The altitudes $BB_1$ and $CC_1$, are drawn in an acute triangle $ABC$. Let $X$ and $Y$ be the points, which are symmetrical to the points $B_1$ and $C_1$, with respect to the midpoints of the sides$ AB$ and $AC$ of the triangle $ABC$ respectively. Let's denote with $Z$ the point of intersection of the lines $BC$ and $XY$. Prove that the line $ZA$ is tangent to the circumscribed circle of the triangle $AXY$ .
Problem
Source: 2021 3nd Final Mathematical Cup Senior Division P2 FMC
Tags: geometry, tangent
30.10.2022 18:48
We see that $B_{1}$ is symmetrical with X through the midpoint of AB, so $AB_{1}BX$ is a rectangle similarly$ AC_{1}CY$ is a rectangle. Let $D$ be the foot of the perpendicular drawn from$ A$ $\Rightarrow$ $A,D,B,X$ are concyclic $\Rightarrow$ $\angle$$XDA$=$\angle$$XBA$=$\angle$$A$ (1) Do the same : $\angle$$YDA$=$\angle$$A$ Combination (1) and (2) $\Rightarrow$ $\angle$$XDY$=2$\angle$$A$ Let $M,N $be the midpoints of $AB $and $AC$ . Since $OM$ is the midline of $ABC, OM$$\|$ $AC$ $\Rightarrow$ OM $\bot$ $AX$ $\Rightarrow$ $OM$ is the midperpendicular of $AX$ Similar proof: $ON$ is the midperpendicular of $AX$ So $O$ is the center of the circumcircle $AXY$ $\Rightarrow$ $\angle$$XOY$=$2( 180$-$\angle$$XAY$)$ =$2( $\angle$$C_{1}CA+$ $\angle$$B_{1}BA-$ $\angle$$A$ )= $2$$\angle$A Thus X,D,O,Y are concyclic $\Rightarrow$ $\overline{\rm ZX}$.$\overline{\rm ZY}$ =$\overline{\rm ZD}$.$\overline{\rm ZO}$ Let $I,J$ are sequence the intersection of $ZO$ with $(AXY)$ $\Rightarrow$ $\overline{\rm ZX}$.$\overline{\rm ZY}$= $\overline{\rm ZI}$.$\overline{\rm ZJ}$=$\overline{\rm ZD}$.$\overline{\rm ZO}$ According to Maclaurin's formula, we have $(ZD,IJ)=-1$ and have $AI$ $\bot$ $ZO$ $\Rightarrow$ $ZA$ is tangent of $(AXY)$
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