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Iranian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Iran

Given an acute-angled triangle $ABC$ with an altitude $AD$ and orthocenter $H$ . Let $E$ be the reflection of $H$ with respect to $A$ . A point $X$ lies on the circumcircle of the triangle $BDE$ such that $DX \parallel AC$ , and similarly a point $Y$ lies on the circumcircle of the triangle $CDE$ such that $DY \parallel AB$ . Prove that the circumcircles of triangles $AXY$ and $ABC$ are tangent to each other.

Solution

Denote by $\Gamma$ the circumcircle of the triangle $ABC$ . Letting $F$ and $G$ be the reflections of $H$ with respect to $AC$ and $AB$ , respectively. It is well-known that $F$ and $G$ lie on $\Gamma$ . Also $EF$ is parallel to $AC$ since $A$ is the midpoint of the segment $EH$ . Therefore $\angle EFB = 90^\circ$ and the pentagon $EFDBX$ is cyclic. Now notice that
$\angle XFB = \angle XDB = \angle ACB = \angle AFB,$

Yielding that the line $FX$ passes through $A$ . Then, since $AF = AH = AE$ we should have $AD = AX$ . Similarly one can show that $G$ , $A$ , and $Y$ are collinear and $AD = AY$ . So $AX = AY$ and $AF = AG$ . This yields that $XY \parallel FG$ and
$\angle XAG = \angle AYX + \angle AXY = \angle AYX + \angle AFG.$
Hence the result follows. ■

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