In the isosceles triangle $ABC$ where $AB = AC$, the point $I{}$ is the center of the inscribed circle. Through the point $A{}$ all the rays lying inside the angle $BAC$ are drawn. For each such ray, we denote by $X{}$ and $Y{}$ the points of intersection with the arc $BIC$ and the straight line $BC$ respectively. The circle $\gamma$ passing through $X{}$ and $Y{}$, which touches the arc $BIC$ at the point $X{}$ is considered. Prove that all the circles $\gamma$ pass through a fixed point.
Problem
Source: Russian TST 2015, Day 7 P2
Tags: geometry, circles
ElevenCubed
23.04.2023 05:38
Let $X'$ denote the other intersection between the ray and the circumcircle $\Omega$ to $BIC$. Let $\gamma'$ denote the circle through $Y$ tangent to $\Omega$ in $X'$. Let $Y'$ denote the intersection between $\gamma$ and $\gamma'$ different from $Y$. I claim that $Y'$ is the midpoint of $BC$, and thus all circles $\gamma$ will pass through the fixed point $Y'$. Let $P \in BC$ denote the pole to the ray wrt $\Omega$. It follows from the radical axis theorem on $\Omega, \gamma, \gamma'$ that $Y' \in BC$. Since $\angle PXY' + \angle PX'Y' = \angle PYX + \angle PYX' = 180^\circ$, quadrilateral $PX'Y'X$ is inscribed. By inversion in $P$, $(Y', \infty, B, C) = (Y, P, C, B) = -1)$ and we are done.
keglesnit
23.04.2023 14:28
Let $K$ be the center of $(BIC)$ and $M$ the midpoint of $BC$. Note that $K$ is the $A$-antipode in $(ABC)$ and hence $AB$ and $AC$ are tangents to $(K)$. Now this implies that $XY$ is the symmedian in triangle $BXC$ so the circle $(XYM)$ is tangent to $(K)$. Thus the desired fix point is $M$.