Let $ABC$ be an acute triangle. Let $P$ be a point such that $PB$ and $PC$ are tangent to circumcircle of $ABC$. Let $X$ and $Y$ be variable points on $AB$ and $AC$, respectively, such that $\angle XPY = 2\angle BAC$ and $P$ lies in the interior of triangle $AXY$. Let $Z$ be the reflection of $A$ across $XY$. Prove that the circumcircle of $XYZ$ passes through a fixed point. (Dominik Burek, Poland)
Problem
Source: 2020 Czech-Polish-Slovak Match p6
Tags: geometry, equal angles, fixed, Fixed point
gabrupro
03.02.2021 13:31
Cool problem Lemma: Given a quadrilateral $ABCD$ and point $P$ in the plane of the quadrilateral, there exists a point $Q$, the isogonal conjugate of $P$ with respect to $ABCD$ if and only if $\measuredangle APB = \measuredangle DPC$. Proof: See this Now we claim that the required fixed point is $Q$, the reflection of $A$ across the midpoint of $BC$. Indeed, by the lemma, there exists a point $Q'$, which is the isogonal conjugate of $P$ with respect to $XBCY$. Then $\measuredangle QBC = \measuredangle ABP = \measuredangle ACB$ and similarly, $\measuredangle QCB = \measuredangle ABC$ . Hence $Q = Q'$ as $ABQ'C$ is a parallelogram and so is $ABQC$. But again by the lemma, $\measuredangle BQC = \measuredangle XQY = 180 - \measuredangle BAC$. Therefore, $\measuredangle XZY = \measuredangle CAB = 180 - \measuredangle BAC = \measuredangle XQY$. Thus $Q$ lies on the circumcircle of $XYZ$. $\blacksquare$
Davsch
03.02.2021 14:25
Reflect the circle over $XY$, then the problem is asking us to show that $(AXY)$ passes through a fixed point. However, this follows from the fact that $X$ and $Y$ move linearly on two lines intersecting at $A$.
gabrupro
03.02.2021 15:43
Davsch wrote: Reflect the circle over $XY$, then the problem is asking us to show that $(AXY)$ passes through a fixed point. However, this follows from the fact that $X$ and $Y$ move linearly on two lines intersecting at $A$. Note that $XY$ is not fixed and hence your solution is wrong