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.