Bulgarian National Mathematical Olympiad

We will need the following lemma.

Lemma. The point $O_1$ lies on the line $H_{1}O$, where $O$ is the center of the circumcircle of $△ABC$. The ratio $H_1{O}_1:O_{1}O$ depends on $\angle ACB$ only.

Proof. Let $K$ and $L$ be the intersection points of $A{H}_1$ and $B{H}_1$ with the circumcircle of $△ABC$. Since $H_1$ and $L$ are symmetric with respect to $AC$, the quadrilateral $H_{1}AL{A}_1$ is a rhombus, as $\angle A{H}_1{A}_1=2\angle A{H}_{1}L=2\gamma$. Analogously, $H_{1}BK{B}_1$ is a rhombus with $\angle B{H}_1{B}_1=2\gamma$. Therefore $H_{1}AL{A}_1$ and $H_{1}BK{B}_1$ are similar, whence $$\frac{H_{1}T}{H_{1}X}=\frac{H_{1}Q}{H_{1}Y}=\frac{1}{1+2\cos 2\gamma }$$ This means that $O_1$ belongs to the line $H_{1}O$ and the ratio $H_1{O}_1:O_{1}O$ depends on $\angle ACB$ only.

It is clear that $C{H}_1{H}_{2}D$ is a parallelogram (it follows from $C{H}_1=C{H}_2=2R\cos \gamma$) and the points $O_1$ and $O_2$ are homothetic to $H_1$ and $H_2$, respectively, with respect to $O$. It follows from the lemma that the ratios of these two homotheties are equal (since they depend on $\gamma$ only). Therefore $O_1{O}_2\parallel H_1{H}_2\parallel CD$, i.e. the line $l_{AB}$ is parallel to $CD$.

We obtain that the sides of $MNPQ$ are parallel to the corresponding sides of $ABCD$. This means that $MNPQ$ is inscribed.