Junior Balkan Mathematical Olympiad

Let $O$, $P$, $K$, $R$ and $T$ be the mid-points of the segments $CD$, $MN$, $CN$, $CH_1$ and $MH_1$, respectively. From $\triangle MNC$ we have that $\overline{PK} = \frac{1}{2}\overline{MC}$ and $PK \parallel MC$.

Analogously, from $\triangle MH_1C$ we have that $\overline{TR} = \frac{1}{2}\overline{MC}$ and $TR \parallel MC$. Consequently, $\overline{PK} = \overline{TR}$ and $PK \parallel TR$. Also $\overline{OK} \parallel \overline{DN}$ (from $\triangle CDN$) and since $\overline{DN} \perp \overline{BC}$ and $\overline{MH_1} \perp \overline{BC}$, it follows that $\overline{TH_1} \parallel \overline{OK}$. Since $O$ is the circumcenter of $\triangle CMN$, $\overline{OP} \perp \overline{MN}$. Thus, $CH_1 \perp MN$ implies $\overline{OP} \parallel CH_1$. We conclude $\triangle TRH_1 \cong \triangle KPO$ (they have parallel sides and $\overline{TR} = \overline{PK}$), hence $\overline{RH_1} = \overline{PO}$, i.e. $\overline{CH_1} = 2\overline{PO}$ and $CH_1 \parallel PO$.

Analogously, $\overline{DH_2} = 2\overline{PO}$ and $DH_2 \parallel PO$. From $\overline{CH_1} = 2\overline{PO} = \overline{DH_2}$ and $CH_1 \parallel PO \parallel DH_2$ the quadrilateral $CH_1H_2D$ is a parallelogram, thus $\overline{H_1H_2} = \overline{CD}$ and $H_1H_2 \parallel CD$. Therefore the area of the quadrilateral $AH_1BH_2$ is $\frac{\overline{AB} \cdot \overline{H_1H_2}}{2} = \frac{\overline{AB} \cdot \overline{CD}}{2} = S$.

Since $MH_1 \parallel DN$ and $NH_1 \parallel DM$, $MDNH_1$ is a parallelogram. Similarly, $NH_2 \parallel CM$ and $MH_2 \parallel CN$ imply $MCNH_2$ is a parallelogram. Let $P$ be the midpoint of the segment $MN$. Then $\sigma_P(D) = H_1$ and $\sigma_P(C) = H_2$, thus $CD \parallel H_1H_2$ and $\overline{CD} = \overline{H_1H_2}$. From $CD \perp AB$ we deduce $A_{AH_1BH_2} = \frac{1}{2} \overline{AB} \cdot \overline{CD} = S$.