Triangle form by perpendicular bisector

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psi241

49 posts

psi241

#1 h Jul 17, 2019, 12:19 PM • 9 Y

Y by v_Enhance, samuel, GioOrnikapa, Adventure10, Rounak_iitr, Tastymooncake2, Funcshun840, cubres, ehuseyinyigit

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$ . A line $\ell$ intersects the lines $AI$ , $BI$ , and $CI$ at points $D$ , $E$ , and $F$ , respectively, distinct from the points $A$ , $B$ , $C$ , and $I$ . The perpendicular bisectors $x$ , $y$ , and $z$ of the segments $AD$ , $BE$ , and $CF$ , respectively determine a triangle $\Theta$ . Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$ .

This post has been edited 1 time. Last edited by psi241, Jan 7, 2020, 5:27 AM

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v_Enhance

6870 posts

v_Enhance

#2 h Jul 17, 2019, 1:02 PM • 22 Y

Y by stroller, JustKeepRunning, Pluto04, samuel, amar_04, Siddharth03, Limerent, v4913, menpo, Lcz, PRMOisTheHardestExam, megarnie, centslordm, GioOrnikapa, motannoir, Adventure10, bhan2025, Rounak_iitr, Tastymooncake2, Funcshun840, JelaByteEngineer, MS_asdfgzxcvb

We let $\ell_a$ , $\ell_b$ , $\ell_c$ denote the reflections of $\ell = \overline{DEF}$ across the perpendicular bisectors of $\overline{AD}$ , $\overline{BE}$ , $\overline{CF}$ .

Claim: The lines $\ell_a$ , $\ell_b$ , $\ell_c$ intersect at a point $T$ on $\omega$ .

Proof. This is direct angle chasing: since $\begin{align*} \measuredangle(\ell_a, \ell_b) &= \measuredangle(\ell_a, \ell) + \measuredangle(\ell, \ell_b) = 2\measuredangle(\ell_a, \overline{AI}) + 2\measuredangle(\overline{BI}, \ell_b) \\ &= 2\measuredangle(\ell_a, \ell_b) + 2\measuredangle(\overline{BI}, \overline{AI}) \\ \implies \measuredangle(\ell_a, \ell_b) &= 2 \measuredangle AIB = \measuredangle ACB \end{align*}$ it follows that $\ell_a$ and $\ell_b$ meet on $\omega$ , as desired. $\blacksquare$

Let line $TI$ meet $\ell$ at $H$ . Then, consider a homothety centered at $T$ which maps $I$ to $H$ . It then maps $\triangle ABC$ to a triangle $\triangle PQR$ , say, whose circumcircle $\gamma$ is tangent to $\triangle ABC$ at $T$ , and whose incenter is $H$ .
$[asy]size(250); defaultpen(fontsize(11pt)); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair T = dir(17); pair I = incenter(A, B, C); pair hom(pair P) { return 1.4*P - 0.4*T; } pair H = hom(I); pair P = hom(A); pair Q = hom(B); pair R = hom(C); pair D = extension(A, I, H, H+dir(I-A)*dir(I-A)/dir(P-A)); pair E = extension(D, H, B, I); pair F = extension(D, E, C, I); filldraw(A--B--C--cycle, invisible, heavycyan); draw(circumcircle(A, B, C), heavycyan); filldraw(P--Q--R--cycle, invisible, heavygreen); draw(circumcircle(P, Q, R), heavygreen); pair U = extension(H, D, P, A); draw(H--T, blue); draw(P--T, pink); draw(Q--T, pink); draw(R--T, pink); draw(E--U, red); draw(A--I, grey); draw(E--I, grey); draw(F--C, grey); draw(P--H, grey); pair Z = hom(dir(40)); pair Y = hom(dir(160)); pair X = hom(dir(270)); draw(X--Y--Z--cycle, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(255)); dot("$C$", C, dir(295)); dot("$T$", T, dir(T)); dot("$I$", I, dir(280)); dot("$H$", H, dir(320)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$D$", D, dir(170)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot(U); dot(Z); dot(Y); dot(X); /* TSQ Source: A = dir 110 B = dir 210 R255 C = dir 330 R295 T = dir 17 I = incenter A B C R280 ! pair hom(pair P) { return 1.4*P - 0.4*T; } H = hom(I) R320 P = hom(A) Q = hom(B) R = hom(C) D = extension A I H H+dir(I-A)*dir(I-A)/dir(P-A) R170 E = extension D H B I F = extension D E C I A--B--C--cycle 0.1 lightcyan / heavycyan circumcircle A B C heavycyan P--Q--R--cycle 0.1 lightgreen / heavygreen circumcircle P Q R heavygreen U .= extension H D P A H--T blue P--T pink Q--T pink R--T pink E--U red A--I grey E--I grey F--C grey P--H grey Z .= hom(dir(40)) Y .= hom(dir(160)) X .= hom(dir(270)) X--Y--Z--cycle red */ [/asy]$
Recall that $\ell \equiv \overline{HD}$ is the reflection of line $\overline{APT}$ across $x$ . As $\overline{HP} \parallel \overline{AD}$ , $x$ is also the perpendicular bisector of $\overline{PH}$ . Similarly, $y$ and $z$ are the perpendicular bisectors of $\overline{QH}$ and $\overline{RH}$ . Since $H$ is the incenter of $\triangle PQR$ , this implies that our perpendicular bisectors determine a triangle with circumcircle $\gamma$ , as the meet at the arc midpoints of $\gamma$ . This is the desired result.

Remark: For me, most of the difficulty in this solution is identifying all the relevant points $T$ , $P$ , $Q$ , $R$ , $H$ , which can be found from an accurately drawn diagram.

This post has been edited 1 time. Last edited by djmathman, Jul 18, 2019, 3:43 PM
Reason: changed diagram a bit

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Dr.Sop

206 posts

Dr.Sop

#3 h Jul 17, 2019, 1:17 PM • 3 Y

Y by nobodyknowswhoIam, Infinityfun, Adventure10

Let $AI, BI, CI$ meet the circumcircle of $ABC$ at $A', B', C'$ . To prove this problem it is enough to verify that the vertices $XYZ$ of $\Theta$ obey the property that $A'X$ , $B'Y$ , $C'Z$ concur on the circumcircle of $ABC$ . Let $X'Y'Z'$ be triangle formed with the perpendicular lines from $D, E, F$ to $AI, BI, CI$ . One can get that $|IX'| = 2|A'X|$ and $IX'||A'X$ . From similar properties and some angle chase we conclude that $A'X, B'Y, C'Z$ meet on the circumcircle of $ABC$ .

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TelvCohl

2312 posts

TelvCohl

#6 h Jul 17, 2019, 1:57 PM • 15 Y

Y by rmtf1111, Hamel, Vietjung, RudraRockstar, Centralorbit, a_simple_guy, Aryan-23, PRMOisTheHardestExam, centslordm, Iceternity, sabkx, Adventure10, Mango247, Tastymooncake2, MS_asdfgzxcvb

Let $T$ be the pole of the Simson line of $\triangle ABC$ with the direction $\perp \ell,$ $T_a, T_b, T_c$ be the intersection of $AI, BI, CI$ with the parallel $\tau$ from $T$ to $\ell$ and let $\triangle XYZ$ be the triangle $\Theta,$ then note that the isogonal conjugate of $T$ WRT $\triangle ABC$ is the point at infinity with the direction $\parallel \ell$ we get $T$ lies on the perpendicular bisector of $BT_b, CT_c,$ so consider the homothety $\mathbf{H}^{B}_{\frac{1}{2}}, \mathbf{H}^{C}_{\frac{1}{2}}$ we conclude that $TX$ passes through the midpoint $M_a$ of arc $BC$ in $\odot (ABC)$ and $\frac{TM_a}{XM_a} = \frac{\text{dist} (I, \tau)}{\text{dist} (I, \ell)}.$

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rmtf1111

698 posts

rmtf1111

#7 h Jul 17, 2019, 7:25 PM • 5 Y

Y by SerdarBozdag, Adventure10, Mango247, Mango247, Tastymooncake2

Let $A_1=y\cap z,$ $B_1=x\cap z,$ and $C_1=x\cap y.$
Let the reflections $l_a$ and $l_b$ of $l$ in $x$ and $y,$ resspectively meet at $Q.$
$\measuredangle{BQA}=\measuredangle(l_b,l_a)=\measuredangle{AIB}-\measuredangle(AI,l_a)-\measuredangle(BI,l_b)=\measuredangle{AIB}-\measuredangle(l,AI)-\measuredangle(l,BI)=-2\measuredangle{BIA}=\measuredangle{BCA}$ Hence, $Q$ lies on $\omega,$ which implies that $l_c,$ the reflection of $l$ in $z$ also passes through $Q.$ Let $\delta(X,m)$ denote the distance from point $X$ to line $m.$
$\delta(C_1, QB)=\delta(C_1,l_b)=\delta(C_1,l)=\delta(C_1,l_a)=\delta(C_1,AQ)$ Thus, $C_1$ lies on the angle bisector of $\angle{AQB}.$ As $A_1B_1C_1$ and the triangle defined by the midpoint of arcs $BC, AC$ and $AB$ are homothetic, this implies that $Q$ is their homothety center, and since $Q$ lies on $\omega \implies$ circle $QA_1B_1C_1$ is tangent to $\omega.$

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jdhua

7 posts

jdhua

#8 h Jul 20, 2019, 7:26 PM • 8 Y

Y by DVDthe1st, Elyson, sabkx, Adventure10, Mango247, Tastymooncake2, Funcshun840, ehuseyinyigit

This was proposed by Mads Christensen, Denmark and lost to G6 by one vote.

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khina

993 posts

khina

#9 h Jul 20, 2019, 11:02 PM • 7 Y

Y by centslordm, MarkBcc168, Adventure10, Mango247, Tastymooncake2, Funcshun840, ehuseyinyigit

wait this would seem really easy for a 3

like this is maximum a 2 - especially considering that the homothety tangency trick has already appeared before in the imo with examples like 2011/6, and once you realise its that kind of problem it's a lot easier

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lminsl

544 posts

lminsl

#10 h Jul 24, 2019, 12:35 PM • 6 Y

Y by stroller, math-lover, smurfcc, Adventure10, Mango247, Tastymooncake2

Wonderful problem

Let $M, N, K$ be the midpoints of the smaller arc $BC, CA, AB$ , respectively.
In order to prove the tangency, we first construct the tangency point:

Main Lemma: Among all points on $\omega$ , there is a unique point $P$ such that $PM:PN:PK=\frac{1}{ID}:\frac{1}{IE}:\frac{1}{IF}.$
Proof. Since such $P$ is unique if it exists, so let us only prove its existence.
We perform an inversion centered at $I$ with radius $1$ . Let $D', E', F'$ denote the image of $D,E,F$ under inversion, respectively. Since $D, E,F$ are collinear, $I, D', E', F'$ are concyclic and moreover, simple angle chasing yields $\triangle D'E'F' \sim \triangle MNK$ . Hence, if $P$ is the point such that $\triangle D'E'F' \cap I \sim \triangle MNK \cap P$ , then $P$ lies on $\omega$ and $P$ satisfies the desired length property, and we are done. $\blacksquare$

Let $x$ meet $NP, KP$ at $Y,Z$ , respectively. A line passing $Y$ and parallel to $MN$ cut $PM$ at $X$ . Then, $\frac{MX}{KZ}=\frac{MX}{NY}\times\frac{NY}{KZ}=\frac{PM}{PN} \times \frac{PN}{PK}=\frac{PM}{PK},$ so $XZ$ is parallel to $MK$ . Also, $\text{dist}(XY, MN)=\text{dist}(YZ, KN) \times \frac{\sin \angle PNM}{\sin \angle PNK}=\frac{ID}{2}\times \frac{PM}{PK}=\frac{IF}{2},$ so $XY$ coincides with $z$ . Similarly, $XZ$ coincides with $y$ , so $\Theta=\triangle XYZ$ .
Since $\triangle XYZ$ and $\triangle MNK$ are homothetic with center $P$ , $\odot(XYZ)$ touches $\Omega$ at $P$ , as desired.

Remark: This proof shows that as $l$ traces with its direction fixed, the tangency point of the circumcircle of $\Theta$ and $\Omega$ is fixed too

This post has been edited 1 time. Last edited by lminsl, Jul 24, 2019, 12:35 PM

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MarkBcc168

1594 posts

MarkBcc168

#11 h Jul 27, 2019, 10:38 AM • 7 Y

Y by jty123, amuthup, flashsonic, sabkx, Adventure10, Tastymooncake2, Funcshun840

What a nice problem! I wished that this was IMO 2018 P2/5 as it's very hard to bash but G6 is clearly more elegant.

Let $AI, BI, CI$ meet $\Omega$ again at $M_A, M_B, M_C$ . Let $X = y\cap z$ , $Y=x\cap z$ , $Z=x\cap y$ . Clearly $\triangle XYZ$ and $\triangle M_AM_BM_C$ are homothetic so lines $XM_A$ , $YM_B$ , $ZM_C$ are concurrent at $T$ .

Let lines through $D,E,F$ perpendicular to lines $AI$ , $BI$ , $CI$ forms a triangle $\triangle A_1B_1C_1$ where $D\in B_1C_1$ , $E\in C_1A_1$ , $F\in A_1B_1$ . The key claim is

Claim: Quadrilaterals $IA_1B_1C_1$ and $TM_AM_BM_C$ are homothetic.

Proof: Both $B_1C_1$ and $M_BM_C$ are perpendicular to $AI$ by definitions so it suffices to show that $IA_1\parallel M_AX$ .

Let $I_A$ be the $A$ -excenter. Notice that projections from points $I_A, X, A_1$ to each of $BI$ and $CI$ are three equally spaced points. Thus $X$ is the midpoint of $I_AA_1$ . Hence $M_AX$ is $I_A$ -midline of $\triangle I_AIA_1$ so we are done.

To finish, note that $\overline{DEF}$ is $I$ -Simson Line of $\triangle A_1B_1C_1$ . Thus by the converse, quadrilateral $IA_1B_1C_1$ is cyclic. By the claim, $TM_AM_BM_C$ is cyclic too so by homothety, these circles are tangent at $T$ .

This post has been edited 1 time. Last edited by MarkBcc168, Jul 27, 2019, 10:39 AM

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Hamel

392 posts

Hamel

#12 h Jul 27, 2019, 11:57 AM • 2 Y

Y by Adventure10, Mango247

jdhua wrote:

This was proposed by Mads Christensen, Denmark and lost to G6 by one vote.

Actually this problem is well known if I am not mistaken.

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TheDarkPrince

3042 posts

TheDarkPrince

#13 h Aug 11, 2019, 5:59 AM • 5 Y

Y by FadingMoonlight, AlastorMoody, rashah76, Adventure10, Tastymooncake2

psi241 wrote:

Let $ABC$ be a triangle with circumcircle $\omega$ and incentre $I$ . A line $\ell$ intersects the lines $AI$ , $BI$ , and $CI$ at points $D$ , $E$ , and $F$ , respectively, distinct from the points $A$ , $B$ , $C$ , and $I$ . The perpendicular bisectors $x$ , $y$ , and $z$ of the segments $AD$ , $BE$ , and $CF$ , respectively determine a triangle $\Theta$ . Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$ .

Solution. Let $M_a$ be midpoint of arc $BC$ and define $M_b,M_c$ similarly. Also, $X=y\cap z$ and define $Y,Z$ similarly. Without loss of generality, let $D$ lie on segment $EF$ . Further let $B_1,B_2$ be feet from $M_b$ on $x$ and $z$ . We have $\angle B_1M_bM_c = \angle M_cM_bM_a$ as $XYZ$ is similar to $M_aM_bM_c$ and $M_bB_1:M_bB_2 = ID/2:IF/2 = ID:IF.$ This gives $\angle YM_bM_c = \angle M_bYB_1 = \angle M_bB_2B_1 = \angle IFD.$ Similarly $\angle M_bM_cZ = \angle DEI$ , so $M_bY\cap M_cZ$ lies on $\odot(M_aM_bM_c)$ and we are done. $~\square$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.770325951955437, xmax = 14.63584922186477, ymin = -3.2441547295906172, ymax = 7.047480792918943; /* image dimensions */ pair M_a = (0.,6.), M_c = (-2.,0.), M_b = (7.,0.), A = (0.,-2.3333333333333326), B = (-2.2,3.0666666666666664), C = (3.6235294117647054,6.560784313725489), D = (0.,1.209721539810738), F = (-1.4068174324789477,0.6920463287745612), I = (0.,2.333333333333333), X = (0.32322002353142404,4.299389006488415), Z = (-1.2971782775518137,-0.5618058967612973), Y = (5.994614077322754,-0.5618058967612973); draw(M_a--M_c--M_b--cycle, linewidth(2.)); /* draw figures */ draw(circle((2.5,1.8333333333333333), 4.85912657903775), linewidth(1.)); draw(M_a--A, linewidth(1.)); draw(M_c--C, linewidth(1.)); draw(B--M_b, linewidth(1.)); draw(F--(1.6021626384635135,1.7992791205121619), linewidth(1.)); draw(D--F, linewidth(1.)); draw(D--(1.6021626384635135,1.7992791205121619), linewidth(1.)); draw(M_c--(1.7029429765660729,-2.959975671154668), linewidth(1.)); draw((1.7029429765660729,-2.959975671154668)--M_b, linewidth(1.)); draw(Z--X, linewidth(1.)); draw(X--Y, linewidth(1)); draw(Y--Z, linewidth(1.)); /* dots and labels */ dot(M_a,dotstyle); label("$M_a$", (-0.20510277581904354,6.1), NE * labelscalefactor); dot(M_c,dotstyle); label("$M_c$", (-2.4,-0.5), NE * labelscalefactor); dot(M_b,dotstyle); label("$M_b$", (7.127886367731051,-0.06281453716107784), NE * labelscalefactor); dot(A,linewidth(4.pt) + dotstyle); label("$A$", (-0.1,-2.8), NE * labelscalefactor); dot(B,linewidth(4.pt) + dotstyle); label("$B$", (-2.5,3.1821524591170527), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C$", (3.692038959907146,6.681626670789546), NE * labelscalefactor); dot((1.6021626384635135,1.7992791205121619),dotstyle); label("$E$", (1.7037013396386818,1.8778029802209413), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (0.06531114053746755,1.368788549432215), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (-1.4776388527908604,0.8120540157570455), NE * labelscalefactor); dot(I,linewidth(4.pt) + dotstyle); label("$I$", (0.16075134631035382,2.323190607161077), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (0.38344515978042176,4.422875134164573), NE * labelscalefactor); dot(Z,linewidth(4.pt) + dotstyle); label("$Z$", (-1.3503852450936789,-1), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (6.1,-0.8581495852684627), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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william122

1576 posts

william122

#14 h Sep 15, 2019, 3:20 AM • 2 Y

Y by centslordm, Adventure10

Denote the reflections $l_A,l_B,l_C$ of $l$ over $x,y,z$ respectively. Defining $l_B\cap l_C=T$ , note $\angle BTC=\angle(l_B,l)+\angle(l,l_C)=2\angle(BI,l)+2\angle(l,CI)=2\angle(BI,CI)$ , from which it is clear that $T$ lies on the circumcircle. Similarly, $l_B\cap l_A$ lies on $\omega$ , so we must have $l_A\cap l_B\cap l_C=T\in\omega$ .

Denoting $y\cap z=A'$ , and $AI\cap \omega=X$ , I claim that $TA'X$ is a line. However, if we look at the triangle formed by $l,l_B,l_C$ , we see that $X$ is the midpoint of its arc, so $TI$ bisects the angle formed by $l_B,l_C$ . Also, $y,z$ are angle bisectors of the other two vertices. Hence, they all concur at $A'$ , the incircle of the triangle in question, and $A'$ must lie on $TX$ as desired.

If we define $Y,Z,B',C'$ similar to $X,A'$ , note that $XY$ and $A'B'$ are both perpendicular to $CI$ , and similar relations give that $XYZ$ and $A'B'C'$ are homothetic. However, we may duplicate the argument above to get that $T$ is their homothetic center. Therefore, $(A'B'C')$ must be tangent to $(XYZT)=\omega$ , as desired.

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Physicsknight

635 posts

Physicsknight

#16 h Oct 19, 2019, 4:57 PM • 1 Y

Y by Adventure10

Let $(I)$ cuts $YZ$ , $ZX$ , $YZ$ at $G$ , $H$ and $J$ respectively.
Let $BH$ cuts $CJ$ at $K$ , by angle chasing we get $\angle BKC=\angle BAC$ . Hence, $K$ lies on $\omega$ . $AG ,BH,CJ$ are concurrent at $K$ , lies on $\omega$ . Let $AG$ , $BH$ and $CJ$ be the reflections of $(I)$ wrt $YZ$ , $ZX$ , $XY$ . Applying Collings' theorem note that $(I)$ passes through the orthocenter $\ell$ of $\triangle XYZ$ . $(I)$ is the Steiner line of $K$ wrt $\triangle XYZ$ . $K$ lies on $\Theta$ . Let $K'$ be the reflection of $K$ wrt $YZ$ . $BI$ cuts $YZ$ at point $P$ , also cuts $\omega$ again $N'$ . Note that $Y$ , $Z$ , $\ell$ , $K'$ lies on a circle. And, $\angle KZY=\angle K'ZY=\angle K'\ell Y=\angle IEF=\angle KBM$ Hence, $K$ , $Z$ , $B$ , $P$ lies on a circle. By Reim's theorem $MN'$ and $MN$ are parallel to $ZY$ , and $N=N'$ . $O$ is the intersection of $AI$ and $\omega$ . $XO$ , $YM$ and $ZN$ are concurrent at $K$ also note that $YZ||MN$ , $ZX||NO$ , $XY||OM$ . Applying homothety with center $K$ , we can prove that $\Theta$ and $\Omega$ are tangent at $K$ $\blacksquare$
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yayups

1614 posts

yayups

#18 h Dec 28, 2019, 8:49 AM • 1 Y

Y by Adventure10

$[asy] unitsize(2inches); pair X=dir(125); pair Y=dir(215); pair Z=dir(-35); pair I=orthocenter(X,Y,Z); pair A=2*foot(I,Y,Z)-I; pair B=2*foot(I,X,Z)-I; pair C=2*foot(I,X,Y)-I; pair L=dir(-80); pair P=(3/5)*X+(2/5)*L; pair E=2*foot(P,B,Y)-B; pair F=2*foot(P,C,Z)-C; pair D=extension(E,F,X,A); draw(X--Y--Z--cycle); draw(circumcircle(X,Y,Z)); draw(X--A); draw(B--Y); draw(C--Z); draw(E--F,dotted); draw(X--L); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$I$",I,dir(I)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$L$",L,dir(L)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy]$
Let $XYZ$ be the arc midpoints of $ABC$ . Note that $I$ is then the orthocenter of $XYZ$ , and $A,B,C$ are the reflections of $I$ over the sides of $XYZ$ . Now, let $PQR$ be the triangle $\Theta$ , so $P=y\cap z$ , $Q=x\cap z$ , $R=x\cap y$ .

Note that $x\parallel YZ$ , $y\parallel XZ$ , $z\parallel XY$ , so the triangle $PQR$ is homothetic to $XYZ$ . Thus, it suffices to show that the center of homothety $XP\cap YQ\cap ZR$ lies on $\omega=(XYZ)$ . To do this end, let $L_X=XP\cap YZ$ , $L_Y=YQ\cap XZ$ , $L_Z=ZR\cap XY$ . It suffices to show that $L_X=L_Y=L_Z$ .

Claim: We have that $L_X$ is the unique point $L$ such that $LY:LZ=IE^{-1}:IF^{-1}$ .

Proof: Using signed distances, we have $\mathrm{dist}(P,XZ)=\frac{\mathrm{dist}(B,XZ)+\mathrm{dist}(E,XZ)}{2}=\frac{\mathrm{dist}(E,XZ)-\mathrm{dist}(I,XZ)}{2}=\frac{IE}{2}.$ Thus, $\frac{LY}{LZ}=\frac{\sin PXY}{\sin PXZ}=\frac{\mathrm{dist}(P,XY)}{\mathrm{dist}(P,XZ)}=\frac{IF}{IE},$ as desired. To finish, note that $LY/LZ=k(LX;YZ)$ for some constant $k$ , and the function from the circle to $\mathbb{RP}^1$ defined by $(LX;YZ)$ is bijective, so $LY/LZ$ is bijective. This completes the proof of the claim. $\blacksquare$

The quantity $IE/IF$ is bijective in the variation of the direction of $\ell$ as well, so to solve the problem, it suffices to show that the set of possible values of $ID^{-1}:IE^{-1}:IF^{-1}$ and $LX:LY:LZ$ are identical as $\ell$ and $L$ vary.

By the law of sines of triangles $IED$ , $IDF$ , $IEF$ , we see that $ID^{-1}:IE^{-1}:IF^{-1} = \sin\angle(XA,\ell):\sin\angle(YB,\ell):\sin\angle(ZC,\ell).$ Let $m$ be a line perpendicular to $\ell$ . Then, we have that $\boxed{ID^{-1}:IE^{-1}:IF^{-1} = \sin\angle(YZ,m):\sin\angle(XZ,m):\sin\angle(XY,m)}.$ Let $T$ be an arbitrary fixed point on $\omega$ . Then, by the extended law of sines, we have $LX:LY:LZ = \sin\angle(TX,TL):\sin\angle(TY,TL):\sin\angle(TZ,TL).$ We see that $\angle XTY=\angle XZY$ and symmetric variants, so there is a fixed rotation sending the triple of directions $(YZ,ZX,XY)$ to $(TX,TY,TZ)$ . Thus, there is some other fixed point $S$ on the circle such that $\boxed{LX:LY:LZ = \sin\angle(YZ,SL):\sin\angle(ZX,SL):\sin\angle(XY,SL)}.$ From the two boxed equations, it is clear that both ratios have the same range as $L$ and $\ell$ vary, so we are done.

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IndoMathXdZ

691 posts

IndoMathXdZ

#20 h Apr 17, 2020, 8:09 AM • 5 Y

Y by RevolveWithMe101, Mathological03, Sugiyem, guptaamitu1, Mango247

Reflect $\ell$ with respect to $x,y,z$ : let them be $\ell_x, \ell_y, \ell_z$ . For obvious reasons, $\ell_x$ passes through $A$ . Similarly, $\ell_y$ passes through $B$ and $\ell_z$ passes through $C$ .
$\textbf{Claim 01.}$ $\ell_x, \ell_y, \ell_z$ concur on $(ABC)$ at point $K$ .
$\textit{Proof.}$ This is because
$\measuredangle (\ell_x, \ell_y) = \measuredangle(\ell_x,\ell) - \measuredangle(\ell_y, \ell) = 2 \measuredangle(AI, \ell) - 2 \measuredangle(BI, \ell) = 2 \measuredangle BIA = \measuredangle BCA$ The other is analogously the same.
Let $x \cap y = Z$ , $y \cap z = X$ , $z \cap x = Y$ .
Denote $d(X,\ell)$ as distance of points $X$ to line $\ell$ . Since $\ell, \ell_y, y$ concur, then we have
$d(X,\ell_y) = d(X,\ell) = d(X,\ell_z)$ This proves that $X$ lies on the angle bisector of $\measuredangle BKC$ .
Now, consider the midpoint arc of $AB, BC, CA: C_0, A_0, B_0$ . Since $A_0$ lies on the angle bisector of $\measuredangle BKC$ . Then we have $K,A_0, X$ being collinear.
This implies that $A_0B_0C_0$ and $XYZ$ are homothetic. But since $A_0 X \cap B_0 Y \cap C_0 Z = K$ . Then $(A_0 B_0 C_0K)$ are cyclic as well, proven.

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lneis1

243 posts

lneis1

#41 h May 24, 2022, 10:20 AM

Y by

Prolly one of the best problems solved so far

Let the $\triangle PQR \equiv \Theta$
$P\equiv y \cap z$ and so on
It is known that the triangle determined by the perpendicular bisectors of $AI, BI, CI$ is the triangle formed by joining midpoints of the minor arcs $BC, CA, AB$ .
Let this triangle be $\triangle LMN$

Hence clearly $\triangle LMN \sim \triangle PQR$

Note that by symmetry it suffices to show that $MQ \cap NR \in \odot (LMN) \equiv \odot (ABC)$

Draw a circle with diameter $MQ$ , and let it intersect $x$ and $z$ at $V$ and $W$ .

Note that due to the above construction, $MW \perp z$ and $MV \perp x$
But $ID \perp x$ and $IF \perp z \implies \angle VWM=\angle DFI$
$\implies \triangle IDF \sim \triangle MVW$

Due to the cyclic quad $\angle VWM=\angle VQM$
We are done since $x \vert \vert MN \implies \angle QMN=\angle DFI$

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NoctNight

108 posts

NoctNight

#42 h Jun 19, 2022, 11:36 AM • 1 Y

Y by PRMOisTheHardestExam

Solution

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gvole

201 posts

gvole

#43 h Mar 20, 2023, 6:56 PM

Y by

NoctNight wrote:

Solution

For there to be an anti-Steiner point, the line has to pass through the orthocenter. Your proof is circular reasoning.

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awesomeming327.

1683 posts

awesomeming327.

#44 h May 21, 2023, 1:08 AM

Y by

https://media.discordapp.net/attachments/932843956335706142/1109648328599281714/Screenshot_2023-05-20_at_18.06.02.png?width=1218&height=1144

Let $A_1$ , $B_1$ , $C_1$ be $y\cap z$ , $z\cap x$ , and $x\cap y$ , respectively. Let $A_2$ , $B_2$ , $C_2$ be $AI$ , $BI$ , $CI$ 's intersection of $\omega$ , respectively. Since $A_1B_1 \parallel A_2B_2$ due to perpendicularity to $CI$ , $\triangle A_1B_1C_1$ is a homothety of $\triangle A_2B_2C_2$ . It suffices to show that the center of homothety, which we will denote as $P$ , is on $\omega$ .

$~$
Let the circle with diameter $A_1A_2$ intersect $z$ and $y$ at $G$ and $H$ , respectively. Similarly, the circle with diameter $B_1B_2$ intersects $x$ and $z$ at $J$ and $K$ , respectively and the circle with diameter $C_1C_2$ intersects $y$ and $x$ at $L$ and $M$ . All angles we write are directed.

$~$
Note that $\angle A_2HA_1=90^\circ$ therefore $A_2H\parallel IE$ . Furthermore, $\angle GA_2H=\angle GA_1H$ . Since $IE$ and $IF$ are perpendicular to the two lines making up that angle, $\angle FIE=\angle GA_2H$ , thus proving a homothety between $\triangle A_2GH$ and $\triangle IFE$ . Analogously, $\triangle B_2JK$ is homothetic to $\triangle IDF$ and $\triangle C_2LM$ to $\triangle IED$ .

$~$
Therefore,
$\begin{align*} \angle B_2PA_2 &= \angle A_2A_1B_1 + \angle A_1B_1B_2 \\ &= \angle A_2A_1G + \angle KB_1B_2 \\ &= \angle A_2HG + \angle KJB_2 \\ &= \angle IEF + \angle FDI \\ &= \angle DIE \\ &= \angle B_2C_2A_2 \end{align*}$ Therefore, $P$ is on $\omega$ as desired.

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HamstPan38825

8857 posts

HamstPan38825

#45 h Mar 4, 2024, 4:02 AM

Y by

Let $\ell_a$ , $\ell_b$ , $\ell_c$ denote the reflections of $\overline{DEF}$ over the three perpendicular bisectors.

Claim. $\ell_a$ , $\ell_b$ , $\ell_c$ concur at some point $K$ on $T$ on $\omega$ .

Proof. Annoying angle chasing; let $\ell_a$ intersect $\omega$ at $K_A$ , and show that $K_A$ , $K_B$ , $K_C$ are consistent. $\blacksquare$

Let the perpendiculars intersect at $P, Q, R$ . Now, let $X = \overline{AK} \cap \overline{DEF}$ , which lies on $\overline{QR}$ ; define $Y$ and $Z$ similarly.

Claim. $R$ is the incenter of triangle $KXY$ .

Proof. Because $\overline{XR}$ and $\overline{YR}$ bisect their respective angles by definition. $\blacksquare$

So $\overline{KR}$ passes through the arc midpoint of $\widehat{AB}$ . But letting $MLN$ be the triangle formed by arc midpoints, the triangles $XYZ$ and $MLN$ are homothetic at $K$ as $K, X, M$ are collinear and their sides are parallel (both perpendicular to $\overline{AI}, \overline{BI}, \overline{CI}$ ). It follows that $K$ takes $(XYZ)$ to $(ABC)$ , hence it is the desired tangency point.

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thdnder

194 posts

thdnder

#46 h Mar 6, 2024, 4:29 PM

Y by

Let $l_A, l_B, l_C$ be the reflections of the line $l$ in the lines $x, y$ and $z$ , respectively. Then we'll prove the following claim:

Claim: $l_A, l_B, l_C$ are concurrent at a point $T$ , which lies on $\omega$ .

Proof. Let $M_A, M_B, M_C$ be the midpoints of arcs $BC, CA$ and $AB$ , respectively. Then it's clear that $I$ is the orthocenter of $M_AM_BM_C$ . Let $T = l_A \cap \omega$ . Then we'll prove that $BT$ is the reflection of $l$ wrt $y$ . Note that
$\begin{align*} \measuredangle(l, M_BB) = \measuredangle(DE, EI) =\\ \measuredangle(DI, EI) + \measuredangle(DE, DI) =\\ \measuredangle(AM_A, BM_B) + \measuredangle(DA, AT) =\\ \measuredangle(M_BM_C, M_AM_C) + \measuredangle(DA, AT) =\\ \measuredangle(M_AM_B, M_BT) + \measuredangle(M_BT, TM_A) =\\ \measuredangle(M_BM_A, M_AT) = \measuredangle(M_BB, BT) \end{align*}$ , so $BT$ coincides with $l_B$ . Thus $l_B$ passes through $T$ , completing the claim. $\blacksquare$

Now let $A' = y \cap z$ and define $B', C'$ similarly. Then the reflections of $T$ across the sides of triangle $A'B'C'$ are lies on $l$ , so by Simson line, $T$ lies on $(A'B'C')$ . Now note that $x \parallel M_BM_C$ , $y \parallel M_AM_C$ and $z \parallel M_AM_B$ , so triangles $M_AM_BM_C, A'B'C'$ are homothetic, and their circumcircles share common point, namely $T$ , hence $\omega, (A'B'C')$ are tangent each other at $T$ , as required. $\blacksquare$

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asdf334

7586 posts

asdf334

#47 h Mar 7, 2024, 4:04 PM

Y by

very interesting problem i actually tried it a while back and never finished.

2018 G5

as a simpler sketch
1) Guess that the desired concurrency point $P$ is such that its Simson line is parallel to $\ell$ . Let $M_A$ be the midpoint of arc $\widehat{BC}$ .
2) Vary $\ell$ of a fixed orientation; in this case $D$ , $E$ , and $F$ move linearly, hence $y\cap z$ moves linearly.
3) Prove that $M_A$ and $P$ both lie on the locus (there exists a line $\ell$ reaching both these points). The first case is trivial ( $D=E=F=I$ ), and the second case requires a bit more work. Done.

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foolish07

24 posts

foolish07

#48 h Apr 28, 2024, 6:23 AM

Y by

TelvCohl wrote:

Let $T$ be the pole of the Simson line of $\triangle ABC$ with the direction $\perp \ell,$ $T_a, T_b, T_c$ be the intersection of $AI, BI, CI$ with the parallel $\tau$ from $T$ to $\ell$ and let $\triangle XYZ$ be the triangle $\Theta,$ then note that the isogonal conjugate of $T$ WRT $\triangle ABC$ is the point at infinity with the direction $\parallel \ell$ we get $T$ lies on the perpendicular bisector of $BT_b, CT_c,$ so consider the homothety $\mathbf{H}^{B}_{\frac{1}{2}}, \mathbf{H}^{C}_{\frac{1}{2}}$ we conclude that $TX$ passes through the midpoint $M_a$ of arc $BC$ in $\odot (ABC)$ and $\frac{TM_a}{XM_a} = \frac{\text{dist} (I, \tau)}{\text{dist} (I, \ell)}.$

I cannot understand this solution, lol

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lnzhonglp

120 posts

lnzhonglp

#49 h Jun 2, 2024, 10:12 PM

Y by

Let $M_a$ , $M_b$ , $M_c$ be the midpoints of arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ respectively, and let $X = y \cap z$ , $Y = x \cap z$ , and $Z = x \cap y$ . The sides of $\triangle M_aM_bM_c$ and $\triangle XYZ$ are parallel, so $\triangle M_aM_bM_c \sim \triangle XYZ$ , so it suffices to prove that their center of homothety lies on $\Omega$ . Let $\ell_a$ , $\ell_b$ , $\ell_c$ be the reflections of $\ell$ across $x$ , $y$ , and $z$ , respectively.

Claim: $\ell_a$ , $\ell_b$ , $\ell_c$ concur at a point $T$ on $\Omega$ .

Proof. We have that $\begin{align*} \measuredangle(\ell_a, \ell_b) &= \measuredangle(\ell_a, \ell) + \measuredangle(\ell, \ell_b) \\ &= 2\measuredangle(AI, \ell) + 2\measuredangle(\ell, BI) \\ &= \measuredangle ACB,\end{align*}$ so $\ell_a$ and $\ell_b$ intersect on $\Omega$ . Similarly, $\ell_b \cap \ell_c$ and $\ell_c \cap \ell_a$ also lie on $\Omega$ . Note that $A \in \ell_a$ , $B \in \ell_b$ , and $C \in \ell_c$ , so either $\ell_a$ , $\ell_b$ , $\ell_c$ all concur on $\Omega$ or else $\ell_a$ , $\ell_b$ , $\ell_c$ must correspond to the sides of $\triangle ABC$ . To prove the second case is impossible, suppose WLOG that $\ell_b = BC$ . Then since $\ell_a$ passes through $A$ , we have $\ell_a = AB$ . Then $\measuredangle(\ell_a, \ell_b) = \measuredangle(AB, BC) = \measuredangle ACB$ , which is impossible because $\measuredangle(AB, BC) = \measuredangle BAC + \measuredangle ACB \neq \measuredangle ACB$ since $0^\circ < \angle BAC < 180^\circ$ . Therefore, $\ell_a$ , $\ell_b$ , $\ell_c$ must concur at a point $T$ on $\Omega$ . $\square$

Now note that $Z$ is the $T$ -excenter of $\triangle TXY$ , so $M_c,$ $Z$ , $T$ are collinear. Similarly, $M_B, Y, T$ and $M_A, X, T$ are also collinear. Therefore, $T$ is the center of the homothety sending $\triangle XYZ$ to $\triangle M_aM_bM_c$ , so $(XYZ)$ is tangent to $\Omega$ at $T$ .

This post has been edited 5 times. Last edited by lnzhonglp, Jun 4, 2024, 12:55 AM

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OronSH

1728 posts

OronSH

#50 h Nov 15, 2024, 11:19 PM

Y by

Let $\triangle MNP$ be the circummidarc triangle and let $\Theta=\triangle XYZ$ such that $XYZ,MNP$ are homothetic (by fact $5$ ) with center $T$ . It suffices to show $T\in\Omega$ .

Fix $D$ and vary $\ell$ with degree $1$ . Then $E,F$ have degree $1$ , so $y,z$ have degree $1$ and $x$ is fixed, so $Y,Z$ have degree $1$ and thus $NY,PZ$ have degree $1$ . Thus if $\mathcal I,\mathcal J$ are the circle points, $(\infty_{NY},\infty_{PZ};\mathcal I,\mathcal J)$ has degree $2$ . Thus to show $T\in\Omega$ we need to check $3$ cases of $\ell$ .

First if $\ell=AI$ and $E=F=I$ then $T=M$ by fact $5$ .

Now let $x$ meet $\Omega$ at $K$ . If $K\in\ell$ then $\measuredangle BEK+\measuredangle KDA=\measuredangle BIA=\measuredangle BNM+\measuredangle NMA=\measuredangle MAC+\measuredangle CBN=\measuredangle MKN=\measuredangle MKK+\measuredangle KKN=\measuredangle MAK+\measuredangle KBN=\measuredangle KBE+\measuredangle DAK.$ Thus $K\in x\implies\measuredangle KDA=\measuredangle DAK\implies\measuredangle BEK=\measuredangle KBE\implies K\in y$ Similarly $K\in z$ , so $X,Y,Z=K$ and $T=K$ .

This works for both possible positions of $K$ , so we have shown $3$ cases so we are done.

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Saucepan_man02

1300 posts

Saucepan_man02

#51 h Feb 12, 2025, 12:55 PM

Y by

Very Nice problem:

Let $\ell_a, \ell_b, \ell_c$ denote the reflections of $\ell$ wrt $AD, BE, CF$ . Let $a, b, c$ denote the perpendicular bisectors of $AD, BC, CF$ respectively.

Claim: $\ell_a, \ell_b, \ell_c$ are concurrent on $\omega$ .
Proof: Let $\ell_b, \ell_c$ intersect at $K$ . Notice that: $\measuredangle(\ell_b, \ell_c) = 2 \measuredangle (b, c) = \measuredangle BAC$ which implies $K \in \omega$ .

Let $L=\ell \cap KI$ . Consider the homothety from $I$ to $L$ , and let $\triangle PQR$ denote the triangle formed after the homothety.
Notice that: $\ell_a = \overline{KPA}, \ell_b = \overline{KQB}, \ell_c = \overline{KRC}$ .
Notice that: $PL \parallel AD$ and $PA, LD$ are symmetric wrt to line $a$ . Thus, they have the same perpendicular bisector and the desired triangle is formed by the midpoints of arcs in $(PQR)$ .

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abeot

123 posts

abeot

#52 h Feb 17, 2025, 10:46 PM • 1 Y

Y by centslordm

Let $M_A$ , $M_B$ and $M_C$ be the arc midpoints of $BC$ , $CA$ and $AB$ in $\omega$ . Let $O_A$ , $O_B$ and $O_C$ be the points such that $O_BO_C$ is the perpendicular bisector of $AD$ , $O_CO_A$ is the perpendicular bisector of $BE$ , and $O_AO_B$ is the perpendicular bisector of $BF$ .

Thus the problem asks to prove $(O_AO_BO_C)$ is tangent to $\omega$ . I will prove the stronger statement that $\triangle M_AM_BM_C$ and $\triangle O_AO_BO_C$ are homothetic under a point on $\omega$ . Notice that $O_BO_C \parallel M_BM_C$ (both are perpendicular to $AI$ ) and similar statements, so the triangles are homothetic.

Claim: Let $\ell_A$ , $\ell_B$ and $\ell_C$ be the reflections of $\ell$ across $O_BO_C$ , $O_CO_A$ and $O_AO_B$ . Then $\ell_A$ , $\ell_B$ , and $\ell_C$ concur at a point on $\omega$ .

Proof. Let $\ell_B$ and $\ell_C$ intersect at $T$ . Directing angles, $\begin{align*} \measuredangle M_BCT &= \measuredangle M_BCM_C + \measuredangle M_CCT \\ &= \measuredangle M_BAM_C + \measuredangle FCT \\ &= \measuredangle M_CIM_B + \measuredangle DFC \\ &= \measuredangle FIE + \measuredangle EFI \\ &= \measuredangle FEI = \measuredangle FEB \\ &= \measuredangle EBT \\ &= \measuredangle M_BBT \end{align*}$ Hence $T$ lies on $\omega$ . Repeating a similar argument for $\ell_A$ and $\ell_B$ shows that $\ell_A$ and $\ell_C$ meet at the point where $\ell_B$ intersects $\omega$ . $\square$

Let this point of concurrence be $T$ . Then we equivalently have that the reflections of $T$ over $O_BO_C$ , $O_CO_A$ and $O_AO_B$ lie on $\ell$ . Then due to Simson, we have $T \in (O_AO_BO_C)$ as well, and moreover $\ell$ passes through the orthocenter of $\triangle O_AO_BO_C$ .

Claim: Let $\ell_1$ be the line through $I$ parallel to $\ell$ . Then the reflections of $T$ over $M_BM_C$ , $M_CM_A$ and $M_AM_B$ lie on $\ell_1$ .

Proof. Note that $I$ is the reflection of $A$ over $M_BM_C$ . We also know $TA$ is the reflection of $\ell$ over $O_BO_C$ .
Since $O_BO_C \parallel M_BM_C$ , it follows that the line through $I$ and the reflection of $T$ over $M_BM_C$ is parallel to $\ell$ , hence it is $\ell_1$ . Now use a similar argument for $M_CM_A$ and $M_AM_B$ . $\square$

Let the reflections of $T$ over $O_BO_C$ , $O_CO_A$ and $O_AO_B$ be $S_A$ , $S_B$ and $S_C$ . Let the reflections of $T$ over $M_BM_C$ , $M_CM_A$ and $M_AM_B$ be $S_A'$ , $S_B'$ and $S_C'$ .

By our claims, we have that $\overline{S_AS_BS_C} = \ell$ and $\overline{S_A'S_B'S_C'} = \ell_1$ and are parallel
Since $M_BM_C \parallel O_BO_C$ and similar, we have that $T$ , $S_A$ and $S_A'$ are collinear and similar equations

Thus there exists a homothety at $T$ that sends $S_A'$ to $S_A$ , $S_B'$ to $S_B$ , and $S_C'$ to $S_C$ . Then this homothety must also send $M_BM_C$ to $O_BO_C$ , $M_CM_A$ to $O_CO_A$ and $M_AM_B$ to $O_AO_B$ .

It follows that this homothety sends $\omega$ to $(O_AO_BO_C)$ . Yet $T$ lies on both circles, hence they must be tangent at $T$ . $\blacksquare$

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EeEeRUT

55 posts

EeEeRUT

#53 h Mar 3, 2025, 8:12 AM

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Let $X,Y,Z$ be intersections of $(y,z)$ , $(z,x)$ , $(x,y)$ , respectively. Denote $M_A$ mid arc $BC$ not containing $A$ . Similarly, $M_B$ , $M_C$ .

Claim: Consider a line $\ell$ . Suppose the reflection of $\ell$ across $BC,CA,AB$ meet at a point $P_{\ell}$ . Then, $P_{\ell},A,B,C$ are cyclic. Moreover, this happens iff $\ell$ pass through the orthocenter of $ABC$ .

Proof : Say $\ell$ intersect $BC,CA,AB$ at $D,E,F$ , respectively. By angle chasing, we get that $A,B,C$ is incenter (or possibly excenter) of $\triangle EFP_{\ell}, \triangle FDP_{\ell}, \triangle DE P_{\ell}$ .
A little more angle chasing yields $P_{\ell}, A,B,C$ concyclic.
Now, say $P,A,B,C$ is cyclic. Note that $\ell$ is a line through the reflection of $P_{\ell}$ wrt $P_{\ell}$ simson line and parallel to the line. Since simson line pass through midpoint of $P_{\ell}$ , $\ell$ pass through $H$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.363571630184413, xmax = 2.859360466545404, ymin = -2.8769366683971946, ymax = 8.494553595619527; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-10.185769710390105,4.976212968577072)--(-5.76009668316386,1.1549270067429658), linewidth(1.2)); draw((-5.76009668316386,1.1549270067429658)--(-11.935829109332868,1.1388892578023546), linewidth(1.2)); draw((-11.935829109332868,1.1388892578023546)--(-10.185769710390105,4.976212968577072), linewidth(1.2)); draw(circle((-8.85030664928438,2.0494282850111416), 3.2170685680498003), linewidth(0.8)); draw((xmin, -2.7287753951224465*xmin-24.610713951762044)--(xmax, -2.7287753951224465*xmax-24.610713951762044), linewidth(0.8) + linetype("4 4")); /* line */ draw(circle((-10.185769710390105,4.976212968577072), 0.6166905790733174), linewidth(0.8) + ccqqqq); draw(circle((-11.935829109332868,1.1388892578023546), 2.3468811784402077), linewidth(0.8) + ccqqqq); draw(circle((-5.76009668316386,1.1549270067429658), 3.457264452438446), linewidth(0.8) + ccqqqq); draw((xmin, -0.2078751828133592*xmin + 3.4887181382774854)--(xmax, -0.2078751828133592*xmax + 3.4887181382774854), linewidth(0.8)); /* line */ draw((xmin, 2.77327428816662*xmin + 27.321473687545485)--(xmax, 2.77327428816662*xmax + 27.321473687545485), linewidth(0.8)); /* line */ draw((xmin, 0.38070332458830347*xmin + 8.194100349068034)--(xmax, 0.38070332458830347*xmax + 8.194100349068034), linewidth(0.8)); /* line */ draw((-11.14658642687847,5.805816829509755)--(-10.185769710390105,4.976212968577072), linewidth(1.2)); /* dots and labels */ dot((-10.185769710390105,4.976212968577072),linewidth(3pt) + dotstyle); label("$A$", (-10.141726205599301,5.045042950677216), NE * labelscalefactor); dot((-11.935829109332868,1.1388892578023546),linewidth(3pt) + dotstyle); label("$B$", (-11.890271256656275,1.21489664836196), NE * labelscalefactor); dot((-5.76009668316386,1.1549270067429658),linewidth(3pt) + dotstyle); label("$C$", (-5.716836688638798,1.2267915126548647), NE * labelscalefactor); dot((-10.18108220431807,3.1711726631001085),linewidth(3pt) + dotstyle); label("$H$", (-10.129831341306398,3.2370235781557284), NE * labelscalefactor); dot((-9.438698417616436,1.1453740522108595),linewidth(3pt) + dotstyle); label("$D$", (-9.20203192645984,1.21489664836196), NE * labelscalefactor); dot((-11.14658642687847,5.805816829509755),linewidth(3pt) + dotstyle); label("$E$", (-11.093315349031668,5.877683451180532), NE * labelscalefactor); dot((-10.549940121115105,4.177703070751983),linewidth(3pt) + dotstyle); label("$F$", (-10.498572134386439,4.248087043052613), NE * labelscalefactor); dot((-7.994485275316855,5.150573226382685),linewidth(3pt) + dotstyle); label("$P$", (-7.9411763114119545,5.223465915070784), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Back to the problem. Let $\ell’$ be a line through $I$ and parallel to $\ell$ . consider $P_{\ell}$ and $P_{\ell’}$ of triangle $XYZ$ and $M_AM_BM_C$ , respectively. We want to claim that points from this $2$ triangle are actually the same. In fact, the line forming these $2$ point are actually the same.
Since $\ell’$ pass through $I$ ( the orthocenter of $M_AM_BM_C$ ), the point $P_{\ell’}$ lies on $(ABC)$ . Let $\ell_A$ be line obtained from reflecting $\ell’$ across $M_BM_C$ .Define $\ell_B, \ell_C$ similarly. Notice that $\ell_A$ pass through $A$ . Reflecting $\ell_A$ across $x$ gives the line $\ell$ . Similarly, reflecting $\ell_B, \ell_C$ across $y,z$ also give us the line $\ell$ . And By the claim, $\ell$ pass through the orthocenter of $XYZ$ and $P_{\ell}, P_{\ell’}$ coincide( They are concurrency point of $\ell_A, \ell_B, \ell_C$ ).

Now, note that $XYZ$ and $M_AM_BM_C$ are homothetic image $\mathcal{H}$ of each other centered at some point $T$ , $\mathcal{H}$ also send $P_{\ell}$ to $P_{\ell’}$ . But these $2$ points coincidences, so we have $T = P_{\ell} = P_{\ell’}$ . That is $(M_AM_BM_C)$ and $(XYZ)$ are tangent.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.03557648958321, xmax = 9.397732402258177, ymin = -7.66970574340795, ymax = 5.705126698237168; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((xmin, 0.24528570141084902*xmin + 3.570684825226503)--(xmax, 0.24528570141084902*xmax + 3.570684825226503), linewidth(0.8) + linetype("4 4")); /* line */ draw(circle((-3.1327105305058716,-0.5703579623504904), 3.9367029737827783), linewidth(0.8)); draw(circle((-3.676641498704823,2.7465112310007282), 0.5755302403505578), linewidth(0.8)); draw((-5.48,2.59)--(-6.137298338084797,-3.1139936301178377), linewidth(1.2)); draw((-6.137298338084797,-3.1139936301178377)--(-0.11598667918599176,-3.0995885783022903), linewidth(1.2)); draw((-0.11598667918599176,-3.0995885783022903)--(-5.48,2.59), linewidth(1.2)); draw((-4.248388132252976,2.812396317206375)--(-3.6752646360412147,2.1709826376127213), linewidth(1.2) + ccqqqq); draw((-3.6752646360412147,2.1709826376127213)--(-3.2578750442618634,3.1413146177532276), linewidth(1.2) + ccqqqq); draw((-3.2578750442618634,3.1413146177532276)--(-4.248388132252976,2.812396317206375), linewidth(1.2) + ccqqqq); draw((-0.2682925886755297,2.130149426698945)--(-7.043533133273599,-0.11969525757980115), linewidth(1.2) + ccqqqq); draw((-7.043533133273599,-0.11969525757980115)--(-3.1232926077586445,-4.507049670691556), linewidth(1.2) + ccqqqq); draw((-3.1232926077586445,-4.507049670691556)--(-0.2682925886755297,2.130149426698945), linewidth(1.2) + ccqqqq); draw((-5.48,2.59)--(-3.1232926077586445,-4.507049670691556), linewidth(1.2)); draw((-6.137298338084797,-3.1139936301178377)--(1.8523945134738202,4.025050712753537), linewidth(1.2)); draw((-7.043533133273599,-0.11969525757980115)--(-0.11598667918599176,-3.0995885783022903), linewidth(1.2)); draw((xmin, 0.24528570141084902*xmin-0.33311245764842573)--(xmax, 0.24528570141084902*xmax-0.33311245764842573), linewidth(0.8) + linetype("4 4") + blue); /* line */ draw((-9.949358586402534,1.1302494257727047)--(-7.043533133273599,-0.11969525757980115), linewidth(1.2)); draw((-3.7697782535487163,3.3144554205608663)--(-7.043533133273599,-0.11969525757980115), linewidth(1.2) + linetype("2 2") + qqwuqq); draw((-3.7697782535487163,3.3144554205608663)--(-0.2682925886755297,2.130149426698945), linewidth(1.2) + linetype("2 2") + qqwuqq); draw((-3.7697782535487163,3.3144554205608663)--(-3.1232926077586445,-4.507049670691556), linewidth(1.2) + linetype("2 2") + qqwuqq); /* dots and labels */ dot((-5.48,2.59),linewidth(3pt) + dotstyle); label("$A$", (-5.418112321279701,2.669207618909772), NE * labelscalefactor); dot((-6.137298338084797,-3.1139936301178377),linewidth(3pt) + dotstyle); label("$B$", (-6.075661614958908,-3.0248894561420716), NE * labelscalefactor); dot((-0.11598667918599176,-3.0995885783022903),linewidth(3pt) + dotstyle); label("$C$", (-0.10175632983079323,-2.996908635134446), NE * labelscalefactor); dot((-5.368390234255033,2.253895461170105),linewidth(3pt) + dotstyle); label("$D$", (-5.306189037249197,2.3334377668182626), NE * labelscalefactor); dot((1.8523945134738202,4.025050712753537),linewidth(3pt) + dotstyle); label("$E$", (1.9128627827182665,4.110219900802499), NE * labelscalefactor); dot((-9.949358586402534,1.1302494257727047),linewidth(3pt) + dotstyle); label("$F$", (-9.895043682499834,1.214204926513232), NE * labelscalefactor); dot((-3.2578750442618634,3.1413146177532276),linewidth(3pt) + dotstyle); label("$Y$", (-3.2076274616772604,3.228824039062287), NE * labelscalefactor); dot((-4.248388132252976,2.812396317206375),linewidth(3pt) + dotstyle); label("$Z$", (-4.186956196944164,2.893054186970778), NE * labelscalefactor); dot((-3.6752646360412147,2.1709826376127213),linewidth(3pt) + dotstyle); label("$X$", (-3.613349366287835,2.2494953037953853), NE * labelscalefactor); dot((-3.7697782535487163,3.3144554205608663),linewidth(3pt) + dotstyle); label("$P$", (-3.7112822398145253,3.3967089651080418), NE * labelscalefactor); dot((-7.043533133273599,-0.11969525757980115),linewidth(3pt) + dotstyle); label("$M_C$", (-6.985038297706748,-0.030941608326114662), NE * labelscalefactor); dot((-0.2682925886755297,2.130149426698945),linewidth(3pt) + dotstyle); label("$M_B$", (-0.21367961386129655,2.2075240722839466), NE * labelscalefactor); dot((-3.1232926077586445,-4.507049670691556),linewidth(3pt) + dotstyle); label("$M_A$", (-3.067723356639131,-4.4239305065233605), NE * labelscalefactor); dot((-4.169697268696029,-1.3558795768714325),linewidth(3pt) + dotstyle); label("$I$", (-4.1170041444251,-1.2760881431654614), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 2 times. Last edited by EeEeRUT, Mar 3, 2025, 9:15 AM

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YaoAOPS

1501 posts

YaoAOPS

#54 h Mar 17, 2025, 4:19 PM • 1 Y

Y by MS_asdfgzxcvb

vectors!!!!!!!!

Let $M_A, M_B, M_C$ be the arc midpoints of $BC, AB, AB$ respectively and let the triangle determined by the bisectors be $XYZ$ . Note that these two triangles are homothetic so it remains to show their homothety point $T$ lies on $\omega$ .
Now consider $\overrightarrow{M_AX}$ . The projections of this onto $BI$ is from the midpoint of $BI$ to the midpoint of $BE$ , so the projection of $2\overrightarrow{M_AX}$ onto $BI$ spans $\overrightarrow{IE}$ and similarly $\overrightarrow{IF}$ onto $CI$ . This vector thus is parallel to $IO_{D}$ where $O_D$ is the circumcircle of $(IEF)$ .
We thus want to show that $\measuredangle M_BTM_C = \measuredangle CIB$ and thus since $\measuredangle M_BTM_C = \measuredangle (M_BY, M_CZ) = \measuredangle O_EIO_F$ we want to show that $\measuredangle O_EI + \measuredangle BI = \measuredangle O_FI + \measuredangle CI.$ Invert about $I$ with arbitrary radius so $IDEF$ is a circle. Then we want to show that $\measuredangle \perp DF + \measuredangle BI = \measuredangle \perp EF + CI \iff \measuredangle DF + \measuredangle BI = \measuredangle EF + \measuredangle CI$ which is just true.

This post has been edited 2 times. Last edited by YaoAOPS, Mar 17, 2025, 4:23 PM

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cursed_tangent1434

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cursed_tangent1434

#55 h Mar 31, 2025, 5:36 AM

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I disagree with the claims that this problem is 'very easy' or 'too easy for a P3/6'. However, it is a fairly routine exercise with the method of moving points. Still finding a elementary solution is by far harder but instructive. The key idea is to work backwards as opposed to forwards, which is quite hard since the homothety is almost immediate to notice so it feels counterintuitive to abandon such progress and work from scratch.

Let $\triangle T_aT_bT_c$ denote the triangle formed by the perpendicular bisectors. We start off with the following key claim.

Claim : The reflections of $\ell$ across the sides $T_aT_b$ , $T_bT_c$ and $T_cT_a$ concur at a point $R$ which lies on $\Omega$ .

Proof : Let $R_{ab}$ denote the intersection of the reflections of $\ell$ across sides $T_bT_c$ ( $\ell_a$ ) and $T_aT_c$ ( $\ell_b$ ) respectively. Note,
$\measuredangle BR_{ab}A = \measuredangle R_{ab}BA + \measuredangle BAR_{ab} = \measuredangle IBA + \measuredangle BAI + \measuredangle EDA + \measuredangle BED = \measuredangle BIA + \measuredangle EID = 2\measuredangle BIA = 2\measuredangle BCA$ which implies that $R_{ab}$ lies on $\Omega$ . A similar argument shows that $\ell_a$ and $\ell_c$ concur on $\Omega$ which implies that all these lines concur at a point $R$ on $\Omega$ as desired.

Further note that this means $R$ is the $\ell-$ Anti Stiener point with respect to $\triangle T_aT_bT_c$ which by the Converse of the Anti-Steiner result implies that the orthocenter $H$ of $\triangle T_aT_bT_c$ lies on $\ell$ and that $R$ lies on $\Theta$ .

We now show that this point $R$ is the center of homothety mapping $\triangle T_aT_bT_c$ to $\triangle M_aM_bM_c$ . For this, we need the following.

Claim : Lines $\overline{T_aM_a}$ , $\overline{T_bM_b}$ and $\overline{T_cM_c}$ concur at $R$ .

Proof : We show that $R$ lies on $\overline{T_aM_a}$ the proof of the others being similar. Note that since $M_a$ is the minor $AB$ arc midpoint, $M_a$ lies on the internal $\angle ARB$ -bisector. Further, let $R_a$ and $R_b$ denote the reflections of $R$ across sides $T_bT_c$ and $T_cT_a$ . Both these points lie on the $R-$ Steiner Line with respect to $\triangle T_aT_bT_c$ which is $\ell$ . Now note,
$\measuredangle T_cRA = \measuredangle DR_aT_c = \measuredangle T_cR_bE = \measuredangle BRT_c$ by reflections and since $T_cR=T_cR_a=T_cR_b$ due to reflections.

This further implies that $T_c$ also lies on the internal $\angle ARB-$ bisector and hence points $R$ , $T_c$ and $T_b$ are collinear as desired.

The triangles $\triangle T_aT_bT_c$ and $\triangle M_aM_bM_c$ have pairwise parallel sides and are hence homothetic, so lines $\overline{T_aM_a}$ , $\overline{T_bM_b}$ and $\overline{T_cM_c}$ concur at the homothety center mapping $\triangle T_aT_bT_c$ to $\triangle M_aM_bM_c$ . Thus, $R$ is indeed this center of homothety, implying that circles $\Theta$ and $\Omega$ are indeed tangent to each other at $R$ .

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