circle and triangle olympiad lamoen

The Lamoen circleDarij Grinberg

Let ABC beanarbitrarytriangle, Ma, Mb, Mc themidpointsof its sidesBC, CA, AB,andS its centroid, i. e. theintersectionof thelinesAMa, BMb andCMc (Fig. 1). Wegetsixtriangles: AMbS, CMbS, CMaS, BMaS, BMcS andAMcS. Thesetriangleshavesomeinterestingproperties. At first, their areasareequal. Theareaof eachoneof thesetriangleswill bedenotedby k.

A

B

C

Ma

Mb

Mc

S

Fig. 1Anotherinterestingproperty, which turnedout to bea theoremof FloorvanLamoen, is

thatthecircumcentersof thesesix trianglesareconcyclic(Fig. 2). Moreprecisely:Theorem 1: Let Ab, Cb, Ca, Ba, Bc, Ac bethecircumcentersof trianglesAMbS, CMbS,

CMaS, BMaS, BMcS, AMcS. Then, Ab, Cb, Ca, Ba, Bc, Ac lie ononecircle (Fig. 2).

A

B

C

Ma

Mb

Mc

S

Ab

AcCb

Ca

BcBa

Fig. 2After hisdiscoverer, I call this circle theLamoen circle of ABC.Hereis a half-syntheticalproof of Theorem1 (Fig. 3). RegardthecircumcentersBa and

Bc; theybothlie on theperpendicularbisectorof thesegmentBS. Hence, BaBc µ BS. On theotherhand, thecircumcentersAb andCb bothlie on theperpendicularbisectorof thesegmentSMb, hence, AbCb µ SMb. For BS andSMb arethesameline, wehaveBaBc 5 AbCb. Analogously, weshowthatAcAb 5 CaBa andCbCa 5 BcAc. Therefore, theoppositesidesof thehexagonAbAcBcBaCaCb arerespectivelyparallel.

A

B

C

Ma

Mb

Mc

S

Ab

Ac CbCa

BcBa

Fig. 3Now wehavethefollowing theorem([1] Aufgabe34; [4] problem109; [5] problem

131):Theorem 2: A hexagon, whoseoppositesidesarerespectivelyparallel, andwhosemain

diagonalsareof equallength, hasa circumcircle.Thus, in orderto showthatthehexagonAbAcBcBaCaCb hasa circumcircle, wemust

prove:

AbBa � AcCa � BcCb.

Wewill calculateAcCa aftertheCosineLaw in triangle AcSCa; but for this aim wemustknow thetwo othersidesandtheoppositeangle. ThesideAcS is thecircumradiusof AMcS; sowehave

k � AS � SMc � McA4 � AcS

� AS � 12 CS � 1

2 c4 � AcS

� AS � CS � c16 � AcS

,

hence

AcS � AS � CS � c16 � k

.

Analogously,

CaS � AS � CS � a16 � k

.

A

B

C

Ma

Mb

Mc

SAc

Ca

Fig. 4Now wewill calculate1AcSCa. (Ourargumentsdependon thearrangementof pointson

Fig. 4, but canbedoneanalogouslyfor otherpositions.) In theisosceles AAcS, wehave

1AcSA � 90° " 121AAcS

� 90° " 1AMcS (centralangle),

andsimilarly1CaSC � 90° " 1SMaC. Thus,

1AcSCa � 1AcSA �1ASC �1CaSC

� �90° " 1AMcS  �1ASC � �90° " 1SMaC � �180° " 1AMcS " 1SMaC  �1ASC

� �180° " 1AMcS " 1SMaC  � �180° " 1McSA � �180° " 1AMcS " 1McSA  � �180° " 1SMaC � 1McAS � �180° " 1SMaC � 1BAMa �1SMaB

� 1BAMa �1AMaB � 180° " *.

Now, wecanapplytheCosineLaw to AcSCa:

AcCa2 � AcS

2 � CaS2 " 2 � AcS � CaS � cos1AcSCa

� AS � CS � c16 � k

2 � AS � CS � a16 � k

2

" 2 � AS � CS � c16 � k

� AS � CS � a16 � k

� cos�180° " * � AS � CS

16 � k

2 � �c2 � a2 " 2ca � cos�180° " *  � AS � CS

16 � k

2 � �c2 � a2 � 2cacos* � AS � CS

16 � k

2 � �2 � BMb 2 (aftera formulafor a trianglemedian)

� AS � CS16 � k

2 � 2 � 32

� BS2

� AS � CS16 � k

2 � �3 � BS 2 � 316

� AS � BS � CSk

2,

therefore

AcCa � 316

� AS � BS � CSk

.

Analogously, onegetsthesameexpressionfor AbBa andBcCb, andtheequationAbBa � AcCa � BcCb is proven!

References[1] H. Dörrie: Mathematische Miniaturen, Wiesbaden1969.[2] D. O. Shkljarskij, N. N. Chenzov, I. M. Jaglom: Izbrannye zadachi i teoremy

elementarnoj matematiki: Chastj 2 (Planimetrija), Moscow1952.[3] D. O. Shkljarskij, N. N. Chenzov, I. M. Jaglom: Izbrannye zadachi i teoremy

planimetrii, Moscow1967.