Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $X_A$ lying on the tangent at $H$ to the circumcircle of triangle $BHC$ is such that $AH=AX_A$ and $X_A \not= H$. Points $X_B,X_C$ are defined similarly. Prove that the triangle $X_AX_BX_C$ and the orthotriangle of $ABC$ are similar.
Problem
Source: Sharygin Geometry Olympiad, Final Round 2016, Problem 2 grade 9
Tags: geometry, similar triangles, circumcircle
anantmudgal09
04.08.2016 16:53
Let $O$ be the circumcenter of triangle $ABC$. Let $DEF$ be the orthotriangle of $ABC$. Let $HX_A$ meet line $BC$ at $T_A$. We see that $\angle AHX_A=\angle THD=90^{\circ}-\angle B+\angle C$. Therefore, $\angle HAX_A=2(\angle B-\angle C)=2\angle HAO$. Since $AH=AX_A$ we conclude that $X_A$ is the reflection of point $H$ in line $AO$. Dilating about $H$ with factor $\frac{1}{2}$, let $Y_A$ be the image of $X_A$. Define $Y_B,Y_C$ analogously. We see that $\angle Y_BY_AY_C=\angle Y_BOY_C=180^{\circ}-\angle BOC=180^{\circ}-2\angle A=\angle EDF$. Similar relations hold and so $\triangle DEF \sim \triangle X_AX_BX_C$.
jslam
05.08.2016 09:36
Denote $H_A, H_B, H_C$ as the feet of the altitudes and let $O$ be the circumcenter of $\triangle{ABC}$.
Through angle chasing, it is easy to show that $AO\perp{HX_A}$, $AO\perp{H_BH_C}$, meaning that $HX_A\parallel{H_BH_C}$. Similarly, $HX_B\parallel{H_AH_C}, HX_C\parallel{H_BH_A}$. Because $AO, BO, CO$ are perpendicular bisectors of $HX_A, HX_B, HX_C$, respectively, $O$ must be the circumcenter of $\triangle{HX_AX_B}, \triangle{HX_BX_C}, \triangle{HX_CX_A}$. It then follows that $O$ is the circumcenter of cyclic quad $HX_AX_BX_C$.
Because of the cyclic quad, $\angle{X_BX_AX_C}=\angle{X_BHX_C}$. If we let $G=HX_C\cap{H_AH_C}$ and let $I=HX_B\cap{H_AH_B}$, then $GHIH_A$ is a parallelogram, so $\angle{X_BHX_C}=\angle{H_BH_AH_C}$. Thus, $\angle{X_BX_AX_C}=\angle{H_BH_AH_C}$. Similarly, by the diagram below, $JHKH_C$ is a parallelogram, and we get that $\angle{X_AX_CX_B}=\angle{H_AH_CH_B}$. It then follows that $\triangle{X_AX_BX_C}\sim\triangle{H_AH_BH_C}$.
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k.vasilev
25.07.2017 01:02
Let the circumcircle of triangle $ABC$ is $\Gamma$ and let $AH \cap \Gamma=\{A,A_1\},BH \cap \Gamma=\{B,B_1\},CH \cap \Gamma=\{C,C_1\} $. We have, that $AX_A=AB_1=AC_1=AH$ therefore $C_1,X_A,H,B_1$ are concyclic. Analogously $A_1,X_B,C_1,H$ and $B_1,X_C,A_1,H$ are concyclic. Consider inversion $I(H,R)$ , where $R$ is arbitrary radius. The image of $\Gamma$ $\Gamma'$ is the circumcircle of triangle $A'B'C'$. By simple angle chasing we obtain $A_{1}',B_{1}',C_{1}'$ are the midpoints of arcs $\overarc {BC} , \overarc {CA}, \overarc {AB}$ in $\Gamma'$ and $X_{A}'$ is the point of intersection of the line trough $H$, which is parallel to $B'C'$ and $B_{1}'C_{1}'$. So the problem now is:
Given $\triangle ABC$ with incenter $I$ and circumcircle $\omega$ . $AI \cap \omega=M_A , M_B, M_C$ are defined analogously. Point $X_A \in M_{B}M_{C}$ is such that $X_A||BC$ . Points $X_B$ and $X_C$ are defined analogously. Prove that $X_A,X_B,X_C$ are collinear.
Proof: By angle chasing we see that $X_{A}A $ is tangent to $\omega$. Consider $T_{A}T_{B}T_{C}$ the triangle of the intersection points of the tangents to $\omega$ at points $A,B,C$. By Desargue's theorem,applied to $\triangle T_{A}T_{B}T_{C}$ and $\triangle M_{A}M_{B}M_{C}$ the problem is equivalent to $M_{A}T_{A}, M_{B}T{B}, M_{C}T_{C}$ are concurrent. But this is obvious since these $3$ lines are just the perpendicular bisectors of $BC,CA,AB$ respectively, which obviously concur at the center of $\omega$. Hence the conclusion follows.
math_pi_rate
21.07.2018 19:09
Beautiful problem, and a super rich configuration. My solution: Let $O$ be the circumcenter of $\odot (ABC)$. Also, let $\triangle H_AH_BH_C$ be the circumcevian triangle of $H$ w.r.t. $\triangle ABC$. Note that as $H$ is the incenter of $\triangle H_AH_BH_C$, by Fact 5 $A$ is the center of $\odot (HH_BH_C)$. As $AH = AX_A$, $X_A$ lies on $\odot (HH_BH_C)$ All other such results hold cyclically. Also, As both $H_BH_C$ and $HX_A$ are antiparallel to $BC$, we have $HX_A \parallel H_BH_C \Rightarrow HX_AH_BH_C$ is an isosceles trapezium. Similarly, we get that $HX_BH_CH_A$ and $HX_CH_AH_B$ are isosceles trapeziums. Now, By an easy angle chase, one gets that $AO$ bisects $\angle X_AAH$. Using, $AX_A = AH$, we get that $OH = OX_A$. Similarly, $OH = OX_B = OX_C \Rightarrow X_A, X_B, X_C, H$ lie on a circle centered at $O$. Thus, $\measuredangle X_BX_AX_C = \measuredangle X_BHX_C =\measuredangle X_CHH_A + \measuredangle H_AHX_B = \measuredangle H_BH_AH + \measuredangle HH_AH_C = \measuredangle H_BH_AH_C$. Similarly, $\measuredangle X_AX_BX_C = \measuredangle H_AH_BH_C$ and $\measuredangle X_BX_CX_A = \measuredangle H_BH_CH_A$ $\Rightarrow \triangle X_AX_BX_C \sim \triangle H_AH_BH_C$. And, As the orthotriangle of $\triangle ABC$ is directly similar to $\triangle H_AH_BH_C$, we are done.
Om245
29.06.2024 10:50
Nice indeed! Take $-\sqrt {HA.HD}$ inversion. $(BHC) \leftrightarrow EF$ and line tangent to $(BHC)$ at $H$ will stay same. As inversion preserve tangency, hence $EF \parallel $ tangent line. Now it's well know that $AO \perp EF$ if $O$ is circumcenter. therefore $AO \perp HX_A$. As $AH=AX_A$ we get $OH=OX_A$ and similarly for others. Hence $H,X_A,X_B,X_C$ lie on same circle with radius $OH$. It can be seen by easy angle chase that $\triangle{X_AX_BX_C} \sim \triangle DEF$
khanhnx
29.06.2024 11:34
Suppose that $O$ is center of $(ABC)$ and $\triangle DEF$ is orthic triangle of $\triangle ABC$. It's easy to see that $H, X_A, X_B, X_C$ lie on a circle with center $O$. So $(X_AX_B, X_AX_C) \equiv (HX_B, HX_C) \equiv (OB, OC) \equiv (DF, DE) \equiv - (DE, DF) \pmod \pi$. Similarly, we have $(X_BX_C, X_BX_A) \equiv - (EF, ED) \pmod \pi$. Then $\triangle X_AX_BX_C \stackrel{-}{\sim} \triangle DEF$