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Problem 1473

AIME late

Geometry Difficulty 6.0 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

The triangle $ABC$ ( $AB > BC$ ) is inscribed in the circle $\Omega$ . On the sides $AB$ and $BC$ , the points $M$ and $N$ are chosen, respectively, so that $AM = CN$ . The lines $MN$ and $AC$ intersect at point $K$ . Let $P$ be the center of the inscribed circle of triangle $AMK$ , and $Q$ the center of the excircle of the triangle $CNK$ tangent to side $CN$ . Prove that the midpoint of the $\operatorname{arc} ABC$ of the circle $\Omega$ is equidistant from $P$ and $Q$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $T$ be the second intersection of two circles $(BMN)$ and $(O)$ . We have
$\angle TAB = \angle TCB, \quad \angle TMB = \angle TNB,$
and $AM = CN$ , so $\triangle TAM \cong \triangle TCN$ . Then $TA = TB$ , which means that $T$ is the midpoint of the arc $BAC$ of circle $(O)$ .

On the other hand, we also have
$\angle TCK = \angle TBA = \angle TNK,$
then $T N C K$ is cyclic. Similarly, $T M A K$ is also cyclic.

Since $AM = CN$ , it is easy to see that $(KAM)$ and $(KCN)$ are equal. So, if we call $D, E$ the midpoints of $\operatorname{arcs} AM$ and $CN$ of circles $(KAM)$ and $(KCN)$ , respectively, then two isosceles triangles $DAM$ and $ECN$ are congruent. But we know that $D, E$ are also the circumcenters of $\triangle PAM$ and $\triangle QCN$ , so $DP = EQ$ .

At last, from $(KAM)$ and $(KCN)$ being equal, we get $TD = TE$ , then $TP = TQ$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.