Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Taiwan

In a scalene triangle $ABC$ with incenter $I$ , the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$ . The tangents to the circumcircle of $\triangle AEF$ at $E$ and $F$ meet at $S$ . Lines $EF$ and $BC$ intersect at $T$ . Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of $\triangle BIC$ .

Solution

Solution: Let $D$ be the foot of the perpendicular from $I$ to $BC$ . Let $X, Y$ be the feet of the perpendiculars from $B, C$ to $CI, BI$ respectively. One can show that $BIFX, CIEY$ are concyclic, hence $X, Y$ lie on line $EF$ . Let $M$ be the midpoint of $BC$ , and let $\omega$ be the circumcircle of $DMXY$ . The original problem is equivalent to proving that $T$ lies on the polar of $S$ with respect to $\omega$ .

Let $K$ be the intersection of $AM$ and $EF$ ; by SL 2005 G6, $K, I, D$ are collinear.

Let $N$ be the midpoint of $EF$ , and $L$ the intersection of $KS$ and $BC$ . From
$-1 = (A, I; N, S) \stackrel{K}{=} (T, L; M, D)$
and
$-1 = (T, D; B, C) \stackrel{I}{=} (T, K; Y, X)$
we get that $T = MD \cap YX$ is the pole of line $KL$ with respect to $\omega$ , as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.