry”asanoun (beingan isometry). Weare thereforeinterestedinwhetheranobjecthassymmetries, notinwhetheritis“symmetric,” aswouldbeusedmorecolloquially.Suppose, forexample, thatourobjectistherectangleshowninFigure 5.1.2 (i). Reflectionin

averticallinethroughthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethisverticallineby L1, andthereflectioninthatlinebyM1. Similarly, reflectioninahorizontallinethroughthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethislineby L2, andthereflectioninthislineby M2. SeeFigure 5.1.2 (ii).Rotationby 180◦ aboutthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethisrotationby R1/2. Arethesetheonlysymmetriesoftherectangle? Notquite. Thereisonemoresymmetry, namelytheidentityisometry. Thesefourisometries, namelyI, R1/2, M1 and M2 areallthesymmetriesoftherectangle.

Letuslookatthesymmetriesofsomeotherobjects, startingwiththeobjectshowninFig-ure 5.1.3 (i). Thesymmetriesofthisfigureare I, R1/3 and R2/3. Hence, comparingthisobjectwiththerectangle, weseethatdifferentobjectscanhavedifferentcollectionsofsymmetries.Ontheotherhand, considertheletterH showninFigure 5.1.3 (ii). ObservethattheletterH haspreciselythesamefoursymmetriesastherectangle, namely I, R1/2, M1 and M2. Wethere-

5.1BasicIdeas 149

(i) (ii)

L1

L2

Figure5.1.2

foreseethattwodifferentobjectscanhavethesamesymmetries. Fromadesignpointofview,clearlytherectangleandtheletter H aredifferent. Fromasymmetrypointofview, however,theyarethesame. Bothpointsofviewareworthconsidering, thoughourconcerninthistextis,needlesstosay, symmetry, notdesign. Finally, considertheobjectshowninFigure 5.1.3 (iii).Colloquially, wemightsaythatthisobject“isnotsymmetric.” Whileexpressingavalidsenti-ment, suchastatementisnotaccuratefromourmathematicalperspective. TheobjectshowninFigure 5.1.3 (iii)doesinfacthaveasymmetry, namelytheidentityisometry I (everyobjecthasI asasymmetry). Thattheobjectappearstobe“notsymmetric” isexpressedpreciselybythefactthat I istheonlysymmetryoftheobject.

(i) (ii) (iii)

HFigure5.1.3

Wehavejustlistedallthesymmetriesofafewdifferentobjects. Inprincipleitispossibletolistall thesymmetriesofanyobject. Someobjectshaveinfinitelymanysymmetries(wewillseeexamplesshortly), sowecannotinpracticemakealistofallthesymmetriesofeveryob-ject. However, wecanstillcollectallthesymmetriesofanobjectintheory, evenifwecannotexplicitlylistthem. Anyobjecthasatleastonesymmetry, namely I, andsothecollectionofsymmetriesofanyobjectdoesindeedexist. Wecallthecollectionofallsymmetriesofanobject


the symmetrygroup oftheobject. Theword“group” usedheredoesnotsimplymeanacol-lectionofthings, butisusedinitstechnicalmathematicalmeaning. Themathematicalconceptofa“group,” discussedinChapter 6, ispartofthemathematicalfieldofabstractalgebra, andhasmanyusesbeyondjustthestudyofsymmetry. Groupshavebeenwidelystudiedbymath-ematicians, andsomefactsfromthetheoryofgroupscanbeusedtoobtainrathersurprisingresultsaboutthesymmetriesofobjectssuchasfriezepatternsandwallpaperpatterns(discussedinSections 5.5 and5.6 respectively).Usingourpreviousexamples, weseethatthesymmetrygroupoftherectangleshowninFig-

ure 5.1.2 (i)is {I, R1/2,M1,M2}, andthesymmetrygroupoftheobjectshowninFigure 5.1.3 (i)is {I, R1/3, R2/3}. Anysymmetrygroupcontainsatleastonesymmetryinit, namelytheidentityisometry. Anobjectthatwouldbedescribedcolloquiallyashaving“nosymmetry,” suchastheobjectshowninFigure 5.1.3 (iii), hasasymmetrygroupthatcontainspreciselyonesymmetry,namelytheidentityisometry.

Exercise 5.1.1. ForeachoftheobjectsshowninFigure 5.1.4, listallsymmetries.

(i) (ii) (iii)

(iv) (v)

Figure5.1.4

WeknowbyProposition 4.6.1 thatanyisometryoftheplaneisatranslation, arotation, areflectionoraglidereflection. Hence, anysymmetryofanobjectisoneofthesefourtypes.Tosaveverbiage, whenanobjecthasasymmetrythatisanon-trivialtranslation, wewillreferto itasa translationsymmetry of theobject; whenanobjecthasasymmetry that isanon-trivialrotation, wewillrefertoitasa rotationsymmetry oftheobject; whenanobjecthasa

5.1BasicIdeas 151

symmetrythatisareflection, wewillrefertoitasa reflectionsymmetry oftheobject; whenanobjecthas a symmetry that is anon-trivial glide reflection, wewill refer to it as a glidereflectionsymmetry oftheobject. Whenwearelookingatthesymmetriesofanobject, wewillrefertotheidentityisometryasthe identitysymmetry oftheobject. Forexample, theobjectinFigure 5.1.3 (i)hasrotationsymmetrybutnoreflectionsymmetry; theobjectinFigure 5.1.3 (ii)hasbothrotationandreflectionsymmetry. Wenotethatobjectsthathavetranslationsymmetryorglidereflectionsymmetryhaveto“goonforever.” InFigure 5.1.5 (i)weseeanobjectthat, ifweassumeitcontinuesindefinitelyinbothdirections, hastranslationsymmetry. Ontheotherhand, justbecauseanobject“goesonforever”doesnotmeanitautomaticallyhastranslationsymmetry; seeFigure 5.1.5 (ii). WewillseefurtherexamplesoftranslationsymmetryandglidereflectionsymmetryinSections 5.5 and5.6.

. . . F F F F F F . . . . . . 1 2 3 4 5 6 . . . (i) (ii)

Figure5.1.5

Wewillneedsomeadditionalterminology. Ifanobjecthasatranslationsymmetry, thenwecanlookatthetranslationvectorofthistranslationsymmetry, andwerefertothistranslationvectorasa translationvectoroftheobject. Ifanobjecthasarotationsymmetry, thenwecanlookatthecenterofrotationofthisrotationsymmetry, andwerefertothiscenterofrotationasa centerofrotationoftheobject. A centerofrotationofaplanarobjectmightbethecenterofrotationforrotationsoftheobjectbyvariousangles. Ifanobjecthasareflectionsymmetry, thenwecanlookatthelineofreflection, andwerefertothislineofreflectionasa lineofreflectionoftheobject. Ifanobjecthasaglidereflectionsymmetry, thenwecanlookatthelineofglidereflection, andwerefertothislineofglidereflectionasa lineofglidereflectionoftheobject.Wesee, therefore, thatagivenplanarobjectmightormightnothavecertaindistinguished

pointsthatarecentersofrotations, anditmightormightnothavecertaindistinguishedlines,someofwhichmightbelinesofreflection, andsomelinesofglidereflection.

Exercise 5.1.2. ForeachoftheobjectsshowninFigure 5.1.6, findandindicatethecentersofrotationandthelinesofreflection.


(i) (ii) (iii)

(iv) (v)

Figure5.1.6

Exercise 5.1.3. ThefollowingquestionsinvolvewordsinEnglishwrittenincapitalletters.Assumethatalllettersareassymmetricaspossible, andthatW isobtainedfromM by 180◦

rotation.

(1) Findfourwordsthathaveahorizontallineofreflection. Findthelongestsuchwordyoucanthinkof.

(2) Findoneormorewordsthathaveaverticallineofreflection. Findthelongestsuchwordyoucanthinkof.

(3) Findoneormorewordsthathavea 180◦ rotationsymmetry. Findthelongestsuchwordyoucanthinkof.

Nowthatwehavethenotionoftranslationvectors, centersofrotations, linesofreflectionandlinesofglidereflectionofanobject, wecanstatethefollowingtechnicalresultabouthowthesedistinguishedvectors, pointsandlinesaretreatedbysymmetriesoftheobject. Wewillusethisresultlater, whenwestudythesymmetriesofcertaintypesofplanarobjects. Thisresultshouldnotbesurprising, becausewhatitsaysintuitivelyisthatasymmetryofanobjecttakesspecialkindsofvectors, pointsandlinestothesamekindsofvectors, pointsandlines, whichisreasonablegiventhatasymmetryofanobjectleavestheobjectlookingunchanged. Weomittheproofofthisproposition.

Proposition 5.1.1. Supposethat P isaplanarobject, andlet S beasymmetryof P.

5.1BasicIdeas 153

1. If X isacenterofrotationof P, then S(X) isacenterofrotationof P.

2. If m isalineofreflectionof P, then S(m) isalineofreflectionof P.

3. If m isalineofglidereflectionof P, then S(m) isalineofglidereflectionof P.

4. If v isatranslationvectorof P, then S(v) isatranslationvectorof P.

A keyideainthestudyofsymmetryisthatwecandomorewithsymmetriesthansimplylistthesymmetriesofeachobject. Giventwosymmetriesofanobject, whicharebothisometriesoftheplane, wecanformthecompositionofthesetwosymmetries(asdiscussedinSection 4.4),toobtainanewisometryoftheplane. Whatmakesthiswholestudyofsymmetriesworkfromamathematicalperspectiveisthatthecompositionoftwosymmetriesofanobjectisinfactalsoasymmetryoftheobject. Wenowformulatethisfactmoreprecisely.

Proposition 5.1.2. Supposethat P and Q aresymmetriesofagivenobject.

1. Q ◦ P isasymmetryoftheobject.

2. P−1 isasymmetryoftheobject.

Demonstration. Suppose that theobject forwhich P and Q aresymmetries iscalled K. Bydefinitionofwhatitmeanstobeasymmetryofanobject, weknowthat P andQ areisometries,andthat P(K) = K and Q(K) = K.

(1). Because P and Q arebothisometries, weknowfromProposition 4.4.1 that Q ◦ P isanisometry. Wealsoobservethat (Q ◦ P)(K) = Q(P(K)) = Q(K) = K. Itfollowsthat Q ◦ P

isasymmetryof K.

(2). Because P isanisometry, weknowfromProposition 4.6.5 that P hasaninverseisometryP−1. Additionally, weknowthat P−1 ◦ P = I, whichmeansthat P−1(P(K)) = I(K). BecauseP(K) = K and I(K) = K (thelatterbecause I istheidentityisometry, whichtakeseveryobjectontoitself), itfollowsthat P−1(K) = K. Itfollowsthat P−1 isasymmetryof K.

Wenotethatitisthisabilitytocombinesymmetriesthatmakestheapproachtosymmetryviaisometriestheonethatisparticularlysuitedtoamathematicaltreatment, anditisthemathe-maticalanalysisofsymmetrythatleadstotheinterestingresultsaboutsymmetrythatwewillseelaterinthischapter. Ifwethinkoftheword“symmetry”inthecolloquialusageasanattributeofanobject, thereforebeinganadjectiveratherthananoun, thenwecouldnotmeaningfullycombinesymmetries.Ourgoalinthischapteristoexplorethesymmetriesofplanarobjects. Themostbasicthing

todoistolookatallpossiblesymmetriesofeachgivenobject, thatis, thesymmetrygroupoftheobject. Becausesymmetriesareisometries, wecanuseourknowledgeofisometriestolearnmoreabouttheobject. Inthissectionwehavediscussedsomegeneralideasaboutsymmetriesofobjects. Insubsequentsectionsinthischapter, wewillrestrictourattentiontovariousspecialtypesofplanarobjects, andineachrestrictedcase, wewillbeabletosaymoredefinitiveresults.Actually, ifallwecoulddowouldbetotakeaplanarobject, andfinditssymmetrygroup,

thatwouldbenice, butnotveryinteresting. Whatwouldbemoreinterestingwouldbetoknow


whetherwecouldfindallpossibletypesofsymmetrygroupsthataplanarobjectcouldhave.Ananalogymightbewithbirdwatching. Wecannotlistallpossibleindividualbirdsfoundinagivenregion, butbirdwatchingguideslistallpossibletypesofbirdsthatcanbefoundintheregion, anddescribevariouscharacteristics(forexample, color, shapeofbeak, etc.) thatcanbeusedtoidentifythetypeofanybirdspottedinthewild. Similarly, wecertainlycannotlistallpossibleobjectsthatcouldeverbedrawnintheplane, becausethereareinfinitelymanydifferentthingsthatcanbepictoriallyrepresented. However, andthisisratherremarkable, inthreeimportantcategoriesofplanarornamentalpatterns(whichbetweenthemencompassmanyoftheornamentalpatternsofinterest), wecanlistallpossibletypesofsymmetrygroupsthatcanariseforeacheachcategory. Thethreecategoriesofplanarpatternswewilldiscussarerosettepatterns, friezepatterns, andwallpaperpatterns, whichwillbetreatedindetailinSections 5.4,5.5 and5.6 respectively. OurdiscussionofisometriesinChapter 4, andourgeneraldiscussionofsymmetrygroupsinthissection, isessentiallyaimedatprovidingusthetoolstounderstandtheclassificationofsymmetrygroupsofrosettepatterns, friezepatternsandwallpaperpatterns.(Itwouldbebeyondthescopeofthisbooktoprovideallthetechnicalmathematicaldetailsforvariousproofsneededfortheanalysisoffriezepatternsandwallpaperpatterns, butwewillbeabletogiveallthedetailsforrosettepatterns, andmanyofthekeyideasfortheothertwocases.)Inordertomakeheadwaywiththeideaofclassifyingobjectsbytheirsymmetries, weneed

toaskwhatitwouldtakeinordertobeabletosaythattwoobjects“havethesametypeofsymmetry”? Wesawanexampleearlierinthissection, namelytherectangleandtheletter H,wheretwodifferentobjectshavethesamesymmetrygroups. Ingeneral, wewillsaythattwoobjectshavethesame symmetrytype iftheyhavethesamesymmetrygroups. Thatis, ifwecanmatchupthetranslationsinonesymmetrygroupwiththetranslationsoftheother, therotationsinonesymmetrygroupwiththerotationsoftheother, andsimilarlyforreflectionsandforglidereflections. (Forthosefamiliarwiththetheoryofgroups, itisnotsufficientsimplytorequirethatthetwosymmetrygroupsbeisomorphic; itisnecessarytohaveanisomorphismbetweenthetwogroupsthattakestranslationstotranslations, rotationstorotations, reflectionstoreflectionsandglidereflectionstoglidereflections. Forthosenotfamiliarwiththetheoryofgroups, donotworryaboutthesetechnicalities.)

5.2 SymmetryofRegularPolygonsI

Inordertogetafeelforsymmetrygroups, westartwiththesymmetrygroupsofregularpolygons.Letusexaminethesymmetriesofanequilateraltriangle, asshowninFigure 5.2.1. NoticeinthefigurethatwelabeledtheverticesofthetrianglebyA, B and C. Theselabelsarenotpartofthetriangle, butaretheresimplytohelpusseetheeffectsofvarioussymmetriesonthetriangle.

Thefirst thingwewanttodois tolistallsymmetriesof theequilateral triangle, that is, allisometriesoftheplanethathavethetrianglelandonitself. Itispermissiblethattheseisometriesinterchangethelettersusedtolabelthevertices, becausetheselettersarenotactuallypartof

5.2SymmetryofRegularPolygonsI 155

A

C B

Figure5.2.1

the triangle, andareonlyusedtohelpuskeeptrackofwhat isgoingon. Forexample, onepermissible isometry isreflectionin thevertical linethroughthemiddleof thetriangle. Thisreflectionleavesthetrianglelookingthesame, thoughitinterchangesvertices B and C (leavingA unmoved). Youmightfindithelpfulatthispointtocutanequilateraltriangleoutofpaper,labeltheverticesasinFigure 5.2.1, andperformtheisometrieswewilldiscuss. Unlessyouhavetransparentpaper, ithelpstowriteonbothsidesofthetrianglewhenyoufirstlabelthevertices.Thetrianglecannothaveanytranslationsymmetryorglidereflectionsymmetry, becauseany

translationorglidereflectionoftheplanewouldmovethetriangleoffitself. Thetrianglecanthereforehaveonlyrotationsymmetryandreflectionsymmetry. Whatweneedtofindarethevariouslinesofreflectionofthetriangle, andthevariouscentersofrotationandanglesofrotationof the triangle. InFigure 5.2.2 are indicated the three linesaboutwhich the trianglecanbereflectedwithoutchangingitsappearance; thesethreelinesaredenoted L1, L2 and L3. Thereflectionsthroughtheselinesaredenoted M1, M2 and M3 respectively. Forexample, ifweapplyM2 tothetriangleaspicturedinFigure 5.2.1, weseethatM2 leavesthevertexlabeled Bunmoved, andinterchangestheverticeslabeled A and C; seeFigure 5.2.3. NotethatthelinesL1, L2 and L3 arenotpartofthetriangle, butratherarefixedreferencelines; theynevermove.

L1

L3

L2

Figure5.2.2


A

C B

C

A B

M2

Figure5.2.3

Theidentityisometry, denoted I, iscertainlyasymmetryofthetriangle. Thereareonlytwonon-identity rotations that leave the triangle lookingunchanged: rotationby 120◦ clockwiseaboutthecenterofthetriangle, androtationby 240◦ clockwiseaboutthecenterofthetriangle.Whataboutrotationby 120◦ or 240◦ counterclockwiseaboutthecenterofthetriangle? Thesearecertainlysymmetriesofthetriangle, butrotationby 120◦ counterclockwisehasthesameneteffectasrotationby 240◦ clockwise, andsimilarlyrotationby 240◦ counterclockwisehasthesameneteffectasrotationby 120◦ clockwise. Asalways, weareonlyinterestedintheneteffectofanisometry, andsoitwouldberedundanttousebothcounterclockwiseandclockwiserotations; wewillsticktotheclockwiseones. Itiseasiertothinkofrotationsbyfractionsofwholeturns, rather thandegrees, sowewillwrite R1/3 and R2/3 todenote theclockwiserotationsby 120◦ and 240◦ respectively. Forexample, ifweapply R1/3 to the triangleaspictured inFigure 5.2.1, weseethat R1/3 takesthevertexlabeledA towherevertex Bwas, takesthevertexlabeled B towherevertex C was, andtakesthevertexlabeled C towherevertex A was; seeFigure 5.2.4.

Wenowhaveacompletelistofallsymmetriesoftheequilateraltriangle, namely I, R1/3, R2/3,M1,M2 andM3. Thislististhesymmetrygroupoftheequilateraltriangle; weletG denotethislist. Ournextstepistoseehowthesymmetriesinthislistcanbecombinedviacomposition.UsingProposition 5.1.2 (1), weknowinprinciplethatifwetakeanytwosymmetriesinG, thentheircompositionwillalsobein G. Considerthefollowingexample. Weknowthat M1 andR1/3 aresymmetriesofthetriangle, andsoM1 ◦ R1/3 mustalsobeasymmetryofthetriangle.Hence M1 ◦ R1/3 mustbein G. Whichofthesixmembersof G isitequalto? Thekeypointisthat M1 ◦ R1/3, thoughformedintwostages, hasasingleneteffect, anditisthisneteffectthatequalstheneteffectofpreciselyoneofthesixmembersof G. RecallthatthecompositionM1 ◦ R1/3 meansfirstdoing R1/3 andthendoing M1. InFigure 5.2.5 weseetheneteffectofperforming the two isometries in thespecifiedorder. It is important to recognize that thetransformation M1 alwaysreferstoareflectionintheline L1 exactlyasshowninFigure 5.2.2,nomatterwhathadbeendonetothetrianglepreviously. (Thelines L1, L2 and L3 arenotparts


A

C B

C

B A

R1/3

Figure5.2.4

ofthetriangle, anddonotmovewhenwerotatetheplane; wealwayswantM1,M2 andM3 tomeanthesamethingsatalltimes.) AnexaminationofFigure 5.2.5 revealsthatM1 ◦ R1/3 leavesthevertexoriginallylabeled B unmoved, anditinterchangestheverticesoriginallylabeled A

and C. Hence, theneteffectof M1 ◦ R1/3 isexactlythesameastheneteffectof M2. Wethereforecanwritetheequation M1 ◦ R1/3 = M2.

M1R1/3

A

C B

C

A B

C

B A

M1 ° R1/3 = M2

Figure5.2.5

WecancomposeanymemberofGwithanymemberofG, andtheresultwillbeamemberofG. WecansummarizeallpossiblecompositionsofmembersofG byconstructinga“multiplica-


tion”table forG, whichistheanalogofthemultiplicationtableswelearninelementaryschool;wewillcallsuchatablea compositiontable. ThecompositiontablefortheequilateraltriangleisshowninTable 5.2.1. If P and Q aremembersof G, wefind Q ◦ P inthetablebylookingattheentrylocatedintherowcontainingQ andthecolumncontaining P. Forexample, tofindR2/3 ◦ M3, welookattheentrylocatedintherowcontaining R2/3 andthecolumncontainingM3; thissquarecontains M1. Hence R2/3 ◦ M3 = M1. Thewayweobtainedthe 36 entriesinthetablewassimplybydirectlycalculatingeachone; thesecalculationscanbedoneeitherbymakingdrawingssimilartowhatisshowninFigure 5.2.5, orbyusingacut-outequilateraltriangle. (WewilllearnamoreefficientwaytoconstructthisoperationtableinSection 5.3.)

◦ I R1/3 R2/3 M1 M2 M3

I I R1/3 R2/3 M1 M2 M3

R1/3 R1/3 R2/3 I M3 M1 M2

R2/3 R2/3 I R1/3 M2 M3 M1

M1 M1 M2 M3 I R1/3 R2/3

M2 M2 M3 M1 R2/3 I R1/3

M3 M3 M1 M2 R1/3 R2/3 I

.

Table5.2.1

A numberofthingscanbeseeninTable 5.2.1. First, thetableisclearlysubdividedintofour3× 3 squares, twoofwhichonlyhaverotations(thinkingof I asarotation), andtheothertwohavingonlyreflections. Thereisanicediagonalpatternineachofthefoursquares. Also, notethateachof thesixmembersof G appearsonceandonlyonceineachrow, andonceandonlyonceineachcolumn. Wealsoseeinthetablethattheorderofcompositionofsymmetriesmatters. Forexample, itisseeninthetablethatM1 ◦ R1/3 = M2 , whereas R1/3 ◦ M1 = M3.Hence M1 ◦ R1/3 = R1/3 ◦ M1.Onefinalpointregardingthesymmetriesoftheequilateraltriangle. InSection 4.6 wedis-

cussedthenotionofaninverseisometry. ByProposition 5.1.2 (2), weknowthattheinverseisometryofeachsymmetryof theequilateral triangle isalsoasymmetryof the triangle. WecanstateveryexplicitlywhattheinverseofeachsymmetryofthetriangleisbyusingProposi-tion 4.6.5. Moreprecisely, wehave I−1 = I, R1/3

−1 = R2/3, R2/3−1 = R1/3 M1

−1 = M1,

M2−1 = M2 and M3

−1 = M3.Everythingthatwehavediscussedconcerningtheequilateraltriangleworkssimilarlyforany

regular n-gon. Justaswehavethreerotations(includingtheidentity)andthreereflectionsforthetriangle, yieldingatotalofsixsymmetries, similarlyaregular n-gonhas n rotationsand n

reflections, yieldingatotalof 2n symmetries. Thesmallestrotationofaregular n-gonis R1/n,andallotherrotationsaremultiplesofthisrotation. Forexample, thesquarewillhavesymmetrygroupwithmembers I, R1/4, R1/2, R3/4,M1,M2,M3 andM4, wherethefourreflectionshavelinesofreflectionasshowninFigure 5.2.6. Thesymmetriesofaregular n-gonare:

I, R1/n, R2/n, R3/n, . . . , R(n−1)/n,M1,M2,M3,M4, . . . ,Mn.


Itdoesnotmakeanysubstantialdifferencehowthelinesofreflectionforaregularpolygonarearranged, but, foruniformity(andlaterconvenience), wewillalwaysassumethattheyarearrangedasinFigure 5.2.6, namelywith L1 vertical, andtheothersincounterclockwiseorder(thisorderwillturnouttobeusefulinSection 5.3). Wecanformthecompositiontableforthesymmetrygroupofeachregular n-gon, analogouslytoTable 5.2.1. Thepatternsthatwesawinthetablefortheequilateraltrianglealsoholdforotherregular n-gons.

L1

L3

L4

L2

Figure5.2.6

Exercise 5.2.1. Listallthesymmetriesofaregularpentagon, andofaregularhexagon.

Exercise 5.2.2. Construct thecomposition table for thesymmetrygroupof thesquare.UselinesofreflectionlabeledasinFigure 5.2.6. Calculateeachentryinthetabledirectly;donotsimplycopy thepatternofTable 5.2.1. (Thepointof thisexerciseis toverifybyactualcalculationthatthecompositiontableforthesquarehasthesamepatternasfortheequilateraltriangle.)


Exercise 5.2.3. Fortheregularoctagon, computethefollowingsymmetries(thatis, expresseachasasinglesymmetry).

(1) R1/8 ◦ R3/8;

(2) R1/4 ◦ R5/8;

(3) R1/8 ◦ M3;

(4) M1 ◦ M5;

(5) (M6)−1;

(6) (R3/8)−1.

5.3 SymmetryofRegularPolygonsII

Supposethatwewantedtocompute R1/3 ◦ M3 ◦ M2 ◦ R1/3 ◦ M1 foranequilateraltriangle.UsingthemethodofSection 5.2, wecouldeitherdoitdirectlybydrawingatriangleanddoingeachisometryoneata time, orwecoulduse thecomposition tablegiven inTable 5.2.1 tocomputethecompositionofthefirsttwoisometries, thenthecompositionoftheresultwiththenextisometry, andsoon. Eitherway, itwouldbeaslightlytediouscalculation, thoughwecoulddoit.Nowsupposewewantedtocompute R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M1 foraregular

100-gon. InprinciplewecouldusethemethodofSection 5.2, butinpracticeitwouldbesotediousthatnoonewouldwanttodoit, becausedrawinga 100-gonwouldbeverydifficult,andmakingacompositiontablefora 100-gonwouldtakealongtime.InthissectionwepresentanalternativeapproachtothematerialwediscussedinSection 5.2,

andthisalternativeapproachwillallowustodocalculationsforaregular 100-gonjustaseasilyaswedoforanequilateraltriangle. Insteadoffiguringoutcompositionsofsymmetriesdirectly,wedevelopan“algebra”ofsymmetriesofregularpolygons. NotonlywillitbeeasiertofillintablessimilartoTable 5.2.1 oncewehavethealgebraicapproach, butwewillseethatessentiallythesamerulesworkforallregular n-gons.Tostart, wewant toexpressall thesymmetriesofaregular n-gonintermsofa fewbasic

symmetries. Thesebasicsymmetrieswillbesortoflikeatoms, outofwhichallothersymmetriesarebuilt. Nomatterwhatthevalueof n is, wewillalwaysusethesamethreesymmetriesasourbuildingblocks. InordertodistinguishthealgebraicapproachofthissectionfromthegeometricapproachofSection 5.2, wewilladoptadifferentnotation, which looksmore likealgebra.(However, whatwearedoinghereisnotthesameasthealgebrawelearninschool—thatdealswithnumbers, whereasherewearedealingwithsymmetries. Numbersandsymmetriesdonotbehaveexactlythesame.)

5.3SymmetryofRegularPolygonsII 161

Supposewehavearegular n-gonforsome n. First, welet 1 denotetheidentitysymmetry,previouslydenotedby I. Next, let r denotethesmallestpossiblenon-trivialclockwiserotationsymmetry, previouslydenoted R1/n. Forexample, in thecaseof theequilateral triangle, wewillhave r = R1/3; inthecaseofthesquare, wewillhave r = R1/4. Hence, thesymbol rdenotesrotationbyadifferentangleforeachdifferentregularpolygon. Inallcases, however,weknowthat r isthesmallestpossiblenon-trivialclockwiserotationsymmetry. Third, let mdenotereflectioninaverticalline, previouslydenoted M1. (Itwouldworkjustaswelltoletm beanyotherreflectionsymmetry, butwewillalwayschoosereflectioninavertical line,sothattherewillbenoambiguityaboutwhichreflectionisreferredtoby m.) Last, insteadofwritingcompositionusingthesymbol ◦, wewillsimplyusethestandardalgebraicnotationformultiplicationtodenotecomposition. Hence, whatweusedtowriteas M1 ◦ R1/n wenowdenote mr.Wecanusesome further standardalgebraicnotationaswell. Tostart, if k isanypositive

integer, wewill let rk meantheproductof r withitself k times. Notethat r1 = r. Wewilllet r−1 denote R−1/n, that is, the smallest possible counterclockwise rotationof the n-gon.

Forconvenience, againfollowingstandardalgebraicpractice, wewilllet r0 = 1, andforanypositiveinteger k, wewilllet r−k = (r−1)k. Thesamesortofnotationappliestoexpressionsoftheform mk.Itturnsoutthatwecanrewritealltheothersymmetriesofaregular n-gonusingthethree

symmetries 1, r and m. Letusstartwiththeequilateraltriangle, beforestatingtheresultmoregenerally. InthenotationweusedinSection 5.2, thesymmetriesoftheequilateraltriangleareI, R1/3, R2/3, M1, M2 and M3. (ItisimportantforwhatfollowsthatthelinesforthesethreereflectionsbeaspicturedinFigure 5.2.2.) Wehavealreadyseenthat I = 1, that R1/3 = r,

andthat M1 = m. Itisstraightforwardtoverifythat R2/3 = R1/32 = r2. Whatabout M2 and

M3? Weproceedasfollows. InTable 5.2.1 wesawthat M1 ◦ M2 = R1/3. HenceweobtainM1 ◦ M1 ◦ M2 = M1 ◦ R1/3. Weknowthat M1 ◦ M1 = I (byProposition 4.6.5 (4)),anditthereforefollowsthat M2 = M1 ◦ R1/3. Switchingtoournewnotation, weseethat

M2 = mr. A similarcalculationshowsthat M3 = mr2; thereaderisaskedtosupplythedetails. (Alternatively, wecouldhavetakenanequilateraltriangle, anddirectlyverifiedaswedidinSection 5.2 thatthecomposition mr hasthesameneteffectas M2, andsimilarlyformr2 and M3.) Wecouldalsohaveexpressed M2 as r2m, and M3 as rm, butforthesakeofuniformitywewillalwayskeeptheletterm ontheleftandtheletter r ontheright. Alltold, thesixsymmetriesoftheequilateraltrianglecanbewrittenas 1, r, r2, m,mr and mr2.A similarcalculationwouldshowthattheeightsymmetriesofthesquarecanbewrittenas

1, r, r2, r3, m, mr, mr2 and mr3. Thesamepatternholds fora regular n-gon, where thecompletelistofsymmetriesinournewnotationisasfollows:

Geometricnotation I R1/n R2/n · · · R(n−1)/n M1 M2 M3 · · · Mn

Algebraicnotation 1 r r2 · · · rn−1 m mr mr2 · · · mrn−1


Onceagain, westressthattheaboveequalitiesholdexactlyaswrittenonlyifweassumethatthelinesorreflectionarearrangedasinFigure 5.2.6, namelywiththelinesofreflectionarrangedconsecutivelyincounterclockwiseorder.Now thatweknowhow towriteour symmetries in termsof 1, r and m, we turn to the

compositionofsymmetries. InSection 5.2, wecomputedthecompositionoftwosymmetriesgeometrically, byseeingtheeffectonthen-gonofeachofthetwoisometriesperformedoneaftertheother. Forexample, inthecaseoftheequilateraltriangle, wecomputed M1 ◦ R1/3 = M2

asshowninFigure 5.2.5, whereweseethetwoisometriesperformedinthespecifiedorder.Thoughstraightforward, suchgeometriccomputationsarequitetedious, andarealsopronetoerror. However, becauseallsymmetriesoftheregularn-gonhavenowbeenrewrittenintermsof1, r andm, oncewecanfigureouthowtocomposethesethreebasicsymmetries, wewillthenhaveaquickmethodforcomposinganytwosymmetriesofthen-gon. Thefollowingpropositionlistssomeofthemostofthebasicrulesforcombining 1, r and m.

Proposition 5.3.1. Let 1, r and m bedefinedasaboveforaregular n-gon.

1. r · 1 = r and r = 1 · r;2. m · 1 = m and m = 1 ·m;

3. rarb = ra+b;

4. mamb = ma+b;

5. meven = 1 and modd = m, where even denotesanyevennumber, and odd denotesanyoddnumber;

6. rn = 1.

Demonstration.

(1). Thesetwoequalitiesareclear, because 1 issimplyanothernotationfortheidentityisom-etry I, andwecanapplyProposition 4.4.2 (1).

(2). ThisissimilartoPart (1).

(3). Recall that ra means theproductof r with itself a times, andsimilarly rb means theproductof r withitself b times. Hence, weseethat rarb meanstheproductof r withitselfb+ a times, whichisthesameas a+ b times. Because ra+b alsomeanstheproductof r withitself a+ b times, wededucethat rarb = ra+b.

(4). ThisissimilartoPart (3).

(5). Recallthat m isareflectionoftheplane. Itthenmustbethecasethat m2 = 1, whichisjustarestatementinourcurrentnotationofProposition 4.6.3 (1). If even denotesapositiveevennumber, then meven = m2m2m2 · · ·m2 = 1. If odd denotesapositiveoddnumber,then modd = m2m2m2 · · ·m2m = m. A similarargumentholdsfornegativeevenandoddnumbers.


(6). Recallthat r isjustanothernotationfor R1/n. ItfollowsfromProposition 4.6.4 (1)thatcomposing R1/n withitself n timesisthesameasa 360◦ rotation, whichequalstheidentityisometry.

NoticethatRules (1)–(4)intheabovepropositionarejustlikestandardalgebraicrulesfornum-bers, whereasRules (5)–(6)arenotatalllikethestandardrulesfornumbers. Hence, althoughournotationusing 1, r and m isreminiscentofstandardalgebra, itisnotthesameasit. Weshouldalwayskeepinmindthatthesymbols 1, r andm asusedhereareshort-handnotationsforvariousisometriesoftheplane, anddonotdenotenumbers. NoticealsothatRules (1)–(5)holdidenticallyforallregularpolygons, whereasRule (6)variesfordifferentvaluesof n. Thatis, forasquareRule (6)is r4 = 1, whereasforaregularpentagonthesameruleis r5 = 1.

Exercise 5.3.1. [UsedinThisSection] Foraregular n-gon, showthat rn−a = r−a foranyinteger a. Inparticular, deducethat rn−1 = r−1.

InProposition 5.3.1 wesawtherulesforcombiningexpressionsinvolvingonly r oronly m.Wehavenotyetseentherulesforcombiningexpressionsinvolvingbothm and r together; wenowturntothismissingcase. Asbefore, letusstartwiththecaseoftheequilateraltriangle.Considerthecomposition rm. Becausethisisthecompositionoftwosymmetriesoftheequi-lateraltriangle, weknowitmustbeequaltoasinglesymmetryoftheequilateraltriangle. Inotherwords, itmustbethecasethat rm equalsoneof 1, r, r2,m,mr ormr2. Ifwecalculatetheneteffectof rm fortheequilateraltriangle, usingadrawingoracut-outtriangle(justaswedidinSection 5.2), wewillseethat rm = mr, andthatinfact rm = mr2 (thereadershouldverifythisequality). Ifwetrythesameresultforthesquare, itwillturnoutthat rm = mr3.Itappears, unfortunately, asifwedonothavethesameresultforthetwodifferentpolygons.However, everythingworksoutnicelyifwerewriteourformulas. Forthecaseoftheequilateraltriangle, observethat r2 = r−1 (useExercise 5.3.1 (2)with n = 3), andtherefore rm = mr−1.Inthecaseofthesquare, wehave r3 = r−1, andthereforewealsohave rm = mr−1. Wenowseeageneralpattern, asstatedinthefirstpartofthefollowingproposition; thesecondpartofthepropositiongeneralizestheresultevenfurther.


1. rm = mr−1;

2. ram = mr−a foranyinteger a.

Demonstration. Thefirstpartoftheproofisgeometric, whereasthesecondpartisalgebraic.

(1). Wewillshowthat rm = mr−1 byapplyingeachof rm and mr−1 toaregular n-gon,andwewillcompare the results. InFigure 5.3.1 weseea regular n-gon, with someof theverticeslabeled. InFigure 5.3.2 weseetheresultofapplyingfirst m andthen r, yieldingtheneteffectofdoing rm tothe n-gon. InFigure 5.3.3 weseetheresultofapplyingfirst r−1 and


thenm, yieldingtheneteffectofdoingmr−1 tothe n-gon. Wethereforeseethattheneteffectofdoing rm and mr−1 isthesame, andthereforethesetwoisometriesareequal.

(2). If a isapositiveinteger, thenwecanapplyPart (1)ofthispropositiontocompute

ram = rr · · · r︸︷︷︸a times

m = rr · · · r︸︷︷︸a−1 times

mr−1 = rr · · · r︸︷︷︸a−2 times

mr−1r−1 = · · · = mr−1r−1 · · · r−1︸︷︷︸a times

= m(r−1)a = mr−a.

If a isnotpositive, thentheaboveargumentdoesn’twork, sowetakethefollowingapproach.Let a beanyinteger. Thenwenotethat mra isthecompositionof m and ra. Whatever a is,weknowthat ra issomerotation, andm isareflection. Hence ra isorientationpreserving, andm isorientationreversing. ByProposition 4.4.3 (2)weseethat mra isorientationreversing.Becauseallthesymmetriesofaregularpolygonarerotationsandreflections, wededucethatmra mustbeareflection. Hence, byProposition 4.6.3 (1)weknowthat (mra)(mra) = 1.Hence mramra = 1. Multiplyingbothsidesontheleftby m andontherightby r−a, weobtain mmramrar−a = m1r−a. Cancellingtheadjacentm’s, andcancelling ra and r−a, weobtain ram = mr−a. (ThisdemonstrationactuallymakesunnecessarythedemonstrationofPart(1), andthedemonstrationofPart(2)when a isapositiveinteger, butthosedemonstrationsareintuitivelymorestraightforward, andsowereworthkeeping.)

DA

B C

Figure5.3.1

Wenowhaveallthealgebraicrulesneededforworkingwith 1, r andm. Usingthesealgebraicrules, wecannoweasilyconstructcompositiontables forthesymmetriesofregular n-gons.Insteadoffiguringouteachentryinthesetablesgeometrically(thatis, bydrawingeachcase, orusingacut-out), aswedidinSection 5.2, wecansimplyuseouralgebraicrules, andessentiallyforgetaboutthegeometry. (Wearenotreallyforgettingthegeometry—thealgebraicrulesforcombining 1, r and m summarize thegeometryof the regularpolygons. Havingdevelopedtheserules, wenolongerneedtokeepgoingbacktothegeometry.) Asanexample, letuslookat thecomposition table for theequilateral triangle. Wehavealreadyseen thiscompositiontableinTable 5.2.1, butletusstartfromscratch. WestartoffwithTable 5.3.1, wherewesee


rm

rm

DA

B C

AD

C B

D

C

B A

Figure5.3.2

mr-1

mr-1

DA

B C

C

A B

DD

C

B A

Figure5.3.3

theoperationtablewithnoentriesfilledin(weputinextralinestomakeiteasiertoview). Wewishtocomputefoursampleentriesinthetable, whichwehavelabeled A, B, C and D.WecomputethefourdesiredentriesintheabovetableusingvariouspartsofProposition 5.3.1

andProposition 5.3.2. (Recallthat, asinSection 5.2, theentrylocatedintherowcontainingQ andthecolumncontaining P is Q ◦ P.) Letusstartwithentry A. Thisentryis theresultofthecomposition r · r, whichclearlyequals r2. Theentry B istheresultofthecompositionm ·mr, whichequals m2r = 1 · r = r. Theentry C istheresultofthecomposition mr · r2,whichequals mrr2 = mr1+2 = mr3 = m · 1 = m. Finally, theentry D istheresultofthecomposition mr2 ·mr, whichequals mr2mr; usingProposition 5.3.2 (2), thislastexpressionequals mmr−2r = m2r−2+1 = r−1, andbecauseweareworkingwithanequilateraltriangle,


· 1 r r2 m mr mr2

1r A

r2

m B

mr C

mr2 D

Table5.3.1

weseethatentry D equals r2. WecanthereforestarttofillinourcompositiontableasshowninTable 5.3.2.

· 1 r r2 m mr mr2

1r r2

r2

m r

mr m

mr2 r2

Table5.3.2

Usingthesamesortsofcalculations, wecaneasilycompletetheentiretable, asshowninTable 5.3.3.

· 1 r r2 m mr mr2

1 1 r r2 m mr mr2

r r r2 1 mr2 m mr

r2 r2 1 r mr mr2 m

m m mr mr2 1 r r2

mr mr mr2 m r2 1 r

mr2 mr2 m mr r r2 1

Table5.3.3

WecannowcompareTable 5.3.3 withTable 5.2.1. Asexpected, thetwotablesareentirelyidentical, exceptforthechangeofnotation. Inotherwords, ifwetakeTable 5.3.3, andreplaceeveryinstanceof 1 with I, everyinstanceof r with R1/3, etc., wewillobtainTable 5.2.1 pre-cisely. Hence, thereisnoessentialdifferencebetweentheresultsofcomputingcompositionsofsymmetriesgeometricallyvs.algebraically, althougheachmethodismoreconvenientorintu-itivelyappealingindifferentsituations. Finally, wementionthatwhatwehavejustdoneforthe


equilateraltrianglecanalsobedoneforanyregular n-gon. Oneoftheadvantagesofthealge-braicapproachisthatthealgebraicrulesarethesameforallregular n-gons(withtheexceptionofProposition 5.3.1 (6)).

Exercise 5.3.2. Construct thecomposition table for thesymmetrygroupof thesquare,analogouslytoTable 5.3.3. Useonlyouralgebraicrules, withoutactuallydrawingasquare.(Calculateeachentryinthetabledirectly; donotsimplycopythepatternofTable 5.3.3.)

Oneuseofthealgebraicapproachtosymmetriesofregularpolygonsisthatitallowsustosimplifycomplicatedexpressionsinvolvingsuchsymmetries. Forexample, supposewearegiventheexpressionmr5m6r3mr forsomeregularpolygon. WecansimplifyitusingtherulesgiveninProposition 5.3.1 andProposition 5.3.2. Recallthat m and r arenotregularnumbers, andthereforewecanuseonlytherulesdiscussedinthissection, andnottheregularrulesforalgebra.Theideaistodothesimplificationonestepatatime. Weproceedasfollows, underscoring,andjustifying, eachsteptaken:

mr3m7r9mr = mr3mr9mr byProposition 5.3.1 (5)

mr3mr9mr = mmr−3r9mr byProposition 5.3.2 (2)

mmr−3r9mr = r−3r9mr byProposition 5.3.1 (5)

r−3r9mr = r6mr byProposition 5.3.1 (3)

r6mr = mr−6r byProposition 5.3.2 (2)

mr−6r = mr−6+1 byProposition 5.3.1 (3)

mr−6+1 = mr−5

Wecanobserveintheabovecalculationsomeideasthatcanbeusedinanysimilarsituation.First, wheneverwehavean m toapower, wecansimplifytheexpression. Second, ifwehaveadjacent letters m, or ifwehaveadjacentpowersof r, wecan simplify. Third, ourgeneralstrategyistomoveallthelettersm totheleft, andalltheletters r totheright, sothateventuallywewillhavearesultthatiseitheroftheform ra forsomeinteger a, oroftheform mra forsomeinteger a. Thewaywemovetheletters m totheleftandtheletters r totherightisbyusingProposition 5.3.2.Intheaboveexample, westoppedwhenweobtainedmr−5. Thereisnopossibilityofsimplify-

ingfurther, giventhatwedonotknowhowmanyedgestheregularpolygonhas. Wenowturntoanotherexample, thistimeforaspecifictypeofregularpolygon. Supposearegiventheexpres-sion mr10m7rm forthesquare. Wecansimplifythisexpressionasfollows, thistimeomittingthejustificationforeachstep(whichthereadershouldsupply), thoughwestillunderscoreeachsteptaken.

mr10m7rm = mr10mrm = mmr−10rm = r−9m = mr9 = mr4r4r = mr.


Observethatitwasonlyintheverylaststepthatweusedthefactthatthepolygonwasasquare;uptillthatpoint, weproceededexactlyaswehaddoneforanarbitrarypolygon.

Exercise 5.3.3. Simplifyeachofthefollowingexpressionsforarbritraryregularpolygons.Ineachcase, theanswershouldbeoftheform 1 or ra or mra forsomeinteger a.

(1) mrmr2.

(2) r5mmrm.

(3) r7m3r2mr.

(4) mr3mr3mr4mr4.

(5) m4rm3rm2rm.

Exercise 5.3.4. Simplifyeachofthefollowingexpressionsfortheregularpolygonindi-cated. Ineachcase, theanswershouldbeoftheform 1 or ra ormra forsomenon-negativeinteger a, where a islessthanthenumberofedgesofthepolygon.

(1) rmr2m fortheequilateraltriangle.

(2) m3r6mrm fortheequilateraltriangle.

(3) mr9m2r forthesquare.

(4) mr4mr3mr2mr fortheregularpentagon.

(5) mr4mr3mr2mr fortheregularhexagon.

Intheaboveexamplesofsimplifyingexpressions, westartedwithanexpressioninalgebraicnotation, andusedouralgebraic rules inorder to simplify. Wecanalsouse thismethod tosimplifyanexpressionwritteninthegeometricnotationofSection 5.2, byfirstconvertingtoalgebraicnotation, thensimplifying, andthenconvertingback. Weconsidertheexamplegivenattheverybeginningofthissection, namelysimplifying R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M29

foraregular 100-gon.

R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M1 = r5 ·mr16 ·mr52 · r3 ·m = r5mr16mr52r3m

= r5mr16mr55m = mr−5r16mr55m = mr11mr55m

= mmr−11r55m = r44m = mr−44 = mr−44r100

= mr56 = M57.

ItwouldbeveryunpleasanttotrytosimplifytheaboveexpressiongeometricallyaswedidinSection 5.2.


Exercise 5.3.5. Simplifyeachofthefollowingexpressionsfortheregularpolygonindi-cated. Ineachcase, theanswershouldbeinthegeometricnotation.

(1) R1/3 ◦ M1 ◦ R2/3 ◦ M3 fortheequilateraltriangle.

(2) M2 ◦ R1/3 ◦ M3 ◦M1 fortheequilateraltriangle.

(3) R1/2 ◦ M1 ◦ M3 ◦ R3/4 forthesquare.

(4) R3/5 ◦ M1 ◦ M5 ◦ R2/5 ◦ M3 fortheregularpentagon.

(5) R1/2 ◦ M1 ◦ M3 ◦ R3/4 fortheregular 60-gon.

(6) R1/2 ◦ M1 ◦ M50 ◦ M3 ◦ R3/4 fortheregular 80-gon.

Wefinishthissectionbymentioningthealgebraicapproachtofindinginversesofsymmetriesofregularpolygons, assummarizedinthefollowingproposition. Part (4)ofthepropositionmightlookasifitwerebackwardsatfirstglance, butitiscorrect, andistheresultofthefactthatordermatterswhencombiningsymmetries.


1. (ra)−1 = r−a = (r−1)a foranyinteger a;

2. m−1 = m;

3. (mra)−1 = mra foranyinteger a.

4. (xy)−1 = y−1x−1 foranysymmetries x and y.

Demonstration.

(1). Recallthat r isanothernotationfor R1/n. Hence ra isanotherwayofwriting Ra/n. Using

Proposition 4.6.5 (3)weseethat(Ra/n

)−1= R−a/n, andthislastexpressionisseentobethe

sameas r−a, whichinturnisthesameas (r−1)a.

(2). Because m is a reflection, the equationweneed to show is simply a restatementofProposition 4.6.5 (4).

(3). AsdiscussedinthedemonstrationofProposition 5.3.2 (2), weknowthatmra isareflec-tion. ThisequationweneedtoshowissimplyarestatementofProposition 4.6.5 (4).

(4). Wecompute

(xy)(y−1x−1) = x(yy−1)x−1 = x · 1 · x−1 = 1.

Itfollowsthat (xy)−1 = y−1x−1.

WenotethatPart (4)of theabovepropositioncanbeextendedtoanynumberofsymme-tries, notjusttwo. Forexample, inthecaseoffoursymmetrieswewouldhave (xyzw)−1 =w−1z−1y−1x−1.


5.4 RosettePatterns

Regularpolygons, thesymmetriesofwhichwestudiedinSections 5.2 and5.3, maybeinter-estingmathematically, buttheyarenotpatternsofgreataestheticinterest. Wewishtoturnourattentiontoaestheticallymoreinteresting—andmathematicallymorecomplicated—objects, of-tenreferredtoasornamentalpatterns. Westartwithrosettepatterns(orsimply, rosettes) whicharethesimplestofthethreetypesofornamentalpatternsthatwewilltreat. A rosettepatternisdefinedtobeanyplanarobjectthathasonlyfinitelymanysymmetries. SeeFigure 5.4.1 forsomeexamplesofrosettepatterns. (Weleaveittothereadertolistallthesymmetriesforeachoftheobjectsinthefigure, thusverifyingthattheyareindeedrosettepatterns.)Thename“rosette”comesfromrosewindowsincathedrals, althoughinfactthehumanfigures

portrayedinarosewindowoftenpreventthewindowfromhavingnon-trivialsymmetry.

(i) (ii) (iii) (iv)

Figure5.4.1

Someauthorsuse the term“finitefigure” insteadof rosettepattern, butwe feel thisnameissomewhatmisleading, becausewhatisfiniteaboutarosettepatternisonlythenumberofsymmetries, not thegeometricnatureof thefigure. Forexample, the infinitecrossshowninFigure 5.4.2 (i)isarosettepattern(ithaseightsymmetries), eventhoughitisnotgeometricallyfinite. Moreover, notallplanarfiguresthatarefiniteinsizearerosettepatterns, forexamplethecircleshowinFigure 5.4.2 (ii). Thecirclecanberotatedaboutitscenterbyanyangle, andsoithasinfinitelymanysymmetries.

Ourultimategoalforrosettepatternsistoclassifythemaccordingtotheirsymmetrygroups.Thatis, wewishtolistallsymmetrygroupsthatariseasthesymmetrygroupsofrosettepatterns,andtobeabletotakeanygivenrosettepattern, andidentifywhichsymmetrygrouponourlistcorrespondstoit—analogoustoacompletefieldguidetothebirdsofNorthAmericathatlistsalltypesofbirdsthatcanbefound, andgivesidentifyingcharacteristicsforeachtypeofbird. Remarkably, wecanmakeacompletefieldguideforrosettepatterns. (Ofcourse, justasthefieldguideforbirdslistsonlyeachtypeofbird, noteachindividualbird, sotooourlistofsymmetrygroupsofrosettepatternsdescribesonlythesymmetriesofrosettepatterns, nottheparticulardesignelements.) IncontrasttoourdiscussionoffriezepatternsandwallpaperpatternsinSections 5.5 and5.6, whereitwouldbebeyondthescopeofthisbooktogiveallthe

5.4RosettePatterns 171

(i) (ii)

Figure5.4.2

mathematicaldetailsofthedemonstrationsofthemainresults, inthecaseofrosettepatternsweareabletodemonstrateallthepropositionsinthissection; thesedemonstrationsaresomewhatlengthy, andtheyarefoundinAppendices CandD.A symmetrygroupisacollectionofsymmetries. Tounderstandwhatcollectionsofsymmetries

ariseassymmetrygroupsofrosettepatterns, wewilllookateachofthefourtypesofisometriesasappliedtorosettepatterns. Westartwithtranslationsandglidereflections. Intuitively, ifarosettepatternhada translationsymmetry, thendoingthetranslationtwicewouldalsobeasymmetry, andthreetimes, fourtimes, etc. wouldallbesymmetries. Itwouldfollowthattheobjecthadinfinitelymanysymmetries, whichcannotbethecaseforarosettepattern. Hence,arosettepatterncannothavetranslationsymmetry. Similarlyforglidereflectionsymmetry. Wethereforehavethefollowingproposition.

Proposition 5.4.1. A rosettepatternhasnotranslationsymmetryandnoglidereflectionsym-metry.

A rosettepatterncanhavereflectionand/orrotationsymmetry. ThepatterninFigure 5.4.1 (i)hasrotationsymmetrybutnoreflectionsymmetry; thepatterninPart (ii)ofthefigurehasrotationandreflectionsymmetry; thepatterninPart (iv)ofthefigurehasnosymmetryotherthantheidentitysymmetry. Westartourdiscussionbyexaminingrotationsymmetryofrosettepatterns.


Whatcanbesaidaboutcentersofrotationofrosettepatterns. Specificially, trytodecidewhetherarosettepatterncanhavemorethanonecenterofrotation. Ifyes, trytodrawsucharosettepattern; ifno, saywhynot.


Ourfirstresultaboutrotationsymmetryofrosettepatternsisthefollowingproposition, whichanswerstheabovequestion.

Proposition 5.4.2. Ifarosettepatternhasrotationsymmetries, allsuchsymmetrieshavethesamecenterofrotation.

Wenowknowthatanyrosettepatternhasatmostonecenterofrotation. Ifthereisacenterofrotationinarosettepattern, canwesaysomethingaboutthepossibleanglesofrotationfortherotationsymmetriesaboutthiscenterofrotation. Itturnsoutthatwecansayagooddealaboutsuchangles. Becausewewillneedasimilaranalysis inour treatmentofwallpaperpatternsinSection 5.6, westateournextpropositioninageneralformthatisnotrestrictedtorosettepatterns.Supposewehaveaplanarobject(notnecessarilyarosettepattern)withacenterofrotation.

Hence, there is at least one rotation symmetryof theobjectwith thispoint as its centerofrotation. Theremightbemorethanonesuchrotationsymmetry; thatis, theremightberotationsymmetriesaboutthiscenterofrotationbyvariousangles. Amongalltheserotationsymmetriesaboutthiscenterofrotation, theremightormightnotbeasmallestclockwiserotationsymmetry(recallthattheterm“rotationsymmetry”alwaysmeansanon-trivialrotation). InFigure 5.4.2 (i)there is a smallest clockwise rotation symmetry, namelyby 90◦; in Figure 5.4.2 (ii) there isnosmallestclockwiserotationsymmetry(rotationsymmetriescanbeusedwitharbitrarilysmallangles). Wenotethatifthereisasmallestclockwiserotationsymmetry, thenrotationbynegativeoftheangleisthesmallestcounterclockwiserotationsymmetry, andvice-versa, soweneedonlyconsiderclockwiserotations.Thesituationswhereacenterofrotationdoesnothaveasmallestrotationsymmetryarecom-

plicatedmathematically, andarenotusefulforus. Bycontrast, thesituationwherecentersofrotationhavesmallestrotationsymmetriesisofgreatinterest. Thecrucialfactisthefollowingproposition, thedemonstrationofwhichisfoundinAppendix C.

Proposition 5.4.3. Let P beaplanarobject. Supposethat A isacenterofrotationof P, andsupposethatthereisasmallestclockwiserotationsymmetryabout A. Thenthereisapositiveinteger n suchthatthefollowingpropertieshold.

1. Thesmallestclockwiserotationsymmetryof P about A is RA1/n.

2. Anyrotationsymmetryof P about A isoftheform[RA1/n

]k= RA

k/n forsomeinteger k.

3. Thecollectionofalltherotationsymmetriesof P is {I, RA1/n, R

A2/n, R

A3/n, . . . R

A(n−1)/n}.

Wecannowdefinesomeveryusefulterminology. Supposethat P beaplanarobject, andsupposethat A isacenterofrotationof P. IfthereisasmallestclockwiserotationsymmetryaboutA, thenbytheabovepropositionweknowthatthesmallestclockwiserotationsymmetryabout A isbyanangleoftheform 360◦/n forsomewholenumber n. Thatis, thesmallestclockwiserotationsymmetryis RA

1/n. Wethensaythatthecenterofrotation A isof order n.(Itismoreconvenienttorefertothenumber n thantothefraction 1/n.) Additionally, if A is


acenterofrotationoforder n, andifthereisanotherrotationsymmetry RAβ about A forsome

angle β, thenProposition 5.4.3 (2) implies that β isan integermultipleof 360◦/n. That is,wehave β = (k · 360◦)/n forsomewholenumber k, whichmeansthat RA

β istheresultofcomposing RA

1/n withitself k times.Wenowreturntoourdiscussionofrosettepatterns. Supposethatarosettepatternhasrotation

symmetries. ByProposition 5.4.2, allsuchsymmetrieshavethesamecenterofrotation, say A.Becausetherosettepatternhasonlyfinitelymanysymmetries, thentheremustbeasmallestclockwiserotationsymmetryabout A. Therefore, asjustdiscussedinthepreviousparagraph,thecenterofrotation A hasorder n forsomewholenumber n. Wethensaythattherosettepatternisof order n. Ifarosettepatternhasnorotationsymmetrywesaythatitisof order 1.Forexample, therosettepatternsinFigure 5.4.1 areoforders 4, 3, 5 and 1 respectively. Everyrosettepatternhasanorder(whichisoneofthenumbers 1, 2, 3, 4, . . .).Wenowturntoreflectionsymmetryofrosettepatterns.


Thinkaboutwhatcanbesaidabouttherelationbetweenlinesofreflectionandcentersofrotationofrosettepatterns.

Thefollowingpropositioncompletelycharacterizestherelationbetweenlinesofreflectionandcentersof rotationof rosettepatterns. Thedemonstrationof thisproposition is found inAppendix D.

Proposition 5.4.4.

1. Ifarosettepatternhasbothreflectionsymmetryandrotationsymmetry, thenalllinesofreflectiongothroughthesinglecenterofrotation.

2. Ifarosettepatternhasmorethanonereflectionsymmetry, thenalllinesofreflectionoftherosettepatterngothroughasinglepoint, andanyrotationsymmetryoftherosettepatternhasthispointasitscenterofrotation.

Putting togetherwhatwehaveseenso far, weknowthat thesymmetrygroupofa rosettepatternhasnotranslationsymmetryorglidereflectionsymmetry; ifithasrotationsymmetries,theyallhavethesamecenterofrotation; ifithasreflectionsymmetryandrotationsymmetry,then all lines of reflection go through the single center of rotation; if it hasmore thanonereflectionsymmetry, alllinesofreflectiongothroughasinglepoint, andthispointisalsothecenterofrotationforallrotationsymmetries. Weknowfurtherthateveryrosettepatternhasanorder, whichisoneof 1, 2, 3, 4, . . .. Onceweknowtheorderofarosettepattern, weknowallthereistoknowaboutitsrotations. Forexample, arosettepatternoforder 5 hasrotations I,R1/5, R2/5, R3/5 and R4/5. Theonlyquestionthatremainsis, therefore, whattypesofreflectionsymmetriesarosettepatterncanhave, onceweknowitsorder.



Supposearosettepatternhasorder n. Thinkabout thepossiblenumbersofreflectionsymmetriestherosettepatterncanhave.

Letuslookatsomeexamples. TherosettepatterninFigure 5.4.3 (i)hassymmetries I, R1/5,R2/5, R3/5 and R4/5; ithasnoreflections. TherosettepatterninFigure 5.4.3 (ii)hassymmetriesI, R1/4, R1/2, R3/4,M1,M2,M3 andM4; thefourlinesofreflectioncorrespondingtothefourreflectionsM1,M2,M3 andM4 aresimilartothefourlinesofreflectionshowninFigure 5.2.6.Notice that in thefirst case thereareno reflections, and in the secondcase thenumberofreflectionsisthesameastheorderoftherosettepattern. Itturnsout(aswillbemadepreciseinProposition 5.4.5 below), thateveryrosettepatternfallsintooneofthesetwopatterns.

(i) (ii)

Figure5.4.3

Tomakeourresultprecise, foreachpositiveinteger n, wedefinethesymmetrygroup Cn tobethecollectionofsymmetries

Cn ={I, R1/n, R2/n, R3/n, . . . R(n−1)/n

}.

Foreachpositiveintegern, wedefinethesymmetrygroupDn tobethecollectionofsymmetries

Dn ={I, R1/n, R2/n, R3/n, . . . R(n−1)/n,M1,M2,M3, . . . ,Mn

}.

(Theletters C andD standfor“cyclic”and“dihedral”respectively, thoughwewillnotbeusingtheseterms. Also, wenotethatthereisnocompletelystandardnotationforthesegroups, andsomeauthorsusenotationthatisdifferentfromours—thoughthenamescyclicanddihedralarequitestandard.) Forexample, wehave

C4 ={I, R1/4, R1/2, R3/4

},

andD3 =

{I, R1/3, R2/3,M1,M2,M3

}.

UsingthealgebraicnotationofSection 5.3, wecanalsowrite Cn as

Cn ={1, r, r2, r3, . . . rn−1

},


and Dn asDn =

{1, r, r2, r3, . . . rn−1,m,mr,mr2,mr3, . . .mrn−1

}.

Wenotethatallthegroups Cn aredifferentfromoneanother, allthegroupsDn aredifferentfromoneanother, andallthegroups Cn aredifferentfromallthegroups Dn.Usingournewnotation, weseethattherosettepatternsinFigure 5.4.3 havesymmetrygroups

oftype C5 and D4 respectively. Itcanbeseen, usingtheideasofSections 5.2 and5.3, thataregular n-gonhassymmetrygroupoftypeDn, andthatDn hassametypeofcompositiontablewesawforaregular n-gon. Thecompositiontablefor Cn issimplytheupperlefthandquarterofthemultiplicationtableforaregular n-gon. Forexample, thecompositiontableforC8, usingthenotationofSections 5.3, isgiveninTable 5.4.1.

· 1 r r2 r3 r4 r5 r6 r7

1 1 r r2 r3 r4 r5 r6 r7

r r r2 r3 r4 r5 r6 r7 1r2 r2 r3 r4 r5 r6 r7 1 r

r3 r3 r4 r5 r6 r7 1 r r2

r4 r4 r5 r6 r7 1 r r2 r3

r5 r5 r6 r7 1 r r2 r3 r4

r6 r6 r7 1 r r2 r3 r4 r5

r7 r7 1 r r2 r3 r4 r5 r6

Table5.4.1

Wecannowstate thecompleteclassificationof symmetrygroupsof rosettepatterns. ThedemonstrationofthispropositionisgiveninAppendix D.

Proposition 5.4.5 (Leonardo’sTheorem). Thesymmetrygroupofarosettepatterniseither Cn

forsomepositiveinteger n, or Dn forsomepositiveinteger n.

Itfollowsthatthecompletelistofallpossiblecollectionsofsymmetriesofrosettespatternsare

C1, C2, C3, C4, . . . , Cn, . . .

D1, D2, D3, D4, . . . , Dn, . . . .

Thereareinfinitelymanysuchgroups, a Cn andaDn foreachinteger n ≥ 1. Thesesymmetrygroupsareknownasthe rosettegroups.TheabovepropositioniscommonlyreferredtoasLeonardo’sTheoreminhonorofLeonardo

daVinci, whoappearstohaveknownthisfact, thoughhedidnothavethemathematicaltoolstoexpressthisknowledgerigorously, nortoprovethattheseareindeedtheonlypossiblecollec-tionsofsymmetriesofrosettepatterns. Leonardo’sinterestinrosettepatternsmayhaveoriginatedinhisinterestinthepossiblesymmetriesofchurcheswithcircularfloorplans. SeeFigure 5.4.4forsomepicturesfromLeonardo’snotebooks.


Figure5.4.4

Leonardo’sTheoremisreallyquiteremarkable, inthatitrulesoutallsortsofcombinationsofsymmetriesasarisingfromrosettepatterns. Forexample, thereisnorosettepatternthesymmetrygroupofwhichhas 5 rotations(includingtheidentity)and 3 reflections; thereisalsonorosettepatternthathasrotationby 1/3 ofawholeturnandby 1/4 ofawholeturn, butbynosmallerrotation.Supposeyouaregivenarosettepattern, forexampletheoneshowninFigure 5.4.5. Howdo

youdetermineitssymmetrygroup? First, figureoutitsorder(order 4 inthecaseofFigure 5.4.5).Thendetermineifithasreflectionsymmetryornot(thereisreflectionsymmetryinthecaseofFigure 5.4.5). Ifthereisreflectionsymmetry, younecessarilyhavethegroup Dn, where n istheorder; ifthereisnoreflectionsymmetry, younecessarilyhavegroup Cn. HencetherosettepatternshowninFigure 5.4.5 hasgroup D4.

Figure5.4.5

Exercise 5.4.1. ForeachoftherosettepatternsshowninFigure 5.4.6, listthesymmetries,andstatewhattypeofsymmetrygroupithas.


(i) (ii) (iii)

(iv) (v) (vi)

(vii) (viii) (ix)

Figure5.4.6

Exercise 5.4.2. Foreachofthefollowingcollectionsofsymmetries, statewhetherornotitisthesymmetrygroupofsomeplanarobject. Ifyes, giveanexampleofanobjectwiththatsymmetrygroup; ifno, explainwhynot.

(1) {I, R1/3, R2/3,M1,M2}.

(2) {I, R1/3, R2/3}.

(3) {I, R1/2, R3/4}.

(4) {I,M1,M2}.

(5) {I,M1}.

(6) {I, R1/2,M1,M2}.


5.5 FriezePatterns

Wenowturntofriezepatterns(orsimply, friezes), anothertypeofornamentalpattern, whichareslightlymorecomplicatedthanrosettepatterns, butwhicharecorrespondinglymoreinterestingaswell. Aswasthecaseforrosettepatterns, inthecaseoffriezepatternswewillalsobeabletostateacompleteclassificationofthesymmetrygroups, analogoustoLeonardo’sTheoremforrosettepatterns, thoughinthepresentcaseitwouldbebeyondthescopeofthisbooktoincludeallthedetailsofthedemonstrationoftheclassificationforfriezepatterns.A friezepattern (alsoknownasa strippattern) isanyplanarobjectthathastranslationsymme-

try, butsuchthatitstranslationsymmetrysatisfiestwoconditions: (1)alltranslationsymmetriesareinparalleldirections; and(2)thereisasmallesttranslationsymmetry(recallthattheterm“translationsymmetry”alwaysmeansanon-trivialtranslation). InFigure 5.5.1, Parts (i)and(ii)arefriezepatterns(notethatinPart (ii)thebasicunitoftranslationistwopeople). Part (iii)isnotafriezepatternbecauseithastranslationsymmetryinnon-paralleldirections(forexample,horizontalandvertical), andPart (iv)isnotafriezepatternbecauseithasnosmallesttranslationsymmetry.

Figure5.5.1

Itisimportanttorecognizethatanyfriezepatternwill“goonforever.” Forexample, thepattern. . . TTTTT . . ., whichweassumeisgoingonforeverinbothdirections, isafriezepattern. Thepattern TTTTT, whichconsistsofpreciselyfive letters T, isnot a friezepattern (though it isarosettepattern, withsymmetrygroup D1). Ifaplanarobjectdoesnotgoonforever, thenitcannotpossiblyhavetranslationsymmetry, andthereforeitcannotbeafriezepattern. However,noteverythingthatgoesonforeverisafriezepattern. Forexample, anon-repeatinginfinitestripofletters, orastraightline, bothgoonforever, butneitherisafriezepattern. Ofcourse, we

5.5FriezePatterns 179

cannotphysicallydrawsomethingthatgoesonforever. Wewillunderstand, however, thateventhoughourpicturesoffriezepatternsdonotgoonforever, weshouldthinkoffriezepatternsasextendingbeyondjustwhatisdrawn, andgoingonforever. Anobjectthatgoesonforeverisnecessarilyamentalconstruct—butsoaremanyotherthingsinbothmathematicsandoutsideofit. Beingamentalconstructisnoliability, atleastfromamathematicalviewpoint. Indeed, itwouldbeapitytolimitourimaginationtoonlythosethingswecanphysicallyconstruct.Foreaseofdiscussion, wewillassumethatallfriezepatternshavebeenpositionedsothatthe

directioninwhichtheycanbetranslatedishorizontal. (ThisisthecaseinFigure 5.5.1.) Anyfriezepattern, nomatterhowitisoriginallydrawn, canberotatedtomakeit“horizontal,” sowearenotlosinganythingbyourassumption.Ourultimategoalforfriezepatternsissimilartoourgoalforrosettepatterns, namelytoclas-

sifyfriezepatternsaccordingtotheirsymmetrygroups. Whatwemeanbythis, attheriskofrepetition, is tolistallsymmetrygroupsthatariseasthesymmetrygroupsof friezepatterns,andtobeabletotakeanygivenfriezepattern, andidentifywhichsymmetrygrouponourlistcorrespondstoit. Asforrosettepatterns, wewillbeginbylookingateachofthefourtypesofisometriesasappliedtofriezepatterns.Westartbylookingat translationsymmetryof friezepatterns. Actually, thereisnothingto

sayhere. Bydefinition, everyfriezepatternmusthavetranslationsymmetry, subjecttocertainrestrictions. Hence, wecannotdistinguishbetweenvarioussymmetrygroups thatarise fromfriezepatternsbyaskingwhetherornottheyhavetranslationsymmetry—theyalldo.Wenextturntorotationsymmetryoffriezepatterns. ThefriezepatterninFigure 5.5.2 (i)hasno

rotationsymmetry. ThefriezepatterninFigure 5.5.2 (ii)hasrotationsymmetryby 180◦ aboutthepointlabeled A, andalsoaboutallpointspointsthatarehalfwaybetweentwoadjacentletters Z inthepattern, andallpointsthatareatthecenterofaletter Z. (AsinSection 4.2, wewillrefertoa 180◦ asahalfturnrotation, orsimplyhalfturn.) Itisnothardtoseethatafriezepatterncannothaverotationsymmetrybyanyangleotherthan 180◦ (oranintegermultipleof180◦), becausethefriezepatternwouldnotlandonitselfifitwererotatedbyanyotherangle.ConsiderthefriezepatternshowninfriezepatterninFigure 5.5.3. Althoughitistruethateachsquareinthefriezepatterncanberotatedby 90◦ aboutitscenter, anditwilllandonitself, sucharotationisnotasymmetryofthefriezepattern, becausewealwaysrotatethewholeplane,notjustonelittlepieceoftheplane, andifwerotatethewholeplaneby 90◦ thenthefriezepatternwillnotlandonitself. Whenitcomestorotationsymmetryforafriezepattern, itiseitherhalfturnsymmetryornothing.Wenotethatifafriezepatternhashalfturnsymmetryaboutonecenterofrotation, thenithas

infinitelymanycentersofhalfturnrotation, obtainedbyapplyingthetranslationsymmetrytotheoriginalcenterofrotation. Further, allcentersofhalfturnrotationmustbeverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).

Next, weturntoreflectionsymmetry. ThefriezepatterninFigure 5.5.4 (i)hasnoreflectionsymmetry; thefriezepatterninFigure 5.5.4 (ii)hasreflectionsymmetryintheverticallinein-dicated(andinotherverticallinesaswell); thefriezepatterninFigure 5.5.4 (iii)hasreflection


(i) (ii)

A

Z Z Z Z Z Z

Figure5.5.2

Figure5.5.3

symmetryinthehorizontallineindicated; thefriezepatterninFigure 5.5.4 (iv)hasreflectionsymmetryinbothverticalandhorizontallines. Itisnothardtoseethatafriezepatterncannothavereflectionsymmetryinalinethatisneitherverticalnorhorizontal, becausethefriezepat-ternwouldnotlandonitselfifitwerereflectedinalinethatisneitherverticalnorhorizontal.Wenotethat ifafriezepatternhasreflectionsymmetryinavertical line, thenitnecessarilyhasreflectionsymmetryininfinitelymanyverticallines, obtainedbyapplyingthetranslationsymmetrytotheoriginalverticallineofreflection. Ontheotherhand, ifafriezepatternhasreflectionsymmetry inahorizontal line, thenthere isonlyonehorizontal lineof reflection,namelythehorizontallinethatisverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).

(i) (ii)

H H H H H HD D D D D D(iii) (iv)

F F F F F F A A A A A A

Figure5.5.4

Finally, weconsiderglidereflectionsymmetry. ThefriezepatterninFigure 5.5.5 (i)hasnoglidereflectionsymmetry; thefriezepatterninFigure 5.5.5 (ii)hasglidereflectionsymmetry, wherethelineofglidereflectionishorizontal, andthetranslationinvolvedtakesa ∪ andmovesitontoanadjacent ∩. Ifafriezepatternhasglidereflectionsymmetry, thenthelineofglidereflection


mustbethehorizontallinethatisverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).

F F F F F F ∩ ∪ ∩ ∪ ∩ ∪ (i) (ii)

H H H H H H(iii)

Figure5.5.5

ThefriezepatterninFigure 5.5.5 (iii)hasglidereflectionsymmetry, butitisfundamentallydifferentfromtheglidereflectionsymmetryofthefriezepatterninFigure 5.5.5 (ii). Anyglidereflectionistheresultofcombiningatranslationandareflection. ForthefriezepatterninFig-ure 5.5.5 (iii), weseethateachofthetranslationandthereflection, thattogetherconstitutetheglidereflectionsymmetry, isitselfasymmetryofthefriezepattern. Bycontrast, thefriezepatterninFigure 5.5.5 (ii)hasnoreflectionsymmetryinahorizontalline, andneitherthetranslationnorthereflection, thattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthefriezepattern. Wecallaglidereflectionsymmetry non-trivial ifneitherthetranslationnorthereflectionthattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthefriezepattern. Wenotethatifafriezepatternhasglidereflectionsymmetry, andhasreflectionsymmetryinahorizontalline, thentheglidereflectionsymmetrymustbetrivial; thereaderisaskedtosupplythedetailsinExercise 5.5.1. Inotherwords, ifafriezepatternhasnon-trivialglidereflectionsymmetry, thenitcannothavereflectionsymmetryinahorizontal line; con-versely, ifafriezepatternhasreflectionsymmetryinahorizontalline, thenithasonlytrivialglidereflectionsymmetry. So, theonlytimetolookfornon-trivialglidereflectionsymmetryiswhenthereisnoreflectionsymmetryinahorizontalline.

Exercise 5.5.1. [UsedinThisSection] Supposethatafriezepatternhasglidereflectionsymmetry, andhasreflectionsymmetryinahorizontalline. Showthattheglidereflectionsymmetrymustbetrivial.

Inthecaseofrosettepatterns, wecouldlistthesymmetrieseachpatternhad, becauseeachlistofsymmetrieswasfinite(bydefinitionofwhatitmeanstobearosettepattern). Wecannotmakesuchlistseasilyforfriezepatterns, becausefriezepatternshaveinfinitelymanysymmetrieseach.However, eventhoughwecannotconvenientlylistsymmetriesinthecaseoffriezepatterns, we


canstillaskwhichtypesofsymmetriescanbecombinedwitheachother. Todoso, weaskthefollowingfourquestionsaboutanygivenfriezepattern.

QuestionA: Istherehalfturnsymmetry?

QuestionB: Istherereflectionsymmetryinaverticalline?

QuestionC: Istherereflectionsymmetryinahorizontalline?

QuestionD: Istherenon-trivialglidereflectionsymmetry?

Giventhateachoftheabovequestionshaseitheryesornoastheanswer, thereare 2·2·2·2 =16 possiblecombinationsofanswerstothesequestions. These 16 casesarelistedinTable 5.5.1.

QuestionsA B C D

1 N N N N2 N N N Y3 N N Y N4 N N Y Y5 N Y N N6 N Y N Y7 N Y Y N8 N Y Y Y9 Y N N N10 Y N N Y11 Y N Y N12 Y N Y Y13 Y Y N N14 Y Y N Y15 Y Y Y N16 Y Y Y Y

Table5.5.1

WhatisinterestingisthatnoteverycombinationlistedinTable 5.5.1 canactuallyoccur. Inotherwords, noteverypossibletypeofsymmetryofafriezepatterncanexistincombinationwitheveryothertypeofsymmetry. Usingsomeofthefactsaboutisometriesthatwehavealreadydiscussed, wewillinfacteliminatethemajorityofthelistedcombinationsofanswerstothefourquestions. Ineachcasethatiseliminated, wewillseethattheanswerstothefourquestionscontradicteachother.



TrytoeliminateasmanyofthecasesinTable 5.5.1 asyoucan; foreachcasethatyoueliminate, statewhynofriezepatterncansatisfythatcombinationofsymmetries. Foreachcasethatyoudonoteliminate, trytofindafriezepatternthathasthatcombinationofsymmetries.

ThecasesthatcanbeeliminatedfromTable 5.5.1 arethefollowing:

(4)NNYY. Thereisreflectionsymmetryinahorizontalline, whichimpliesthattheonlyglidereflectionsymmetryistrivial(asmentionedinourdiscussionofglidereflectionsymmetriesoffriezepatterns). Becausethiscasedoeshaveanon-trivialglidereflectionsymmetry, wehaveacontradiction.

(6)NYNY. There is reflectionsymmetry inavertical lineandaglidereflectionsymmetry. Ifthefirstof these isometries is followedby the second, thenbyExercise 4.6.7 weknow theresultingisometryisahalfturnsymmetry. Becausethiscasehasnohalfturnsymmetry, wehaveacontradiction.

(7)NYYN. Thereisreflectionsymmetryverticallineandreflectionsymmetryinahorizontalline. Ifthefirstofthesereflectionsisfollowedbythesecond, thenbyProposition 4.6.3 (3)weknowtheresultingisometryisahalfturnsymmetry. Becausethiscasehasnohalfturnsymmetry,wehaveacontradiction.

(8)NYYY. ThiscaseisjustlikeCase (7).

(10)YNNY. Thereisahalfturnsymmetryandaglidereflectionsymmetry. Thecenterofrotationofthehalfturnsymmetrymustbeonthelineofglidereflection. Ifthehalfturnisfollowedbytheglidereflection, thenbyExercise 4.6.6 wededucethattheresultingisometryisareflectionsymmetryinaverticalline. Becausethiscasehasnoreflectionsymmetryinaverticalline, wehaveacontradiction.

(11)YNYN. There isahalfturn symmetryand reflection symmetry inahorizontal line. Thecenterofrotationofthehalfturnsymmetrymustbeonthehorizontallineofreflection. Ifthehalfturnisfollowedbythereflectioninthehorizontalline, thenbyExercise 4.6.4 wededucethattheresultingisometryisreflectionsymmetryinaverticalline. Becausethiscasehasnoreflectionsymmetryinaverticalline, wehaveacontradiction.

(12)YNYY. ThiscaseisjustlikeCase (4).

(13)YYNN. Thereisahalfturnsymmetryandreflectionsymmetryinaverticalline. Ifthehalfturnisfollowedbythereflectioninaverticalline, thenbyExercises 4.6.4 and4.6.5 wededucethattheresultingisometryiseitherareflectionsymmetryinahorizontallineoraglidereflectionsymmetry. Becausethiscasehasneitherofthesesymmetries, wehaveacontradiction.

(16)YYYY. ThiscaseisjustlikeCase (4).


Therearesevencombinationsofanswersthatwehavenoteliminated, namelyCases (1), (2),(3), (5), (9), (14)and(15). Infact, eachofthesecombinationsdoesarisefromafriezepattern,aswillbeseenveryshortly. Moreover, all friezepatterns thathave thesameanswers to thefourquestionshavethesamesymmetrygroups, andeachofthesesevencombinationsofan-swerscorrespondstoadifferentsymmetrygroup. (Theproofofthesefactsusessomeadvancedmathematicsthatisbeyondthescopeofthisbook.) Insum, therearepreciselysevensymmetrygroupsoffriezepatterns. Thesesevensymmetrygroups, knownasthe friezegroups, areoftendenotedwiththesymbols f11, f12, f1m, f1g, fm1, fmm and fmg. (Therationaleforthesesymbolsisasfollows: the f standsforfrieze; thefirstsymbolafterthe f is 1 ifthereisnore-flectionsymmetryinaverticalline, andis m ifthereis; thesecondsymbolafterthe f is 1 ifthereisnoothersymmetry, ism ifthereisreflectionsymmetryinahorizontalline, is g ifthereisnon-trivialglidereflectionsymmetry, andis 2 ifthereishalfturnsymmetry.) Wesummarizetheclassificationofthesymmetrygroupsoffriezepatterns, andgiveanexampleofeachoftheseventypes, inthefollowingproposition.

Proposition 5.5.1 (ClassificationofFriezePatterns). ThesymmetrygroupofanyfriezepatternisoneofthesevengroupslistedinTable 5.5.2.

QuestionsName A B C D Example

f11 N N N N F F F F F F F F

f1g N N N Y D ∪ D ∩ D ∪ D ∩f1m N N Y N D D D D D D D D

fm1 N Y N N T T T T T T T T

f12 Y N N N S S S S S S S S

fmg Y Y N Y ∪ ∩ ∪ ∩ ∪ ∩ ∪ ∩fmm Y Y Y N O O O O O O O O

Table5.5.2

InSection 5.4 wenotonlystatedthetypesofsymmetrygroupsthatcouldariseforrosettepatterns, namelythe Cn and Dn groups, butweexplicitlylistedallthemembersofeachofthesegroups; forexample, westatedthat

Cn ={1, r, r2, r3, . . . rn−1

}.

Canwegiveasimilarexplicitdescriptionofeachofthesevenfriezegroups? Intheorywecoulddoso, thoughitismorecomplicatedthaninthecaseoftherosettegroups, becauseeachrosettegroupisfinite, whereaseachfriezegroupisinfinite. Consider, forexample, thefriezegroup f11,whichisthesymmetrygroupoffriezepatternsthathavenosymmetryotherthantranslation, forexample · · · FFFFF · · · . Let t denotethesmallestpossibletranslationsymmetrytotherightof


thisfriezepattern. Thenthecollection f11 ofallsymmetriesofthisfriezepatternis

f11 ={· · · t−3, t−2, t−1, 1, t, t2, t3, · · ·

}.

Wecanthinkof t as t1, and 1 as t0. Althoughwewillnotwriteacompositiontablefor f11,becausesuchatablewouldbeinfinite, wecanexplicitlydescribehowtocombineanytwosymmetriesin f11 bytherule tatb = ta+b.

Exercise 5.5.2. Listthesymmetriesinthefriezegroup f1m, similarlytothewaywelistedthesymmetriesin f11. Onceagainlet t denotethesmallestpossibletranslationsymmetrytotherightofthisfriezepattern, andlet h denotereflectioninahorizontalline.

LetusnowuseTable 5.5.2 toanalyzethesymmetriesofthefriezepatterninFigure 5.5.6.Wefirstaskifthefriezepatternhashalfturnsymmetry. Inthiscasetheanswerisyes; thereadershouldfindacenterofrotationforahalfturnsymmetry. Thenextquestioniswhetherthereisreflectionsymmetryinverticallines. Theanswerisyes; thereadershouldfindaverticallineofreflection. Next, weaskwhetherthereisreflectionsymmetryinahorizontalline. Theanswerisno. Finally, weaskifthefriezepatternhasnon-trivialglidereflectionsymmetry. Theanswerisyes; thereadershouldfindthenon-trivialglidereflection. WethereforehaveanswersYYNYtoQuestionsA,B,C,D.Itfollowsthatthefriezepatternhassymmetrygroup fmg.

Figure5.5.6

Observethatthesituationforfriezepatternsisverydifferentfromrosettepatternsinthefollow-ingcrucialway: thereareinfinitelymanydistinctrosettegroups, butonlysevenfriezegroups.Thisisanamazingfact. Inasense, thegreatergeometriccomplexityoffriezepatternsrestrictshowtheycanbeconstructed. Theredoesnotseemtobeanysimpleintuitivereasonforthenumberoffriezegroups, namelyseven; itsimplycomesoutofthemathematicaldetails.

Exercise 5.5.3. ForeachofthefriezepatternsshowninFigure 5.5.7, statetheanswerstoQuestionsA–D,andstatewhatsymmetrygroupithas.


Figure5.5.7

Exercise 5.5.4. Findandphotocopy 7 friezepatterns, allwithdifferentsymmetrygroups.Foreachofthefriezepatternsyoufind, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.

5.6WallpaperPatterns 187

Exercise 5.5.5. ThemathematicianJohnH.Conwayhascomeupwiththefollowingde-scriptivenamesforthesevenfriezegroups, eachonebasedonaformofbodilymotion:hop, jump, step, sidle, spinninghop, spinningjumpandspinningsidle. Performeachtypeofmotion(partoftheproblemisfiguringoutwhateachmotionis), andlookatyourfoot-prints. Thefootprintsfromeachmotionformafriezepattern. MatchupthesefootprintfriezepatternswiththesevenlistedinTable 5.5.2.

5.6 WallpaperPatterns

Thelast, andmostinteresting, ofourthreetypesofornamentalpatternsarewallpaperpatterns.Thoughwallpaperpatternsaremorecomplicatedtechnicallythanfriezepatterns, heretoowewillbeabletostateacompleteclassificationofthesymmetrygroupsthatarise. Onceagainitwouldbebeyondthescopeofthisbooktoincludeallthedetailsofthedemonstrations.A wallpaperpattern isanyplanarobjectthathastranslationsymmetrysubjecttotwocondi-

tions: (1)thetranslationsymmetriesarenotallinparalleldirections; and(2)thereisasmallesttranslationsymmetryinanypossibledirectionforwhichthereistranslationsymmetry. Addi-tionally, weassumethatateverycenterofrotationofthewallpaperpatternthereisasmallestclockwiserotationsymmetry. InFigure 5.6.1, Parts (i)and(ii)arewallpaperpatterns. Part (iii)isnotawallpaperpatternbecausealltranslationsymmetriesareinparalleldirections, andPart (iv)isnotawallpaperpatternbecauseithasnosmallesttranslationsymmetryintheverticaldirec-tion. Thislastexampleshowsthatnoteverythingyoumightputonawalliscalleda“wallpaperpattern”inthetechnicalsense.

Justasafriezepatternhadto“goonforever”inordertohavetranslationsymmetry, thesameholdsforawallpaperpattern, exceptthatwallpaperpatternsgoonforeverinalldirections, notjustone. Ofcourse, anypicturewedrawofawallpaperpatternwillnotgoonforever, butthatissimplytheresultofourhumanlimitations. Wewillunderstand, however, thateventhoughourpicturesofwallpaperpatternsdonotgoonforever, weshouldthinkofwallpaperpatternsasextendingbeyondjustwhatisdrawn, andgoingonforever.Onecontrastbetweenhowwedrawwallpaperpatternsandfriezepatternsisthatfriezepat-

ternswerealwaysdrawnhorizontally(forconvenience), whereasforawallpaperpatternthereisnooneparticulardirectionthatcanbesingledoutandmadehorizontal.Ourultimategoal forwallpaperpatterns is just likeourgoal for friezepatterns, namelyto

classifywallpaperpatternsaccordingtotheirsymmetrygroups. Weproceedverymuchaswedidwithfriezepatterns, namelyfirstexaminingeachofthefourtypesofisometriesasappliedtowallpaperpatterns. Aswithfriezepatterns, wedonotneedtosayanythingabouttransla-tionsymmetryofwallpaperpatterns, becauseeverywallpaperpatternmusthave translationsymmetry, subjecttocertainrestrictions.


Figure5.6.1

Letusstartbyexaminingrotationsymmetryofwallpaperpatterns. Aswith friezepatterns,awallpaperpatternmightormightnothaverotationsymmetry. ThewallpaperpatterninFig-ure 5.6.2 (i)hasnorotationsymmetry(otherthantheidentity); thewallpaperpatterninFig-ure 5.6.2 (ii)hasrotationsymmetryby 120◦ orby 240◦ aboutthepointslabeled A, B and C

(andaboutallsimilarpoints); thewallpaperpatterninFigure 5.6.2 (iii)hasrotationsymmetryby 90◦, 180◦ or 270◦ aboutthepointslabeled X and Y (andallsimilarpoints), androtationsymmetryby 180◦ aboutthepointlabeled Z (andallsimilarpoints). Wethereforeseethatincontrasttofriezepatterns, whererotationsymmetrycanonlybeby 180◦, forwallpaperpatternsrotationsymmetrycanbebyavarietyofangles; moreover, differentcentersofrotationinthesamewallpaperpatterncanhavedifferentanglesofrotation.

IfwelookatthewallpaperpatternshowninFigure 5.6.2 (iii), weseethreecentersofrotation,labeled X, Y and Z respectively. Ofcourse, thewallpaperpatternhasothercentersof rota-tion, besidesthethreethatarelabeled. Indeed, becausewallpaperpatternsrepeatthemselvesinfinitely, ifawallpaperpatternhasonecenterofrotation, thenithasinfinitelymanycentersofrotation. Itwould, therefore, besillyforustoattempttofindliterallyallthecentersofrotation


AB

C

X

Y Z

(i) (ii)

(iii)

Figure5.6.2

ofagivenwallpaperpattern. Whatwecanhopetofindareallthe“genericallydifferent”typesofcentersofrotationofawallpaperpattern. Wemakethisconceptpreciseasfollows.Supposewearegivenawallpaperthathascentersofrotation. Wesaythat twocentersof

rotationofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonecenterofrotationtotheother(thesymmetrycouldbeanyofthefourtypesofisometries). InFigure 5.6.3, weseefourcentersofrotationlabeledA,B,C, andD. ThepointsAand B areequivalentcentersofrotation, becausereflectionintheverticallinehalfwaybetweenthemisasymmetryofthewallpaperpatternthattakes A to B. Ontheotherhand, notwoofthepointsA, C andD areequivalent, becausenosymmetryofthewallpaperpatterntakesoneofthemtoanother.

Ingeneral, foranycenterofrotationofawallpaperpattern, wecanlookforallthecentersofrotationthatareequivalenttoit; allsuchcentersofrotationwillinfactbeequivalenttoeachotheraswell. Wecallsuchacollectionofequivalentcentersofrotationan equivalenceclass of


A B

DC

Figure5.6.3

centersofrotation. Forexample, inFigure 5.6.3, theequivalenceclassofthecenterofrotationC consistsofallpointsthatareinthemiddlesofallthe“bricks”outofwhichthepatternisbuilt. Foranywallpaperpattern, itcanbeshownthatthecollectionofallitscentersofrotationcanbebrokenupintoafinitenumberofequivalenceclasses, whichwillbedisjointfromeachother. Now, withthisnotionofequivalenceclasses, wecanstatemorepreciselywhatitmeanstofindallthe“genericallydifferent”typesofcentersofrotationofawallpaperpattern. Givenawallpaperpattern, whatwewanttofindispreciselyonecenterofrotationperequivalenceclass. Forexample, thecentersofrotationlabeled X, Y and Z inFigure 5.6.2 (iii)areexactlyonerepresentativefromeachequivalenceclassofcentersofrotationforthiswallpaperpattern.Assuch, wecansayinformallythatwehavefound“allthecentersofrotation”ofthepattern.

Exercise 5.6.1. Foreachwallpaperpatterns shown inFigure 5.6.4, findand labelonecenterofrotationperequivalenceclass(ifthereareany).

Tomakesenseofrotationsymmetryofwallpaperpatterns, weneedtorecallfromSection 5.4thenotionofacenterofrotationofanobjecthavingorder n. Becauseweareassumingthatateverycenterofrotationofawallpaperpatternthereisasmallestclockwiserotationsymmetry,theneverycenterofrotationofawallpaperpatternhassomeorder n, thoughdifferentcentersofrotationofagivenwallpaperpatterncanhavedifferentorders. Forexample, thepoints A,B and C inFigure 5.6.2 (ii)areallcentersofrotationoforder 3, andthepoints X, Y and Z inFigure 5.6.2 (iii)arecentersofrotationoforders 4, 4 and 2 respectively.Foragivenwallpaperpattern, wecanfinditscentersof rotation(that is, wecanfindone

centerofrotationperequivalenceclass). Eachofthesecentersofrotationhasanorder.


(i) (ii)

(iii) (iv)

Figure5.6.4


Trytofigureoutwhichnumberscanoccurasordersofcentersofrotationofwallpaperpatterns. Areallnumberspossible, oronlysome? Ifthelatter, whichnumbersoccur?

WeknowfromtheexamplesinFigure 5.6.2 that 2, 3 and 4 canoccurasordersofcentersofrotationofwallpaperpatterns. Arethereanyothernumberspossible? Forinstance, isitpossibletohaveawallpaperpatternwithacenterofrotationoforder 5? Howabout 7, or 583? Itturnsout, quiteremarkably, thatthereareveryfewpossibleordersforcentersofrotationofwallpaperpatterns, aswenowstate.

Proposition 5.6.1. Everycenterofrotationofawallpaperpatternhasorder 2, 3, 4 or 6.

A rigorousproofoftheabovepropositionusesgrouptheory, andisbeyondthescopeofthisbook. Aninformaldiscussionofwhythisresultistruecanbefoundin[Wey52, pp.101–103]. ToseehowremarkableProposition 5.6.1 is, noteinparticularthatitmeansthattherecannotbeawallpaperpatternwithacenterofrotationoforder 5. Simplyarrangingpentagonsinaninfinite


grid, asshowninFigure 5.6.5 (i)willnotyieldawallpaperpatternwithcenterofrotationoforder5, becauseweneedtorotatetheentireplane, notjustonepentagonatatime. InFigure 5.6.5 (ii)weseeanexampleofacleverIslamicdesignthatincorporatespentagons, andmightthereforegiveanillusionoforder 5 centersofrotation, butitisonlyanillusion.

Figure5.6.5

Givenawallpaperpattern, wecanlookforthedifferentcentersofrotationthatithas. Eachcenterofrotation, ifthereareany, hasanorderthatis 2, 3, 4 or 6. Wenowwanttodefinetheorderforthewholewallpaperpattern. Ifthewallpaperpatternhasnocentersofrotation, thenwesaythatthewallpaperpatternisof order 1. Ifthewallpaperpatternhascentersofrotation, wesaythatthewallpaperpatternisof order n if n isthehighestorderfoundamongthecentersofrotationofthewallpaperpattern. Forexample, thewallpaperpatterninFigure 5.6.2 (i)isoforder1; thewallpaperpatterninFigure 5.6.2 (ii)isoforder 3; thewallpaperpatterninFigure 5.6.2 (iii)isoforder 4.


Exercise 5.6.2. FindtheorderofeachwallpaperpatternshowninFigure 5.6.4.

Exercise 5.6.3. Supposewearegivenawallpaperpattern. Supposefurtherthat, amongitscentersofrotation, thewallpaperpatternhasanorder 2 centerofrotationandanor-der 3 centerofrotation. Fromthispartialinformation, canyoudeterminetheorderofthewallpaperpattern? Ifyes, whatistheorder, andwhy?

Weturnnexttoreflectionsymmetry. Aspreviouslymentioned, incontrasttofriezepatterns,wherewedistinguishedbetweenverticalandhorizontallinesofreflection, forwallpaperpat-ternsthereisnosuchdistinction, becauseawallpaperpatterngoesonforeverinalldirections,sowecannotisolateonedirectionas“horizontal.” ThewallpaperpatterninFigure 5.6.6 (i)hasnoreflectionsymmetry; thewallpaperpatternsinFigure 5.6.6 (ii)and(iii)bothhavereflectionsymmetry, inthelinesindicated(andinallsimilarlines). Ifawallpaperpatternhasreflectionsymmetryinsomelineofreflection, thenitwillnecessarilyhaveinfinitelymanylines, obtainedbyapplyingthetranslationsymmetryofthewallpaperpatterntotheoriginallineofreflection.AlthoughthewallpaperpatternsinFigure 5.6.6 (ii)and(iii)bothhavereflectionsymmetry, thereisonemajordifferencebetweenthereflectionsymmetryofthesetwopatterns, namelythatallthelinesofreflectionforPart (ii)areparallel, whereasthelinesofreflectionarenotallparallelforPart (iii).

Ifawallpaperpatternhasonelineofreflection, thenithasinfinitelymanylinesofreflection. Aswasthecasewithcentersofrotation, wewouldliketofindallthe“genericallydifferent”typesoflinesofreflectionofawallpaperpattern; onceagain, weusetheconceptofequivalence.Supposewearegivenawallpaperthathaslinesofreflection. Wesaythattwolinesofreflectionofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonelineofreflectiontotheother. InFigure 5.6.7, weseethreelinesofreflectionlabeledm, nand k. Thelines m and n areequivalentlinesofreflection, becauserotationinthecenterofrotation A showninFigure 5.6.3 isasymmetryofthewallpaperpatternthattakes m to n. Ontheotherhand, thelines m and k arenotequivalent, becausenosymmetryofthewallpaperpatterntakes m to k.

Givenawallpaperpattern, andalineofreflectionforthiswallpaperpatterns, wecanlookforallthelinesofreflectionthatareequivalenttoit. Wecallsuchacollectionofequivalentlinesofreflectionan equivalenceclass oflinesofreflection. Forexample, inFigure 5.6.7, theequivalenceclassof the lineof reflection m consistsofallvertical linesof reflectionof thewallpaperpattern(thatis, allverticallinesthatareboundariesbetweenthe“bricks”). Givenawallpaperpattern, whatwewanttofindispreciselyonelineofreflectionperequivalenceclass.Forexample, thelinesofreflectionshowninFigure 5.6.6 (iii)areexactlyonerepresentative


(i) (ii)

(iii)

Figure5.6.6

m n

k

Figure5.6.7

fromeachequivalenceclassoflinesofreflectionforthiswallpaperpattern. Assuch, wecansayinformallythatwehavefound“allthelinesofreflection”ofthepattern.


Exercise 5.6.4. ForeachwallpaperpatternshowninFigure 5.6.4, findandlabelonelineofreflectionperequivalenceclass(ifthereareany).

Wenowlookatglidereflectionsymmetryforwallpaperpatterns. Asforfriezepatterns, weareinterestedonlyinnon-trivialglidereflectionsymmetry, that is, glidereflectionsymmetrysuchthatneitherthetranslationnorthereflection, thattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthewallpaperpattern. Justaswasthecaseforfriezepatterns,ifawallpaperpatternhasalineofglidereflectionthatisalsoalineofreflection, thentheglidereflectionsymmetryinthatlineistrivial. Moreover, itturnsoutthatinawallpaperpattern, anylineofreflectionisautomaticallyalineofglidereflection(foratrivialglidereflectionsymmetry);thereaderisaskedtosupplythedetailsinExercise 5.6.5. Puttingtheseobservationstogether,weseethattofindanon-trivialglidereflectionsymmetry, weneedtofindalineofglidere-flectionthatisnotalineofreflection. Wecallsuchlinesofglidereflection non-triviallinesofglidereflection. ThewallpaperpatterninFigure 5.6.8 (i)hasnoglidereflectionsymmetry; thewallpaperpatterninFigure 5.6.8 (ii)hasnon-triviallinesofglidereflectionasindicated(notethattheverticallinesthroughthemiddleoftheletters M aretriviallinesofglidereflection).

(i) (ii)

Figure5.6.8

Exercise 5.6.5. [Used inThis Section] Suppose that awallpaperpatternhas a lineofreflection. Show that this lineof reflectionmustalsobea lineofglide reflection (foratrivialglidereflectionsymmetry). ThisexerciseusesideasfromAppendix C.

Itissometimestrickyinpracticetofindnon-triviallinesofglidereflectioninwallpaperpat-terns, certainlytrickierthanitistofindcentersofrotationandlinesofreflection. Linesofglidereflectionoftentendtobe“inbetween”featuresof thewallpaperpattern, forexampleasin


Figure 5.6.8 (ii). Moreprecisely, ifalineofglidereflectionisparalleltolinesofreflection, thenitmustbehalfwaybetweenthelinesofreflection. SeeExercise 5.6.6 fordetails. Exercise 5.6.7discussestherelationbetweenalineofglidereflectionandcentersofrotationnotonit. Note,however, thatlinesofglidereflectionneednotbeparalleltoanylinesofreflection, andcaninfactintersectlinesofreflection; thereaderisaskedtosupplyanexampleinExercise 5.6.8.

Exercise 5.6.6. [UsedinThisSection] Supposethatawallpaperpatternhasanon-triviallineofglidereflectionthatisparalleltolinesofreflection. Showthatthelineofglidereflec-tionishalfwaybetweenthelinesofreflection. ThisexerciseusesideasfromAppendix C.

Exercise 5.6.7. [UsedinThisSection] Supposethatawallpaperpatternshasanon-triviallineofglidereflection, andithasacenterofrotationthatisnotonthelineofglidereflection.Showthatthereisanothercenterofrotationatthesamedistancefromthelineofglidereflection, buton theother side (thoughnotdirectly across from theoriginal centerofrotation). ThisexerciseusesideasfromAppendix C.

Exercise 5.6.8. [UsedinThisSection] Findanexampleofawallpaperpatternthathasnon-triviallinesofglidereflectionandhaslinesofreflection, andsuchthatthenon-triviallinesofreflectionintersectsomelinesofreflection. Thereissuchanexampleamongthewallpaperpatternsshownsofarinthissection, thoughitslinesofreflectionandlinesofglidereflectionarenotshown.

Justaswehavethenotionofequivalentcentersofrotation, andequivalentlinesofreflec-tion, wehavethesamenotionfornon-triviallinesofglidereflection. Supposewearegivenawallpaperthathasnon-triviallinesofglidereflection. Wesaythattwonon-triviallinesofglidereflectionofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonelineofglidereflectiontotheother. InFigure 5.6.9, weseethreenon-triviallinesofglidereflectionlabeled a, b and c. Thelines a and b areequivalentlinesofglidereflection,becausereflectionintheverticallinehalfwaybetween a and b isasymmetryofthewallpaperpatternthattakes a to b. Ontheotherhand, thelines a and c arenotequivalent, becausenosymmetryofthewallpaperpatterntakes a to c.

Givenawallpaperpattern, andanon-triviallineofglidereflectionforthiswallpaperpatterns,wecanlookforallthenon-triviallinesofglidereflectionthatareequivalenttoit. Wecallsuchacollectionofequivalentnon-triviallinesofglidereflectionan equivalenceclass ofnon-triviallinesofglidereflection. Forexample, inFigure 5.6.9, theequivalenceclassofthelineofglidereflection a consistsofallverticalnon-triviallinesofglidereflectionofthewallpaperpattern


a b

c

Figure5.6.9

(thatis, allverticallinesthatarehalfwaybetweentheverticallinesthroughtheedgesofthe“bricks”). Foragivenwallpaperpattern, wewanttofindpreciselyonenon-triviallineofglidereflectionperequivalenceclass.

Exercise 5.6.9. For eachwallpaper pattern shown in Figure 5.6.4, find and label onenon-triviallineofglidereflectionperequivalenceclass(ifthereareany).

Inthecaseoffriezepatterns, afterdiscussingthedifferenttypesofsymmetriesthatcouldoccur,wewereledtofourquestionsconcerningthesetypesofsymmetries. Wefollowasimilarplanforwallpaperpatterns, thoughwithoneadditionalquestionthatdoesnothaveananalogamongthefourquestionswehadforfriezepatterns. Uptillnowwehavelookedseparatelyateachofthefourtypesofisometriesastheycanoccurassymmetriesofwallpaperpatterns. Wenowneedtoaskonequestionconcerninghowthesedifferenttypesofsymmetriesinteract. Supposeawallpaperpatternhasbothrotationsymmetryandreflectionsymmetry. Thewallpaperpatternmustthereforehavebothcentersofrotationandlinesofreflection.


Supposethatawallpaperpatternhasbothcentersofrotationandlinesofreflection. Mustallthehighestordercentersofrotationbeonlinesofreflection?

Theanswertotheabovequestionisthatinsomewallpaperpatternsthehighestordercentersofrotationareallonlinesofreflection, andinotherwallpaperpatternstheyarenot. (Centersofrotationthatarenotthehighestorderarenotofusetousforourpresentpurpose.) Forexample,inthewallpaperpatterninFigure 5.6.2 (iii), thehighestordercentersofrotationarethepoints


labeledX and Y (theyareorder 4), andboththesepointsareonlinesofreflection. Bycontrast, inthewallpaperpatterninFigure 5.6.3, thehighestordercentersofrotationarethepointslabeledA, C and D (theyareallorder 2); wedonotneedthepoint B, becauseitisequivalentto A.Thepoints C and D areonlinesofreflection, butthepoint A isnotonalineofreflection.Itturnsoutthatwenowhaveeverythingweneedtoclassifywallpaperpatternsaccordingto

theirsymmetries. Aswasthecasewithfriezepatterns, wecannotconvenientlylistallthesym-metriesofwallpaperpatterns, butwecanstillaskwhichtypesofsymmetriescanbecombinedwitheachother. Inparticular, weaskthefollowingfivequestionsaboutanygivenwallpaperpattern.

QuestionA: Whatistheorderofthewallpaperpattern? (Answer: 1, 2, 3, 4, or6.)

QuestionB: Istherereflectionsymmetry? (Answer: YesorNo.)

QuestionC: Istherereflectionsymmetryinnon-parallellines? (Answer: YesorNo.)

QuestionD: Areallhighestordercentersofrotationonlinesofreflection? (Answer: YesorNo.)

QuestionE: Isthereglidereflectioninnon-triviallinesofglidereflection? (Answer: YesorNo.)

Itcanbeseenthatthereare 5 · 2 · 2 · 2 · 2 = 80 possiblecombinationsofanswerstothesequestions. Wewillnotlistall 80 here. Aswithfriezepatterns, itturnsoutthatmostofthese 80casescannotactuallyoccur. Hence, noteverypossibletypeofsymmetryofawallpaperpatterncanexistincombinationwitheveryothertypeofsymmetry. Wewillnotgooverthedetailsofhowtoeliminatethecasesthatcannotoccur; todosowouldbebeyondthescopeofthisbook.Somecasesaresimpletoeliminate, however, andarelefttothereaderinExercises 5.6.10 and5.6.11.

Exercise 5.6.10. [UsedinThisSection] Showthatnowallpaperpatterncanhaveanswers1, yesandyestoQuestionsA–C,regardlessofwhattheanswerstoQuestionsD andE are.(Wecanthereforeeliminatethecombinationsofanswers 1YYNNN, 1YYNNY, 1YYNYN,1YYNYY, 1YYYNN, 1YYYNY, 1YYYYN, 1YYYYY.)

Exercise 5.6.11. [UsedinThisSection] Showthatnowallpaperpatterncanhaveanswers3, yesandnotoQuestionsA–C,regardlessofwhattheanswerstoQuestionsD andE are.Similarly, showthatnowallpaperpatterncanhaveanswers 4, yesandno, oranswers 6,yesandno, toQuestionsA–C.ListallthecombinationsofanswerstoQuestionsA–E thatcanthereforebeeliminated.

AfteralltheimpossiblecombinationsofanswerstoQuestionsA–E areeliminated, itturnsoutthatthereare 17 combinationsofanswersthatdooccur. Moreover, allwallpaperpatternsthat


havethesameanswershavethesamesymmetrygroups, andeachofthese 17 combinationsofanswerscorrespondstoadifferentsymmetrygroup. (Theproofofthesefactsusessomeadvancedmathematicsthatisbeyondthescopeofthisbook.) Insum, thereareprecisely 17 symmetrygroupsofwallpaperpatterns, knownasthe wallpapergroups. Thewallpapergroupsareoftendenotedwiththesymbols p1, pg, pm, cm, p2, pgg, pmg, cmm, pmm, p3, p31m, p3m1,p4, p4g, p4m, p6 and p6m. (Thereareothersetsofsymbolsthatvariousauthorsuse, butthesymbolswehaveusedseemtobethemostcommon; asforthesymbolsusedtodenotethefriezegroups, thereisarationaleforthewallpapergroupsymbols, butitisnotworthdwellingupon.) Thereasonforthenumber 17 isnomoreintuitivelyobviousthanthereasonthattherearepreciselysevenfriezegroups; inbothcasesitcomesoutofthemathematicalanalysis. Wesummarizetheclassificationofthesymmetrygroupsofwallpaperpatternsasfollows.

Proposition 5.6.2 (ClassificationofWallpaperPatterns). Thesymmetrygroupofanywallpaperpatternisoneofthe 17 groupslistedinTable 5.6.1.

QuestionsName A B C D Ep1 1 N N N Npg 1 N N N Ypm 1 Y N N Ncm 1 Y N N Yp2 2 N N N Npgg 2 N N N Ypmg 2 Y N N Ycmm 2 Y Y N Ypmm 2 Y Y Y Np3 3 N N N Np31m 3 Y Y N Yp3m1 3 Y Y Y Yp4 4 N N N Np4g 4 Y Y N Yp4m 4 Y Y Y Yp6 6 N N N Np6m 6 Y Y Y Y

Table5.6.1

Anexampleofeachofthe 17 typesofwallpaperpatternsisgiveninFigures 5.6.10 and5.6.11(thefirstofthesefiguresshowsallthewallpaperpatternsoforders 1 and 2, andthesecondofthefiguresshowsallthewallpaperpatternsoforders 3, 4 and 6).


p1 pg pm

cm p2 pgg

pmg cmm pmm

Figure5.6.10

Thoughthe 17 wallpapergroupsweretreatedmathematicallyonlyinthelate19thcentury,theyseemtohavebeenknownonsomeintuitivelevelearlier. Forexample, wallpaperpatternsforallthesegroupscanbefoundintheAlhambra inGranada, Spain, whichwasbuiltduringthe9th–14thcenturies. SeeFigure 5.6.12 foronesuchpattern. TheAlhambrawasbuiltbytheArabrulerswhocontrolledpartofSpainatthetime. BecauseIslam forbidstheuseofrepresen-tationalpictures, Muslimartistsexcelledatgeometricdesigns. (ItshouldbenotedthatArabicculture wasgenerallymoreadvancedmathematicallythantheEuropeancultureduringtheMid-dleAges. Moreover, duringtheRenaissance, theEuropeanslearnedmuchGreekmathematicsthroughArabictranslations. TheArabicculturehasnotalwaysbeengiventhecredititdeservesinthesematters. Itisnotclear(totheauthor, anyway)whetherthedesignersoftheAlhambraac-tuallyknewexplicitlythattherewereseventeendifferenttypesofsymmetryconfigurationsthatawallpaperpatterncouldhave, orwhethertheyweresimplysogoodatdesigninggeometricpatternsthattheymanagedtofindallofthembyaccident. Asasidenote, theDutchartistM.C.Escher wasinspiredtomakehisownrepeating, interlockingfiguresaftervisitingtheAlhambra,ashisnotebooksshow. Itisclaimedin[Wey52]thattheancientEgyptianshadfoundwallpaperpatternsofall 17 types; manyothercultures, includingChinaandvariouspeoplesinAfrica,alsoexcelatgeometricdesign.

LetusnowuseTable 5.6.1 toanalyzethesymmetriesofthewallpaperpatterninFigure 5.6.13.First, wefindthecentersofrotation. ThepointsA, B and C inFigure 5.6.13 arethreecentersof


p3 p31m p3m1

p4 p4g p4m

p6 p6m

Figure5.6.11

rotation, andallothersareequivalenttothese. Allthesecentersofrotationhaveorder 2, sothewallpaperhasorder 2. Next, weaskifthepatternhasreflectionsymmetry. Theanswerisyes.Thenextquestioniswhetherthereisreflectionsymmetryinnon-parallellines. Becausetherearebothhorizontalandverticallinesofreflection, theanswerisyes. Fourth, weaskifallhighestordercentersofrotationareonlinesofreflection. Allthreeof A, B and C arehighestorder(inthiscaseorder 2), butbecause C isnotonalineofreflection, theanswertothisquestionisno.Finally, weaskwhetherthewallpaperpatternhasglidereflectionsymmetryinnon-triviallinesofglidereflection. Theanswerisyes, asthereadershouldverify. LookingatTable 5.6.1 leadsustoconcludethatthewallpaperpatternhassymmetrygroup cmm.

Exercise 5.6.12. ForeachofthewallpaperpatternsshowninFigure 5.6.14, statethean-swerstoQuestionsA–E,andstatewhatsymmetrygroupithas.


Figure5.6.12

Figure5.6.13


Figure5.6.14


Exercise 5.6.13. Findandphotocopy 4 wallpaperpatterns, allwithdifferent symmetrygroups. Foreachofthewallpaperpatternsyoufind, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.

Exercise 5.6.14. Draw 4 wallpaperpatterns, allwithdifferentsymmetrygroups. (IfyouarealsodoingExercise 5.6.13, thenmakesurethewallpaperpatternsyoudrawhavedifferentsymmetrygroupsthantheonesyoufoundandphotocopied.) Foreachofthewallpaperpatternsyoudraw, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.

5.7 ThreeDimensionalSymmetry

Havingsofardiscussedthesymmetryofplanarobjectsinthischapter, weturnbrieflytoalookatsymmetryofthreedimensional objects(thatis, spatialobjects). Thestudyofsymmetryofthreedimensionalobjectsisinmanywayssimilartothestudyofsymmetrywehaveseenforplanarobjects, thoughitismorecomplicated, andwewillmentiononlyafewideas, andwillnotgiveathoroughtreatmentaswedidforplanarobjects. Forsomeinterestingissuesconcerningspatialobjects, see[Wey52].Justas thestudyofsymmetryofplanarobjects isbasedonthenotionof isometriesof the

plane, thestudyofthesymmetryofthreedimensionalobjects(whichwewillrefertoas“threedimensionalsymmetry”)isbasedonisometriesofthreedimensionalspace. Assuch, athoroughtreatmentof threedimensionalsymmetrywouldcommencewithanexaminationofallpos-sibletypesofisometriesofthreedimensionalspace. Ratherthangivingacompletetreatmentofisometriesinthreedimensionalspace, whichwouldbeverylengthy, wewilllookatafewexamplesofsymmetriesofthreedimensionalobjects, startingwiththesymmetriesofthecube,analogouslytowhatwedidinSection 5.2, wherewelookedatthesymmetriesoftheregularpolygons. Wepointout, withoutgoingintothedetails, thatallthebasicideasaboutisometriesandsymmetriesthatholdfor theplanehaveanalogsforthreedimensionalspace; forexam-ple, thecompositionoftwosymmetriesofathreedimensionalobjectisstillasymmetryoftheobject, etc.InFigure 5.7.1 weseeacube, withitsverticeslabeled(justaswelabeledtheverticesofregular

polygonsinSection 5.2). Aswithregularpolygons, therearenotranslationorglidereflectionssymmetriesof thecube(thoughother threedimensionalobjectscanhavesuchsymmetries).Clearlytheidentityisometryofthreedimensionalspace, denoted I asintheplanarcase, isasymmetryofthecube. Letusnowtrytofindallthenon-trivialrotationsymmetriesofthecube.Intheplane, eachrotationisperformedaboutapoint, calledthecenterofrotation, whichisfixedbytherotation. Inthreedimensionalspace, bycontrast, eachrotationisperformedaround

5.7ThreeDimensionalSymmetry 205

aline, calledthe axisofrotation. Thereisoneslightcomplicationinvolvingrotationinthreedimensionalspace, however. Intheplane, weusedtheconventionthatrotationbyapositiveangleistakentobeclockwise. Wecouldadoptthisconventionbecausewecanalldistinguishbetweenclockwiseandcounterclockwiserotations. Inthreedimensionalspace, supposewewanttorotateaboutagivenaxisofrotationbyagivenpositiveangle. Inwhichdirectionshouldwerotate? Therearetwopossibilitiesforrotatingbythegivenpositiveangleaboutthegivenaxisofrotation, andweneedtofindawaytospecifywhichoneistobeused. Themethodforsolvingthisproblemisthateveryaxisofrotationwillbegivenadirection, specifiedbyanarrowhead, asseenforexampleonline a inFigure 5.7.2. Wethenadopttheconventionthatwewillconsiderclockwiserotationaboutthelinetobethedirectionofrotationthatappearsclockwisewhenwelookfromthetailofthearrowtowardtheheadofthearrow. Wesaythatsuchrotationfollowstherighthandrule. Thatis, weconsiderclockwiserotationaboutalinewithanarrowheadtobethedirectiongivenbycurlingthefingersofyourrighthand, whenyouplaceyourthumbparalleltotheaxisofrotation, andinthedirectionofthearrowhead. Thisrighthandruleisusedregularlyinphysics.Using theaboveconsiderations in thecaseof thecube, wesee inFigure 5.7.2 a rotation

symmetryof thecube, namelyrotationby 1/4 turnaround the line labeled a, which is theverticallinethroughthecenterofthecube. Noticethattherotationisclockwisewhenviewedfromabove thecube, which is looking in thedirectionof thearrowhead shownon line a.Rotationby 1/2 and 3/4 around line a are also symmetriesof the cube. Wedenote thesesymmetriesby Ra

1/4, Ra1/2 and R

a3/4 respectively. Theseareallthenon-trivialrotationsymmetries

aroundline a.

B C

DAF G

HE

Figure5.7.1


Trytofindasmanyrotationsymmetriesofthecubeaspossible.

Whataretheotheraxesofrotationofthecube? Therearetwomorethatareverysimilarto a,namelythe“front-to-back”horizontalline b throughthecenterofthecubethatisperpendiculartothesquare ADHE, andthe“left-to-right”horizontalline c throughthecenterofthecube


B C

DA

a

F G

HE

A B

CDE F

GH

R1/4a

Figure5.7.2

thatisperpendiculartothesquare ABFE. Then Rb1/4, R

b1/2, R

b3/4, R

c1/4, R

c1/2 and Rc

3/4 areallnon-trivialrotationsymmetriesofthecube.Therearealsoothertypesofaxesofrotationofthecube. Thelines a, b and cwerethroughthe

centersofopposingsquarefaces. Therearealsoaxesofrotationthroughmidpointsofopposingedges. Forexample, inFigure 5.7.3 (i)weseethelinethatgoesthroughthemidpointsof AB

and HG, pointinginthedirectionofthemidpointof HG; thislineisdenoted d. Then Rd1/2

isasymmetryofthecube. (Observethat Rd1/4 and Rd

3/4 arenotsymmetriesofthecube.) There

arefiveothersimilaraxesofrotation: theline e thatgoesthroughthemidpointsof BC andEH; theline f thatgoesthroughthemidpointsof CD and E F; theline g thatgoesthroughthemidpointsofAD and FG; theline h thatgoesthroughthemidpointsofAE and CG; andtheline i thatgoesthroughthemidpointsof DH and BF; inallcasesthelinespointinthedirectionofthemidpointofthesecondlistededge. Hence Re

1/2, Rf1/2, R

g

1/2, Rh1/2 and Ri

1/2 aresymmetriesofthecube.

B C

DA

d j

F G

HE

B C

DAF G

HE

(i) (ii)

Figure5.7.3


Therearealsoaxesofrotationthroughopposingverticesofthecube. Forexample, inFig-ure 5.7.3 (ii)weseethelinethatgoesthroughtheverticesA andG, pointinginthedirectionofG; thislineisdenoted j. Then Rj

1/3 and Rj

2/3 aresymmetriesofthecube. Therearethreeothersimilaraxesofrotation: theline k thatgoesthrough B and H; theline l thatgoesthrough C

and E; andtheline m thatgoesthrough D and F; inallcasesthelinespointinthedirectionofthesecondlistedvertex. Hence Rk

1/3, Rk2/3, R

l1/3, R

l2/3, R

m1/3 and Rm

2/3 aresymmetriesofthecube. Wenowhaveacompletelistofrotationsymmetriesofthecube.Next, weturntoreflectionsymmetriesofthecube. Intheplane, wereflectedinaline, called

thelineofreflection. Inthreedimensionalspace, wereflectinaplane, calledthe planeofre-flection. Thatreflectionofthreedimensionalspaceisinaplaneisquitereasonableintuitively—mirrorsareplanes!


Trytofindasmanyreflectionsymmetriesofthecubeaspossible.

ReferringtothecubeshowninFigure 5.7.1, itisevidentthatthecubehasreflectionsymmetryintheplanethatgoesthroughthecenterofthecubeandisparalleltothetop(ABCD)andthebottom(EFGH). Callthisplane p, anddenotereflectioninthisplaneby Mp. Therearetwoothersimilarplanesofreflection: theplane q thatgoesthroughthecenterofthecubeandisparalleltotheleftside(ABFE)andtherightside(DCGH); andtheplane r thatgoesthroughthecenterofthecubeandisparalleltothefront(ADHE)andtheback(BCGF). ThenMq andMr aresymmetriesofthecube.Thereisanothercollectionofplanesofreflectionofthecube. Forexample, let s denotethe

planecontainingtheedges AB and GH. ThenMs isareflectionsymmetryofthecube. Therearefiveothersimilarplanesofreflection: theplane t containingtheedges BC and EH; theplane u containingtheedges CD and E F; theplane v containingtheedges AD and FG;theplane w containingtheedges AE and CG; andtheplane x containingtheedges BF andDH. HenceMt,Mu,Mv,Mw andMx aresymmetriesofthecube. Wenowhaveacompletelistofreflectionsymmetriesofthecube.Havewenowfoundallthesymmetriesofthecube? Itmightatfirstappearasifwedoknow

allthesymmetriesofthecube, giventhatweknowallthereflectionandrotationsymmetriesofthecube, andweknowthattherearenotranslationorglidereflectionsymmetries. However, inthreedimensionalspace, thecompletelistoftypesofisometriesisnotjusttranslations, rotations,reflectionsandglidereflections. Itturnsoutthattherearetwoadditionaltypesofisometriesinthreedimensionalspace, called rotaryreflections and screws. Bothofthesetypesofisometriesaresimilartoglidereflections, inthattheyaresingleisometriesthataredescribedintermsoftwo-stepprocesses. A rotaryreflectionistheresultoffirstrotatingaroundanaxisofrotation,andthenreflectinginaplanethatisperpendiculartotheaxisofrotation; ascrewistheresultoffirst rotatingaroundanaxisof rotation, and then translating inadirectionparallel to theaxisofrotation. (See[Mar82, Section16.1]forathoroughdiscussionoftheisometriesofthreedimensionalspace, includingthethreedimensionalanalogofProposition 4.6.1.)


Thecubedoesnothaveanyscrewsymmetries, butitdoeshaverotaryreflectionsymmetries.Forexample, considerthecomposition Mp ◦ Ra

1/4. Thiscompositioniscertainlyasymmetryof thecube, being thecompositionof twosymmetries. However, byusingdrawings similartoFigure 5.7.2, itcanbeverifiedthatthiscompositionisnotequaltoanyoftherotationorreflectionsymmetrieswehavelistedforthecube(suchaverificationwouldentailcomparingtheneteffectofMp ◦ Ra

1/4 withtheneteffectsofeachoftherotationsandreflectionsthatwehavefound; weleavethedetailstothereader). Hence, thecomposition Mp ◦ Ra

1/4 isasymmetryofthecube, andisnotequaltoanyothertheothersymmetriesthatwehaveseensofar. Forconvenience, weusethefollowingnotation: If α isanangle, if a isalineinthreedimensionalspace, andif p isaplanethatisperpendicularspace, welet Ca

α,m denotetherotaryreflectionthatconsistsoffirstdoingtherotation Ra

α, andthendoingthereflection Mm. Hence, wewriteCa

1/4,p asanabbreviationforMp ◦ Ra1/4. Therearesixothersimilarrotaryreflectionsymmetries

ofthecubethatcanbeobtainedbycompositionsofrotationandreflectionssymmetriesofthecube, andtheseare Ca

1/2,p, Ca3/4,p, C

b1/4,r, C

b3/4,r, C

c1/4,q, C

c3/4,q.

Thereadermighthavenoticedthatwedidnot include Cb1/2,r and Cc

1/2,q in theabovelistof rotary reflectionsymmetriesof thecube. These twocompositionsare indeedvalid rotaryreflectionsymmetriesofthecube, butitturnsoutthattheyarebothequalto Ca

1/2,p. (Again, thereadercanverifythatthesethreecompositionshavethesameneteffects.) Actually, theneteffectofthesethreecompositionsisaparticularlynicesymmetryofthecube. SeeFigure 5.7.4 forthecomposition Ca

1/2,p. Observethattheneteffecttakeseachvertex, andmovesittothelocationdiametricallyoppositeitwithrespecttothecenterofthecube. Let O denotethecenterofthecube. TheisometrythattakeseverypointinthreedimensionalspaceandsendsittothepointdiametricallyoppositeitwithrespecttoO iscalled inversion inO. Let JO denotethisisometry.Fromnowon, insteadofwriting Ca

1/2,p wewillwrite JO. Itturnsoutthat JO canbeobtained

in sixadditionalwaysas rotary reflections; eachof Mu ◦ Rd1/2, Mv ◦ Re

1/2, Ms ◦ Rf1/2,

Mt ◦ Rg

1/2, Mx ◦ Rh1/2 and Mw ◦ Ri

1/2 isequalto JO.

Wearestillnotfinishedlookingforrotaryreflectionsymmetriesofthecube. Certainly, onecanobtainarotaryreflectionsymmetryofthecubebycomposingarotationsymmetryofthecubewithareflectionsymmetryofthecube(aslongastheplaneofreflectionisperpendiculartotheaxisofrotation). However, notallrotaryreflectionsymmetriesofthecubeareobtainedthatway. Itisalsopossibletoformarotaryreflectionsymmetryofthecubewheretherotaryre-flectionisthecompositionofarotationandareflection, neitherofwhichaloneisasymmetryofthecube, buttheircompositionis. (A similarphenomemonoccurredwhenwestudiedglidere-flectionsymmetryoffriezepatternsandwallpaperpatterns.) Forexample, let j denotetheplanecontainingthecenterofthecubethatisperpendiculartotheline j (showninFigure 5.7.3 (ii)).Thenneither Rj

1/6 nor Mj isasymmetryofthecube, butthecomposition Mj ◦ Rj

1/6, abbre-

viatedasbeforeby Cj

1/6,j, isinfactasymmetryofthecube. Theneteffectofthissymmetryis

showninFigure 5.7.5. Therearesevenothersimilarrotaryreflectionsymmetriesofthecube,andtheseare Cj

5/6,j, Ck

1/6,k, Ck

5/6,k, Cl

1/6,l, Cl

5/6,l, Cm

1/6,m, Cm5/6,m.


B C

DAF G

HE

D A

BCH E

FG

H E

FGD A

BC

R1/2

Mp

R1/2

aMp C

1/2, p° =

a

a

Figure5.7.4

B C

DAF G

HE

F B

CGE A

DH

R1/6

jM j C

1/6, j° =j

Figure5.7.5

Wenow, finally, haveacompletelistofsymmetriesofthecube:

I, Ra1/4, R

a1/2, R

a3/4, R

b1/4, R

b1/2, R

b3/4, R

c1/4, R

c1/2, R

c3/4, R

d1/2, R

e1/2,

Rf1/2, R

g

1/2, Rh1/2, R

i1/2, R

j

1/3Rj

2/3, Rk1/3, R

k2/3, R

l1/3, R

l2/3, R

m1/3, R

m2/3,

Mp,Mq,Mr,Ms,Mt,Mu,Mv,Mw,Mx, JO, Ca1/4,p, C

a3/4,p,

Cb1/4,r, C

b3/4,r, C

c1/4,q, C

c3/4,q, C

j

1/6,j, C

j

5/6,j, Ck

1/6,k, Ck5/6,k, C

l1/6,l, C

l5/6,l, C

m1/6,m, C

m5/6,m.


Clearly, thislistofsymmetriesismuchlarger, andmorecomplicated, thanthelistofsymmetriesofthesquare, whichisthetwodimensionalanalogofthecube(thecubehas 48 symmetries,versus 8 forthesquare). Nonetheless, weseethatforthreedimensionalobjectsitispossibletoformcompletelistsofsymmetries; inotherwords, wecanformthesymmetrygroups ofthreedimensionalobjectsjustaswedidforplanarobjects. Moreover, wecanformthecompositionsofsymmetriesofanobjectinthreedimensionalspace, andinprinciplewecouldformcompositiontables forthreedimensionalobjectsjustaswedidforregularpolygonsinSections 5.2 and5.3. Inpracticeformingsuchanoperationtablewouldbeverytimeconsuming—forthecubewewouldhavea 48×48 table, whichwouldhave 2304 entries—andsowewillnotactuallyconstructsuchtables. InExercise 5.7.1 thereaderisaskedtocomputethecompositionsofvarioussymmetriesofthecube; thesecalculationscomputesomeoftheentriesofthecompositiontableforthecube. Thebottomlineisthatsymmetryofthreedimensionalobjectscanbestudiedsimilarlytothestudyofplanarobjects, butthreedimensionalobjectsareagoodbitmorecomplicated.

Exercise 5.7.1. Forthecube, computethefollowingsymmetries(thatis, expresseachasasinglesymmetry).

(1) Rj

1/3 ◦ Ra1/4.

(2) Mp ◦ Rc1/4.

(3) Ms ◦ Mp.

(4) Mq ◦ Ca1/4,p.

Exercise 5.7.2. Foreachofthefollowingpolyhedra, listallofitssymmetries. Usepicturesorwordstodescribetheaxesofrotationandplanesofsymmetry.

(1) A pyramidoverasquare.

(2) A prismoveranequilateraltriangle, wherethesidesarerectangles, butnotsquares.

(3) A regulartetrahedron.

(4) A regularoctahedron.

Exercise 5.7.3. Howmanysymmetriesdoestheprismoveraregular n-gonhave? Assumethat the sidesof theprismare rectangles, butnot squares. (Youdonotneed to list thesymmetries, justcountthem.)


Exercise 5.7.4. Whatistherelationbetweenthesymmetriesofaconvexpolygonandthesymmetriesofitsdual?

Intheplane, westudiedthesymmetryofvariousclassesofobjects: regularpolygons, rosettepatterns, friezepatternsandwallpaperpatterns. Arethereanalogsforsuchclassesofobjectsinthreedimensionalspace? Theanswerisdefinitelyyes. A particularlyinterestingclassofob-jectsinthreedimensionalspaceistheanalogofwallpaperpatterns, thatis, patternsinthreedimensionalspacethathavetranslationsymmetryinatleastthreedifferentdirections(wherenotalldirectionsareinasingleplane). Suchpatternsarecalled crystals, becauseofthefactthatthemoleculesinchemicalcrystals, suchassalt(NaCl), alignthemselvesinalattice-likeformthatcorrespondsexactlytothenotionofhavingtranslationsymmetryinthreedifferentdirectionsinthreedimensionalspace. Thestudyofchemicalcrystalsiscalledcrystallography,andthesymmetrygroupsofmathematicalcrystalsarecalledthe crystallographicgroups (alsoknownasthe spacegroups). Thecrystallographicgroupscanbeclassifiedanalogouslytotheclassificationoffriezegroupsandwallpapergroups, althoughtheclassificationismuchmorecomplicated. Thereare 230 crystallographicgroups(incontrastto 17 wallpapergroups). Thecrystallographicgroupswerefirstcompletelyclassifiedin1891byEvgrafStepanovichFedorov(1853-1919)andArthurSchoenflies (1853-1928), eachworkingindependentlyoftheother. See[Sen90]formoredetailsaboutthecrystallographicgroups.Although the symmetrygroupof thecube ismuch largerandmorecomplicated than the

symmetrygroupofthesquare, thereisonesimilaritybetweenthesetwosymmetrygroupsthatwecanobserve. Inthesymmetrygroupof thesquare, half thesymmetriesarerotations(weconsider I tobeatrivialrotation), andhalfarereflections. AswesawinSection 5.4, thisequalsplitbetweenrotationsandreflectionsholdsforallrosettegroupsthathavereflectionsymme-try. Noticeinparticularthatrotationspreserveorientation andreflectionsreverseorientation.Hence, foranyrosettegroup, halfthesymmetriesareorientationpreservingandhalfareorien-tationreversing. Now, inthecaseofthecube, thesymmetrygroupcontainsnotonlyrotationsandreflections, butalsorotaryreflections. However, itstillisthecaseforthecubethathalfofitssymmetriesareorientationpreserving(therotations), andhalfareorientationreversing(thereflectionsandtherotaryreflections). Infact, itturnsoutthatanyfinitesymmetrygroupforanobjectinthreedimensionalspace(andactuallyinanydimensionalspace), eitherconsistsofallorientationpreservingsymmetries, orhalfitssymmetriesareorientationpreservingandhalfareorientationreversing. ThedemonstrationofthisfactisoutlinedinExercise 5.7.5.


Exercise 5.7.5. [UsedinThisSection] Ourgoalistoshowthatforanyfinitesymmetrygroupforanobjectinthreedimensionalspace, preciselyoneofthefollowingsituationsholds:eitherallthesymmetriesareorientationpreserving, orhalfthesymmetriesareorientationpreservingandhalfareorientationreversing. Wewillmakeuseofthefollowingtwofactsaboutisometriesthatwehaveseenfortheplane, andwhichinfactholdtrueinthree(orhigher)dimensionalspace; wewillnotbeabletodemonstratethesetwofacts—thatwouldrequiremoretechnicalitiesthanweareusing. First, theanalogofProposition 4.4.3 holdsinthreedimensions. Second, everyisometryhasaninverse.Suppose G isafinitesymmetrygroupforanobjectinthreedimensionalspace. Theargu-menthasanumberofsteps, mostofwhichhavesomethingforthereadertodo.

(1) Supposethat A, B and C aresymmetriesin G, andthat A = B. Showthat C ◦A = C ◦ B.

(2) If G hasallorientationpreservingsymmetries, thenthereisnothingtodemonstrate,soassumefromnowonthatnotallsymmetriesinG areorientationpreserving. Showthat G hasbothorientationpreservingandorientationreversingsymmetries.

(3) Let {A1, A2, . . . , An} denotetheorientationpreservingsymmetries in G, andlet{B1, B2, . . . , Bm} denotetheorientationreversingsymmetriesin G, where n andm aresomepositiveintegers. Ourgoalistoshowthat n = m, whichwillimplythat G hasthesamenumberoforientationpreservingsymmetriesandorientationreversingsymmetries.

(4) Considerthecollectionofsymmetries {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An}. Showthatthesesymmetriesarealldistinct.

(5) Showthatallthesymmetries {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An} areorientationreversing.

(6) Deduce that every oneof {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An} is contained in{B1, B2, . . . , Bm}.

(7) Deducethat n ≤ m.

(8) Usesimilarideastoshowthatallthesymmetries {B1 ◦ B1, B1 ◦ B2, . . . , B1 ◦ Bm}

aredistinct, andallarecontainedin {A1, A2, . . . , An}. Deducethat m ≤ n.

(9) Becausewehaveseenthat n ≤ m andthat m ≤ n, itfollowsthat n = m, whichiswhatweneededtoshow.

Finally, wearenowinapositiontoclarifysomethingleftunfinishedinSection 3.3, wherewediscussedthesemi-regularpolyhedra. Inparticular, welistedallsuchpolyhedra(inPropo-sition 3.3.1), andwementionedthatallexceptoneofthem(thepseudorhombicuboctahedron)satisfiedastrongerpropertycalledvertextransitivity. WecouldnotdefinethispropertyinSec-


tion 3.3, butwenowhavethenecessarytoolforthedefinition, namelysymmetry. A polyhedronissaidtobe vertextransitive if, givenanytwovertices v and w ofthepolyhedron, thereisasymmetryofthepolyhedronthattakes v to w.For example, we claim that theprismover a regular pentagon is vertex transitive. In Fig-

ure 5.7.6 weseeaprismoveraregularpentagon. Toshowthatthisprismisvertextransitive, weneedtoshowthatforanytwoverticesoftheprism, thereisasymmetryoftheprismtakingonevertextotheother. ConsidertheverticeslabeledA and B, asseeninFigure 5.7.6. Observethatrotationby 1/5 ofaturnaroundtheverticallinethroughthecenteroftheprismisasymmetryoftheprism, andthissymmetrytakesvertex A tovertex B. Rotationby −1/5 ofaturntakes B toA. TotakevertexA tovertex C, asseeninthefigure, weneedtherotaryreflectionobtainedbyfirstrotatingby 2/5 ofaturnaroundtheverticallinethroughthecenteroftheprism, andthenreflectingintheplanethatisparalleltothetopandbottompentagons, andishalfwaybetweenthem. Usingtheseideas, itisseenthatforanytwoverticesoftheprism, thereisasymmetryoftheprismtakingonevertextotheother. Hencethisprismisvertextransitive.

A B

C

Figure5.7.6

Itcanbeshownthatallofthesemi-regularpolyhedraotherthanthepseudorhombicubocta-hedronarevertextransitive; weomitthedetails. Bycontrast, thepseudorhombicuboctahedronisnotvertextransitive. InFigure 5.7.7 (i)weseethepseudorhombicuboctahedron, withthreeofitsverticeslabeled. Thereis, forexample, nosymmetryofthepseudorhombicuboctahedronthat takesvertex A tovertex B. Hence, thepseudorhombicuboctahedron isnotvertex tran-sitive. (Thatdoesnot, however, meanthatnovertexofthepseudorhombicuboctahedroncanbetakenbyasymmetrytoanothervertex; forexample, rotationby 1/4 turnaroundtheverti-callinethroughthecenterofthepseudorhombicuboctahedroninFigure 5.7.7 (i)takesvertexA tovertex C.) Bywayofcomparison, observethat fortherhombicuboctahedronshowninFigure 5.7.7 (ii), reflectionin thehorizontalplanethroughthecenterof thepolyhedronisasymmetryoftherhombicuboctahedronthattakesvertex A tovertex B.

Some textsadd thepropertyof vertex transitivity to thedefinitionof semi-regular (thoughwedonot), andiftheydo, thentheydonotconsiderthepseudorhombicuboctahedrontobesemi-regular, andtheyhaveonly13Archimedeansolids.


AA

B B

CC

(i) (ii)

Figure5.7.7

6Groups

6.1 Thebasicidea

AtthestartofChapter 4 wereadaquotebyHermanWeyl whichended:

Toacertaindegreethisschemeistypicalforalltheoreticknowledge: Webeginwithsomegeneralbutvagueprinciple(symmetryinthefirstsense), thenfindanimportantcasewherewecangivethatnotionaconcreteprecisemeaning(bilateralsymmetry), andfromthatcasewegraduallyriseagaintogenerality, guidedmorebymathematicalconstructionandabstractionthanbythemiragesofphilosophy; andifweareluckyweendupwithanideanolessuniversalthantheonefromwhichwestarted. Gonemaybemuchofitsemotionalappeal, butithasthesameorevengreaterunifyingpowerintherealmofthoughtandisexactinsteadofvague.”

Inthepresentchapter, thelastinourbook, wenowindeedrisetothelevelofmathematicalgenerality, andunifyingpower, towhichWeylwasreferring. Atfirstitmightnotbeapparentwhatthematerialinthischapterhastodowithsymmetry, butwewillmaketheconnectionclearinourverylastsection, Section 6.6.Inthischapterwewilldiscussthemathematicalconceptofagroup. Unlikethecolloquialus-

ageofthisword, toamathematicianagroupisaverypreciselydefinedconcept, aswillbeseenbelow. Thoughatfirstglancegroupsappeartobeveryabstract, likeallworthwhileabstractiontheyarebasedonconcreteexamples. Indeed, itistheextremelybroadrangeofexamplesofgroupsthathaveledthegroupconcepttobeconsideredverycentraltomodernmathematics.Thetheoryofgroups, thoughlessthan200yearsold, ishighlydeveloped, withnewdiscoveriesbeingmadeallthetime. Groupsareextremelyusefulineverythingfromgeometrytochemistry;inparticular, groupsarevital to thestudyof symmetry, and it is for this reason thatwedis-cussthemhere. Further, themethodologyofgrouptheoryepitomizestheabstractapproachof

216 6. Groups

modernmathematics(asspearheadedearlierinthiscenturybythegreatmathematicianEmmyNoether), andthischapter’sexcursionintotheabstractshouldbeseenasatasteofwhatmanymathematiciansdotoday.Considertheintegers −3,−2,−1, 0, 1, 2, 3 . . .. Wewillusethestandardabbreviation Z to

denote the setof integers. Theword“set” is simply thecommonlyusedmathematical termtomeana“collection”of things, in thiscasenumbers, thoughasetcouldcontainanytypeofobject, not justnumbers. (The letterZ,by theway, stands for theGermanwordZahlen,whichmeansnumbers.) Ifallwecoulddowith the integerswouldbe towrite themdown,theywouldbeentirelyuseless. Whatmakestheintegerssousefulisthatwecancombinethem,viaaddition, subtraction, multiplicationanddivision. Actually, subtractionisjustdoingaddition“backwards,” anddivisionisjustmultiplication“backwards,” sowereallyneedtoconsideronlyadditionandmultiplication. (Whatdoes 5 − 3 mean? Itmeansthenumberthatyouaddto 3toget 5, namely 2.) Wewillconsidereachofthetwooperations, additionandmultiplication,separately. Eachoftheseoperationsisreferredtoasa binaryoperation,inthatittakestwothings(inthiscasenumbers)asinputs, andgivesoneoutput(inthiscase

alsoanumber).Whatpropertiescanweascribetotheoperationofadditionasappliedtotheintegers? Some

ofthesepropertiesmayseemsoobviousastobehardlyworthmention, buttheirvaluewillbeapparentlateron. (Itmightbethesimplicityofthesepropertiesthatcausedmathematicianstotakesolongtofocusinonthem.) First, wenotethatifwetakeanytwointegersandaddthem,wegetanotherinteger. Wecallthispropertythe closure propertyoftheintegerswithrespecttoaddition. Toappreciatetheworthofthisproperty, notethatifyoutakeanytwointegersanddivideonebytheother, youwillmostlikelynotgetaninteger, forexample 3 dividedby 2.Next, supposeyouwanttoaddanythreeintegers, forexample 2, 3 and 7. Becausewecan

formallyaddonlytwointegersatatime, wehavetogrouptheintegers 2, 3 and 7 withparen-thesestoprescribetheorderofaddition. Ifwekeepthesethreeintegersinthegivenorder, weseethattherearetwowaysofgroupingthem, namely (2 + 3)+ 7 and 2+ (3+ 7). Theformersaystoadd 2 and 3 first, andthentoadd 7 totheresult; thelattersaystoadd 3 and 7 first,andthentoadd 2 totheresult. Ofcourse, wegetthesamefinalanswerinbothcases, inthat(2 + 3) + 7 = 5 + 7 = 12 and 2 + (3 + 7) = 2 + 10 = 12. Indeed, becausewegetthesameanswerbothways, itissafetodroptheparenthesesandsimplywrite 2 + 3 + 7, lettingeachpersondotheadditionanywayshechooses. Wecanstatethispropertymoregenerallybysayingthatforanythreeintegers a, b and c, wealwayshave (a+ b) + c = a+ (b+ c).Wecallthispropertythe associative propertyfortheintegerswithrespecttoaddition.Ifyouwereaskedtochosethesinglemostimportantinteger, whichwouldyouchoose? Al-

though each personmay have a personal favorite number, mathematically the uncontestedleader of the pack is the number zero. Of itsmany properties, 0 is the only number that,whenaddedtoanyothernumber, leavestheothernumberunchanged. Forexample, wehave5 + 0 = 5. Toputthismoregenerally, foranyinteger a, wehave a + 0 = a and 0 + a = a.Wecallthispropertyof 0 the identity propertyfortheintegerswithrespecttoaddition.

6.1Thebasicidea 217

Onewayofobtaining 0 isbyaddingany integer and itsnegative. For example, wehave5 + (−5) = 0. Moregenerally, foranyinteger a, wehave a+ (−a) = 0 and (−a) + a = 0.Notethattheseequationsholdwhether a ispositive, negativeor 0. Theessentialpointhereisthatforanyinteger a, thereisanotherinteger, namely −a, that“cancels a out.” Wecallthispropertythe inverses propertyoftheintegerswithrespecttoaddition.Althoughtheabovefourpropertiesoftheintegersandadditionarethemostcrucialonesfor

ourpurpose, there isonemorepropertywewillmention, which, thoughwellknown, turnsouttobelesscentralthanthepropertiesmentionedsofar. Thisadditionalproperty, calledthecommutative property, saysthattheorderofadditiondoesnotmatter. Forexample, wehave5 + 3 = 3 + 5. Ingeneral, foranytwointegers a and b, wealwayshave a+ b = b+ a.To summarize, we see that the integers togetherwith the operation addition, symbolized

(Z,+), satisfythefourfundamentalpropertiesofclosure, associativity, identityandinverses, aswellastheadditionalpropertyofcommutativity. Theintegerswithadditionsatisfyanumberofotherpropertiesaswell, butafterlookingatmanyothermathematicalsystems, mathematiciansfoundthatthesefourpropertiesareextremelyprevalentinmanyseeminglyunrelatedfields, fromgeometrytoquantummechanics, andhavethereforechosentofocusonthesefourproperties.Letuslookatsomeothermathematicalsystems, toseeiftheysatisfythesamepropertiesas

theintegerswithaddition. Thenextmostobviousexampletoconsideristheintegerswiththeoperationmultiplication, abbreviated (Z, ·). Weneedtocheckwhetherthefourpropertiesofclosure, associativity, identityandinverses, aswellasthecommutativeproperty, holdfor (Z, ·).Letusstartwithclosure. Itiscertainlythecasethatifwemultiplyanytwointegerswegetaninteger, so theclosurepropertyholds. It isalsonothard tosee that theassociativepropertyholds, thatis, foranythreeintegers a, b and c, itisalwaystruethat (a · b) · c = a · (b · c).Whatabouttheidentityproperty? Weneedtofindaspecialmemberoftheintegersthatplaysthesamerolewithrespecttomultiplicationthat 0 doesforaddition; inotherwords, weneedanumberso thatmultiplyingby itdoesnotchangeanything. Certainly thenumber 1 is theintegerwewant. If a isanyinteger, then 1 · a = a and a · 1 = a. Hence 1 istheidentityfortheintegerswithmultiplication, andsotheidentitypropertyholdsfor (Z, ·). Next, weneedtoverifywhethertheinversespropertyholdsfor (Z, ·). Thismeansthatforeveryinteger, weneedtofindanotherintegerthatcancelsitoutbymultiplication, yielding 1. Letustrythisfortheinteger 2. Thereiscertainlyanumberthatcancels 2 outwithrespecttomultiplication, namely1/2, because 2 ·(1/2) = 1 and (1/2) ·2 = 1. Thereisamajorproblemhere, however, becausewearedealingwiththeintegers, and 1/2 isnotaninteger. Thereiscertainlynoothernumberthatcancels 2 outwithrespecttomultiplication, sowehavetoconcludethat 2 doesnothaveamultiplicativeinverseintheintegers. (Thenumber 2 doeshaveanadditiveinverse, namely−2,butthatdoesnothelpushere.) Hence, weseethat (Z, ·) doesnothavetheinversesproperty.Therefore, eventhough (Z, ·) satisfiesthefirstthreepropertiesthat (Z,+) satisfies, itdoesnotsatisfythefourthproperty. Itisnothardtoseethatthecommutativepropertyholdsfor (Z, ·),thatis, foranytwointegers a and b, itisalwaystruethat a · b = b · a.Theproblem thatoccuredwith (Z, ·) might suggest toyouwherewecanfindsomething

thatdoessatisfyallfiveproperieswithrespecttomultiplication. Thenumber 1/2 isnotinthe

218 6. Groups

integers, butitisafraction, sowhydon’twelookatthesetofallfractions, denoted Q. (Theletter Q standsforquotient.)


Trytofigureoutwhether (Q, ·) satisfytheclosure, associative, identity, inversesandcom-mutativeproperties.

Justaswith (Z, ·), itisnothardtoseethat (Q, ·) satisfiestheclosure, associativityandidentityproperties(onceagain 1 istheidentitywithrespecttomultiplication). Butthistime, unliketheintegers, itseemsthattherearemultiplicativeinverses. Foranyfraction, thefractionthatcancelsitoutbymultiplicationisjustthereciprocaloftheoriginalfraction. Forexample, thereciprocalof 5/3 is 3/5, andsureenough (5/3) · (3/5) = 1 and (3/5) · (5/3) = 1. Soitappearsasif(Q, ·) hastheinversesproperty. Almost, butthereisstillonelittleglitch. Thenumber 0 canbeconsideredasafraction, say 0/1. Unfortunately, thefraction 0/1 hasnoreciprocal, becausewewouldwanttouse 1/0, butthatisnotallowedbecausewecannotdivideby 0. Itfollowsthatthefraction 0/1 doesnothaveamultiplicativeinverse. However, thenumber 0 istheonlyproblem, becauseanyfractionthatdoesnotequal 0 doeshaveareciprocal. Wewillbypasssthisproblemcausedby 0 asfollows. LetususethesymbolQ∗ todenotethesetoffractionswiththenumber 0 removed. Then, ifweputalltheabovereasoningtogether, weseethat (Q∗, ·) doessatisfythefourpropertiesofclosure, associativity, identityandinverses. Moreover, becausetheorderofmultiplicationoftwonumbersdoesnotmatter, forexample 4 · 7 = 7 · 4, weseethat(Q∗, ·) alsosatisfiesthecommutativeproperty.

Exercise 6.1.1. Determinewhichofthefivepropertiesofclosure, associativity, identity,inversesandcommutativityaresatisfiedbyeachofthefollowingsystems.

(1) Theevenintegerswithaddition.

(2) Theoddintegerswithaddition.

(3) Allrealnumberswithaddition.

(4) Allrealnumberswithmultiplication.

6.2 Clockarithmetic

Sofarwehavebeenconcernedwithvarioussetsofnumbers, suchasintegersandfractions. Allthesesetshavebeeninfinite. Wenowwishtoexamineamathematicalsystemthatisfiniteinsize. Thismathematicalsystemisbasedontheideaof“clockarithmetic,” whichyoumayhaveseen; ifyouhavenot, itwillbesufficientthatyouhaveseenaclock. Allourreferencestotime

6.2Clockarithmetic 219

willbebasedontheAmerican 12 hoursystem(thoughwewillignorea.m.vs.p.m.), asopposedtothe 24 hoursystemusedmanyotherplacesaroundtheworld(andtheU.S.military); eithertimesystemwouldworkforourpurpose, andwehavesimplychosenoneofthemonceandforalltoavoidanyambiguity.Sayitis 2 o’clock, andyouwanttoknowwhattimeitwillbein 3 hours. A sillyquestion, you

maybethinking, becausethetimeinthreehoursissimply 2 + 3 = 5 o’clock. Right, butnowsupposeitis 7 o’clock, andyouwanttoknowwhatthetimewillbein 6 hours. Youcouldgo7 + 6 = 13, butyouwouldn’tsay 13 o’clock, becausethereisnosuchthing; youwouldsay1 o’clock, ofcourse, andyouwouldberight. Howdidyouget 1 o’clock? Yousubtracted 12from 13, because 13wasgreaterthan 12, andthereforetoolarge. Now, supposeitis 11 o’clock,andyouwanttoknowwhattimeitwillbeafter 30 hours(again, ignoringa.m.andp.m.). Youwouldstartbygoing 11 + 30 = 41, butonceagainthisistoolarge, becauseyoucannothave41 o’clock. Theonly“o’clocks”youcanhavearefrom 1 to 12 (roundingofftowholehours,aswearedoing). Therefore, youwanttotake 41 and“bringitdown”tobetween 1 and 12. Todothis, youwanttosubtractfrom 41 asmanycopiesof 12 asyoucan. Thebestyoucandoissubtract 3 times 12, whichis 36. Now, wecompute 41 − 36 = 5, soifyoustartat 11 o’clockandgoanother 30 hours, youendupat 5 o’clock.

Exercise 6.2.1. Ifyoustartat 7 o’clock, andgoanother 20 hours, whattimewillitbe?

Letusnowlookmorecarefullyatwhatwejustdid; wewilldropthe“o’clocks”forconve-nience. Thereweretwothingswewantedtoaccomplish, whichseemedsomewhatatoddswitheachother: ontheonehand, wewantedtorestrictourselvestotheintegers 1 through 12. Ontheotherhand, wewantedtobeabletoaddnumbers, whichtookusoutsideofthe 1 to 12range. Toresolvetheproblem, wetookanynumberthatwasoutsideofthe 1 to 12 range, andreduceditrepeatedlyby 12 untilwewerebackinthedesiredrange. Forexample, wereduced13 to 1 bysubtracting 12, andwereduced 41 to 5 bysubtracting 3 times 12. Inotherwords,weareessentiallyconsidering 13 and 1 asequivalent(fromthepointofviewofclocks), and 41and 5 areconsideredequivalent.Wearethereforeledtoanewnotion, called congruencemod 12. Wesaythattwointegers

are congruentmod 12 iftheydifferbyanintegermultipleof 12. Therefore, wesaythat 13 and1 arecongruentmod 12, andthat 41 and 5 arecongruentmod 12. Ontheotherhand, thenumbers 17 and 3 arenotcongruentmod 12 becausetheirdifferenceis 14, whichisnotanintegermultipleof 12. Thenumbers 5 and 29 arecongruentmod 12, becausetheirdifferenceis 5− 29 = −24 = (−2) · 12.

220 6. Groups

Exercise 6.2.2. Whichofthefollowingpairsofnumbersarecongruentmod 12?

(1) 15 and 3;

(2) 9 and 57;

(3) 7 and −5;

(4) 11 and 1;

(5) 0 and 12.

Forthesakeofbrevity, weintroducethefollowingnotation. Ifintegers a and b arecongruentmod 12, wewritethis a ≡ b (mod 12). Forexample, wehave 41 ≡ 5 (mod 12). If a and b

arenotcongruentmod 12, wewrite a ≡ b (mod 12). Forexample, wehave 3 ≡ 7 (mod 12).Fromtheclockexample, wenoticedthatanyintegerwhatsoevercouldbereducedbymulti-

plesof 12 untilwhatisleftissomewherefrom 1 through 12. Hence, ifweareinterestedonlyinintegersmod 12, thenweneedtoconsideronly 1 through 12, becauseanythingelsecanbereducedtooneofthesenumbers. Foreaseofuselateron, wewillmakeonesmallchangeatthispoint. Insteadofconsideringtheintegersfrom 1 to 12, wewillswitchtotheintegersfrom 0 to11. Thischangehasnosubstantialeffect, because 0 ≡ 12 (mod 12). Anyintegerwhatsoevercanbereducedbymultiplesof 12 untilwhatisleftisanintegerfrom 0 through 11. Inotherwords, forany integerwhatsoever, there isanother integer, this time from 0 to 11, which iscongruentmod 12 totheoriginalinteger; moreover, thereisonlyonesuchintegerfrom 0 to 11.Insymbols, foranyinteger a, thereisauniqueinteger x from 0 to 11 sothat x ≡ a (mod 12).Forexample, ifwelet a = 13, then x = 1, because 1 ≡ 13 (mod 12). If a = 35, then x = 11,because 11 ≡ 35 (mod 12). If a = 12, then x = 0, because 0 ≡ 12 (mod 12). Notethatthenumber a neednotbepositive. If a = −4, then x = 8, because 8 ≡ (−4) (mod 12).Additionally, notethatif a = 7, then x = 7 aswell, because 7 isalreadybetween 0 to 11.

Exercise 6.2.3. Foreachinteger a givenbelow, findtheinteger x from 0 to 11 sothatx ≡ a (mod 12).

(1) a = 18;

(2) a = 41;

(3) a = −17;

(4) a = 3.

Wearenowledtothefollowingmethodforconstructinganewmathematicalsystem, whichsimplyencapsulateswhatwedowhenwetelltime. Oursystemwillhavetwelveobjects, de-

6.2Clockarithmetic 221

noted 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. ThecollectionofthesetwelveobjectswillbedenotedZ12. Weputthe“hat”ontheseobjectstoindicatethat, althoughtheycorrespondtotheintegersfrom 0 through 11, theydonotbehaveexactlyliketheintegerstowhichtheycorrespond. Thedifferenceisinhowweaddtheelementsin Z12. Letusstartwithsomeexamples. Toadd 3 and4 iseasy; itissimply 3+4 = 7. Ontheotherhand, wecannotsaythat 6+8 is 14, becausethereisnosuchthingas 14 in Z12. So, asonaclock, whatwedoistoreduce 14 byintegermultiplesof 12. Moreconcisely, wewanttofindanintegerfrom 0 to 11 thatiscongruentmod 12 to 14.Thenumberisclearly 2, andsowesay 6+8 = 2. Ingeneral, if a and b aretwonumbersin Z12,tofind a+ b wefirstfind a+b asusual; if a+b isfrom 0 to 11, weputahatoverit, andthatisouranswer; if a+ b islargerthan 11, wefindaninteger x from 0 to 11 sothat x ≡ (a+ b)

(mod 12), andthenlet a + b = x. Forexample, wehave 3 + 7 = 10, and 7 + 9 = 4, and2 + 0 = 2. Itshouldbeclearthatalthoughweusetheusual“+”signtodenote“addition”inZ12, thisoperationisnotthesameasstandardadditionofintegers, becausewereducemod 12.(Itwouldbesensibletoputa“hat”onthe + signthatweusefor Z12, similarlytothehatweputon 0, 1, 2, . . ., 11, butitisnotstandardtodoso, andwewillstickwithstandardnotation.)

Exercise 6.2.4. Calculatethefollowing.

(1) 4 + 5;

(2) 7 + 8;

(3) 5 + 11;

(4) 0 + 3.

Weareinterestedinthesystem (Z12,+). Onehelpfultoolforunderstandingthissystemisadevicethathelpeduslearnmultiplicationaschildren, namelymultiplicationtables, suchasTable 6.2.1, whichshowsmultiplicationupto 10.Thistablesummarizesexplicitlyallpossiblemultiplicationsbetweenintegersfrom 1 to 10.

Forexample, tofind 3 · 7, lookintherowlabeled 3, andthecolumnlabeled 7, andintheintersectionofthisrowandthiscolumnwefind 21, justasexpected.Wecan, similarly, makeanadditiontablefor (Z12,+), showninTable 6.2.2. Forexample,

tofind 4 + 9, lookintherowlabeled 4, andthecolumnlabeled 9, andintheintersectionofthisrowandthiscolumnwefind 1, whichis 4 + 9. Noticethecyclicpatterninthetable.Wenowaskwhether (Z12,+) thesamefiveproperties(discussedinSection 6.1)that (Z,+)

satisfies. Theclosurepropertyholdsfor (Z12,+), becausetheway+wasdefinedforZ12 insuresthataddinganytwoelementsinZ12 yieldsanotherelementinZ12. Asforassociativity, withabitofthoughtitisnothardtoseethatbecausethestandardadditionfortheintegersisassociative,theadditionof Z12 isalsoassociative; wewillomitthedetails.Toseethattheidentitypropertyholds, wenotethat 0 playstheroleofanidentityelement,

becauseforany a in Z12, itisseenthat a+ 0 = a and 0 + a = a.

222 6. Groups

· 1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 102 2 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 506 6 12 18 24 30 36 42 48 54 607 7 14 21 28 35 42 49 56 63 708 8 16 24 32 40 48 56 64 72 809 9 18 27 36 45 54 63 72 81 9010 10 20 30 40 50 60 70 80 90 100

Table6.2.1

Whatabout inverses? Youmight thinkatfirst that therecannotbe inverses, because therearenonegativenumbersin Z12. Butnegativeisarelativeterm(dependingonyourset, youroperationandyourzero), andinfactthereareinversesin (Z12,+). Letusstartwith 1. What,ifanything, is its inverse in (Z12,+)? Inotherwords, is thereanelement in Z12 that, whenaddedto 1, yields 0. Recallingthat 0 ≡ 12 (mod 12), weseethatthenumberthatcancels1 outisprecisely 11, because 1 + 11 = 12, andtherefore 1 + 11 = 0. Itisnothardtoseethateveryelementin Z12 hasaninversewithrespecttoaddition. Forexample, theinverseof5 is 7, because 5 + 7 = 12, andtherefore 5 + 7 = 0. Hence, theinversespropertyholdsfor(Z12,+). Itisnothardtoseethatthecommutativepropertyalsoholdsfor (Z12,+), forexample5 + 6 = 6 + 5. Wehavethereforeverifiedthat (Z12,+) satisfiesthesamefivepropertiesas(Z,+).

Exercise 6.2.5. Findtheinverseswithrespecttoadditionof 3, 6, 8 and 0 in Z12.

Wecanalsomakeamultiplicationtablefor Z12, showninTable 6.2.3.Notice that themultiplication table for Z12 doesnot have the same simplepattern along

upward-slopinglinesasdidtheadditiontablefor Z12. Moreover, notethatintheadditiontable,eachof 0, 1, . . . , 11 appearsonceandonlyonceineachrowandineachcolumn; thispropertydoesnotholdforthemultiplicationtable. Alltold, multiplicationfor Z12 isnotaswellbehavedasaddition. SeeExercise 6.2.6 fordetails.

Exercise 6.2.6. Whichofthefiveproperties(closure, associativity, identity, inverses, com-mutativity)holdsfor (Z12, ·)? Forthoseelementsof Z12 thathaveinverseswithrespecttomultiplication, statewhattheirinversesare.

6.3TheIntegersMod n 223

+ 0 1 2 3 4 5 6 7 8 9 10 11

0 0 1 2 3 4 5 6 7 8 9 10 11

1 1 2 3 4 5 6 7 8 9 10 11 0

2 2 3 4 5 6 7 8 9 10 11 0 1

3 3 4 5 6 7 8 9 10 11 0 1 2

4 4 5 6 7 8 9 10 11 0 1 2 3

5 5 6 7 8 9 10 11 0 1 2 3 4

6 6 7 8 9 10 11 0 1 2 3 4 5

7 7 8 9 10 11 0 1 2 3 4 5 6

8 8 9 10 11 0 1 2 3 4 5 6 7

9 9 10 11 0 1 2 3 4 5 6 7 8

10 10 11 0 1 2 3 4 5 6 7 8 9

11 11 0 1 2 3 4 5 6 7 8 9 10

Table6.2.2

6.3 TheIntegersMod n

InSection 6.2 webasedourdiscussiononthenumbertwelvebecauseofourfamiliaritywithclocks. Wecan, however, repeatthewholeprocedurewithanyotherpositiveintegerreplacing12. Chooseanypositiveinteger n. Wesaythatanytwointegersare congruentmod n iftheydifferbysomeintegermultipleof n. Insymbols, suppose a and b areintegers. Wesaythat aand b arecongruentmod n, written a ≡ b (mod n), if a−b = kn forsomeinteger k (whichcouldbepositive, negativeorzero). If a and b arenotcongruentmod n, wewrite a ≡ b

(mod n). Forexample, saywechoose n = 5. Then 17 and 2 arecongruentmod 5, written17 ≡ 2 (mod 5), because 17 − 2 = 15 = 3 · 5, whichisanintegermultipleof 5. Also, wehave 3 ≡ 11 (mod 4) because 3−11 = −8 = (−2) ·4. However, wehave 13 ≡ 2 (mod 9),because 13− 2 = 11, whichisnotamultipleof 9.

Exercise 6.3.1. Whichofthefollowingaretrue, andwhicharefalse?

(1) 3 ≡ 9 (mod 2);

(2) 7 ≡ (−1) (mod 8);

(3) 4 ≡ 11 (mod 3);

(4) 0 ≡ 24 (mod 6).

(5) 9 ≡ 9 (mod 5).

224 6. Groups

· 0 1 2 3 4 5 6 7 8 9 10 11

0 0 0 0 0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7 8 9 10 11

2 0 2 4 6 8 10 0 2 4 6 8 10

3 0 3 6 9 0 3 6 9 0 3 6 9

4 0 4 8 0 4 8 0 4 8 0 4 8

5 0 5 10 3 8 1 6 11 4 9 2 7

6 0 6 0 6 0 6 0 6 0 6 0 6

7 0 7 2 9 4 11 6 1 8 3 10 5

8 0 8 4 0 8 4 0 8 4 0 8 4

9 0 9 6 3 0 9 6 3 0 9 6 3

10 0 10 8 6 4 2 0 10 8 6 4 2

11 0 11 10 9 8 7 6 5 4 3 2 1

Table6.2.3

Foreachpositiveinteger n greaterthanorequalto 2, wecanformasystemcalled Zn com-pletelyanalogouslytothewayweformed Z12. Wewillobtainonesuchsystemforeachinteger2, 3, 4, . . . (Weskipover thecase n = 1, becausethat turnsout tobeuseless.) Letusstartwiththeexampleof n = 8. Analogouslytowhatwedidwith 12, weseethatfor n = 8, anyintegercanbereducedbymultiplesof 8 untilwhatisleftissomewherefrom 0 through 7. Inotherwords, foranyinteger, thereisauniqueintegerfrom 0 to 7 thatiscongruentmod 8 totheoriginalinteger. Insymbols, foranyinteger a, thereisauniqueintegerfrom 0 to 7, denoted x,sothat x ≡ a (mod 8). Forexample, ifwelet a = 10, then x = 2, because 2 ≡ 10 (mod 8).Nowsupposewestartwith a = 1950. Inthiscase, wecouldproceedbysubtracting 8, andthenanother 8, andthenagainandagain, asmanytimesasareneeded, untilweareleftwithsomenumberfrom 0 to 7. Thatwouldwork, butwouldbeverytedious. A bettermethodwouldbetodivide 1950 by 8. Wewouldthenseethatthequotientis 243, andtheremainderis 6;thatis, weseethat 1950/8 = 243 + (6/8). Itfollowsthat 1950 = 243 · 8 + 6, andhencethat6 − 1950 = (−243) · 8. Wethereforeseethat 6 ≡ 1950 (mod 8), andthereforewecanusex = 6. Wenote, asbefore, thatifwestartwithanumber a thatisalreadyfrom 0 to 7, then x

isjust a itself.


Exercise 6.3.2. Foreachinteger a givenbelow, findtheinteger x from 0 to 7 sothat x ≡ a

(mod 8).

(1) a = 15;

(2) a = 54;

(3) a = 1381;

(4) a = −2;

(5) a = 3;

(6) a = 8.

Asbefore, theset Z8 andwillhave 8 members, denoted 0, 1, 2, 3, 4, 5, 6 and 7. Weaddelementsof Z8 asbefore, exceptthatthistimewereducebymultiplesof 8. Forexample, in Z8

wehave 2 + 3 = 5 andwehave 5 + 4 = 1, wherethelatterholdsbecause 5 + 4 = 9, and1 ≡ 9 (mod 8).

Exercise 6.3.3. Calculatethefollowingin Z8.

(1) 4 + 1;

(2) 3 + 7;

(3) 0 + 3.

Justaswedidfor (Z12,+), wecanformanadditiontablefor (Z8,+), showninTable 6.3.1.Noticethesamediagonalpatternasbefore.

+ 0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 0

2 2 3 4 5 6 7 0 1

3 3 4 5 6 7 0 1 2

4 4 5 6 7 0 1 2 3

5 5 6 7 0 1 2 3 4

6 6 7 0 1 2 3 4 5

7 7 0 1 2 3 4 5 6

Table6.3.1

226 6. Groups

Weaskwhether (Z8,+) satisfiesthesamefivepropertiesas (Z,+), andtheanswerisyes.Theclosure, associativeandidentitypropertiesholdfor (Z8,+) justastheydidfor (Z12,+).Weobservethat 0 isonceagaintheidentityelement. Theinversepropertyalsoholds, althoughalittlecautionmustbetaken, becausetheinversesin Z8 arenotthesameasin Z12. Forexample,theinverseof 1 in Z8 is 7, because 1+7 = 8, andso 1 + 7 = 0. Thiscontrastswiththeinverseof 1 in Z12, whichis 11. Thecommutativepropertyalsoholdsfor (Z8,+).

Exercise 6.3.4. Findtheinverseswithrespecttoadditionof 2, 4, 5 and 0 in Z8.

Justas (Z8,+) satisfiesthefivepropertiesofclosure, associativity, identity, inversesandcom-mutativity, sodoes (Zn,+) foranypositiveinteger n, where n ≥ 2. Thesystem (Zn,+) iscalledthe groupofintegersmod n withtheoperationaddition. Noticethat (Zn,+) haspre-cisely n members.

Exercise 6.3.5. Considerthesystem (Z6,+).

(1) Listtheelementsofthissystem.

(2) In (Z6,+), whatare 5 + 2 and 4 + 1?

(3) Constructtheadditiontablefor (Z6,+).

(4) Findtheinverseswithrespecttoadditionof 2, 4, 5 and 0 in Z6.

Exercise 6.3.6. Observethatintheadditiontablefor (Z8,+), showninTable 6.3.1, alltheentriesonthedownwardsslopingdiagonalareevennumbers. Willthesamefactholdintheadditiontableforany (Zn,+)? Ifyes, explainwhy. Ifnot, describewhatdoeshappenonthedownwardsslopingdiagonalfor (Zn,+) ingeneral, andexplainyouranswer.

Havinglookedat (Zn,+), letusnowturnto (Zn, ·). Considerthecaseof (Z5, ·). Themulti-plicationtablefor (Z5, ·) isshowninTable 6.3.2.NoticethatTable 6.3.2 doesnotsatisfythesimplepatternalongupward-slopinglinesasin

Tables 6.2.2 and6.3.1.Itisseenthat (Z5, ·) doessatisfytheclosure, associative, andidentityproperties(with 1 as

identity), andthecommutativepropertyaswell. However, theinversespropertyisnotsatisfied,because 0 hasnoinversewithrespecttomultiplication. Toseethisfact, wenotethataninversefor 0 wouldbesome x in Z5 suchthat 0 · x = 1 and x · 0 = 1. A lookatthetableshowsthatnosuch x exists. Wecanremedythisproblemjustaswedidfor Q inSection 6.1, bydropping


· 0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

Table6.3.2

theproblematic 0. Letususethesymbol Z∗5 todenote Z5 with 0 removed. Wethenobtainthe

multiplicationtablefor (Z∗5, ·), shown inTable 6.3.3. Weleaveittothereadertoverifythat

(Z∗5, ·) satisfiesallofourproperties.

· 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

Table6.3.3

Unfortunately, whatworksfor (Z∗5, ·) doesnotworkforall (Z∗

n, ·). Forexample, weseeinTable 6.3.4 themultiplication table for Z∗

6. Notallfivepropertieshold for (Z∗6, ·). First, the

closurepropertydoesnothold; forexample, weseethat 2 · 3 = 0, but 0 isnotin Z∗6. The

associativepropertyholds, asdoestheidentityproperty(withidentity 1), andthecommutativepropertyholds. Theinversespropertydoesnothold; forexample, itisseenthat 2 doesnothaveaninversewithrespecttomultiplication.

· 1 2 3 4 5

1 1 2 3 4 5

2 2 4 0 2 4

3 3 0 3 0 3

4 4 2 0 4 2

5 5 4 3 2 1

Table6.3.4

228 6. Groups


Istheresomethingaboutthenumber 5 thatmakes (Z∗5, ·) satisfyallourproperties, and

aboutthenumber 6 thatmakes (Z∗6, ·) failtosatisfytheclosureandidentityproperties?

Ingeneral, what is itaboutan integer n thatwoulddeterminewhetherornot (Z∗n, ·)

satisifiesallfiveproperties? (Youwillmostlikelynotbeabletodemonstrateyouranswerrigorously, unlessyouknowmoreaboutnumbersthanweareassuming, buttrytomakeaneducatedguess.)

Itturnsoutthattherelevantdifferencebetween 5 and 6 thatleadsto (Z∗5, ·) satisfyingourfive

propertiesbut (Z∗6, ·) notsatisfyingallfiveistheissueofprimenumbers vs.compositenumbers.

A positiveintegerisa primenumber ifitsonlypositivefactorsare 1 anditself. A positiveintegerthatisnotprimeiscalledcomposite. Forexample, thenumbers 2, 3, 5 and 7 areprime, whereas6 iscomposite, havingfactors 1, 2, 3, and 6. Itturnsout, thoughthisisfarfromobvious, that(Z∗

n, ·) satisfiesallfivepropertiesifandonlyif n isaprimenumber. Theproofusesfactsaboutprimenumbers.

Exercise 6.3.7. Constructthemultiplicationtablefor (Z8, ·).

6.4 Groups

Intheprevioussectionsofthischapter, wesawanumberofmathematicalsystemsthatsatisfiedthesamepropertiesofclosure, associativity, identityandinverses. (Allthesystemsdiscusseduptillnowalsosatisfiedthecommutativityproperty, butwewillseesystemsthatdonotsatisfythispropertyinalittlewhile.) Mathematicianshaveinfactfoundsomanysystemsthatsatisfythesesamefourfundamentalproperties(thoughnotnecessarilycommutativity), thattheydecidedtogiveallsuchsystemsanamesothattheycouldbestudiedtogether, andcommonpropertiescouldbefound. Wewillcallanysystemsatisfyingthesefourpropertiesagroup.Letusphrasethisconceptmoreprecisely. Firstofall, agroupisasetofobjects, whichmay

benumbers(asinthecaseoftheintegers), butwhichcouldbeotherthingsaswell. Supposethat G isasetofobjects. Themembersof G willbereferredtoas elements of G. Next, thissetofobjects G needsa binaryoperation, whichcombineselementsof G bytakinganytwoelementsof G asinputs, andfortheseinputsgivesauniqueoutput. Now, supposethat ∗ isabinaryoperation. Thatmeansthatforanytwoelements g and h inG, wecancombinethemtogetasinglethingdenoted g ∗ h. Thenotation g ∗ h ismeanttobesimilartothenotationsforadditionandmultiplication, ourtwomostfamiliarbinaryoperations. Wewilldenoteaset Gtogetherwithanoperation ∗ bythepair (G, ∗). Oneexampleofsuchapair (G, ∗) isthepair(Z,+). A pair (G, ∗) iscalleda group ifitsatisfiestheabovementionedfourproperties, whichwenowstateintheirmostgeneralform:

6.4Groups 229

Closureproperty: If g and h arein G, then g ∗ h isin G.

Associativeproperty: If g, h and k arein G, then (g ∗ h) ∗ k = g ∗ (h ∗ k).Identityproperty: ThereisadistinguishedelementinG, calledan identityelement anddenotede, sothatif g isin G, then e ∗ g = g and g ∗ e = g.

Inversesproperty: If g is in G, there isanelement g ′ in G, calledan inverse of g, so thatg ∗ g ′ = e and g ′ ∗ g = e.

Some, thoughnotall, groupsalsosatisfythefollowingproperty:

Commutativeproperty: If g and h arein G, then g ∗ h = h ∗ g.A groupthatalsosatisfiesthecommutativepropertyiscalledan abelian group. (Itwouldbe

entirelyreasonabletocallsuchagroupa“commutativegroup;” however, thatisnotstandardterminology. ThetermabeliangroupisinhonoroftheNorwegianmathematicianNielsAbel.Thischoiceofnamehasgivenrisetothefollowingwellknownjoke(wellknownamongmath-ematicians, atleast). Question: whatispurpleandcommutative? Answer: anabeliangrape.)

Ifwegobackandconsidertheexampleswehaveseensofar, nowusingtheterminologyofgroups, weseethat (Z,+) isanabeliangroup, asis (Zn,+) foreachpositiveinteger n, wheren ≥ 2. Ontheotherhand, thesystem (Z, ·) isnotagroup, becausewesawthatitdidnotsatisfytheinversesproperty. Wealsosawthat (Z∗

n, ·) isagrouppreciselyif n isaprimenumber.

Exercise 6.4.1. Whichofthefollowingsystemsisagroup? Whichisanabeliangroup?

(1) Theevenintegerswithaddition.

(2) Theoddintegerswithaddition.

(3) Allrealnumberswithaddition.

(4) Allrealnumberswithmultiplication.

Thereisonematterweneedtoclarifyrightawayaboutthedefinitionofgroups. Inthestate-mentoftheidentityproperty, wementioned“anidentityelement,” andinthestatementoftheinversesproperty, wementioned“aninverse.” Coulditbethecasethatagrouphasmorethanoneidentityelement, orthatanelementinagrouphasmorethanoneinverse? Intuitivelythatsoundsunlikely, andthefollowingpropositionshowsthatourintuitioniscorrect.

Proposition 6.4.1. Supposethat (G, ∗) isagroup.

1. Thegroup G hasauniqueidentityelement.

2. If g isanelementof G, then g hasauniqueinverse.

Demonstration. WeshowPart (1), leavingPart (2)tothereaderinExercise 6.4.2.

230 6. Groups

(1). Wefollowthestandardmathematicalapproach toshowing thatsomething isunique,whichistosupposethattherearetwoofthething, andthenshowthatthetwothingsareinfactequal. Inparticular, supposethat e and c arebothidentityelementsof G. Then

e = e ∗ c = c,

whereinthefirstequalitywearethinkingof c asanidentityelement, andinthesecondequalitywearethinkingof e asanidentityelement. Itfollowsthat e = c, andthereforetheidentityelementof (G, ∗) isunique.

Bytheaboveproposition, wecannowreferto“theidentityelement”ofagroup, and“theinverse”ofeachelementofthegroup.

Exercise 6.4.2. [UsedinThisSection] DemonstrateProposition 6.4.1 (2).

Groupscomeintwobasicvarieties, infinite orfinite, dependingonhowmanyelementsareinthegroup. Forexample, thegroup (Z,+) isaninfinitegroup, andeach (Zn,+) isafinitegroup.Althoughtheinfinitegroupswehavedealtwithsofarmayseemmorenatural(forexample, theintegers), finitegroupsareofteneasiertoworkwithmathematically. Letuslookatsomefurtherexamplesoffinitegroups.Recall that thegroup (Zn,+) hasprecisely n elements in it. Because thisworks foreach

positiveinteger n with n ≥ 2, weseethatthereisagroupofeachpossiblefinitesize. (Thereisalsoagroupwithoneelement, namelythesetwiththesingleelement 0, andwiththeoperationaddition.) However, therearemanyotherfinitegroupsbesidesthegroups (Zn,+), thoughmanyofthemaremorecomplicatedtoconstruct. Someofthefollowingexamplesoffinitegroupsmayappearsomewhatarbitrary. Wheredidthesegroupscomefrom? Sometimestrialanderrorwasused, thoughifthatweretheonlymethod, notonlywouldthatberathertedious, butitwouldberatherunappealing. Therearevarioussystematicwaysofconstructingfinitegroups,someofwhichareextremelycomplex, yieldinghugegroupswithnicknamessuchas“monster”(seriously).To see someexamplesoffinitegroups, recall thatagroup ingeneral is a setof elements

G, togetherwithanoperation ∗, subject to fourproperties. Whendealingwith the familiaroperationsofadditionandmultiplication, allwehadtodowasnametheseoperations, andeveryoneknewexactlywhatweweretalkingabout. Inunfamiliarsituation, whenwecannotrefertoanoperationbysimplygivinganamewithwhicheveryonewouldbefamiliar, wewillreturntotheideaofthemultiplicationtablementionedpreviously. Todescribeagroup, wewillfirstdescribeasetG, andthendescribeanoperation ∗ usinga“multiplication”table, whichwewillcallan operationtable fromnowon.Weuseoperationtablesjustasweusedadditiontablesfor (Z,+) and (Z,+) previously. Whatwillbenewnowisthatinsteadofusinganoperationtabletogiveapictorial representation ofabinaryoperationwithwhichwearealreadyfamiliar,nowwewill define newbinaryoperationsbygivingoperation tables for them. Ifwedefine

6.4Groups 231

defineabinaryoperation ∗ bygivingitsoperationtable, thentofind a ∗ b, wesimplylookattheentryintheoperationtableintherowcontaining a andthecolumncontaining b.Letusconstructatwo-elementgroupviaanoperationtable. Westartwithaset, labeled T ,

whichcontainstwoelements, called r and s; wecanabbreviatethisbywriting T = {r, s}. Wethenspecifyabinaryoperation ∗ bygivingTable 6.4.1.

∗ r s

r r s

s s r

Table6.4.1

FromTable 6.4.1 wesee, forexample, that r ∗ s = s and s ∗ s = r. Wewant toverifywhether (T, ∗) isagroup. Now, wedonotknowwhether r and s aremeanttodenotenumbers,orperhapshouses, orsomethingelse; wealsodonotknowthat ∗ “means,” inthewaythatweknowwhatadditionandmultiplicationofnumbersmeans. So, isitpossibletoverifywhether(T, ∗) isagroupusingonlytheoperationtablegivenfor ∗? Theanswerisyes—everythingthatcanbeknownabout ∗ iscontainedinitsoperationtable.Theclosurepropertycertainlyholdsfor (T, ∗), becauseanytwoelementsintheset T yieldan

elementof T whencombinedby ∗; thisfactisevidentbecausealltheentriesinTable 6.4.1 arethemselvesin T . Ingeneral, aslongasallentriesinanoperationtablearethemselveselementsoftheoriginalset, thentheclosurepropertywillnecessarilyhold.Tochecktheassociativityof ∗ weneedtocheckmanycases. Inprinciple, onewouldhave

tocheckeverypossiblewaytocombine threeelementsof T (repeatsallowed), tosee if theassociativityruleholds. Forexample, does (r ∗ s) ∗ s equal r ∗ (s ∗ s)? Usingtheoperationtable, weseethat (r ∗ s) ∗ s = s ∗ s = r, and r ∗ (s ∗ s) = r ∗ r = r, whichiswhatwehadhopedfor. Tocheckwhetherassociativityholdsfor (T, ∗), wehavetodoallotherpossiblecases, whichwouldbequitetedious. A casebycasecheckdoesshowthatassociativityholdsinthisexample; wewillnotgothroughthedetails. Becausecheckingforassociativityissotedious(andnotveryenlightening), thereadercanassumetheassociativepropertyforanyoperationtablesweusefromnowon(noteverypossibleoperationtablesatisfiestheassociativeproperty,butwearenotinterestedinthosethatdonot).Wedefinitelycannotassumethetwootherpropertiesofgroups, however, andsoweneedto

verifywhethertheidentityandinversespropertiesholdfor (T, ∗). Toverifywhethertheidentitypropertyholds, weneedtofindanelementof T thatisanidentityelementwithrespectto ∗.Thatis, wewanttofindanelementof T thatbehaveswithrespectto ∗ justasthenumber 0behaveswithrespecttoadditionofnumbers. Lookingatthetable, weseethat r ispreciselysuchanelement, because r ∗ r = r, and r ∗ s = s and s ∗ r = s. Hence, theidentitypropertyholdswithidentityelement r.Toverifytheinversesproperty, wehavetofindaninverseforeachelementof T , wherean

inverseofanelementissomethingthatcancelsitout, whichmeansthattheelementanditsinversecombinetoyieldtheidentityelement. (Ofcourse, ifasetandbinaryoperationdonot

232 6. Groups

haveanidentityelement, thequestionofinversesismoot.) Bylookingatthetable, weseethatr ∗ r = r and s ∗ s = r. Inotherwords, theelements r and s areeachtheirowninverses.Itmayseemsomewhatstrangethatsomethingisitsowninverse, thatis, itcancelsitselfout,butthereisnothinginvalidhere. Consequently, theinversespropertyholdsfor (T, ∗). Puttingthisalltogether, weseethat (T, ∗) isagroup. Doesthecommutativepropertyhold? Notethatr ∗ s = s and s ∗ r = s, andso r ∗ s = s ∗ r. Becausethisistheonlypossiblepairofelementstocheckforcommutativity, anditworks, weseethat (T, ∗) satisfiesthecommutativeproperty,andhenceisanabeliangroup.Thatwasafairbitofefforttoverifythat (T, ∗) wasagroup, butifwewanttobesurethat ∗

asgiveninTable 6.4.1 yieldsagroup, wecannotavoidthateffort, becausenoteveryoperationtableyieldsagroup. Infact, ifyourandomlywritedownanoperationtable, itishighlyunlikelythatitwillyieldagroup. Asanexample, letustakethesameset T = {r, s} asbefore, butletusdefineadifferentbinaryoperation, denoted • thistime, givenbyTable 6.4.2.

• r s

r r s

s s s

Table6.4.2

Wewanttoverifywhetherornot (T, •) isagroup. Thereadermayverifythat (T, •) onceagainsatisfiestheclosure, associativityandidentityproperties, with r againplayingtheroleoftheidentity. Butwhataboutinverses? Fromthetableweseethat r•r = r, so r isitsowninverse.Ontheotherhand, lookingatthetabledoesnotyieldaninversefor s. Weseethat r • s = s

and s • s = s, soneither r nor s canbetheinverseof s. Itfollowsthat (T, •) doesnotsatisfytheinversesproperty, andisthereforenotagroup. So, notalloperationtableswork.Letustryafewmoreoperationtables. ConsiderthesetV = {x, y, z,w}withbinaryoperation

⊕ givenbyTable 6.4.3.

⊕ x y z w

x w z y x

y z w x y

z y x w z

w x y z w

Table6.4.3

Is (V,⊕) agroup? Becausealltheentriesinthetableareintheset V , theclosurepropertyholds. Asmentioned, wecanassume theassociativityproperty. As for the identityproperty,noticethat w playstheroleofanidentityelement, because w combinedwithanythingisthatsamething. Observethatthecolumnunderw intheoperationtableisthesameasthecolumnattheleftendofthetable; similarly, therowtotherightof w isthesameastherowatthetopofthetable. Thisphenomenonholdspreciselybecause w istheidentityelement, andthis

6.4Groups 233

methodcanbeusedtofindidentityelements(iftheyexist)quicklyinanyoperationtable. Forinverses, itisseenthateachelementisitsowninverse(thiswillnotbethecaseforallfinitegroups, sodon’tjumptoanyconclusionshere). Hence, allthepropertiesofagrouphold, and(V,⊕) isindeedagroup. Itisalsothecasethatthecommutativepropertyholdsfor (V,⊕), Toseethis, youcouldtryallpossibilities. Forexample, weseethat x ⊕ y = z and y ⊕ x = z,sothat x ⊕ y = y ⊕ x; similarlyfortheothercases. Hence (V,⊕) isanabeliangroup. Thereisaneasierwaytoseethatcommutativityholdsfor (V,⊕). Noticethattheoperationtablefor(V,⊕) issymmetricaboutitsdownwardslopingdiagonal. Ifyouthinkaboutit, youwillseethatingeneral, foranygroup, thistypeofsymmetryoftheoperationtablewillholdpreciselyifagroupsatisfiesthecommutativeproperty.

Exercise 6.4.3. Foreachcollectionofobjectsandoperationtableindicatedbelow, answerthefollowingquestion:

(a) Istheclosurepropertysatisfied?

(b) Isthereanidentityelement? Ifso, whatisit?

(c) Whichelementshaveinverses? Forthosethathaveinverses, statetheirinverses? (Ifthereisnoidentityelement, thisquestionismoot.)

(d) Isthecommutativepropertysatisfied?

(e) Assumingthattheassociativepropertyholds, dothecollectionofobjectsandgivenoperationformagroup? Iftheyareagroup, isitanabeliangroup?

(1) Theset V = {x, y, z,w} withbinaryoperation ⋄ givenbyTable 6.4.4.

(2) Theset K = {m,n, p, q, r} withbinaryoperation ⋆ givenbyTable 6.4.5.

(3) Theset M = {1, s, t, a, b, c} withbinaryoperation ⊙ givenbyTable 6.4.6.

(4) Theset W = {e, f, g, h,w, x, y, z} withbinaryoperation ∗ givenbyTable 6.4.7.

⋄ x y z w

x z w y x

y x y z w

z y z w x

w z w x z

Table6.4.4

Wehaveseensomeexamplesoffinitegroupsgivenbyoperationtables, andotherexamples(namelythegroups (Zn,+))thatwerenotgivenbyoperationtables. However, anyfinitegroup

234 6. Groups

⋆ m n p q r

m n p q m r

n p r m n p

p q m r p n

q m n p q r

r r p n r q

Table6.4.5

⊙ 1 s t a b c

1 1 s t a b c

s s t 1 b c a

t t 1 s c a b

a a c b 1 t s

b b a c s 1 t

c c b a t s 1

Table6.4.6

hasanoperationtable, evenifthatisnothowthegroupwasinitiallydescribed; nomatterhowabinaryoperationisdefined, wecanalwayswriteoutanoperationtablesimplybyseeingwhatthebinaryoperationdoestoeachpairofelementsofthegroup.


Lookattheexamplesofoperationtablesthatwehaveseensofarthatyieldgroups. Canyouseeanynicefeaturesofthewaythattheelementsarearrangedinthesetables?

Thefollowingpropositionstatesaverynicefeatureofoperationtablesoffinitegroups.

Proposition 6.4.2. Supposethat (G, ∗) isafinitegroup. Intheoperationtableforthegroup,eachelementofthegroupappearsexactlyonceineachrow, andonceineachcolumn.

Demonstration. SupposetothecontrarythatasingleelementofG appearstwiceinonerow. Inparticular, supposethatthissameelementappearstwiceintherowcorrespondingtotheelementa; supposethatitappearsinthecolumnscorrespondingtoelements b and c. Itfollowsthata∗b = a∗c. Because (G, ∗) isagroup, weknowthat a hasaninverse, say a ′. Wededucethata ′ ∗ (a ∗b) = a ′ ∗ (a ∗ c), andhencebyassociativityweseethat (a ′ ∗a) ∗b = (a ′ ∗a) ∗ c.If e istheidentityelementofthegroup, thenitfollowsthemeaningofinverseelementsthate ∗ b = e ∗ c. By themeaningof the identityelement, wededuce that b = c. However,weassumedthatthecolumnscorrespondingto b and c aredistinctcolumns, andsowehavearrivedata logical impossibility. Theonlywayoutof thissituationis toconcludethateachelementof G appearsatmostonceineachrow. Inordertofillupeachrowintheoperation

6.4Groups 235

∗ e f g h w x y z

e e f g h w x y z

f f g h e x y z w

g g h e f y z w x

h h e f g z w x y

w w x y z e f g h

x x y z w f g h e

y y z w x g h e f

z z w x y h e f g

Table6.4.7

tablewithelementsof G, itmustbethecasethateveryelementof G appearsatleastonceineachrow. ThefinalconclusionisthateachelementofG appearsexactlyonceineachrow. Thesameideawillworkforcolumnsinsteadofrows, andwewillskipthedetails.

Wenotethateventhougheachelementofthegroupappearsexactlyonceineachrow, andonce in each column, of its operation table, it is definitely not the case that anyoperationtablethatsatisfiesthispropertyyieldsagroup. ThereaderisaskedtofurnishanexampleinExercise 6.4.4.

Exercise 6.4.4. [UsedinThisSection] Findanexampleofafinitesetwithabinaryoperationgivenbyanoperationtable, suchthateachelementofthesetappearsexactlyonceineachrow, andonceineachcolumn, andyetthesetwiththisbinaryoperationisnotagroup.

Exercise 6.4.5. Let C betheset C = {k, l,m}. Constructanoperationon C, bymakinganoperationtable, whichturns C intoagroup.

Whatdoesitmeantosaythattwogroupsaredifferent? Certainly, thegroup (T, ∗) givenbyTable 6.4.1 isdifferentfromthegroup (V,⊕) givenbyTable 6.4.3, becausetheformerhastwoelements(namely r and s), whereasthelatterhasfourelements(namely x, y, z and w).Nowconsidertheset Q = {E, F,G,H} withbinaryoperation ⊞ givenbyTable 6.4.8.Thereadercanverifythat (Q,⊞) isindeedanabeliangroup. Wenowcomparethegroups

(V,⊕) and (Q,⊞). Technicallytheyaredistinctgroups, havingdifferentelementsandoperationtables, butintuitivelytheyappeartobe“essentiallythesame.” Moreprecisely, observethatwe

236 6. Groups

⊞ F G H E

F E H G F

G H E F G

H G F E H

E F G H E

Table6.4.8

canobtainTable 6.4.8 fromTable 6.4.3 bythefollowingsubstitutions:

x 7−→ F

y 7−→ G

z 7−→ H

w 7−→ E.

ThefactthatTable 6.4.8 isobtainedfromTable 6.4.3 bythissubstitutionmeansthatnotonlydotheelementsof V correspondtotheelementsofQ, buttheoperation ⊕ correspondstotheoperation⊞. Hence, wesaythat (V,⊕) and (Q,⊞) areessentiallythesameinthatthesecondgroupisobtainedfromthefirstsimplybyrenamingtheelementsoffirstgroupandrenamingthebinaryoperation. Wenote, moreover, that it isacceptable foroneoperation table tobeobtainedfromanotherbysubstitution, evenifonetablehastoberearranged. Forexample, if⊞ hadinitiallybeengivenbyTable 6.4.9, thattoowouldbeessentiallythesameas ⊕. Ifonegroupcanbeobtainedfromanotherbyrenamingtheelementsandthebinaryoperation, andpossiblyrearrangingtheoperationtable, thenwesaythetwogroupsare isomorphic.

⊞ E F G H

E E F G H

F F E H G

G G H E F

H H G F E

Table6.4.9

Clearly, twogroupswithdifferentnumbersofelementscannotbeisomorphic. Ontheotherhand, not all groups of the same size are isomorphic. For example, consider the set Z ={a, b, c, d} withbinaryoperation ⋆ givenbyTable 6.4.10.Again, thereadercanverifythat (Z, ⋆) isanabeliangroup. However, weclaimthat (Z, ⋆)

isnot isomorphic to (Q,⊞) (andhencenot to (V,⊕) either). Themostdirectway toshowthat (Q,⊞) and (Z, ⋆) arenotisomorphicwouldbetotryeverypossiblewayofrenamingtheelementsofQ as a, b, c and d, andthenobservingthatweneverobtainTable 6.4.10 for ⋆ fromTable 6.4.8 for ⊞, evenafterrearranging. Suchaverificationwouldbequitetedious. Themoreappealingwaytoverifythattwogroupsarenotisomorphicistofindsomepropertyofoneof

6.5Subgroups 237

⋆ a b c d

a a b c d

b b c d a

c c d a b

d d a b c

Table6.4.10

thegroupsthatdoesnotholdfortheothergroup, butsuchthatthepropertywouldbepreservedbyrenamingandrearranginganoperationtable. Forexample, weobservethatinTable 6.4.10,whichhasidentityelement a, eachof a and c isitsowninverses, but b and d arenottheirowninverses(theyareinversesofeachother). Bycontrast, inTable 6.4.8, whichhasidentityelement E, weobservethateachof F, G, H and E isitsowninverses. Hence, in (Z, ⋆) twoelementsaretheirowninverses, whereasin (Q,⊞) allfourelementsaretheirowninverses. Itfollowsthatthesetwogroupscouldnotpossiblybeisomorphic.

Exercise 6.4.6. Isthegroup (Z4,+) isomorphictoeitherof (Z, ⋆) or (Q,⊞)? If (Z4,+)isisomorphictooneofthesetwogroups, demonstratethisfactbyshowinghowtorenametheelementsof (Z4,+) appropriately.

6.5 Subgroups

Oneinterestingphenomenoninthetheoryofgroupsistheideaofasubgroup. Considerthegroup (Z,+). Insidethesetofintegers Z isthesetofevenintegers

E = {. . . ,−6,−4,−2, 0, 2, 4, 6, . . .}.

Thesystem (E,+), isitselfagroup. (YouwereaskedtoverifythisfactinExercise 6.4.1 (1); thepointisthataddingtwoevennumbersgivesanevennumber, sotheclosurepropertyholds,andtheotherpropertiescanbeverifiedsimilarly.) Hence, weseethat (E,+) isbothagroupinitsownright, andit isalsocontainedinthelargergroup (Z,+). Wesaythat (E,+) isasubgroupof (Z,+). Ingeneral, acollectionofelementsofagroup forma subgroup if thiscollection, togetherwiththeoperationoftheoriginalgroup, formagroupintheirownright.Noteverycollectionofelementsofagroupsformsasubgroup. Forexample, considerthesetofoddnumbers, denoted O. Itturnsoutthat (O,+) isnotagroup, becausetheclosurepropertydoesnothold; toseethis, notethatthesumoftwooddnumbersisanevennumber, notanoddnumber.

238 6. Groups

Exercise 6.5.1.

(1) Let T bethesetofallintegermultiplesof 3, thatis, theset

T = {. . . ,−9,−6,−3, 0, 3, 6, 9, . . .}.

Is (T,+) subgroupof (Z,+)?

(2) Let V bethesetofallperfectsquareintegersandtheirnegatives, thatis, theset

V = {. . . ,−16,−9,−4,−1, 0, 1, 4, 9, 16 . . .}.

Is (V,+) subgroupof (Z,+)?

It turnsout thatagroupmayhavemanysubgroups, or itmayhavevery few. Everygroupcontainswhatiscalledthe trivialsubgroup, whichisthesubgroupconsistingofnothingbuttheidentityelement. Thistrivialsubgrouphasonlyoneelementinit, whichmaymakeitseemlessthanexciting, butitisreallyavalidgroup. Forexample, thetrivialsubgroupof (Z,+) isjusttheoneelementset {0}, togetherwiththeoperationofaddition. Notethat 0 + 0 = 0, sotheclosurepropertyholdsforthistrivialgroup; theotherpropertiesofgroupscanalsobeverifiedfor {0}. Everygroupalsocontainsatleastoneothersubgroup, namelyitself. Wedidnotrequirethatasubgrouphavefewerelements thantheoriginalgroup. Ofcourse, whatwearereallyinterestedinaresubgroupsthatarenottheentireoriginalgroup. A subgroupthatisnotequaltotheoriginalgroupiscalleda propersubgroup. Thequestionnowbecomeswhetherthereareanyproper, non-trivialsubgroupsinagivengroup.Letusstartwiththeexampleof (Z8,+), theoperationtableforwhichisgiveninTable 6.3.1.

Wewanttofindasubcollectionofelementsof (Z8,+) thatformagroupbythemselves. Be-cause theoperation + isassociative forall theelementsof Z8, it iscertainlyassociative foranysubcollectionofelements. Asaresult, wewillnothavetoworryaboutassociativitywhenlookingforsubgroupsof (Z8,+), orsubgroupsofanythingelseforthatmatter. Ontheotherhand, wedohavetoworryaboutclosure, identityandinverses. Becausethesubcollectionsof(Z8,+) thatwearelookingformustsatisfytheidentityproperty, theymustcontain 0. So, weneedtofindasubcollectionof Z8 = {0, 1, 2, 3, 4, 5, 6, 7} thatcontains 0, andthatsatisfiestheclosureproperty, andsuchthatforanyelementofthesubcollection, itsinversewillbeinthesubcollection. A goodwaytotrytofindsuchasubcollectionistochoosesomeelements, andconstructtheoperationtable.Letustrythesubcollection A = {0, 1, 3, 7} of Z8, chosenrandomly. Theoperationtablefor

theseelements, showninTable 6.5.1, wasobtainedbydeletingalltheunnecessaryrowsandcolumnsoftheoperationtablefor (Z8,+).Aninspectionoftheoperationtablefor A revealsthat (A,+) isnotasubgroupof (Z8,+).

First, itdoesnotsatisfytheclosureproperty, because, forexample, weseethat 1 + 3 = 4, and

6.5Subgroups 239

+ 0 1 3 7

0 0 1 3 7

1 1 2 4 0

3 3 4 6 2

7 7 0 2 6

Table6.5.1

yet 4 isnotinthesubcollection A. Ingeneral, ifalltheentriesintheoperationtableforthesubcollectionarethemselvesinthesubcollection, thentheclosurepropertyholds; conversely,ifsomeoftheentriesintheoperationtableforthesubcollectionarenotinthesubcollection,thentheclosurepropertydoesnothold. Further, notethat 3 doesnothaveaninversein A,becausethereisnothingin A which, whenaddedto 3, yields 0. (Theelement 3 doeshaveaninversein (Z8,+), namely 5, but 5 isnotin A.) So, ifwewanttofindsubgroups, weneedtochooseoursubcollectionsmorecarefully.Letusnowtrythesubcollection B = {0, 2, 4, 6}. Theoperationtablefor (B,+) isshownin

Table 6.5.2.

+ 0 2 4 6

0 0 2 4 6

2 2 4 6 0

4 4 6 0 2

6 6 0 2 4

Table6.5.2

Thistimethingslookmorepromising. Noticethatalltheelementsofthetablearefromthesubcollection B, sothattheclosurepropertyholds. Theelement 0 isinthecollection, sotheidentitypropertyholds. Asforinverses, notethat 0 isitsowninverse, that 4 isitsowninverse,andthat 2+ 6 = 0 and 6+ 2 = 0, sothat 2 and 6 areinversesofeachother. Hencetheinversespropertyholds. Becausetheassociativitypropertyisautomatic, asmentionedabove, weseethat (B,+) isindeedasubgroup.Arethereanyotherpropersubgroupsof (Z8,+)? Twothatareeasytofindare C = {0} (the

trivialsubgroup)and D = {0, 4}. Itisnothardtoverifythat C and D areindeedsubgroupsbyexaminingtheiroperationtables. Arethereanyothersubgroups? Wecouldexamineeachpossiblesubcollectionof Z8 aswedidsubcollection A above, butthatwouldbeverytedious.Ingeneral, itisnoteasytofindallsubgroupsofagivengroup, butthereisonefactthatis

veryusefulinreducingtheworkincheckingthevarioussubcollections. Thisresult, knownasLagrange’sTheorem, isasfollows. A proofofthistheoremisbeyondthescopeofthistext.

Proposition 6.5.1 (LaGrange’sTheorem). Supposethat (G, ∗) isafinitegroup. Supposethat Hisasubgroupof G. Thenthenumberofelementsin H dividesthenumberofelementsin G.

240 6. Groups

Inotherwords, ifyouhaveafinitegroupandyouarelookingforsubgroups, youcanruleoutanysubcollectionwherethenumberofelementsdoesnotdividethenumberofelementsoftheoriginalgroup. However, justbecauseasubcolletiondoeshaveanacceptablenumberofelementsdoesnotmeanthatthesubcollectionisnecessarilyasubgroup.LetusapplyLagrange’sTheoremtothegroup (Z8,+). Thisgrouphas 8 elements. Theonly

numbersthatdivide 8 are 1, 2, 4 and 8. Wecanignore 1 and 8, becausetheonlysubgroupwithoneelementisthetrivialsubgroup {0}, andtheonlysubgroupwith 8 elementsisthewholeof(Z8,+). Hence, byLagrange’sTheorem, alltheproper, non-trivialsubgroupsof (Z8,+) musthave 2 or 4 elements. Inparticular, therecanbenosubgroupsof (Z8,+) witheither 3, 5, 6 or7 elements. Ontheotherhand, noteverysubcollectionof (Z8,+) with 2 or 4 elementsisasubgroup. Forexample, thesubcollection A discussedabovehad 4 elements, andyetwasnotasubgroup. Infact, itturnsoutthattherearenoothersubgroupsof (Z8,+) otherthan B, CandD givenabove(weomitthedetails, thoughitcouldbeverifieddirectly, albeittediously, bycheckingallsubcollectionsof Z8 witheither 2 or 4 elements).Next, letusfindallsubgroupsofthegroup (Z7,+). Thisgrouphas 7 elements. ByLagrange’s

Theoremanysubgroupwouldhave tohaveanumberofelements thatdivides 7. But 7 isaprimenumber; thatis, therearenonumbersthatdivide 7 exceptitselfand 1. A subgroupwith7 elementswouldjustbethewholegroup (Z7,+), andasubgroupwith 1 elementwouldhavetobethetrivialsubgroup. Inotherwords, weseefromLagrange’sTheoremthat (Z7,+) hasnoproper, non-trivialsubgroup. Thissamereasoningappliestoanyfinitegroupthathasaprimenumberofelements.

Exercise 6.5.2. Thegroup (Z36,+) has 36 elements. Howmanyelementscouldasubgroupof (Z36,+) possiblyhave?

Exercise 6.5.3. Which, if any, of the following subcollections of Z6 are subgroups of(Z6,+)? UseLagrange’sTheorem, andconstructoperationtablesaswedidforsubgroupsof (Z8,+).

(1) A = {0, 3};

(2) B = {0, 2};

(3) C = {0, 1, 4};

(4) D = {0, 2, 4}.

(5) E = {0, 1, 2, 3}.

6.6SymmetryandGroups 241

Exercise 6.5.4. Let (M,⊙) beas inExercise 6.4.3 (3). Which, ifany, of the followingsubcollectionsof M aresubgroupsof (M,⊙)?

(1) E = {1, s};

(2) F = {1, a};

(3) C = {1, s, t};

(4) D = {1, a, b, c}.

Exercise 6.5.5. Findasmanypropersubgroupsasyoucanof (Z12,+). Theoperationtablefor (Z12,+) isgiveninTable 6.2.2.

6.6 SymmetryandGroups

Althoughthestudyofsymmetryappearstobe“geometric”innature, andthestudyofgroupsappearstobe“algebraic,” infactsomeofthesameideasappearinbothfields. Forexample,recallLeonardo’sTheorem(Proposition 5.4.5)aboutrosettepatterns, whichstatedthatthesym-metrygroupofarosettepatterniseitherCn forsomepositiveinteger n, orDn forsomepositiveinteger n. The Cn groupsshouldlookveryfamiliarafterourdiscussionoftheintegersmod n

inSection 6.3. Forexample, wesaw inTable 5.4.1 theoperationtablefor (C8, ·). Comparethatoperationtablewiththeoperationtablefor (Z8,+), showninTable 6.3.1. Itisseenfromthesetwooperationtablesthat (C8, ·) and (Z8,+) areisomorphicgroups; simplyreplace 1by 0, replace r by 1, replace r2 by 2, replace r3 by 3, etc. Thesameideashowsthatforeachpositiveinteger n, thegroup (Cn, ·) isisomorphictothegroup (Zn,+). Wethereforeseethatthesamebasicobjectcanariseinthestudyofgeometryandthestudyofalgebra. Geometryandalgebraare, wesee, notasunrelatedasonemightthinkafterseeingthetwofieldsstudiedratherseparatelyintypicalhighschoolcourses.Tounderstand the relationbetweensymmetryandalgebramoreexplicitly, recall the term

“symmetrygroup” thatwestartedusinginSection 5.1, thoughatthetimewesimplyusedthistermtorefertothecollectionofallsymmetriesofagivenobject. Nowthatwehavediscussedthegeneralconceptofgroups(whichisinherentlyanalgebraicconcept), weneedtoaskwhethera“symmetrygroup”aspreviouslydefinedisindeedagroupaswehavenowdefinedit. Theanswer, notsurprisinglygivenourchoiceofterminology, isyes.

Proposition 6.6.1. Supposethat K isaplanarobject. LetG denotethecollectionofallsymme-triesof K. Then (G, ◦) isagroup.

242 6. Groups

Demonstration. TheclosurepropertiesfollowsfromProposition 5.1.2 (1). Theassociativeprop-ertyfollowsfromProposition 4.4.2 (2). Theidentityof (G, ◦) istheidentitysymmetry I. TheinversespropertiesfollowsfromProposition 5.1.2 (2).

Anotherexampleofasymmetrygroupthatalsoarisesnaturallyinalgebraisthefriezegroupf11, whichwasdiscussedinSection 5.5. Thefriezegroup f11 isthesymmetrygroupoffriezepatternsthathavenosymmetryotherthantranslation, forexample · · · FFFFF · · · . AsstatedinSection 5.5, wehave

f11 ={· · · t−3, t−2, t−1, 1, t, t2, t3, · · ·

},

where t denotesthesmallestpossibletranslationsymmetrytotherightofthefriezepattern. Wecanthinkof t as t1, and 1 as t0, andwecancombineanytwosymmetriesin f11 bytheruletatb = ta+b. Itcannowbeobservedthatthegroup (f11, ◦) isisomorphictothegroup (Z,+),where 1 in f11 correspondsto 0 in Z, where t correspondsto 1, where t2 correspondsto 2,where t3 correspondsto 3, etc.Observethatsymmetrygroupsarenotnecessarilyabelian, forexamplethesymmetrygroup

ofanequilateraltriangle. WementionthattheanalogofProposition 6.6.1 forthreedimensional(andhigher)objectsalsoholds(andforthesamereasons), butwewillnotgointodetailshere.Also, weshouldnotethatalthougheverysymmetrygroupofaplanarobjectisagroup, noteverygrouparisesasthesymmetrygroupofaplanarobject(seeExercise 6.6.1).

Exercise 6.6.1. [Used in This Section] Show that the group (W, ∗) given in Exer-cise 6.4.3 (4)isnotthesymmetrygroupofanyplanarobject. Theideaisasfollows. Giventhat W isfinite, if itwere the symmetrygroupofaplanarobject, itwouldhave tobethesymmetrygroupofa rosettepattern (because thoseareprecisely theplanarobjectswithfinitesymmetrygroups). ByLeonardo’sTheorem (Proposition 5.4.5), weknowthatanyrosettepatternhassymmetrygroupeither Cn or Dn forsomepositiveinteger n. Findreasonstoshowwhy (W, ∗) isnotisomorphictoanyofthe Cn or Dn groups.

Nowthatweknowthatsymmetrygroupsareindeedgroups, variousideasaboutgroupscanbeusedtogainabetterunderstandingofsymmetry. Indeed, mathematicallycompleteproofsofProposition 5.5.1 andProposition 5.6.2, inwhichwestatedtheclassificationoffriezepatternsandwallpaperpatternsrespectively, arebasedonsomeideasfromgrouptheorythatarebeyondthescopeofthistext.Oneconceptfromthetheoryofgroupsthatcanbeappliedtosymmetrygroupsisthenotion

ofsubgroups. (Indeed, subgroupsplayanimportantroleontheproofsreferredtointhepreviousparagraph.) Giventhesymmetrygroupofanobject, wecanaskwhichcollectionsofsymmetriesoftheobjectformsubgroups. Forexample, letusexaminethesymmetrygroupofthesquare,theoperationtableforwhichweseeinTable 6.6.1.

6.6SymmetryandGroups 243

· 1 r r2 r3 m mr mr2 mr3

1 1 r r2 r3 m mr mr2 mr3

r r r2 r3 1 mr3 m mr mr2

r2 r2 r3 1 r mr2 mr3 m mr

r3 r3 1 r r2 mr mr2 mr3 m

m m mr mr2 mr3 1 r r2 r3

mr mr mr2 mr3 m r3 1 r r2

mr2 mr2 mr3 m mr r2 r3 1 r

mr3 mr3 m mr mr2 r r2 r3 1

Table6.6.1

AnexaminationofTable 6.6.1 showsthatthisgrouphasninepropersubgroups, asfollows:

{1},

{1, r2},

{1,m},

{1,mr},

{1,mr2},

{1,mr3},

{1, r, r2, r3},

{1, r2,m,mr2},

{1, r2,mr,mr3}.

Wefoundthesesubgroupsbytrialanderror, thoughwemadeuseofLaGrange’sTheorem, whichsaidthatweonlyneededtolookforsubgroupswith 1, 2 or 4 elements.

Exercise 6.6.2. Foreachofthefollowingobjects, findallpropersubgroupsofitssymmetrygroup.

(1) Theequilateraltriangle.

(2) Theregularpentagon.

Are thereanygeneral rules forfinding subgroupsof symmetrygroups? Forexample, doesthecollectionofalltranslationsymmetriesformasubgroup? Whataboutthecollectionofallrotationsymmetries? Whataboutthecollectionofallrotationandallreflectionsymmetries?Thesecondandthirdofthesecollectionsofsymmetriesarenotalwayssubgroups(andexamplewillbegivenshortly), butthefirstalwaysis, asshownbythefollowingproposition.

244 6. Groups

Proposition 6.6.2. Supposethat K isaplanarobject. LetG denotethecollectionofallsymme-triesof K. Thenthefollowingsubcollectionsof G aresubgroupsof (G, ◦):

1. Alltranslationsymmetriesof A;

2. Alltranslationandrotationsymmetriesof A.

Demonstration.

(1). Let T denote thecollectionofall translationsymmetriesof K. WeknowfromPropo-sition 4.6.2 that thecompositionofanytwotranslations isa translation, andweknowfromProposition 5.1.2 (1)thatthecompositionofanytwosymmetriesofanobjectisasymmetry.Puttingthesetwofactstogether, wededucethat (T, ◦) satisfiestheclosureproperty. Theasso-ciativepropertyfor (T, ◦) isautomaticallytrue, becauseitistruefor ◦ ingeneral(seePropo-sition 4.4.2 (2)). Next, wecanthinkoftheidentityisometry I astranslationby 0, andso I isin T . Therefore (T, ◦) satisfiestheidentityproperty. WededucefromProposition 4.6.5 (2)thattheinverseofanytranslationisatranslation, andweknowfromProposition 5.1.2 (2)thattheinverseofanysymmetryofanobjectisasymmetry. Puttingthesetwofactstogether, wededucethat (T, ◦) satisfiestheinversesproperty. Alltold, weseethat (T, ◦) isagroupinitsownright,andhenceitisasubgroupof (G, ◦).

(2). ThispartisverysimilartoPart (1), andthedetailsarelefttothereader.

Theabovepropositiongivestwoverysimpletypesofsubgroupsofsymmetrygroups, thoughthereareother subgroupsaswell. Thecollectionof all rotation symmetries isnot alwaysasubgroup—itdependsupontheobject. Forarosettepattern, forexample, thecollectionofallrotationsymmetriesisasubgroup; thereaderisaskedtosupplythedetailsinExercise 6.6.4.Bycontrast, forafriezepatternthathashalfturnrotationsymmetry, thecollectionofallrotationsymmetriesisnotasubgroup, becausethecompositionoftwohalfturnrotationsaboutdifferentcentersofrotationisatranslation, andthereforetheclosurepropertyisnotsatisfied.

Exercise 6.6.3. Iseachofthefollowingcollectionofsymmetriesalwaysasubgroupofthesymmetrygroupofaplanarobject. Explainyouranswers.

(1) Thecollectionofallreflectionsymmetries.

(2) Thecollectionofalltranslationandallhalfturnrotationsymmetries.

(3) Thecollectionofallrotationandallreflectionsymmetries.

Exercise 6.6.4. [UsedinThisSection] Showthatforarosettepattern, thecollectionofallrotationsymmetriesisasubgroupofthesymmetrygroup.

SuggestionsforFurtherReading

Therearemanyexcellenttextsthatyoumightwishtoreadtofurtheryourstudyofthematerialdiscussedinthisbook(thoughsomeofthesetextsshouldbeapproachedwithanunderstandingof their strengthsandweaknesses). What follows isavery idiosyncraticallyannotated listofvariousbooksyoumightconsiderforfurtherreading, arrangedbythechaptersinthistext.

GeometryBasics

• Euclid, “TheElements”(3vols.), Dover, 1956.

Oneof thegreatestworksofWesternCivilization, andoneof themore tediousaswell. It’sunquestionablytruethatourliveswouldbeverydifferenttodayifthisbookhadnotbeenwritten,butthat’snoreasontoattempttoreadthewholething. ItiswellworthknowingwhatEuclidwastryingtodo, howhedidit, andwhetherornothesucceeded, butitdoesn’ttakeallthreevolumestogetthat. Lookitover, inanycase. Thisusedtoberequiredreadingforeverypersonclaimingtobeeducated. Unfortunately, lessEuclidinschoolshasnotbeenreplacedbyotherkindsofgeometry.

• RobinHartshorne, “Geometry: EuclidandBeyond,” Springer-Verlag, NewYork, 2000.

OneofthemostimpressivemathematicstextbooksI haverecentlyseen. Thistext, meantasacompaniontoEuclid’s“TheElements,” doesnotsummarizeEuclid, butratherexplainswhathisconceptualunderstandingwasandhowitdiffersfromourcontemporaryapproach, andshowshowEuclidcanbebroughtmathematicallyuptodate. Thoughmostofthebookisaimedatanaudienceofjuniororseniorlevelcollegemathematicsmajors(and, inparticular, makesuseofabstractalgebra), muchofthediscussionofEuclideangeometryinthefirsttwochaptersis

246 SuggestionsforFurtherReading

accessibletoabroaderaudience, andwellworththepriceofhavingtoskipoversometechni-calities. HartshornehasdoneanastonishingjoboffiguringEuclidout, makingthisasubstantialbookwithequallysubstantialrewards.

Polygons

• MarthaBoles&RochelleNewman, “TheGolden-Relationship, Book1,” PythagoreanPress, 1987.

A workbook that actually has you get your hands dirtywith geometric constructions usingstraightedgeandcompass. InbetweentheproblemsandprojectsareveryreadablediscussionsoftheGoldenRatio, Fibonaccinumbersandthelike. Somereadersmayfindthephilosophicalexpositionabitflaky, butit’sworthwadingthroughitforthesakeofthehands-onapproach.Besides, howcanyougowrongwithabookthathasarecipefor“FibonacciFudge”?

• TheodoreA.Cook, “TheCurvesofLife,” Dover, 1914.

Morethanyoueverwantedtoknowaboutspirals, fromrams’hornstospiralstaircases. Thefirstandlastfewchaptersareworthreading; thestuffinbetween(whichisafairbit)makesforfunbrowsing.

• MatilaGhyka, “TheGeometryofArtandLife,” Dover, 1977.

Inspiteof thebroadtitle, mostof thebookfocusesontheGoldenRatioandrelatedtopics.Someofthematerialisgood, thoughabittechnical; otherpartsofthebookarespeculative(toputitpolitely), concerningvariousesoterictheoriestheauthorappearstobelieve. Interestingreadingifyoucandealwithit.

• H.E.Huntley, “TheDivineProportion,” Dover, 1970.

A rhapsodyabouttheGoldenRatio(a.k.a.theDivineProportion), andbeautyinmathematicsingeneral. Someofthematerialisphilosophical, somefairlytechnical. It’sworthpickingbitsandpiecesoutofthisbook.

• RobertLawlor, “SacredGeometry,” Crossroad, 1982.

Greatpictures, andallkindsofesoterictheories—withlotsofgeometricalconstructionsthrownin. Youwillhavetodecideforyourselfwhat’sgoingonhere, becauseI amnotsure.


Polyhedra

• PeterR.Cromwell, “Polyhedra,” CambridgeUniversityPress, Cambridge, 1997.

A lovelytextforthenon-specialist. Thereisawealthofhistoricalinformationonthestudyofpolyhedra, wonderfulillustrations, andanexcellentchoiceoftopics, rangingfromsuchstan-dardsasthePlatonicsolidstolesswellknown(toapopularaudience)gemssuchasDescartes’TheoremonangledefectsandConnelly’sflexiblesphere. Theonerealdrawbackisthelackofexercisesforthereader, devaluingthisbookasatextbook, butwellworthreadingnonetheless.

• MarjorieSenechalandGeorgeFleck, “ShapingSpace,” Birkhäuser, Boston, 1988.

Thisbookistheproceedingsfromaconferenceonvariousaspectsofpolyhedraandrelatedtopics, whichmightsounddulluntilyoutakealookatit—lookingthroughthisbookmakesmewishthatI hadbeenatthatconference! Thoughafewofthearticlesarequitetechnical, manyareaimedatageneralaudience, includinganicehistoryofthestudyofpolyhedra. Thebookisverywellillustrated. I wouldn’tnecessarilyrecommendbuyingthisoneunlessyouareahardcorepolyhedrafan, butitiswellworthabrowse.

HigherDimensions

• EdwinA.Abbott, “Flatland,” Dover, 1952(orothereditions; alsoavailableontheweb).

A minorclassic, withheavyemphasisonbothwords. “Flatland”recountstheadventuresofASQUARE,wholivesina2dimensionalworld. Thefirstpartofthebook, asatireoftheVictoriansocietyinwhichAbbottlived, describestheracistandsexistsocialorderinwhichourherolives.ThesecondpartofthebookdescribesA SQUARE’S encounterwithlowerandhigherdimen-sionalbeings, thusintroducingthereadertosomeimportantideasaboutthefourthdimensionandhigher. Neithergreatwritingnorbrilliantmathematics, “Flatland”straddlesthefencesowellthatitsplaceinthecanonisassured. (Becarefulwiththeintroductionstovariouseditionsof“Flatland”—theonebyBaneshHoffmannintheDoveredition, andtheonebyIsaacAsimovintheHarperCollinsedition, bothentirelymissthepointofthebook.)

• DionysBurger, “Sphereland,” PerennialLibrary(Harper&Row), 1965.

A modern sequel to “Flatland,” introducingmanymathematical ideas recognizedas impor-tantsincetheadventofEinstein’stheoryofrelativity(whichpost-dates“Flatland”by25years).“Sphereland”waswrittenbyamathematician, whichshowsinboththewellchosenmathemat-icaltopics, andthelessthangrippingnarrativestyle. Thoughmathematicallymoresubstantialthan“Flatland,” itlacksthelatter’ssatiricalbite.

• RudyRucker, “TheFourthDimension: aGuidedTouroftheHigherUniverses,” HoughtonMifflin, 1984.


A funbookcoveringalotofseriousmaterial, andsomeratheresotericstuff toboot. Ruckermakeshigherdimensions, relativityandgeometryenjoyableandsurprisinginawaynooneelsecan. Lotsofgoodproblemsandpuzzles, andgreatquotesandillustrations. Cometoyourownconclusionsaboutthemorespeculativestuff—I’msureRudywouldn’thaveitanyotherway.

• Thomas F. Banchoff, “Beyond theThird Dimension,” ScientificAmerican Library, NY1990.

I wishI hadwrittenthisone, thoughitisjustaswellthatI didn’t, becauseitishardtoimaginethatanyoneelsewouldhavecomeclose todoing itaswellasBanchoff (aserious researchmathematicianwithagenuineinterestinreachingabroadaudience). Thisbookissuchacare-fullythoughtoutandbeautifullyillustratedtreatmentofhigherdimensionsthatitcouldmakeafinecoffeetablebook, thoughdon’tletthatfoolyou—thisbookdiscussesseriousstuff. Theexcellentchoiceof topics range fromunfoldingand slicinghigherdimensionalcubes (withgreatcomputergraphics)toperspectiveandscaling. Afterreadingtheclassics“Flatland”and“Sphereland,” thiswouldbeanexcellentnextplacetowhichtoturnifyouwanttoknowmoreabouthigherdimensions.

• A.K.Dewdney, “ThePlaniverse,” PoseidonPress, 1984.

A verydetailedexplorationofwhata2-dimensionalworldcouldreallybelike, wrappedinasomewhatsillynarrative. Theemphasisisnotonmathematics(asinFlatland), butonphysics,biologyandtechnologyin2dimensions. Whatwoulda2-dimensionalsailboatlooklike? Howwould2-dimensionalintestineskeepfromsplittingacreatureintwo? It’sallquitefun, thoughabitmorethanyoumightwanttoknow.

• MichioKaku, “Hyperspace,” Anchor, 1994.

A rhapsodyaboutthelatesttheoriesofphysics(forexample, stringtheory, paralleluniverses,wormholesandthelike), andtheirrelationtomathematics, especiallythestudyofhigherdi-mensions. Writtenbyaphysicist, ithastheadvantageofaninsider’sviewofthelatestphysicaltheories, andthedisadvantageofaphysicistsviewofmathematics—whichtothismathemati-cianseemsabitdistorted. Thefirstfewchaptersonhigherdimensionscontainsomeinterestinghistoricaldiscussionoftheriseofpopularinterestinthesubject, butthemathematicalideascanbefoundtreatedbetterelsewhere. Ifyouwanttolearnaboutphysics, thenbyallmeansreadthisbook.

• CharlesH.Hinton, “SpeculationsontheFourthDimension,” Dover, 1980.

Probablyamustforhard-core4thdimensionfans, butnotnecessarilyforanyoneelse. Hinton,amathematicianobsessedwiththe4thdimension, wroteavarietyofessaysand“Flatland”stylestoriesthathavebeenexcerptedandcollectedbyRudyRuckerinthisvolume. Thefictionat-temptstobemorescientificthan“Flatland”(havingadifferentsortof2dimensionalworld, andanticipatingthelaterbook“ThePlaniverse”), butthenarrativeistedious, andimbuedwithHin-ton’smysticalideas. Theessaysarefineinpart, but, aswiththefiction, thereisbetterelsewhere.


SymmetryofPlanarObjectsandOrnamentalPatterns

• HermanWeyl, “Symmetry,” Princeton, 1952.

A classicbyoneofthegreatmathematiciansofthe20thcentury. Thecaliberofthephilosophicalandhistoricaldiscussionsreflectthestatureoftheauthor. Weyldoeslapseintosomeoverlytechnicalpassages, buttheyarewellworthwadingthroughfortherest. Greatillustrationsaswell.

• Farmer, David, “GroupsandSymmetry,” AMS,1996

Theideaisofthisbookisgreat: anexpositionoflovelymathematicaltopicsincludingsymmetry,ornamentalpatternsandgroups, aimedatnon-mathematicians, donenotbylecturingbutbybriefdiscussioncombinedwithlotsof‘tasks’forthereadertoexplore. Unfortunately, thewritingisattimesawkward, thechoiceofterminologyisonoccasionunfortunate, theorganizationispoorandthe‘tasks’varyfromtrivialtoextremelyhardwithnowarning. A fewextrarevisionswouldhavehelped. A well-meaningbookthatdoesnotquiteliveuptoitspromise.

• GeorgeE.Martin, “TransformationGeometry,” SpringerVerlag, 1982.

A verytechnicalbookappropriateforpeoplewithatminimumsomeCalculuslevelmathemat-ics(thoughCalculusperseisnotrequired). Thebookhasaverynicetreatmentoffriezeandwallpapergroups, tilings, andprojectivegeometry. Thisonedemandsseriousstudy.

Tilings

• BrankoGrünbaum&G.C.Shephard, “TilingsandPatterns,” W.H.Freeman, NY,1987.

Theultimatereferenceonthemathematicaltheoryoftilingsandotherplanarornamentalpat-terns, thismassivebookwillsurelybethedefinitivesourceintheforeseeablefuture. Thoughmostofthetextismathematicallysophisticated, thelovelyintroductionisaccessibletoall, andthepicturesandfiguresthroughoutthetextaregreat. I wouldnotrecommendbuyingthisoneunlessyouareplanningaseriousstudyofthesubject, butitiswellworthlookingthrough.

Bibliography

[Bar01] ArthurBaragar, ASurveyofClassicalandModernGeometries, withComputerActiv-ities, PrenticeHall, UpperSaddleRiver, NJ, 2001.

[Euc56] Euclid, TheThirteenBooksoftheElements, trans. ThomasL. Heath, Dover, NewYork,1956.

[Fed82] P. J. Federico, DescartesonPolyhedra, Springer-Verlag, NewYork, 1982.

[Gre93] MarvinJ. Greenberg, EuclideanandNon-EuclideanGeometries, 3rded., W. H. Free-man, NewYork, 1993.

[Har00] RobinHartshorne, Geometry: EuclidandBeyond, Springer-Verlag, NewYork, 2000.

[Joh66] NormanW. Johnson, Convexpolyhedrawithregularfaces, Canad. J. Math. 18 (1966),169–200.

[Jus61] NortonJuster, ThePhantomTollbooth, Epstein&Carroll, NewYork, 1961.

[Loo40] ElishaLoomis, ThePythagoreanProposition, EdwardsBrothers, AnnArbor, 1940.

[Mar75] GeorgeMartin, TheFoundationsofGeometryandtheNon-EuclideanPlane, Springer-Verlag, NewYork, 1975.

[Mar82] , TransformationGeometry: An Introduction to Symmetry, Springer-Verlag,NewYork, 1982.

[Mey99] WalterMeyer, GeometryanditsApplications, AcademicPress, SanDiego, 1999.

252 Bibliography

[Sen90] MarjorieSenechal, CrystallineSymmetries: AnInformalMathematicalIntroduction,AdamHilger, Bristol, 1990.

[Tru87] RichardTrudeau, TheNon-EuclideanRevolution, Birkhäuser, Boston, 1987.

[WW98] EdwardWallaceandStephenWest, RoadstoGeometry, 2nded., PrenticeHall, UpperSaddleRiver, NJ, 1998.

[Wey52] HermanWeyl, Symmetry, PrincetonUniversityPress, Princeton, 1952.

polygons, polyhedra, patterns & beyond

Documents