IMO Team Selection Contest

The orthocenter $H$ of the triangle $ABC$ is located on the circumcircle of the triangle $CDE$, because $\angle HDC + \angle HEC = 90^\circ + 90^\circ = 180^\circ$ (see fig. 35). Let $\alpha = \angle BAC$; then also $\angle CHE = 90^\circ - \angle ECH = \alpha$. Therefore $\angle MPE = 180^\circ - \angle CPE = 180^\circ - \angle CHE = 180^\circ - \alpha$, from which we get that points $A, M, P, E$ are concyclic. Analogously we see that points $B, M, P, D$ are concyclic.

Point $M$ is the circumcenter of the right triangle $ABE$. Therefore $|ME| = |MA|$ and $\angle MEA = \alpha$, due to which also $\angle MPA = \alpha$. But since $\angle MQB = \angle CQB = \angle CAB = \alpha$, we must have $AP \parallel BQ$. Analogously $BP \parallel AQ$.

Figure 35

In conclusion we get that $APBQ$ is a parallelogram with diagonals $AB$ and $PQ$. As the diagonals of a parallelogram divide each other in half, the desired claim follows.

Similarly to the previous solution we show that points $A, M, P, E$ are located on one circle. Let $\vec{u} \cdot \vec{v}$ be the dot product of vectors $\vec{u}$ and $\vec{v}$. Then
$$\overrightarrow{AC} \cdot \overrightarrow{BC} = \overrightarrow{AC} \cdot \overrightarrow{EC} = \overrightarrow{MC} \cdot \overrightarrow{PC} = \overrightarrow{MC} \cdot (\overrightarrow{MC} - \overrightarrow{MP}) = \overrightarrow{MC} \cdot \overrightarrow{MC} - \overrightarrow{MC} \cdot \overrightarrow{MP}.$$

On the other hand,
$$\begin{align*} \overrightarrow{AC} \cdot \overrightarrow{BC} &= (\overrightarrow{MC} - \overrightarrow{MA}) \cdot (\overrightarrow{MC} - \overrightarrow{MB}) = \\ &= \overrightarrow{MC} \cdot \overrightarrow{MC} - \overrightarrow{MC} \cdot \overrightarrow{MB} - \overrightarrow{MA} \cdot \overrightarrow{MC} + \overrightarrow{MA} \cdot \overrightarrow{MB} = \\ &= \overrightarrow{MC} \cdot \overrightarrow{MC} - \overrightarrow{MC} \cdot (\overrightarrow{MA} + \overrightarrow{MB}) + \overrightarrow{MA} \cdot \overrightarrow{MB}. \end{align*}$$

In conclusion $\overrightarrow{MC} \cdot \overrightarrow{MP} = \overrightarrow{MC} \cdot (\overrightarrow{MA} + \overrightarrow{MB}) - \overrightarrow{MA} \cdot \overrightarrow{MB}$. But as $M$ is the midpoint of $AB$, we have $\overrightarrow{MA} + \overrightarrow{MB} = \vec{0}$, and due to choice of $Q$, we also have $\overrightarrow{MA} \cdot \overrightarrow{MB} = \overrightarrow{MC} \cdot \overrightarrow{MQ}$. Therefore $\overrightarrow{MC} \cdot \overrightarrow{MP} = -\overrightarrow{MC} \cdot \overrightarrow{MQ}$, meaning that

$$\overrightarrow{MC} \cdot (\overrightarrow{MP} + \overrightarrow{MQ}) = 0.$$

As $\overrightarrow{MP}$, $\overrightarrow{MQ}$ and $\overrightarrow{MC}$ have the same direction, the equality is true only if $\overrightarrow{MP} + \overrightarrow{MQ} = \vec{0}$. Therefore $|MP| = |MQ|$.