Ukrainian Mathematical Olympiad

Let $A{A}_1$ be the altitude of triangle $ABC$. Consider the case when the lines $BQ$ and $CP$ intersect at some point $T$. We need to prove that the point $T$ lies on the line $A{A}_1$. Suppose first that points $A$ and $P$ lie on the same side of line $BC$, and points $A$ and $Q$ lie on different sides of line $BC$ (see figure).

As is known, the points symmetric to the orthocenter of a triangle with respect to the midpoints of its sides lie on the circumcircle of this triangle: indeed, if in triangle $ABC$ the point $F$ is symmetric to the orthocenter $H$ with respect to the midpoint $N$ of side $AC$, then $\angle AFC=\angle AHC=180^{\circ }-\angle ABC$. Further, since $FC\parallel AH$, we have $\angle FCB=90^{\circ }$, and points $B$ and $F$ are diametrically opposite. Hence, $HQ\perp BT$. Similarly, $HP\perp CT$.

The quadrilaterals $HQTP$ and $BQ{A}_{1}H$ are cyclic. Therefore, we have: $\angle QBC=\angle QH{A}_1$, $\angle QHT=\angle QPC$. Since $\angle QPC=\angle QBC$, it follows that $\angle QHT=\angle QH{A}_1$, which completes the proof.

If the chord $PQ$ and point $A$ lie on different sides of line $BC$, then $\angle QBC=\angle QH{A}_1$, $\angle QHT=\angle QPT$, $\angle QPT=\frac{\angle QP}{2}+\frac{\angle PC}{2}=\angle QBC$. Other cases of the arrangement of points $P$ and $Q$ on the circumcircle of triangle $ABC$ are considered similarly.

If the lines $BQ$ and $CP$ are parallel, it is not difficult to prove that the point $H$ lies on the segment $MN$, points $P$ and $F$ coincide, and therefore $AH\parallel PC$.

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Alternative solution.

Let $\omega$ be the circumcircle of triangle $ABC$, and ${\omega }_B$ and ${\omega }_C$ be the circles constructed on diameters $HB$ and $HC$ respectively. Since $\angle HPC=90^{\circ }$, we have $P\in {\omega }_C$, and the line $CP$ is the radical axis of circles $\omega$ and ${\omega }_C$. Similarly, $BQ$ is the radical axis of circles $\omega$ and ${\omega }_B$. Further, the line $A{A}_1$ is the radical axis of circles ${\omega }_B$ and ${\omega }_C$. Therefore, the statement of the problem follows from the theorem on the radical axes of three circles.

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Alternative solution.

(This solution was proposed by olympiad participant Anna Mitrushchenkova.)

The case when points $P$ and $F$ coincide, i.e., $AH\parallel PC$, was considered in Solution 1. Let $Y$ be such a point that $H$ is the midpoint of segment $AY$. Then $BY\parallel PH$, and $CP\perp BY$. Similarly, $BQ\perp CY$. Moreover, $AH\perp BC$. The required result follows from the concurrency of the altitudes of triangle $BCY$.