Mongolian Mathematical Olympiad 46

Let us denote the feet of the perpendicular from $M$ to $AC$ by $R$. Then $P$, $Q$ and $R$ are collinear (Simson's line).

If $\angle CAM = \varphi_1$, $\angle ACM = \varphi_2$ and $\angle ABC = \beta$, then $\varphi_1 + \varphi_2 = \beta$.

We first show that $S_{PHQ} = S_{PMQ}$. Since
$$S_{PMQ} = S_{BPMQ} - S_{BPQ} \text{ and } S_{PHQ} = S_{BPQ} - S_{BHQ} - S_{BHP},$$
it suffices to prove
$$S_{BPQ} = \frac{S_{PQMP} + S_{BHQ} + S_{BHP}}{2} = S_{PMQ} + S_{BHQ} + S_{BHP} \quad (1).$$
$\frac{BM}{\sin(\gamma + \varphi_2)} = 2R$ and $MQ = BM \cdot \sin \varphi_1$. From this $MQ = 2R \sin \varphi_1 \sin(\gamma + \varphi_2)$. And similarly $MP = 2R \sin \varphi_2 \sin(\alpha + \varphi_1)$, $BQ = 2R \cos \varphi_1 \sin(\gamma + \varphi_2)$, $BP = 2R \cos \varphi_2 \sin(\alpha + \varphi_1)$ (here $\sin(\alpha + \varphi_2) = \sin(\gamma + \varphi_2)$).

$HA_1 = BH \cos \gamma = 2R \cos \beta \cos \gamma$, $HC_1 = 2R \cos \alpha \cos \beta$. From this the (1) equality is equivalent to $\sin \beta(BP \cdot BQ - MP \cdot MQ) = HA_1 \cdot BQ + HC_1 \cdot BP$.
$$\begin{aligned} & \sin \beta(4R^2 \sin(\alpha + \varphi_1) \sin(\gamma + \varphi_2))(\cos \varphi_1 \cos \varphi_2 - \sin \varphi_1 \sin \varphi_2) = \\ & = 4R^2 \cos \beta(\cos \gamma \cos \varphi_1 \sin(\gamma + \varphi_2) + \cos \alpha \cos \varphi_2 \sin(\alpha + \varphi_1)) \\ & \sin \beta \sin(\alpha + \varphi_1) = \cos \gamma \cos \varphi_1 + \cos \alpha \cos \varphi_2 \\ & \sin \beta \sin \alpha \cos \varphi_1 + \sin \beta \cos \alpha \sin \varphi_1 = \cos \varphi_1(\sin \alpha \sin \beta - \cos \alpha \cos \beta) + \end{aligned}$$
$$+ \cos \alpha (\cos \beta \cos \varphi_1 + \sin \beta \sin \varphi_1) = \cos \alpha (\sin \beta \sin \varphi_1 + \cos \beta \cos \varphi_1) \\ \sin \beta \sin \alpha \cos \varphi_1 = \sin \alpha \sin \beta \cos \varphi_1$$
So $S_{PHQ} = S_{PMQ}$, implies that $PQ$ intersects $HM$ at the midpoint of $HM$.