AoPS - Problem

Source: Junior Olympiad of Malaysia 2013 P5

Tags: geometry

navi_09220114

20.07.2015 14:34

Consider a triangle $ABC$ with height $AH$ and $H$ on $BC$. Let $\gamma_1$ and $\gamma_2$ be the circles with diameter $BH,CH$ respectively, and let their centers be $O_1$ and $O_2$. Points $X,Y$ lie on $\gamma_1,\gamma_2$ respectively such that $AX,AY$ are tangent to each circle and $X,Y,H$ are all distinct. $P$ is a point such that $PO_1$ is perpendicular to $BX$ and $PO_2$ is perpendicular to $CY$. Prove that the circumcircles of $PXY$ and $AO_1O_2$ are tangent to each other.

$BX \parallel AO_1$ since they are both perpendicular to $XH$. Since $PO_1 \perp BX$ so $AO_1 \perp PO_1$. Similarly, $AO_2 \perp PO_2$. Therefore, the circumcenter of $AO_1O_2$ is the midpoint of $AP$. On the other hand, since $XH \perp O_1P$ so $\angle PO_1O_2= \angle XHO_1$ or $\angle PAO_2= \tfrac 12 \angle XAH$. This follows $AP$ is the bisector of $\angle XAY$. Note that $AX=AY=AH$ so $AP$ is perpendicular bisector of $XY$. Hence, the circumcenter of $PXY$ will lie on $AP$. This means circumcenters of $PXY$ and $AO_1O_2$ and $P$ are collinear. Hence, the circumcircles of $PXY$ and $AO_1O_2$ are tangent to each other.