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Geometry Difficulty 8.7 Shortlist Prove it Balkan Mathematical Olympiad

Let $ABC$ be a triangle with circumcircle $\omega$ , circumcenter $O$ , and orthocenter $H$ . Let $K$ be the midpoint of $AH$ . The perpendicular to $OK$ at $K$ intersects $AB$ and $AC$ at $P$ and $Q$ , respectively. The lines $BK$ and $CK$ intersect $\omega$ again at $X$ and $Y$ , respectively. Prove that the second intersection of the circumcircles of triangles $KPY$ and $KQX$ lies on $\omega$ .

Solution

Claim 1. $PK = KQ$ .
Proof of Claim 1. Let $L$ and $N$ be the midpoints of $AB$ and $AC$ , respectively. Since $L$ and $K$ are midpoints of $AB$ and $AH$ , then $LK \parallel BH$ and so $LK \perp AC$ . Since also $LO \perp AB$ , then $\angle KLO = \angle BAC = \alpha$ . Also, $\angle OLP = 90^\circ = \angle OKP$ , so the quadrilateral $OKLP$ is cyclic and therefore $\angle KPO = \angle KLO = \alpha$ . Similarly, $\angle KQO = \alpha$ . Therefore, the triangle $OPQ$ is isosceles, and since $OK \perp PQ$ then $K$ is the midpoint of $PQ$ . $\square$

Claim 2. The intersection of $YP$ and $AK$ lies on $\omega$ .
Proof of Claim 2. Let $D$ be the other point of intersection of $AK$ with $\omega$ . Since $OK \perp PQ$ , then $K$ is the midpoint of the chord $\ell$ of $\omega$ through $P, Q$ . Since $AD$ and $YC$ intersect at $K$ , by the Butterfly theorem the points $P' = YD \cap \ell$ and $Q = AC \cap \ell$ are equidistant from $K$ . Thus $P' = P$ and $YP \cap AK = D \in \omega$ . $\square$

Now let $A'$ be the other point of intersection of $AO$ with $\omega$ and let $S$ be the other point of intersection of $A'H$ with $\omega$ .

Claim 3. $PQ$ is the perpendicular bisector of $HS$ .
Proof of Claim 3. We have $\angle HSA = \angle A'SA = 90^\circ$ so $S$ lies on the circle $\omega'$ with diameter $AH$ centered at $K$ . So $KS = KH$ . Since $O$ and $K$ are the circumcenters of $\omega$ and $\omega'$ respectively, and $AS$ is their common chord, then $OK \perp AS$ . But we also have $HS \perp AS$ and $PQ \perp OK$ , thus $HS \perp PQ$ . Since $KS = KH$ , then $K$ belongs on the perpendicular bisector of $HS$ and the result follows. $\square$

From Claim 3 we have $\angle KSP = \angle KHP$ . From Claim 1 and the fact that $KP = KQ$ we have that $AQPH$ is a parallelogram and so (using Claim 2 as well)
$\angle KHP = \angle KAC = \angle DAC = \angle DYC = \angle PYK.$

Since $\angle KSP = \angle KYP$ we get that $S$ belongs on the circumcircle of triangle $KPY$ . Similarly it belongs to the circumcircle of triangle $KQX$ and therefore the result follows.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.