paul yiumath.fau.edu/yiu/oldwebsites/geometry2007spring/... · 2007. 1. 5. · contents 1 the...

Elementary Geometry

Paul Yiu

Department of MathematicsFlorida Atlantic University

Spring 2007

Contents

1 The Pythagorean theorem 11.1 The hypotenuse of a right triangle . .. . . . . . . . . . . . . . 11.2 The Pythagorean theorem. . . . . . . . . . . . . . . . . . . . . 31.3 Integer right triangles . .. . . . . . . . . . . . . . . . . . . . . 3

2 The equilateral triangle 52.1 Construction of equilateral triangle . .. . . . . . . . . . . . . . 52.2 The medians of an equilateral triangle .. . . . . . . . . . . . . . 62.3 The center of an equilateral triangle . .. . . . . . . . . . . . . . 72.4 The circumcircle and incircle of an equilateral triangle . . . . . . 72.5 Trigonometry . . . . . .. . . . . . . . . . . . . . . . . . . . . 82.6 Some interesting examples on equilateral triangles .. . . . . . . 10

3 The square 133.1 Construction of a square on a segment. . . . . . . . . . . . . . 133.2 The diagonals of a square. . . . . . . . . . . . . . . . . . . . . 143.3 The center of the square .. . . . . . . . . . . . . . . . . . . . . 153.4 Some interesting examples on squares. . . . . . . . . . . . . . 15

4 Some basic principles 174.1 Angle properties . . . . .. . . . . . . . . . . . . . . . . . . . . 17

4.1.1 Parallel lines . . .. . . . . . . . . . . . . . . . . . . . . 174.1.2 Angle sum of a triangle . . . . .. . . . . . . . . . . . . . 184.1.3 Angle properties of a circle . . .. . . . . . . . . . . . . . 18

4.2 Tests of congruence of triangles . . . .. . . . . . . . . . . . . . 194.2.1 Construction of a triangle with three given elements . . . 194.2.2 Congruence tests .. . . . . . . . . . . . . . . . . . . . . 19

iv CONTENTS

5 Circumcircle and incircle 235.1 Circumcircle . .. . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.1.1 The perpendicular bisector locus theorem .. . . . . . . . 235.1.2 Construction of circumcircle . . . .. . . . . . . . . . . . 245.1.3 Circumcircle of a right triangle . . .. . . . . . . . . . . . 24

5.2 The incircle . .. . . . . . . . . . . . . . . . . . . . . . . . . . 255.2.1 The angle bisector locus theorem .. . . . . . . . . . . . 255.2.2 Construction of incircle. . . . . . . . . . . . . . . . . . . 255.2.3 The incircle of a right triangle . . .. . . . . . . . . . . . 26

5.3 Tangents of a circle . . . . .. . . . . . . . . . . . . . . . . . . 265.3.1 Tangent at a point on the circle . . .. . . . . . . . . . . . 265.3.2 The tangents from a point to a circle . . .. . . . . . . . 26

5.4 Some interesting examples on the circumcircle and incircle of atriangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6 Uses of congruence tests 296.1 Isosceles triangles . . . . . .. . . . . . . . . . . . . . . . . . . 296.2 Chords of a circle . . . . . .. . . . . . . . . . . . . . . . . . . 306.3 Parallelograms .. . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.3.1 Properties of a parallelogram . . . .. . . . . . . . . . . . 316.3.2 Quadrilaterals which are parallelograms . .. . . . . . . . 326.3.3 Special parallelograms. . . . . . . . . . . . . . . . . . . 32

6.4 The midpoint theorem and its converse . .. . . . . . . . . . . . 336.4.1 The midpoint theorem. . . . . . . . . . . . . . . . . . . 336.4.2 The converse of the midpoint theorem . . .. . . . . . . . 346.4.3 Why are the three medians concurrent? . .. . . . . . . . 34

6.5 Some interesting examples .. . . . . . . . . . . . . . . . . . . 35

7 The regular hexagon, octagon and dodecagon 377.1 The regular hexagon . . . . .. . . . . . . . . . . . . . . . . . . 377.2 The regular octagon . . . . .. . . . . . . . . . . . . . . . . . . 387.3 The regular dodecagon . . .. . . . . . . . . . . . . . . . . . . 39

8 Similar triangles 418.1 Tests for similar triangles . .. . . . . . . . . . . . . . . . . . . 418.2 The parallel intercepts theorem . . . . . .. . . . . . . . . . . . 438.3 The angle bisector theorem .. . . . . . . . . . . . . . . . . . . 448.4 Some interesting examples of similar triangles . .. . . . . . . . 45

CONTENTS v

9 Tangency of circles 499.1 External and internal tangency of two circles . . . .. . . . . . . 499.2 Three mutually tangent circles . . . . .. . . . . . . . . . . . . . 509.3 Some interesting examples of tangent circles . . . .. . . . . . . 50

10 The regular pentagon 5310.1 The golden ratio . . . . .. . . . . . . . . . . . . . . . . . . . . 5310.2 The diagonal-side ratio of a regular pentagon . . . .. . . . . . . 5410.3 Construction of a regular pentagon with a given diagonal . . . . 5510.4 Various constructions of the regular pentagon . . . .. . . . . . . 5610.5 Some interesting constructions of the golden ratio .. . . . . . . 58

11 The regular decagon 6111.1 Constructions . . . . . .. . . . . . . . . . . . . . . . . . . . . 6111.2 Some relations among sides of regular polygons inscribed in a

circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

12 Construction of regular n-gons 6512.1 Gauss’ theorem . . . . .. . . . . . . . . . . . . . . . . . . . . 6512.2 Construction of a regular15-gon . . . . . . . . . . . . . . . . . 6512.3 Construction of a regular17-gon . . . . . . . . . . . . . . . . . 66

13 Regular solids 6913.1 The faces of a regular solid . . . . . .. . . . . . . . . . . . . . 6913.2 Edge lengths of inscribed regular solids. . . . . . . . . . . . . . 70

14 The regular tetrahedron 7114.1 Construction. . . . . . . . . . . . . . . . . . . . . . . . . . . . 7114.2 Measurements . . . . . .. . . . . . . . . . . . . . . . . . . . . 71

14.2.1 The height of a regular tetrahedron . . . . . .. . . . . . . 7114.2.2 The circumsphere of a regular tetrahedron . .. . . . . . . 72

14.3 The angle between a line and a plane .. . . . . . . . . . . . . . 7314.4 The angle between an edge and a face of a regular tetrahedron . . 7314.5 The angle between two planes . . . . .. . . . . . . . . . . . . . 7414.6 The dihedral angle of a regular tetrahedron . . . . .. . . . . . . 7414.7 Inscribed regular tetrahedron . . . . .. . . . . . . . . . . . . . 7514.8 The anglesΩ andΨ . . . . . . . . . . . . . . . . . . . . . . . . 76

vi CONTENTS

15 The cube 7715.1 The face diagonals and the space diagonals of a cube . . . . . . . 7715.2 The angle between a space diagonal and a face . .. . . . . . . . 7715.3 Inscribed cube .. . . . . . . . . . . . . . . . . . . . . . . . . . 78

15.3.1 Inscribed cube with two faces parallel to the equator . . . 7815.3.2 Inscribed cube with a vertex at the north pole . . . . . . . 78

16 The square pyramid and the regular octahedron 7916.1 The square pyramid . . . . .. . . . . . . . . . . . . . . . . . . 7916.2 Dihedral angle between a face and the base. . . . . . . . . . . . 7916.3 Dihedral angle between two adjacent equilateral triangles . . . . 8016.4 Inscribed regular octahedron. . . . . . . . . . . . . . . . . . . 80

16.4.1 Inscribed regular octahedron with a vertex at the north pole 8016.4.2 Inscribed regular octahedron with a face parallel to the

equator .. . . . . . . . . . . . . . . . . . . . . . . . . . 80

17 The regular icosahedron 8317.1 Construction . .. . . . . . . . . . . . . . . . . . . . . . . . . . 8317.2 Dihedral angle of the regular icosahedron .. . . . . . . . . . . . 84

18 The regular dodecahedron 8518.1 Euclid’s construction of the regular dodecahedron. . . . . . . . 85

18.1.1 Two golden points above a mid-line of a square . . . . . . 8518.1.2 The pentagonal faces .. . . . . . . . . . . . . . . . . . . 85

18.2 Dihedral angle between two adjacent faces of a regular dodeca-hedron . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . 87

18.3 Inscribed regular dodecahedron with a vertex at the north pole . . 88

Chapter 1

The Pythagorean theorem

1.1 The hypotenuse of a right triangle

This chapter is on the famous Pythagorean theorem. We certainly all know thatthis is an important relation on the sides of aright triangle. The side oppositeto the right angle is called thehypotenuseand is the longest among all threesides. The other two are called thelegs. If we denote the legs bya andb, and thehypotenuse byc, the Pythagorean theorem says that

a2 + b2 = c2.

A

B

Cb

ac

Why is this relation true?Suppose you have a cardboard whose length and breadth are 3 inches and

4 inches. How long is its diagonal? Your young students have not known thePythagorean theorem yet. Help themcalculate this length without throwing thePythagorean theorem to them.

Take two identical copies of the cardboard and cut each one along a diagonal.In this way there are 4congruent right triangles. Arrange these 4 triangles in thisway:

2 The Pythagorean theorem

4 1

23

In this figure, the 4 right triangles bound a large square outside and their hy-potenuses bound another smaller square inside.

We calculate areas.(1) The outer side square has side length3 + 4 = 7 units. Its area is72 = 49

square units.(2) The 4 right triangles make up 2 rectangular cardboards each with area

3× 4 = 12 square units. The 4 triangles have total area2× 12 = 24 square units.(3) The inside square therefore has area

49 − 24 = 25

square units. Each of its sides has length5 units.This method indeed applies to any right triangle with given legs.

Square on hypotenuse= square on sum of legs

− 2 × rectangle with given legs as sides.

b a

b

a

a b

b

ac

c c

c

4 1

23

Why is the inside region a square?

1.2 The Pythagorean theorem 3

1.2 The Pythagorean theorem

Rearrange the “puzzle” in§1.1 in another form:

4

1

2

3a

a

b

b

Here, we see the same 4 right triangles inside the same outer square. Insteadof the inside square, we now have two smaller squares, each built on a leg of theright triangle. This gives the famousPythagorean theorem:

a2 + b2 = c2.

1.3 Integer right triangles

If we take any two numbersp andq and form

a = 2pq, b = p2 − q2, c = p2 + q2, (1.1)

then we havea2 + b2 = c2.

In particular, ifp andq are integers, then so area, b, andc.Sometimes, integer right triangles constructed from formula (1.1) can be re-

duced. For example, withp = 3 andq = 1, we obtain(a, b, c) = (6, 8, 10), whichclearly is simply the right triangle(3, 4, 5) magnified by factor2. We say that aninteger right triangle(a, b, c) is primitive if a, b, c do not have common divisor(other than1). Every integer right triangle is a primitive one magnified by an in-teger factor. A right triangle constructed from (1.1) is primitive if we choose theintegersp, q of different parity1 and without common divisors.

1This means that one of them is even and the other is odd.

4 The Pythagorean theorem

Here are all primitive integer right triangles withp, q < 10:

p q a b c

2 1 4 3 53 2 12 5 134 1 8 15 174 3 24 7 255 2 20 21 295 46 16 57 27 47 68 18 38 58 79 29 49 8

How many distinct integer right triangles are there whose sides are not morethan 100?

Chapter 2

The equilateral triangle

Notations(O) circle with centerOO(A) circle with centerO, passing throughAO(r) circle with centerO, radiusr

2.1 Construction of equilateral triangle

An equilateral triangle is one whose three sides are equal in length. The very firstproposition of Euclid’sElements teaches how to construct an equilateral triangleon a given segment.

A B

C

If A andB are the endpoints of the segments, construct the circlesA(B) andB(A) to intersect at a pointC. Then, triangleABC is equilateral.

An equilateral triangle is alsoequiangular. Its three angles are each equal to60.

6 The equilateral triangle

2.2 The medians of an equilateral triangle

LetABC be an equilateral triangle, andM the midpoint ofBC. The segmentAMis amedianof the triangle. SinceABM ≡ ACM by thetest, this lineAM is also(i) thebisector of angleA,(ii) the altitude on the sideBC (sinceAM ⊥ BC), and therefore(iii) the perpendicular bisectorof the segmentBC.

B C

A

M

This also means that one half of an equilateral triangle is a right triangle withangles30 and60. We call this an30 − 60 − 90 right triangle.

In a30− 60− 90 right triangle, the hypotenuse is twice as long as the shorterleg. The two legs are in the ratio

√3 : 1.

2

1

√3

2.3 The center of an equilateral triangle 7

2.3 The center of an equilateral triangle

Consider two medians of the equilateral triangle intersecting atO.

B C

A

M

NO

TrianglesAON andBOM are both30 − 60 − 90 right triangles. Therefore,OA = 2 ×ON andOB = 2 × OM .However,OA+OM andOB +ON have the same length.This meansOM = ON , 1

andOM = 13× AM , ON = 1

3× BN . We have shown thatany two medians of

an equilateral triangle intersect at a point which divides each of them in the ratio2 : 1. This is the same pointO for all three medians, and is called thecenter ofthe equilateral triangle.

2.4 The circumcircle and incircle of an equilateraltriangle

Since the three medians of an equilateral triangle have the same length, the centerO is equidistant from the vertices. It is the center of thecircumcircle of theequilateral triangle. The same centerO is also the center of theincircle of theequilateral triangle.

1AM = BN ⇒ OA+OM = OB+ON ⇒ 2×ON+OM = 2×OM+ON ⇒ OM = ON .


B C

A

O

B C

A

O

2.5 Trigonometry

For an acute angleθ, the trigonometric ratiossin θ, cos θ andtan θ are defined by

sin θ =a

c, cos θ =

b

c, tan θ =

a

b,

where(a, b, c) are the sides of a right triangle containing an acute angleθ withopposite lega.

If any one of the three ratios is known, then the other two can be found. Why?

(1) sin2 θ + cos2 θ = 1.

(2) tan θ = sin θcos θ

.

(3) sin(90 − θ) = cos θ.

The precise determination of the trigonometric ratios of an angle is in generalbeyond the scope of elementary geometry. There are, however, a few specialangles whose trigonometric ratios can be determined precisely.

2.5 Trigonometry 9

c

b

a

θ

A C

B

θ 30 60

sin θ

cos θ

tan θ


2.6 Some interesting examples on equilateral trian-gles

(1) For an arbitrary pointP on theminor arcBC of the circumcircle of an equi-lateral triangleABC, AP = BP + CP .

B C

A

P

Q

Proof. If Q is the point onAP such thatPQ = PC, then triangleCPQ is equi-lateral since . Note that∠ACQ = ∠BCP .Thus,ACQ ≡ BCP by the test. From this,AQ = BP ,andAP = AQ+QP = BP + CP .

(2) Consider a triangleABC whose angles are all less than120. Constructequilateral trianglesBCX, CAY , andABZ outside the triangle.

LetF be the intersection of the circumcircles of the equilateral triangles. Notethat ∠CFA = ∠AFB = 120. It follows that∠AFB = 120, andF lies onthe circumcircle of the equilateral triangleBCX as well. Therefore,∠XFC =∠XBC = 60, and∠XFC + ∠CFA = 180. The three pointsA, F , X arecollinear. Thus,

AX = AF + FX = AF +BF + CF

by (1) above. Similarly,B, F , Y are collinear, so doC, F , Z, andBY , CZ areeach equal toAF +BF + CF .

Theorem 2.1 (Fermat point). If equilateral triangles BCX , CAY , and ABZare erected on the sides of triangle ABC, then the lines AX , BY , CZ concur ata point F called the Fermat point of triangleABC. If the angles of triangleABCare all less than 120 (so that F is an interior point), then the segments AX , BY ,CZ have the same length AF +BF + CF .

2.6 Some interesting examples on equilateral triangles 11

F

Z

X

Y

A

B C

(3∗) Napoleon’ theorem. If equilateral trianglesBCX, CAY , andABZ areerected on the sides of triangleABC, the centers of these equilateral trianglesform another equilateral triangle.

Z

X

Y

A

B C

Chapter 3

The square

3.1 Construction of a square on a segment

Given a segmentAB, construct(1) an equilateral triangleABP ,(2) an equilateral triangleAPQ (different fromABP ),(3) the segmentPQ and its midpointM ,(4) the rayAM ,(5) the circleA(B) to intersect this ray atD,(6) the circlesB(A) andD(A) to intersect atC.ThenABCD is a square.

A B

CD

PQM

This construction gives anoriented square. If you apply it toAB andBA,you will get two different squares on the two sides ofAB.

14 The square

3.2 The diagonals of a square

A diagonal of a square divides it into two congruent right triangles. Each is anisosceles right triangleor a45 − 45 − 90 triangle. If each side of a square hasunit length, a diagonal has length

√2.

1

1√

2

1

√2

1

θ 45

sin θ

cos θ

tan θ

The diagonals of a square are perpendicular to each other. They divide thesquare into 4 isosceles right triangles.

3.3 The center of the square 15

3.3 The center of the square

The intersection of the diagonals is thecenter of the square. It is the commoncenter of the circumcircle and the incircle of the square.

3.4 Some interesting examples on squares

(1) If ABCbCa andACBcBa are squares erected externally on the sidesAB andAC of a triangleABC, the triangleACaBa has the same area as triangleABC.

Bc

Ba

Ca

Cb

A

B C

(2∗) The perpendicular fromA toBaCA passes through the midpoint ofBC.Likewise, those fromB toCbAb and fromC toAcBc pass through the midpointsof CA andAB respectively. These three perpendiculars meet at the centroidG oftriangleABC.

16 The square

G

Bc

Ba

Ca

Cb

Ab Ac

A

B C

(3∗) The midpoint ofBcCb does not depend on the position of the vertexA. Itis also the same as the center of the square erected on theBC, on the same sideasA.

Bc

Ba

Ca

Cb

A

B C

M

Chapter 4

Some basic principles

4.1 Angle properties

4.1.1 Parallel lines

Consider two parallel lines1, 2, and a transversalL.

1

2

L

α

β

γ δ

(1) Thecorrespondinganglesα andβ are equal.(2) Thealternate anglesβ andγ are equal.(3) Theinterior anglesβ andδ are supplementary,i.e., β + δ = 180.The converses of these statements are also true. In other words, if any one of

(1), (2), (3) holds, then the lines1 and2 are parallel.

18 Some basic principles

4.1.2 Angle sum of a triangle

The angle sum of a triangle is always180. This is because when a side of thetriangle is extended, the external angle formed is equal to the sum of the tworemote internal angles.

A

B Cβ

α

4.1.3 Angle properties of a circle

(1) LetP be a point on themajor arcAB of a circle(O).

∠AOB = 2∠APB.

P

A B

O

Q

P

A B

O

Q

PA B

O

(2) LetP be a point on theminor arcAB of a circle(O).

∠AOB + 2∠APB = 360.

(3) The opposite angles of a cyclic quadrilateral are supplementary.(4) The angle contained in a semicircle is a right angle.(5) Equal chords subtend equal angles at the center.

4.2 Tests of congruence of triangles 19

P

A BO

A

B D

O

C

4.2 Tests of congruence of triangles

4.2.1 Construction of a triangle with three given elements

A triangle has six elements: three sides and three angles. Consider the construc-tion of a triangle given three of its six elements. The triangle is unique (up to sizeand shape) if the given data are in one of the following patterns.

(1) SSSGiven three lengthsa, b, c, we construct a segmentBC with lengtha, and the two circlesB(c) andC(b). These two circles intersect at two points if(and only if)b + c > a [triangle inequality]. There are two possible positions ofA. The resulting two triangles are congruent.

(2) SAS Given b, c, and angleA < 180, the existence and uniqueness oftriangleABC is clear.

(3) ASA or AAS These two patterns are equivalent since knowing two of theangles of a triangle, we easily determine the third (their sum being180). GivenB, C anda, there is clearly a unique triangle providedB + C < 180.

(4) RHS Givena, c andC = 90.

4.2.2 Congruence tests

These data patterns also provide the valid tests of congruence of triangles. Two tri-anglesABC andXY Z are congruent if theircorrespondingelements are equal.Two triangles are congruent if they have three pairs of equal elements in one thefive patterns above. The five valid tests of congruence of triangles are as follows.


(1) SSS: ABC ≡ XY Z if

AB = XY, BC = Y Z, CA = ZX.

A

B C

X

Y

Z

(2) SAS: ABC ≡ XY Z if

AB = XY, ∠ABC = ∠XY Z, BC = Y Z.

A

B C

X

Y

Z

(3) ASA: ABC ≡ XY Z if

∠BAC = ∠Y XZ, AB = XY, ∠ABC = ∠XY Z.

A

B C

X

Y

Z

(4) AAS. We have noted that this is the same asASA: ABC ≡ XY Z if

∠BAC = ∠Y XZ, ∠ABC = ∠XY Z, BC = Y Z, .

4.2 Tests of congruence of triangles 21

A

B C

X

Y

Z

(5) RHS. The ASS is not a valid test of congruence. Here is an example. Thetwo trianglesABC andXY Z are not congruent even though

∠BAC = ∠Y XZ, AB = XY, BC = Y Z.

A X

B Y

C Z

However, if the equal angles are right angles, then the third pair of sides areequal:

AC2 = BC2 −AB2 = Y Z2 −XY 2 = XZ2,

andAC = XZ. The two triangles are congruent by the SSS test. Without repeat-ing these details, we shall simply refer to this as theRHS test.ABC ≡ XY Zif

∠BAC = ∠Y XZ = 90, BC = Y Z, AB = XY.

A X

B Y

C Z

The congruence tests for triangles form a paradigm for proofs in euclideangeometry. We shall illustrate this with numerous examples in Chapter 6. Here, wegive only one example, theconverseof the Pythagorean theorem.


Theorem 4.1 (Converse of Pythagoras’ theorem).If the lengths of the sides ofABC satisfy a2 + b2 = c2, then the triangle has a right angle at C.

ac

b

a

bC X

B Y

A Z

Proof. Consider a right triangleXY Z with ∠Z = 90, Y Z = a, andXZ = b.By the Pythagorean theorem,XY 2 = Y Z2 + XZ2 = a2 + b2 = c2 = AB2. Itfollows thatXY = AB, andABC ≡ XY Z by the test, and∠C = ∠Z = 90.

Chapter 5

Circumcircle and incircle

5.1 Circumcircle

Theperpendicular bisector of a segment is the line perpendicular to it throughits midpoint.

5.1.1 The perpendicular bisector locus theorem

Theorem 5.1. A point P is equidistant from B and C if and only if P lies on theperpendicular bisector of BC.

P

B CM

Proof. LetM be the midpoint ofBC.(⇒) If PB = PC, thenPBM ≡ PCM by the test. This

means∠PMB = ∠PMC andPM ⊥ BC. The pointP is on the perpendicularbisector ofBC.

(⇐) If P is on the perpendicular bisector ofBC, thenPMB ≡ PMC bythe test. It follows thatPB = PC.

24 Circumcircle and incircle

5.1.2 Construction of circumcircle

Given a triangleABC, there is a circle(O) through the three vertices. The centerO of the circle, being equidistant fromB andC, must lie on the perpendicularbisector ofBC. For the same reason, it also lies on the perpendicular bisectors ofCA andAB. This is called thecircumcenter of triangleABC.

O

A

B C

5.1.3 Circumcircle of a right triangle

To construct a circle through the vertices of a right triangleABC, mark the mid-pointM of the hypotenuseAB. The circleM(C) passes throughA andB, and isthecircumcircle of the right triangle.

A B

C

M

5.2 The incircle 25

5.2 The incircle

5.2.1 The angle bisector locus theorem

The distance from a pointP to a line is the distance betweenP and itspedal(orthogonal projection) on.

P

K

A H

Theorem 5.2. A point P is equidistant from two lines if and only if P lies on thebisector of an angle between the two lines.

Proof. LetA be the intersection of the two lines, andH,K the pedals of a pointP on these lines.

(⇒) If PH = PK thenAPH ≡ APK by the test. Itfollows that∠PAH = ∠PAK, andP lies on the bisector of angleHAK.

(⇐) If ∠PAH = ∠PAK, thenAPH ≡ APK by thetest. It follows thatPH = PK.

5.2.2 Construction of incircle

The incircle of a triangle is the one which is tangent to each of the three sidesof the triangle. Its center is the common point of the angle bisectors, and can belocated by constructing two of them. Thisincenter I is equidistant from the threesides.

I

A

B C


5.2.3 The incircle of a right triangle

If d is the diameter of the incircle of a right triangle, then

a+ b = c+ d.

C

B

Ab

a cd

5.3 Tangents of a circle

5.3.1 Tangent at a point on the circle

A tangent to a circle is a line which intersects the circle at only one point. Givena circleO(A), the tangent to a circle atA is the perpendicular to the radiusOAatA.

A

O

A

O

B

PM

5.3.2 The tangents from a point to a circle

If P is a point outside a circle(O), there are two lines throughP tangent to thecircle. Construct the circle withOP as diameter to intersect(O) at two points.These are the points of tangency.

The two tangents have equal lengths sinceOAP ≡ OBP by thetest.

5.4 Some interesting examples on the circumcircle and incircle of a triangle27

5.4 Some interesting examples on the circumcircleand incircle of a triangle

(1) Let s = 12(a + b + c) be thesemiperimeterof triangleABC. The lengths of

the tangents from the vertices to the incircle are as follows.

tangent fromA =s− a,tangent fromB =s− b,tangent fromC =s− c.

C A

B

b

a c

r

rr

(2) The following rearrangement shows that for an arbitrary triangle, the radiusof its incircle is given by

r =area of triangle

semi-perimeter of triangle.

r


(3∗) Construct the circumcircle(O) of triangleABC and select an arbitrarypointP on it. Construct thereflectionsof P in the sidelines of the given triangle.These three reflection points always lie on a straight line. Furthermore, asPvaries on(O), the line passes through a fixed point.

P

H

A

B C

(4∗) Construct the circumcircle(O) and incircle(I) of a triangleABC. Selectan arbitrary pointP on (I) and construct the perpendicular toIP atP . This is atangent to(I). Let it intersect the circle(O) atY andZ.

Construct the tangents fromY andZ to (I). These two tangents always inter-secton the circumcircle(O).

I

P

X

Y

ZO

A

B C

Chapter 6

Uses of congruence tests

6.1 Isosceles triangles

An isosceles triangle is one with two equal sides.

Proposition 6.1. A triangle is isosceles if and only if it has two equal angles.

B

A

CM B

A

CD

Proof. (⇒) Let ABC be a triangle in whichAB = AC. LetM be the midpointof BC. ThenABM ≡ ACM by the test. From this concludethat∠BAM = ∠CAM .

(⇐) LetABC be a triangle in which∠B = ∠C. Construct the perpendicularfrom A to BC, and let it intersectBC atD. ThenABD ≡ ACD by the

test. From thisAB = AC and the triangle is isosceles.

30 Uses of congruence tests

6.2 Chords of a circle

(1) LetM be a point on a chordBC of a circle(O). M is the midpoint ofBC ifand only ifOM ⊥ BC.

O

MB C

Proof. (⇒) If M is the midpoint ofBC, thenOMB ≡ OMC by thetest. Therefore,∠OMB = ∠OMC andOM ⊥ BC.

(⇐) If OM ⊥ BC, thenOMB ≡ OMC by the test. Itfollows thatBM = CM , andM is the midpoint ofBC.

(2) Equal chords of a circle are equidistant from the center. Conversely, chordsequidistant from the center are equal in length.

Proof. SupposeBC andPQ are equal chords of a circle(O), with midpointsM andN respectively. ThenBM = PN , andOBM ≡ OPN by the

test. Therefore,OM = ON , and the chords are equidistant fromthe centerO.

Conversely, ifOM = ON , thenOBM ≡ OPN by thetest. It follows thatBM = PN , andBC = 2 · BM = 2 · PN = PQ.

6.3 Parallelograms 31

6.3 Parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides.

6.3.1 Properties of a parallelogram

(1) The opposite sides of a parallelogram are equal in length. The opposite anglesof a parallelogram are equal.

A B

CD

Proof. Consider a parallelogramABCD. Construct the diagonalBD. Note thatABD ≡ CDB by the test. It follows thatAB = CD andAD = CB, and∠BAD = ∠DCB. Similarly,∠ABC = ∠CDA.

(2) The diagonals of a parallelogram bisect each other.

M

A B

CD

Proof. Consider a parallelogramABCD whose diagonalsAC andBD intersectatM . Note thatAD = CB andAMD ≡ CMB by the test.It follows thatAM = CM andDM = BM . The diagonals bisect each other.


6.3.2 Quadrilaterals which are parallelograms

(1) If a quadrilateral has two pairs of equal opposite sides, then it is a parallelo-gram.

A B

CD

Proof. LetABCD be a quadrilateral in whichAB = CD andAD = BC. Con-struct the diagonalBD. ThenABD ≡ CDB by the test. Itfollows that andAD//BC. Also, andAB//DC. Therefore,ABCD is a parallelogram.

(2) If a quadrilateral has one pair of equal and parallel sides, then it is a paral-lelogram.

A B

CD

Proof. SupposeABCD is a quadrilateral in whichAB//DC andAB = CD.ThenABD ≡ CDB by the test. It follows thatAD = CBandABCD is a parallelogram since .

6.3.3 Special parallelograms

A quadrilateral is(1) arhombus if its sides are equal,(2) arectangle if its angles are equal (and each90),(3) asquare if its sides are equal and its angles are equal.

6.4 The midpoint theorem and its converse 33

These are all parallelograms since(1) a rhombus has two pairs of sides, and(2) a rectangle has two pairs of sides.Clearly, then a square is a parallelogram.

(1) A parallelogram is a rhombus if and only if its diagonals are perpendicularto each other.

(2) A parallelogram is a rhombus if and only if it has an angle bisected by adiagonal.

(3) A parallelogram is a rectangle if and only if its diagonals are equal inlength.

6.4 The midpoint theorem and its converse

6.4.1 The midpoint theorem

Given triangleABC, letE andF be the midpoints ofAC andAB respectively.The segmentFE is parallel toBC and its length is one half of the length ofBC.

A

B C

FE

D

Proof. ExtendFE toD such thatFE = ED. Note thatCDE ≡ AFE bythe test. It follows that(i) CD = AF = BF ,(ii) ∠CDE = ∠AFE, andCD//BA. Therefore,CDFB is a parallelogram, andFE//BC. Also,BC = FD = 2FE.


6.4.2 The converse of the midpoint theorem

Let F be the midpoint of the sideAB of triangleABC. The parallel throughFtoBC intersectsAC at its midpoint.

A

B C

FE

D

Proof. Construct the parallel throughC to AB, and extendFE to intersect thisparallel atD. Then,CDFB is a parallelogram, andCD = BF = FA. It followsthatAEF ≡ CED by the test. This means thatAE = CE, andE is the midpoint ofAC.

6.4.3 Why are the three medians concurrent?

LetE andF be the midpoints ofAC andAB respectively, andG the intersectionof themediansBE andCF .

Construct the parallel throughC toBE, and extendAG to intersectBC atD,and this parallel atH. .

A

B C

FE

D

G

H

6.5 Some interesting examples 35

By the converse of the midpoint theorem,G is the midpoint ofAH, andHC =2 ·GE

JoinBH. By the midpoint theorem,BH//CF . It follows thatBHCG is aparallelogram. Therefore,D is the midpoint of (the diagonal)BC, andAD isalso a median of triangleABC. We have shown that the three medians of triangleABC intersect atG, which we call thecentroid of the triangle.

Furthermore,

AG =GH = 2GD,

BG =HC = 2GE,

CG =HB = 2GF.

The centroidG divides each median in the ratio2 : 1.

6.5 Some interesting examples

(1) Why are the three altitudes of a triangle concurrent?LetABC be a given triangle. Through each vertex of the triangle we construct

a line parallel to its opposite side. These three parallel lines bound a larger triangleA′B′C ′. Note thatABCB ′ andACBC ′ are both parallelograms since each hastwo pairs of parallel sides. It follows thatB ′A = BC = AC ′ andA is themidpoint ofB ′C ′.

H

X

Y

Z

A

B C

A′

B′C′


Consider the altitudeAX of triangleABC. Seen in triangleA′B′C ′, this lineis the perpendicular bisector ofB ′C ′ since it is perpendicular toB ′C ′ through itsmidpointA. Similarly, the altitudesBY andCZ of triangleABC are perpendic-ular bisectors ofC ′A′ andA′B′. As such, the three linesAX, BY , CZ concur ata pointH. This is called theorthocenter of triangleABC.

(2∗) The butterfly theorem. LetM be the midpoint of a chordPQ of a circle(O). AB andCD are two chords passing throughM . JoinBC to intersectPQ atX andAD to intersectPQ atY . ThenMX =MY .

MP Q

O

A

B

C

D

Y X

Chapter 7

The regular hexagon, octagon anddodecagon

A regular polygon ofn sides can be constructed (with ruler and compass) if it ispossible to divides the circle inton equal parts by dividing the360 at the centerinto the same number of equal parts. This can be easily done forn = 6 (hexagon)and8 (octagon).1 We give some easy alternatives for the regular hexagon, octagonand dodecagon.

7.1 The regular hexagon

A regular hexagon can be easily constructed by successively cutting out chords oflength equal to the radius of a given circle.

1Note that if a regularn-gon can be constructed by ruler and compass, then a regular2n-goncan also be easily constructed.

38 The regular hexagon, octagon and dodecagon

7.2 The regular octagon

(1) Successive completion of rhombi beginning with three adjacent45-rhombi.

(2) We construct a regular octagon by cutting from each corner of a givensquare (of side length2) an isosceles right triangle of (shorter) sidex. This means2 − 2x : x =

√2 : 1, and2 − 2x =

√2x; (2 +

√2)x = 2,

x =2

2 +√

2=

2(2 −√2)

(2 +√

2)(2 −√2)

= 2 −√

2.

P

Q

A B

CD

x 2 − 2x x

x

O

P

Q

A B

CD

ThereforeAP = 2−x =√

2, which is half of the diagonal of the square. ThepointP , and the other vertices, can be easily constructed by intersecting the sidesof the square with quadrants of circles with centers at the vertices of the squareand passing through the centerO.

7.3 The regular dodecagon 39

7.3 The regular dodecagon

(1) Trisection of right angles.

(2) A regulardodecagoncan be formed from4 equilateral triangles inside asquare.8 of its vertices are the intersections of the sides of these equilateral trian-gles, while the remaining4 are the midpoints of the sides of the squares formedby the vertices of the equilateral triangles inside the square.

An easy dissection shows that the area of the regular dodecagon is34

of that ofthe (smaller) square containing it.

40 The regular hexagon, octagon and dodecagon

(3) Successive completion of rhombi beginning with five adjacent30-rhombi.

This construction can be extended to other regular polygons. If an angle ofmeasureθ := 360

2n

can be constructed with ruler and compass, beginning with

n − 1 adjacentθ-rhombi, by succesively completing rhombi, we obtain a regular2n-gons tesellated by

(n− 1) + (n− 2) + · · ·+ 2 + 1 =1

2n(n− 1)

rhombi.

Chapter 8

Similar triangles

8.1 Tests for similar triangles

Two triangles are similar if their corresponding angles are equal and their corre-sponding sides proportional to each other,i.e., ABC ∼ XY Z if and onlyif

∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z,and

BC

Y Z=CA

ZX=AB

XY.

It is enough to conclude similarity of two triangles when they have three pairsof

(1) AAA : if the triangles have two (and hence three) pairs of equal angles.

A

B C

X

YZ

42 Similar triangles

(2) proportionalSAS: ifCA

ZX=AB

XYand∠A = ∠X.

A

B C

X

YZ

(3) proportionalSSS: ifBC

Y Z=CA

ZX=AB

XY.

A

B C

X

YZ

8.2 The parallel intercepts theorem 43

8.2 The parallel intercepts theorem

The intercepts on two transversals between a set of parallel lines are proportionalto each other: if0, 1, 2, . . . , n−1, n are parallel lines andL1 andL2 are twotransversals intersecting these lines atA0, A1,A2, . . . ,An−1, An, andB0,B1,B2,. . . ,Bn−1,Bn, then

A0A1

B0B1

=A1A2

B1B2

= · · · =An−1AnBn−1Bn

.

A0 B0

A1 B1

A2 B2

An−1 Bn−1

An Bn

C0

C1

Cn−1

L1L2

0

1

2

n−1

n

Proof. ThroughB1, B2, . . . ,Bn construct lines parallel toL1, intersecting0, 1,. . . , n−1 atC0, C1, . . . ,Cn−1 respectively. Note that(i) B1C0 = A1A0,B2C1 = A2A1, . . . ,BnCn−1 = AnAn−1, and(ii) the trianglesB0B1C0,B1B2C1, . . . ,Bn−1BnCn−1 are similar.It follows that

B1C0

B0B1

=B2C1

B1B2

= · · · =BnCn1

Bn−1Bn.

Hence,A0A1

B0B1=A1A2

B1B2= · · · =

An−1AnBn−1Bn

.


8.3 The angle bisector theorem

The bisector of an angle divides the opposite side in the ratio of the remaining twosides,i.e.,if AX bisects angleA, then

BX

XC=AB

AC.

A

B CX

D

Proof. Construct the parallel throughB to the bisector to intersect the extensionof CA atD. Note that

∠ABD = ∠BAX = ∠CAX = ∠CDB = ∠ADB.

This means that triangleABD is isosceles, andAB = AD. Now, from theparallelism ofAX andDB, we have

AX

XB=DA

AC=AB

AC.

8.4 Some interesting examples of similar triangles 45

8.4 Some interesting examples of similar triangles

(1) The altitude from the right angle vertex divides a right triangle into two trian-gles each similar to the given right triangle.

A B

C

D

Proof. LetABC be a triangle with a right angle atC. If D is a point onAB suchthatDC ⊥ AB, thenACD ∼ CBD ∼ ABC.

Here are some interesting consequences.

(a) FromACD ∼ CBD, we haveAD

CD=CD

BD. It follows that

CD2 = AD · CD.

(b) FromCBD ∼ ABC, we haveBD

CB=BC

ABso that

BC2 = BD · AB.

Similarly, fromACD ∼ ABC, we haveAD

AC=AC

ABso that

AC2 = AD · AB.

It follows that

BC2 +AC2 = BD ·AB +AD ·AB = (BD +AD) ·AB = AB ·AB = AB2.

This gives an alternative proof of the Pythagorean theorem.(a) and (b) above give simple constructions of geometric means.


Construction 8.1. Given a pointP on a segmentAB, construct(1) a semicircle with diameterAB,(2) the perpendicular toAB atP , intersecting the semicircle atQ.

ThenPQ2 = AP × PB.

A BO P

Q

A BO P

Q

θ

θ

(2) The radius of the semicircle is the geometric mean ofAP amdBQ.

A

B

O

Q

P

This follows from the similarity of the right trianglesOAP andQBO. If

r = OA = OB, thenr

QB=AP

r. From this,r2 = AP · BQ.

8.4 Some interesting examples of similar triangles 47

(3∗) Heron’s formula for the area of a triangle.

I

I′

Z′

s− b s− c s− a

r′

r

Y ′ Y

B

C A

TheA-excircle of triangleABC is the circle which is tangent to the sideBCand to the extensions ofAB andAC. If Y ′ andZ ′ are the points of tangency withAC andAB, it is easy to see thatAY ′ = AZ ′ = s, the semiperimeter. Hence,CY ′ = s− b. If the incircle touchesAC atY , we have known from Example 1 of§5.4 thatCY = s− c andAY = s− a.

From these we can find the radiir of the incircle, andr′ of theA-excircle quiteeasily. From the similarity of trianglesAIY andAI ′Y ′, we have

r

r′=s− as.

Note that alsoICY ∼ CI ′Y ′, from which we haver

s− b =s− cr′. and

r · r′ = (s− b)(s− c).Multiplying these two equations together, we obtain

r2 =(s− a)(s− b)(s− c)

s.

From Example 2 of§5.4, the area of triangleABC is given by = rs. Hence,we have the famousHeron formula

=√s(s− a)(s− b)(s− c).

Chapter 9

Tangency of circles

9.1 External and internal tangency of two circles

Two circles(O) and(O′) are tangent to each other if they are tangent to a line at the same lineP , which is a common point of the circles. The tangency isinternal or external according as the circles are on the same or different sides ofthe common tangent.

O O′ O O′P

P

The line joining their centers passes through the point of tangency.The distance between their centers is the sum or difference of their radii, ac-

cording as the tangency is external or internal.

50 Tangency of circles

9.2 Three mutually tangent circles

Given three non-collinear pointsA,B, C, construct(i) triangleABC and its incircle, tangent toBC, CA, AB respectively atX, Y ,Z,(ii) the circlesA(Z),B(Y ) andC(X).These are tangent to each other externally.

C A

B

Z

Y

X

9.3 Some interesting examples of tangent circles

(1) In each of the following diagrams, the shaded triangle is similar to the3−4−5triangle.

9.3 Some interesting examples of tangent circles 51

52 Tangency of circles

(2∗) Given triangleABC, letD, E, F be the midpoints of the sidesBC, CA,AB respectively. The circumcircle of triangleDEF is always tangent internallyto the incircle of triangleABC.

I

D

EF

A

B C

Chapter 10

The regular pentagon

10.1 The golden ratio

The construction of the regular pentagon is based on the division of a segmentin thegolden ratio. 1 Given a segmentAB, to divide it in the golden ratio is toconstruct a pointP on it so that the area of the square onAP is the same as thatof the rectangle with sidesPB andAB, i.e.,

AP 2 = AB · PB.

A construction ofP is shown in the second diagram.

A BP BA

M

P

1In Euclid’sElements, this is called division into theextreme and mean ratio.

54 The regular pentagon

SupposePB has unit length. The lengthϕ of AP satisfies

ϕ2 = ϕ+ 1.

This equation can be rearranged as

(ϕ− 1

2

)2

=5

4.

Sinceϕ > 1, we have

ϕ =1

2

(√5 + 1

).

Note thatAP

AB=

ϕ

ϕ+ 1=

1

ϕ=

2√5 + 1

=

√5 − 1

2.

This explains the construction above.

10.2 The diagonal-side ratio of a regular pentagon

Consider a regular pentagonACBDE. It is clear that the five diagonals all haveequal lengths. Note that(1) ∠ACB = 108,(2) triangleCAB is isosceles, and(3) ∠CAB = ∠CBA = (180 − 108) ÷ 2 = 36.

In fact, each diagonal makes a36 angle with one side, and a72 angle withanother.

A

E D

BP

C

10.3 Construction of a regular pentagon with a given diagonal 55

It follows that(4) trianglePBC is isosceles with∠PBC = ∠PCB = 36,(5) ∠BPC = 180 − 2 × 36 = 108, and(6) trianglesCAB andPBC are similar.

Note that triangleACP is also isosceles since(7) ∠ACP = ∠APC = 72. This means thatAP = AC.

Now, from the similarity ofCAB andPBC, we haveAB : AC = BC : PB.In other wordsAB · AP = AP · PB, orAP 2 = AB · PB. This means thatPdividesAB in the golden ratio.

10.3 Construction of a regular pentagon with a givendiagonal

Given a segmentAB, we construct a regular pentagonACBDE with AB as adiagonal.(1) DivideAB in the golden ratio atP .(2) Construct the circlesA(P ) andP (B), and letC be an intersection of thesetwo circles.(3) Construct the circlesA(AB) andB(C) to intersect at a pointD on the sameside ofBC asA.(4) Construct the circlesA(P ) andD(P ) to intersect atE.

ThenACBDE is a regular pentagon withAB as a diameter.

A

E D

BP

C


10.4 Various constructions of the regular pentagon

(1) To construct a regular pentagon with vertices on a given circleO(A),(i) construct a radiusOP perpendicular toOA,(ii) mark the midpointM of OP and join it toA,(iii) bisect the angleOMA and let it intersectOA atQ,(iv) construct the perpendicular toOA atQ to intersect the circle atB andE,(v) markC,D on the circle such thatAB = BC andAE = ED.

ThenABCDE is a regular pentagon.

E

D C

B

A

PM

Q

O

(2) A compass-only costruction of the regular pentagon:2 construct(i) (O) = O(A),(ii) A(O) to markB andF on (O),(iii) B(O) to markC,(iv) C(O) to markD,(v) D(O) to markE,(vi) A(C) andD(AC) to intersect atX,(vii) Y = C(OX) ∩E(OX) inside(O),(viii) P = A(OY ) ∩ (O),(ix) Q = P (A) ∩ (O),(x) D(AP ) to intersect(O) atR andS (so thatR is betweenQ andD),(xi) S(R) to intersect(O) atT .

ThenAQRST is a regular pentagon inscribed in(O).

2Modification of a construction given by E. P. Starke,3 of a regular pentagon in 13 steps usingcompass only.

10.4 Various constructions of the regular pentagon 57

A

BC

D

E F

O

X

Y

P

Q

R

S

T

(3∗) Another construction of an inscribed regular pentagon. LetO(A) be agiven circle.(i) Construct an isosceles triangleAXY whose height is5

4of the radius of the

circle.(ii) ConstructA(O) intersectingO(A) atB ′ andC ′.(iii) Mark P = AX ∩OB′ andQ = AY ∩OC ′.(iv) JoinP andQ by a line intersecting the given circle atB andC.

ThenAB andAC are two sides of a regular pentagon inscribed in the circle.

C′B′

YX

QPCB

O

A


10.5 Some interesting constructions of the golden ra-tio

(1) Construction of36, 54, and72 angles. Each of the following constructionsbegins with the division of a segmentAB in the golden ratio atP .

36A

B

C

P36 36

36A

B

C

D

P

54

54

AB

C

P72

72

(2∗) Odom’s construction. LetD andE be the midpoints of the sidesAB andAC of an equilateral triangleABC. If the lineDE intersects the circumcircle ofABC atF , thenE dividesDF in the golden ratio.

A

B

D

C

EF

10.5 Some interesting constructions of the golden ratio 59

(3) Given a segmentAB, erect a square on it, and an adjacent one with baseBC. If D is the vertex aboveA, construct the bisector of angleADC to intersectAB atP . ThenP dividesAB in the golden ratio.

A B C

D

P

Proof. If AB = 1, then

AP : PC =1 :√

5,

AP : AC =1 :√

5 + 1

AC : 2AB =1 :√

5 + 1

AB : AP =√

5 + 1 : 2

=ϕ : 1.

This means thatP dividesAB in the golden ratio.

(4) Hofstetter’s parsimonious construction.4 Given a segmentAB, construct(i) C1 = A(B),(ii) C2 = B(A), intersectingC1 atC andD,(iii) the lineAB to intersectC1 atE (apart fromB),(iv) C3 = E(B) to intersectC2 at F (so thatC andF are on opposite sides ofAB),(v) the segmentCF to intersectAB atG.

Then the pointG divides the segmentAB in the golden section.

4K. Hofstetter, Another 5-step division of a segment in the golden section,Forum Geom., 4(2004) 21–22.


A B

C

D

E

F

G

C1

C2

C3

(5) Hofstetter’s rusty compass construction.5

(i) ConstructA(B) andB(A) intersecting atC andD.(ii) JoinCD to intersectAB at its midpointM .(iii) ConstructM(AB) to intersectB(A) atE on the same side ofAB asC.(iv) JoinDE to intersectAB atP .

ThenP divides the segment in the golden ratio.

A B

C

D

M P

E

5K. Hofstetter, Division of a segment into golden ratio with ruler and rusty compass,ForumGeom., 5 (2005) 133–134.

Chapter 11

The regular decagon

11.1 Constructions

(1) Successive completion of rhombi beginning with four adjacent36-rhombi.

(2) Another construction.

62 The regular decagon

11.2 Some relations among sides of regular polygonsinscribed in a circle

Given a circle(O), we denote bysn the length of a side of a regularn-gon in-scribed in the circle. Note that the radius of the circle is equal tos6.

Theorem 11.1 (Euclid). s6 + s10 : s6 = ϕ : 1

Proof. Consider a regular pentagonABCDE inscribed in a circle(O). LetF bethe midpoint of the arcAB (so thatAF is a side of a regular decagon inscribed in(O)). ExtendAF andOB to intersect atG.

B

C D

E

A

F

G

O

(1) TriangleOAF is isosceles, with angles(36, 72, 72).(2) TriangleGOA is also isosceles with base angles72.Therefore,F dividesGA in the golden ratio.Note also thatGF = FO = s6.

Theorem 11.2 (∗Euclid). s25 = s26 + s210.

Proof. LetABCDE be a regular pentagon inscribed in a circle(O). LetF be themidpoint of the arcAB so thatAF is a side of a regular decagon inscribed in thesame circle. LetP be the midpoint of the arcAF (so thatAP is a side of a regular20-gon inscribed in the circle). JoinOP to intersectAB atH.

(1) SinceF is the midpoint of the arcAB, triangleFAB is isosceles. TheradiusOP is the perpendicular bisector ofAF . SinceH lies on this radius, it isequidistant fromA andF . The triangleHAF is also isosceles. The two triangles

11.2 Some relations among sides of regular polygons inscribed in a circle 63

B

C D

E

A

F

P

O

H

FAB andHAF , sharing a common base angle atA, are similar. Therefore,AB : AF = AF : AH, and

AB × AH = AF 2 = s210.

(2) TriangleOAB is clearly isosceles, with angles72, 54, and54. In trian-gleHBO, clearly∠B = 54. Also, ∠BOH = 54, being 3

4of ∠BOA. There-

fore, it is also isosceles with the same base angle as triangleOAB. From thesimilarity ofOAB andHOB, we haveAB : OB = OB : HB, and

AB ×HB = OB2 = s26.

(3) The result follows from noting that

AB × AH + AB ×HB = AB2 = s25.

64 The regular decagon

Chapter 12

Construction of regular n-gons

12.1 Gauss’ theorem

The construction problem of regularn-gons was solved by Carl Friedrich Gauss(1777-1855), who first discovered in 1796 a construction of the regular17-gon,and later established the following theorem.

Theorem 12.1 (Gauss).A regular n-gon can be constructed with ruler and com-pass if and only if n is an integer of the form 2k · p1 · p2 · · · pk, where p1, p2, . . . pkare distinct prime numbers of the form 2m + 1.

If a number2m + 1 is prime, thenm = 2n for some integern. We writeFn = 22n

+ 1. The first 5 values ofFn are all prime numbers.

n 0 1 2 3 4Fn 3 5 17 257 65537

It is not known if there are other prime numbers of the formFn. For example,Leonhard Euler (1707-1783) has shown that

F5 = 225

+ 1 = 232 + 1 = 4294967297 = 641 × 6700417.

12.2 Construction of a regular15-gon

LetA1A2A3A4A5 be a regular pentagon inscribed in a circle(O). Construct equi-lateral trianglesA1A2B5, A2A3B5, A3A4B1, A4A5B2, A5A1B3 with the extra

66 Construction of regular n-gons

vertices inside the circle. Extend the sides of these equilateral triangles to inter-sect the circle again. The10 intersections, together the vertices of the regularpentagon, form the vertices of a regular15-gon inscribed in the circle.

A1

A2

A3

A4

A5

B4

B5

O B1

B2

B3

12.3 Construction of a regular17-gon

The following construction1 of a regular17-gon makes use of Gauss’ computation

cos360

17=

1

16

(√17 − 1 +

√34 − 2

√17

)

+1

8

√17 + 3

√17 −

√34 − 2

√17 − 2

√34 + 2

√17.

Consider a circle(O) with two perpendicular diametersPQ andRS.(1) On the radiusOR, mark a pointA such thatOA = 1

4OP .

(2) Construct the internal and external bisectors of angleOAP to intersect the line

1J. J. Callagy, The central angle of the regular17-gon,Math. Gaz., 67 (1983) 290–292.

12.3 Construction of a regular17-gon 67

OP atB andC respectively.(3) MarkD andE onPQ such thatCD = CA andBE = BA.(4) Mark the midpointM of QD.(5) MarkF onOS suchMF =MQ.(6) Construct the semicircle onOE and mark a pointG on it such thatOG = OF .(7) MarkH onOP such thatEH = EG, and construct the perpendicular toOPatH to intersect the circle atP1.

Then∠POP1 = 36017

, andP1 is a vertex of the regular17-gon adjacent toPon the circle.

A

BC OPQ

D EM

S

FG

H

P1

R

68 Construction of regular n-gons

Chapter 13

Regular solids

13.1 The faces of a regular solid

A regular solid is a convex solid whose faces are congruent regular polygons.Euclid has shown in hisElements that there are only five regular solids, and hegave explicit constructions of these solids.

Theorem 13.1. The faces of a regular solid can only be equilateral triangles,squares, or regular pentagons.

Proof. Around each vertex of a regular solid there are at least three faces, and thesum of the angles must be smaller than360. Therefore, each angle in the regularpolygonal faces is smaller than360

3

= 120. These faces can only be equilateraltriangles, squares, or regular pentagons.

If the faces are equilateral triangles, there are either 3, 4, or 5 triangles aroundeach vertex. If these are squares or regular pentagons, then there must be 3 facesaround each vertex.

Faces around a vertex regular solid(1) 3 equilateral triangles regular tetrahedron(2) 4 equilateral triangles regular octahedron(3) 5 equilateral triangles regular icosahedron(4) 3 squares cube(5) 3 regular pentagons regular dodecahedron

70 Regular solids

13.2 Edge lengths of inscribed regular solids

Given a sphere, we inscribe a regular solid of each type above. The length of anedge of the solid is related to the radius of the sphere as follows.

solid length of an edge

(1) regular tetrahedron st = 2√

63R

(2) regular octahedron so =√

2R

(3) regular icosahedron si =√

2(√

5−1)√5R

(4) cube sc = 2√

33R

(5) regular dodecahedronsd = 2√

33ϕR =

√3(√

5−1)12

R

The diagram below is a vertical section of a sphere with the edge lengths ofthe regular solids inscribed in the sphere.

sd

sc = ϕ · sd

so

st

si

N

S

O

C

T

D

I

Chapter 14

The regular tetrahedron

14.1 Construction

The regular tetrahedron has4 faces, which are congruent equilateral triangles. Ithas four vertices and six edges. An easy way to construct a regular tetrahedron isto fold an equilateral triangle along segments joining the midpoints of its sides.

CD D

D

A B

14.2 Measurements

14.2.1 The height of a regular tetrahedron

Let ABCD be a regular tetrahedron. We shall assume each edge of lengtha.Consider the equilateral triangleABC as the base of the pyramid, with center

72 The regular tetrahedron

O. The vertexD is vertically aboveO. In the right triangleADO, AD = a,AO = 1√

3a. Therefore,

DO2 = AD2 −AO2 = a2 − a2

3=

2

3a2 =

6

9a2.

The height of the regular tetrahedron ish = DO = 13

√6a.

A

D

C

B

O

M

K

Ω

14.2.2 The circumsphere of a regular tetrahedron

LetR be the radius of the sphere containing the four vertices of the regular tetrahe-dron. Suppose each side of the tetrahedron has lengtha. Its height ish = 1

3

√6a.

The radius of the circle containingABC is r = 13

√3a. If the circumsphere has

centerK and radiusR, then

AK = R, OK = h−R, AO = r,

and

R2 =(h− R)2 + r2 = h2 − 2hR+R2 + r2;

2hR =h2 + r2,

R =h2 + r2

2h=a2

2h=

32h2

2h=

3

4h.

The center of the circumsphere of a regular tetrahedron divides an altitude inthe ratio3 : 1.

14.3 The angle between a line and a plane 73

14.3 The angle between a line and a plane

The angle between a line and a plane is the angle between the line and its or-thogonal projection on the plane. If a line intersects a planeΠ at a pointA, wechoose a pointP on and construct from it the perpendicular toΠ to intersect theplane atQ. Q is called theorthogonal projection of P onΠ. The lineAQ is theorthogonal projection of onΠ, and∠PAQ is the angle between andΠ.

A

Q

P

Π

14.4 The angle between an edge and a face of a reg-ular tetrahedron

The angleΩ between an edge of a regular tetrahedron and a face not containing itis given by

cos Ω =r

a=

1√3.

This is approximately54.74.


14.5 The angle between two planes

Consider two planesΠ1 andΠ2 intersecting in a line. LetA be a point on. Oneach plane draw a line throughA perpendicular to. The angle between the twolines is the angle between the planesΠ1 andΠ2.

A Π1

Π2

14.6 The dihedral angle of a regular tetrahedron

The dihedral angleΨ of a regular tetrahedron is the angle between two faces. Thisis equal to angleOMD, whose cosine is clearly1

3. This is approximately70.53.

14.7 Inscribed regular tetrahedron 75

A

D

C

B

O

M

K

14.7 Inscribed regular tetrahedron

Note that if the height of a regular tetrahedron ish, then its circumradius isR =34h. This givesh = 2

3· 2R. Therefore, in a given sphere, we can inscribe a regular

tetrahedron by choosing an arbitrary point as a vertex. Regard this vertex as thenorth pole. The remaining three vertices are on the latitude circle which is twothirds of the diameter below the north pole.

st

N

S

O

T

Ψ

(1) If st is the length of a side of a regular tetrahedron inscribed in the sphere,

st =2√

2√3R.

(2) The latitude containing the opposite face of the north pole is90−Ψ south,


wherecos Ψ = 13.

14.8 The anglesΩ and Ψ

1

√2

√3

Ω ≈ 54.74

1

2√

23

Ψ ≈ 70.53

Ω

Chapter 15

The cube

15.1 The face diagonals and the space diagonals ofa cube

A cube has6 faces,8 vertices, and12 edges. The diagonals of a cube are of twokinds: the face diagonals and the space diagonals. If each side of a cube has lengtha, then(i) a face diagonal has length

√2a, and

(ii) a space diagonal has length√

3a.

Ω

The radius of the sphere containing the vertices of the cube isR =√

32a.

15.2 The angle between a space diagonal and a face

The angle between a space diagonal and a face of a cube is90 − Ω.

78 The cube

15.3 Inscribed cube

There are two natural ways to inscribe a cube in a given sphere.

15.3.1 Inscribed cube with two faces parallel to the equator

If two opposite faces are parallel to the equator, here is a vertical section throughtwo face diagonals on the latitude circles. The northern latitude circle can be deter-mined by erecting an equilateral triangle onEW (two antipodes on the equator).The center of the triangle is the center of the latitude circle.

O

Ω

N

S

OEW

Ω

sc

st

N

S

O

C

Ψ

T

Ω

15.3.2 Inscribed cube with a vertex at the north pole

If we put one vertex at the north pole, then the opposite vertex is at the south pole.The remaining6 vertices form antipodal two equilateral triangles on two latitudecircles. These two latitude circles divide the diameterNS into three equal parts.

Chapter 16

The square pyramid and the regularoctahedron

16.1 The square pyramid

Let ABCD be a square andE a point vertically above the centerO of ABCDsuch thatEAB, EBC, ECD, andEDA are equilateral triangles. The height ofE aboveO is the same asOA. This means that each slant edge of the squarepyramid makes and angle45 with the square base.

A B

CD

O

E

16.2 Dihedral angle between a face and the base

Note that in the right triangleEMO, OE : OM =√

2 : 1. This means that∠EMO = Ω.

80 The square pyramid and the regular octahedron

A B

CD

O

E

M

Ω

16.3 Dihedral angle between two adjacent equilat-eral triangles

If we take an identical copy of the square pyramind and identify the square faces,we obtain a solid with8 congruent equilateral triangles as faces. This is a regularoctahedron. The dihedral angle of each edge is clearly2Ω. This is the same forthe angle between two equilateral triangle faces of the square pyramid.

16.4 Inscribed regular octahedron

16.4.1 Inscribed regular octahedron with a vertex at the northpole

If a regular octahedron has a vertex at the north pole, its opposite vertex is at thesouth pole. The other four vertices form a square on the equator. If the sphere hasradiusR, the length of a side of an inscribed regular octahedron isso =

√2R.

16.4.2 Inscribed regular octahedron with a face parallel to theequator

If a face of a regular octahedron is parallel to the equator, say an equilateral tri-angleABC on the northθ-latitude circle, its opposite faceA′B′C ′ is an antipodalequilateral triangle at the southθ-latitude circle. The orthogonal projectionsX,Y ,Z ofA,B,C on the south latitude circle are such thatA′ZB′XC ′Y is a regularhexagon.

16.4 Inscribed regular octahedron 81

so

N

S

O

If the radius of the latitude circle isr, then(i) each side of the regular hexagonA′ZB′XC ′Y has lengthr,(ii) each side of the equilateral triangle (and hence each edge of the octahedron)has length

√3r,

and (iii) the separation of the two latitude circles is given by√

(√

3r)2 − r2 =√2r.

These are the same two latitudes to inscribe a cube with two faces parallel tothe equator.

r

√3r

C′ X

A

Ω

r

√2

2r

A

O

Ω

N

S

OEW

A

Ω

82 The square pyramid and the regular octahedron

Chapter 17

The regular icosahedron

17.1 Construction

In Book XIII of the Elements, Euclid shows how to inscribe a regular icosahedronwhose faces are congruent equilateral triangles with vertices on a given sphere.

si

N

S

I

O

In a vertical great circle of the sphere, inscribe a rectangle of dimensions2 : 1.Place a regular pentagon on the latitude circle at levelI along the longer side ofthe rectangle. These vertices, their antipodes on the sphere, and the north andsouth poles, form the12 vertices of a regular icosahedron inscribed in the sphere.There are20 faces and 30 edges. If the sphere has radiusR, each side of theinscribed regular icosahedron has length

si =

√2(√

5 − 1)√5

R.

84 The regular icosahedron

17.2 Dihedral angle of the regular icosahedron

Let A1A2A3A4A5 be a regular pentagon on the latitude circle throughI. If Mis the midpoint of the sideNA1, then the angleA2MA5 is the dihedral angle ofthe two faces containing the edgeNA1. If each side of the regular pentagon haslengths, thenA2A5 = ϕ · s. The dihedral angle is equal to angleA2PA5 in thediagram below. This is approximately138.183.

A1

P

A2 A5

Chapter 18

The regular dodecahedron

18.1 Euclid’s construction of the regular dodecahe-dron

18.1.1 Two golden points above a mid-line of a square

Given a square with centerO, letA andA′ be the midpoints of two opposite sidesof the square. We callAA′ a mid-line of the square. Divide the segmentOA(respectivelyOA′) in the golden ratio at a pointB (respectivelyB ′), and erecta square on the segmentOB (respectivelyOB ′), in a plane perpendicular to thesquare. Denote the vertex of the square opposite toO by P (respectivelyP ′). WechooseP andP ′ on the same side of the square, and call them the two goldenpoints above the mid-lineAA′ of the square.

A

B O B′A′

P K P ′

18.1.2 The pentagonal faces

Consider a cube with a mid-line on each of its six faces:

86 The regular dodecahedron

Faces Mid-linetop and bottom left-rightleft and right front-rearfront and rear top-bottom

Construct two golden points above the indicated mid-line of each face, allexternally of the cube. Euclid showe that the8 vertices of the cube, together withthe 12 golden points constructed above the faces, form the vertices of a regulardodecahedron.

Theorem 18.1 (Euclid). The two vertices common to the top and the front faces,the two golden points above the left-right mid-line of the top face, and the “top”golden point above the top-bottom mid-line of the front face form a planar regularpentagon.

XY

A B O

P K P ′

C

D

O′

Q

Proof. AssumeOB = 1 so thatOA = ϕ. The lengths of the various segmentsare very easily related to these two.

(1) We first show that the triangleXQY and the quadrilateralXY P ′P arecoplanar. Note that the pointsQ,D C,OK,K are on the same vertical plane, and(i) tanOCK = OK

OC= 1

ϕ= ϕ− 1,

(ii) tanDQC = DCDQ

= ϕ−11

= ϕ− 1.This shows thatQ, C,K are along the same line, and the two polygons are copla-nar.

18.2 Dihedral angle between two adjacent faces of a regular dodecahedron87

(2) Note that

XP 2 =XB2 +BP 2 = AB2 +OC2 +BP 2

=OA2 + AB2 +OB2

=ϕ2 + (ϕ− 1)2 + 12

=2ϕ2 − 2ϕ+ 2

=2(ϕ+ 1) − 2ϕ+ 2 = 4.

This meansXP = 2. By symmetry,Y P ′ = XQ = Y Q = 2 = PP ′.Therefore, the pentagonQXPP ′Y is equilateral.

(3) A similar calculation shows thatPY = XY = XP ′. Therefore, triangleXPP ′ andY PP ′ are isosceles triangles with top angles108.

(4) Now it is easy to see that∠PXQ andP ′Y Q are also108.This completes the proof thatQXPP ′Y is a regular pentagon.

Theorem 18.1 holds if we replace the front face replaced by the rear face.These two regular pentagons share a common edge, namely, the segment joiningthe two golden points above (the left-right mid-line of) the top face. Similarly,above the indicated mid-line of each face, there is a pair of regular pentagons.This gives a regular solid with12 faces which are regular pentagons. It has30edges.

Corollary 18.2. The length of a side of a regular dodecahedron inscribed in agiven sphere is

sd =1

ϕ· sc,

where sc is the length of a side of an inscribed cube.

18.2 Dihedral angle between two adjacent faces of aregular dodecahedron

The dihedral angle between two adjacent faces of a regular dodecahedron is2Φ ≈116.565. In the diagram below,CC ′ is the front-rear mid-line of the top face, andK is the midpoint of the segmentPP ′ above the left-right mid-line of the sameface.

88 The regular dodecahedron

1

2ϕC C′

K

O

ΦΦ

18.3 Inscribed regular dodecahedron with a vertexat the north pole

Suppose each face of an inscribed dodecahedron is a regular pentagon of sidelengths and diagonald. If one vertex of a regular dodecahedron is at the northpole, its opposite vertex is at the south pole, the remaining18 vertices lie on 4latitude circles, the3 neighbors of the north pole on one latitude, and6 on a lowerlatitude each at a distanced from the north pole.

sd

ϕ · sd

N

S

O

C

T

D

(1) On the latitude circle at the level ofC, which is one-third of the diameterbelow the north pole, inscribe a hexagon whose alterate sides have two lengthsin the golden ratio. This hexagon can be constructed as follows. First inscribean equilateral triangle in the circle. Divide the sides of the triangle in the goldenratio. Join each vertex to the point of division on its opposite side to meet thecircle again. These six points define the hexagon on the latitude circle.

(2) On the latitude circle (levelD) of radius 23R, construct an equilateral tri-

angle whose vertices are on the meridians of the midpoints of the longer sides ofthe hexagon on the latitude at levelC.

18.3 Inscribed regular dodecahedron with a vertex at the north pole 89

C

(3) The 9 points on these two latitudes circles, together with the antipodalpoints, and the north and south poles, form the vertices of a regular dodecahedroninscribed in the given sphere.

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