FORUM GEOM ISSN 1534-1178
The M-Configuration of a Triangle
Alexei Myakishev
Abstract. We give an easy construction of pointsAa,Ba,Caon the sides of a triangleABCsuch that the figure M pathBCaAaBaCconsists of 4 segments of equal lengths. We study the configuration consisting of the three figures M of a triangle, and define an interesting mapping of triangle centers associated with such an M-configuration.
1. Introduction
Given a triangleABC, we consider pointsAa on the lineBC,Baon the half lineCA, andCaon the half lineBAsuch thatBCa=CaAa=AaBa=BaC. We shall refer toBCaAaBaC as Ma, because it looks like the letter M when triangle ABC is acute-angled. See Figures 1a. Figure 1b illustrates the case when the triangle is obtuse-angled. Similarly, we also have Mb and Mc. The three figures Ma, Mb, Mcconstitute the M-configuration of triangleABC. See Figure 2.
Aa
Ba
Ca
A
B C
Figure 1a
A
B C
Aa
Ca
Ba
Figure 1b
Aa
Ba
Ca
Ab
Bb
Cb
Ac
Bc
Cc
A
B C
Figure 2
Proposition 1. The linesAAa,BBa,CCaconcur at the point with homogeneous barycentric coordinates
1
cosA : 1
cosB : 1 cosC
.
Proof. Letla be the length of BCa = CaAa = AaBa = BaC. It is clear that the directed length BAa = 2lacosB andAaC = 2lacosC, andBAa : AaC = cosB : cosC. For the same reason, CBb : BbA = cosC : cosA and ACc : CcB = cosA: cosB. It follows by Ceva’s theorem that the linesAAa,BBa,CCa
concur at the point with homogeneous barycentric coordinates given above.1
Publication Date: June 30, 2003. Communicating Editor: Paul Yiu.
The author is grateful to the editor for his help in the preparation of this paper.
1This point appears in [3] asX92.
Remark. Since2lacosB+2lacosC=a= 2RsinA, whereRis the circumradius of triangleABC,
la= a
2(cosB+ cosC) = RsinA
cosB+ cosC = RcosA2
cosB−2C. (1) For later use, we record the absolute barycentric coordinates ofAa,Ba,Cain terms ofla:
Aa =2la
a (cosC·B+ cosB·C), Ba=1
b(la·A+ (b−la)C), (2) Ca=1
c(la·A+ (c−la)B). 2. Construction of Ma
Proposition 2. LetA be the intersection of the bisector of angleAwith the cir- cumcircle of triangleABC.
(a)Aais the intersection ofBCwith the parallel toAAthrough the orthocenter H.
(b) Ba (respectively Ca) is the intersection of CA (respectively BA) with the parallel toCA(respectivelyBA) through the circumcenterO.
O
Aa
Ba
Ca
A
B C
H
A
Figure 3a
A
B C
Aa
Ca
Ba O
H
A
Figure 3b
Proof. (a) The line joiningAa= (0 : cosC: cosB)toH = a
cosA: cosbB : coscC has equation
0 cosC cosB
cosaA b
cosB c
cosC
x y z
= 0.
This simplifies to
−(b−c)xcosA+a(ycosB−zcosC) = 0. It has infinite point
(−a(cosB+ cosC) :acosC−(b−c) cosA: (b−c) cosA+acosB)
=(−a(cosB+ cosC) :b(1−cosA) :c(1−cosA)).
It is clear that this is the same as the infinite point(−(b+c) : b :c), which is on the line joiningAto the incenter.
O
Aa
Ca
Ba
A
B C
A M Y Z
Figure 4
(b) LetM be the midpoint ofBC, andY,Z the pedals ofBa,CaonBC. See Figure 4. We have
OM =a
2cotA=la(cosB+ cosC) cotA, CaZ=lasinB,
MZ=a
2 −lacosB =la(cosB+ cosC)−lacosB =lacosC.
From this the acute angle between the lineCaOandBChas tangent ratio CaZ−OM
MZ =sinB−(cosB+ cosC) cotA cosC
=sinBsinA−(cosB+ cosC) cosA cosCsinA
=−cos(A+B)−cosCcosA
cosCsinA = cosC(1−cosA) cosCsinA
=1−cosA
sinA = tanA 2.
It follows thatCaOmakes an angle A2 with the lineBC, and is parallel toBA. The same reasoning shows thatBaOis parallel toCA.
3. Circumcenters in the M-configuration
Note that∠BaAaCa = ∠A. It is clear that the circumcircles ofBaAaCaand BaACaare congruent. The circumradius is
Ra= la 2 sinπ
2 −A2 = la
2 cosA2 = R
2 cosB−2C (3) from (1).
Proposition 3. The circumcircle of triangleABaCacontains (i) the circumcenter O of triangle ABC, (ii) the orthocenter Ha of triangle AaBaCa, and (iii) the midpoint of the arcBAC.
Proof. (i) is an immediate corollary of Proposition 2(b) above.
I
Aa
Ca
Ba
Ha
A
B C
O A
Figure 5
(ii) LetHabe the orthocenter of triangleAaBaCa. It is clear that
∠BaHaCa=π−∠BaAaCa=π−∠BAC =π−∠CaABa.
It follows that Ha lies on the circumcircle of ABaCa. See Figure 5. Since the triangleAaBaCais isosceles,BaHa=CaHa, and the pointHalies on the bisector of angleA.
(iii) LetAbe the midpoint of the arcBAC. By a simple calculation,∠AAO=
π2−12|B−C|. Also,∠ACaO = π2 +12|B−C|.2This shows thatAalso lies on
the circleABaOCa.
The pointsBa andCaare therefore the intersections of the circle OAA with the sidelinesACandAB. This furnishes another simple construction of the figure Ma.
2This isC+A2 ifC≥BandB+A2 otherwise.
Remarks. (1) If we take into consideration also the other figures Mb and Mc, we have three trianglesABaCa,BCbAb,CAcBcwith their circumcircles intersecting atO.
(2) We also have three triangles AaBaCa, AbBbCb,AcBcCc with their ortho- centers forming a triangle perspective withABCat the incenterI.
Proposition 4. The circumcenter Oa of triangle AaBaCa is equidistant from O andH.
Aa
Ca
Ba
A
B C
O
H N
Q P
H
M
X Y
Z
Figure 6
Proof. Construct the circle through OandH with center Qon the lineBC. We prove that the midpointP of the arcOHon the opposite side ofQis the circum- centerOaof triangle AaBaCa. See Figure 6. It will follow thatOais equidistant fromO andH. LetN be the midpoint of OH. Suppose the line P Qmakes an angleϕwithBC. LetX,Y, andM be the pedals ofH,N,Oon the lineBC.
SinceH,X,Q,N are concyclic, and the diameter of the circle containing them isQH = sinNXϕ = 2 sinRϕ. This is the radius of the circleOP H.
By symmetry, the circleOP H contains the reflectionHofHin the lineBC.
∠HHP = 1
2∠HQP = 1
2∠HQN = 1
2∠HXN = 1
2|B−C|.
Therefore, the angle between HP andBC is π2 − 12|B −C|. It is obvious that the angle betweenAaOaandBCis the same. But from Proposition 2(a), the angle betweenHAaandBCis the same too, so is the angle between the reflectionHAa andBC. From these we conclude thatH,Aa,OaandP are collinear. Now, letZ be the pedal ofP onBC.
AaP = P Z
cos12(B−C) = QPsinϕ
cos12(B−C) = R
2 cos12(B−C) =Ra. Therefore,P is the circumcenterOaof triangleAaBaCa.
Applying this to the other two figures Mb and Mc, we obtain the following re- markable theorem about the M-configuration of triangleABC.
Theorem 5. The circumcenters of triangles AaBaCa,AbBbCb, andAcBcCc are collinear. The line containing them is the perpendicular bisector of the segment OH.
Aa
Ba
Ca
Ab
Bb
Cb
Ac
Bc
Cc
A
B C
Oa
Ob
Oc
O
H
N
Figure 7
One can check without much effort that in homogeneous barycentric coordi- nates, the equation of this line is
sin 3A
sinA x+sin 3B
sinB y+sin 3C sinC z= 0. 4. A central mapping
LetP be a triangle center in the sense of Kimberling [2, 3], given in homoge- neous barycentric coordinates (f(a, b, c) : f(b, c, a) : f(c, a, b))where f = fP satisfies f(a, b, c) = f(a, c, b). If the reference triangle ABC is isosceles, say, withAB=AC, thenP lies on the perpendicular bisector ofBC and has coordi- nates of the form(gP : 1 : 1). The coordinate gdepends only on the shape of the isosceles triangle, and we express it as a function of the base angle. We shall call g =gP the isoscelized form of the triangle center functionfP. LetP∗ denote the isogonal conjugate ofP.
Lemma 6. gP∗(B) = 4 cosgP(B2B) .
Proof. IfP = (gP(B) : 1 : 1)for an isosceles triangleABCwithB =C, then P∗=
sin2A
gP(B) : sin2B : sin2B
=
4 cos2B gP(B) : 1 : 1
sincesin2A= sin2(π−2B) = sin22B = 4 sin2Bcos2B. Here are some examples.
Center fP gP
centroid 1 1
incenter a 2 cosB
circumcenter a2(b2+c2−a2) −2 cos 2B orthocenter b2+c12−a2 −2 coscos 2B2B
symmedian point a2 4 cos2B
Gergonne point s−1a 1−coscosBB Nagel point s−a 1−coscosBB Mittenpunkt a(s−a) 2(1−cosB) Spieker point b+c 1+2 cos2 B
X55 a2(s−a) 4 cosB(1−cosB)
X56 sa−2a 1−cos4 cos3BB
X57 s−aa 1−cos2 cos2BB
Consider a triangle center given by a triangle center function with isoscelized formg=gP. The triangle center of the isosceles triangleCaBAais the pointPa,b with coordinates (g(B) : 1 : 1) relative toCaBAa. Making use of the absolute barycentric coordinates ofAa,Ba,Cagiven in (2), it is easy to see that this is the point
Pa,b=
g(B)la
c : g(B)(c−la)
c + 1 +2la
a cosC: 2la a cosB
. The same triangle center of the isosceles triangleBaAaCis the point
Pa,c=
g(C)la b : 2la
a cosC: g(C)(b−la) b +2la
a cosB+ 1
. It is clear that the linesBPa,bandCPa,cintersect at the point
Pa=
g(B)g(C)la2
bc : 2g(B)la2cosC
ca : 2g(C)la2cosB ab
= (ag(B)g(C) : 2bg(B) cosC : 2cg(C) cosB)
=
ag(B)g(C)
2 cosBcosC :bg(B)
cosB : cg(C) cosC
.
Figure 8 illustrates the case of the Gergonne point.
In the M-configuration, we may also consider the same triangle center (given in isoscelized formgP of the triangle center function) in the isosceles triangles . These are the pointPb,c,Pb,a,Pc,a,Pc,b. The pairs of linesCPb,c,APb,aintersect- ing atPbandAPc,a,BPc,bintersecting atPc. The coordinates ofPb andPccan be
Aa
Ba
Ca
Pa,b
Pa,c
Pa
A
B C
Figure 8
written down easily from those ofPa. From these coordinates, we easily conclude that thatPaPbPc is perspective with triangleABCat the point
Φ(P) =
agP(A)
cosA : bgP(B)
cosB : cgP(C) cosC
= (gP(A) tanA:gP(B) tanB:gP(C) tanC). Proposition 7. Φ(P∗) = Φ(P)∗.
Proof. We make use of Lemma 6.
Φ(P∗) =(gP∗(A) tanA:gP∗(B) tanB:gP∗(C) tanC)
=
4 cos2A
gP(A) tanA: 4 cos2B
gP(B) tanB : 4 cos2C gP(C) tanC
=
sin2A
gP(A) tanA : sin2B
gP(B) tanB : sin2C gP(C) tanC
=Φ(P)∗.
We conclude with some examples.
P Φ(P) P∗ Φ(P∗) = Φ(P)∗ incenter incenter
centroid orthocenter symmedian point circumcenter
circumcenter X24 orthocenter X68
Gergonne point Nagel point X55 X56
Nagel point X1118 X56 X1259 =X1118∗
Mittenpunkt X34 X57 X78=X34∗
For the Spieker point, we have Φ(X10) =
tanA
1 + 2 cosA : tanB
1 + 2 cosB : tanC 1 + 2 cosC
=
1
a(b2+c2−a2)(b2+c2−a2+bc) :· · ·:· · ·
.
This triangle center does not appear in the current edition of [3].
Remark. ForP =X8, the Nagel point, the pointPahas an alternative description.
Antreas P. Hatzipolakis [1] considered the incircle of triangleABC touching the sidesCAandAB atY andZ respectively, and constructed perpendiculars from Y,ZtoBCintersecting the incircle again atYandZ. See Figure 9. It happens thatB,Z,Pa,bare collinear; so areC,Y,Pa,c. Therefore,BZandCYintersect atPa. The coordinates ofYandZ are
Y=(a2(b+c−a)(c+a−b) : (a2+b2−c2)2: (b+c)2(a+b−c)(c+a−b)), Z=(a2(b+c−a)(a+b−c) : (b+c)2(c+a−b)(a+b−c) : (a2−b2+c2)2).
Aa
Ba
Ca
A
B X C
Z
Y
Z
Y Pa,b
Pa,c
Pa
Figure 9
The linesBZandCYintersect at
Pa=
a2(b+c−a) :(a2+b2−c2)2
c+a−b :(a2−b2+c2)2 a+b−c
=
a2(b+c−a)
(a2−b2+c2)2(a2+b2−c2)2 : 1
(c+a−b)(a2−b2+c2)2 : 1
(a+b−c)(a2+b2−c2)2
.
It was in this context that Hatzipolakis constructed the triangle center X1118 =
1
(b+c−a)(b2+c2−a2)2 :· · ·:· · ·
. References
[1] A. P. Hatzipolakis, Hyacinthos message 5321, April 30, 2002.
[2] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1 – 285.
[3] C. Kimberling, Encyclopedia of Triangle Centers, May 23, 2003 edition available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
Alexei Myakishev: Smolnaia 61-2, 138, Moscow, Russia, 125445 E-mail address:alex geom@mtu-net.ru
