Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.3 National olympiad Prove it Ukraine

In triangle $ABC$ , $H$ is the orthocenter and $AK$ is an altitude. Circle $w$ goes through points $A$ and $K$ and intersects sides $AB$ and $AC$ at points $M$ and $N$ respectively. The line through point $A$ parallel to $BC$ intersects the circumscribed circles of triangles $AHM$ and $AHN$ a second time at points $X$ and $Y$ respectively. Prove that $XY = BC$ .

Solution

Let $Z$ be the point of intersection of circle $w$ with line $BC$ , then $AZ$ is the diameter of $w$ (see Fig. 35).

Really, if $K = Z$ , $w$ is tangent to $BC$ , so, as $AK \perp BC$ , the center of $w$ is on $AK$ . If $K \neq Z$ , then $\angle AKZ = 90^\circ$ and $AZ$ is the diameter of $w$ . Then $\angle AMZ = 90^\circ$ .

Let line $MZ$ intersect the circumscribed circle of $\triangle XAH$ a second time at point $S$ , then $\angle AMS = 90^\circ$ . Also $\angle SHA = \angle SMA$ , then $SH \perp AH$ . So $AX \parallel SH \parallel BC$ , it also follows that $XAHS$ is a rectangle. So, $AX = SH$ .

Also notice that $CH \perp AB$ . Apart from that, $ZM \perp AB$ . So $HC \parallel SZ$ . Then $SHCZ$ is a parallelogram, and so $SH = ZC$ . Then $AX = SZ$ . Also, $AY = BZ$ , so $XY = BC$ .

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