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#1 h Mar 29, 2025, 5:58 PM

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Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$ , with $AB < AC$ . The incircle $(I)$ touches the sides $BC$ , $CA$ , and $AB$ at $D$ , $E$ , and $F$ , respectively. A line through $I$ , perpendicular to $AI$ , intersects $BC$ , $CA$ , and $AB$ at $X$ , $Y$ , and $Z$ , respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$ ). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$ ). Let $P$ be the midpoint of the arc $BAC$ of $(O)$ . The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$ , respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$ . Prove that $AT \perp BC$ .

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#2 h Mar 30, 2025, 2:21 PM

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bump!!!!

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#3 h Mar 30, 2025, 7:48 PM • 1 Y

Y by hanzo.ei

I dont understand why there are so many unnecessary points defined, but oh well
Let $S$ be the midpoint of $BC$ , $R$ be the midpoint of $AH$ where $H$ is the foot of the altitude, $AI\cap BC=P,$ , let $V$ be the point on $I$ such that $SV=SD$ . Let $AV\cap BC=Q$ . It is well known that $\angle AVD=90^\circ$ Let $U=AI\cap DV$ , and $AT\cap BC = H'$
We first claim that $V$ lies on $(DKL)$ , and that the center is the midpoint of $BC$ . First, consider inversion about $I$ . $K,L$ are inverses, because $L$ is mapped to $AI\cap (CDEI)$ which is $K$ by Iran lemma. This implies that $(DKL)$ is orthogonal to $I$ , so the center of $DKL$ lies on $BC$ as $ID\perp BC$ . Also by Iran lemma, $BL\perp AI, CK\perp AI\implies BL\parallel CK$ . So if we now consider the perpendicular bisector of $KL$ it passes through $S$ . Thus, $S$ is the center of $(DKL)$ . Now, by definition, $V$ lies on this circle aswell. Again, it is well known that $Q$ is the $A$ -extouch point, so it also lies on $(DKL)$ . Note that $(L,K;D,V)=-1$ , so $T$ lies on $DV$ , and $(T, O;D,V)=-1$ . Now, as
$-1=(T,O;D,V) \overset{A}{=}(H',P;D,Q)$ and as its well known that $R,I,Q$ lie on a line, we also have:
$-1=(H,A;\infty_{\perp BC},R)\overset{I}{=}(H,P;D,Q)$ It follows that $H=H'$ as desired.
I believe that most of the well known lemmas i stated throughout the solution are in EGMO chapter 4

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