Let $\gamma_A, \gamma_B, \gamma_C$ be excircles of triangle $ABC$, touching the sides $BC$, $CA$, $AB$ respectively. Let $l_A$ denote the common external tangent to $\gamma_B$ and $\gamma_C$ distinct from $BC$. Define $l_B, l_C$ similarly. The tangent from a point $P$ of $l_A$ to $\gamma_B$ distinct from $l_A$ meets $l_C$ at point $X$. Similarly the tangent from $P$ to $\gamma_C$ meets $l_B$ at $Y$. Prove that $XY$ touches $\gamma_A$.
Problem
Source: XVII Sharygin Correspondence Round, P14
Tags: geometry, excircle
amar_04
02.03.2021 19:29
Animate $P$ on $\ell_A$. $P\mapsto X$ and $P\mapsto Y$ are homographies $\implies X\mapsto Y$ is a homography from $\ell_B$ to $\ell_C$. Also $X\cancel{\equiv}Y$ for any input of $P$ on $\ell_A$, if that were true, $\gamma_B,\gamma_C$ would be internally tangent. So by dual of steiner conic theorem $XY$ is tangent to a fixed conic tangent to $\ell_B,\ell_C$. $XY\equiv BC,AB,AC$ when $P=\ell_A\cap BC, \ell_A\cap AB, \ell_A\cap AC$ respectively. Hence, $XY$ is tangent to the conic inscribed in $\{\ell_B,\ell_C,BC,CA,AB\}$ which is $\gamma_A$. Thus, $XY$ is tangent to $\gamma_A$. $\quad\blacksquare$
mathaddiction
05.03.2021 16:28
Moving points for the win! 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Now suppose $BC$ intersect $MN$ at $C_1$. We can now introduce $P$ to the picture (the diagram is on the next page). applying Brianchon's theorem to $OYPQAB$ we have $PB,YA,OQ$ are concurrent at a point $R$. Suppose $CY$ intersect $MN$ at $U$. Then we have the following projections: $$(P,B_1;Q,S)\overset{B}{\to}(R,O;Q,BS\cap OQ)\overset{A}{\to}(Y,O;A_1,M)\overset{C}{\to}(U,C_1;N,M)$$We can do similar manipulations to the other side. By applying Brianchon's theorem to $PXTCAS$, $PC$,$XA$ and $TS$ are concurrent at a point $W$. Then, $$(P,B_1;Q,S)\overset{C}{\to}(W,T;QN\cap ST,S)\overset{A}{\to}(X,T;N,V)\overset{B}{\to}(BX\cap MN,C_1,N,M)$$Comparing them we have $BX\cap MN=U$, hence $MN,CY,BX$ are concurrent. By the converse of Brianchon's theorem we have $YX$ is tangent to $\gamma_A$ as desired. 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meysam1371
11.01.2022 16:56
My solution with moving point method is presented: We show centers of $\gamma_B$ and $\gamma_C$ with $I_{B}$ and $I_{C}$ respectively. $P_A$, $P_B$ and $P_C$ are intersection points of $l_A$ with $BC$, $CA$ and $AB$ respectively. Assume that $l_B$ and $l_C$ intersect with $l_A$ at points $D$ and $E$. And finally, as in the figure, $Q$ and $R$ are intersections of $l_B$ with $AB$, and $l_C$ with $AC$ respectively. By Brianchon's theorem, tangency of $XY$ to $\gamma_A$ is equivalent to concurrency of $BX$, $CY$ and $QR$. So we prove it instead. Consider $P$ is a moving point on $l_A$. So degree of $P$ is $1$. $\gamma_B$ is incircle (or excircle) of triangle $\triangle EPX$ and so angle $\angle PI_{B}X$ is constant. Degree of $I_B$ is zero, so degree of $I_BP$ is $1$. $I_BX$ is obtained by rotation of $I_BP$ around $I_B$ with constant angle $\angle PI_{B}X$ counterclockwise. Thus, degree of $I_BX$ equals $1+0=1$. On the other hand, degree of $l_C$ is zero, so its intersection with $I_BX$, $X$, has degree $1$. With similar argument we conclude that degree of $Y$ equals $1$. Clearly degree of $QR$ is zero. Thus concurrency of $BX$, $CY$ and $QR$ is a problem of degree $1+1+0=2$. So it is enough to prove it in just three special cases. And finally we recommend three cases when $P$ coincides with $P_A$, $P_B$ and $P_C$ which are trivial. Proof is complete! $\blacksquare$
guptaamitu1
02.06.2022 15:15
Does anyone have a solution without using projective maps stuff? Tough, the official solution also uses that technique only, a elementary proof still might exist.
axolotlx7
02.03.2024 20:44
Let me post another solution that doens't use MMP Lemma (Cross ratio of four tangents). Let $c$ be a conic, and let $T_c$ denote the set of tangents to $c$. Then for $\ell, \ell_1, \ell_2, \ell_3, \ell_4 \in T_c$, we have \[ (\ell \cap \ell_1, \ell \cap \ell_2; \ell \cap \ell_3; \ell \cap \ell_4) = (c \cap \ell_1, c \cap \ell_2; c \cap \ell_3, c \cap \ell_4)_c. \]In other words, $(\ell_1, \ell_2; \ell_3; \ell_4)_c$ is well-defined (the dual concept of cross ratio of four points on a conic). Proof of Lemma. When $c$ is a circle, the statement is readily provable with inversion about $c$. Here we provide a projective proof for the general case. Let $P_i = \ell_i \cap c$ for $i = 1, 2, 3, 4$, $P = \ell \cap c$, $A_i = \ell_i \cap \ell$, $B_1 = AA_1 \cap P_2P_3$, $B_4 = AA_4 \cap P_2P_3$, $C_1=PP_1 \cap P_2P_3$, $C_4=PP_4 \cap P_2P_3$, and $Q$ be the point on $c$ such that $(P,Q;P_2,P_3)_c=-1$. Note that $A_1A$ is the polar of $C_1$, so $(B_1,C_1;P_2,P_3)=-1$ and hence $B_1P_1$ passes through $Q$. Similarly $B_4P_4$ passes through $Q$. Now \[ (A_1,A_2;A_3,A_4) \overset{A}= (B_1,P_2;P_3,B_4) \overset{Q}= (P_1,P_2;P_3,P_4)_c \]as desired. Back to the problem. Let $H_A$ be the exsimilicenter of $\gamma_B$ and $\gamma_C$, and define $H_B$, $H_C$ similarly. By Monge, $H_A, H_B, H_C$ are collinear. Now let $P_B = AC \cap l_A$, and $l_B$ meet $BC$ and $AB$ at $J_B$ and $K_B$ respectively, and define $P_C, J_C, K_C$ similarly. As \[ (H_C,X;K_C,J_C) = (P_CH_C,PX;P_BK_C,H_AJ_C)_{\gamma_B} = (P_C,P;P_B,H_A) \]and similarly $(H_B,Y;K_B,J_B) = (P_B,P;P_C,H_A)$, we get $(H_C, CX \cap H_BH_C; H_B, H_A) \overset{C}= (H_C,X;K_C,J_C) = (P_C,P;P_B,H_A) = (K_B,Y;H_B,J_B) \overset{B}= (H_C, BY \cap H_BH_C; H_B, H_A)$, so $BY$ and $CX$ concur on $H_BH_C$. Now Brianchon on $H_CBCH_BYX$ finishes.