Balti Tee 1995 · Ülesanne 18

If $|A B|=|B C|$, the points $M, H, P$ and $Q$ coincide and the circle degenerates to a point. We will assume that $|A B|<|B C|$, so that $P$ lies inside the triangle $A B C$, and $Q$ lies outside of it.

Let the line $A P$ intersect $B C$ at $P_{1}$, and let $C Q$ intersect $A B$ at $Q_{1}$. Then $|A P|=\left|P P_{1}\right|$ (since $\triangle A P B \cong$ $\left.\triangle P_{1} P B\right)$, and therefore $M P \| B C$. Similarly, $M Q \| A B$. Therefore $\angle A M Q=\angle B A C$. We have two cases:

(i) $\angle B A C \leq 90^{\circ}$. Then $A, H, P$ and $B$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $\angle B A C=\angle H M Q$. Therefore $H, P, M$ and $Q$ lie on a circle.

(ii) $\angle B A C>90^{\circ}$. Then $A, H, B$ and $P$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $180^{\circ}-\angle H A B=\angle B A C=\angle H M Q$, and therefore $H, P, M$ and $Q$ lie on a circle.

Figure 3