NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Without loss of generality, we may assume that $AB < AC$ and $AP < AQ$. We shall prove that angles $BAC$ and $PMQ$ have parallel sides, which trivially implies the conclusion.

First approach. Consider the tangency points $E \in AC$ and $F \in AB$ on the inscribed circle. Let $P'$ and $Q'$ be the intersection points of line $AI$ with $DE$ and $DF$, respectively. We intend to prove that $MP' \parallel AC$ and $MQ' \parallel AB$, and afterwards that $P \equiv P'$ and $Q \equiv Q'$.
Notice that $\angle P'IB = 180^\circ - \angle AIB = 90^\circ - \frac{1}{2}\angle ACB$. As the triangle $CED$ is isosceles with base $DE$, we have $\angle P'DC = 90^\circ - \frac{1}{2}\angle ACB$, so $\overline{P'IB} \equiv \overline{P'DC}$. Hence, the quadrilateral $BDP'I$ is cyclic, so $BP' \perp AI$.
Let $N$ be the midpoint of $AB$. As $P'AB$ is a right triangle, we have $P'N = NA = NB$, so $\overline{AP'N} \equiv \overline{P'AN} \equiv \overline{P'AC}$, which leads to $NP' \parallel AC$. Also, we have $MN \parallel AC$, since $MN$ is a midline of triangle $ABC$. It follows that the points $M, N, P'$ are collinear and $MP' \parallel AC$. Hence $\overline{DP'M} \equiv \overline{DEC} \equiv \overline{CDE}$ (because $CDE$ is isosceles), so triangle $MDP'$ is also isosceles, with $MD = MP'$.

A similar reasoning shows that $MQ' \parallel AB$ and $MD = MQ'$. Therefore, $MP' = MQ' = MD$, which means that $M$ is the circumcenter of triangle $DP'Q'$, which finishes our proof.

Another approach. Let $R$ be the foot of the bisector of angle $BAC$. Then $MP = MQ = MD = \frac{a}{2} - (p-b) = \frac{b-c}{2}$, so
$$\frac{PM}{AC} = \frac{b-c}{2b}.$$
On the other hand, $RC = \frac{ab}{b+c}$ and $RM = RC - MC = \frac{ab}{b+c} - \frac{a}{2} = \frac{a(b-c)}{2(b+c)}$, so
$$\frac{RM}{RC} = \frac{b-c}{2b} = \frac{PM}{AC}.$$
Since the triangles $RMP$ and $RCA$ have one common (obtuse) angle $R$, the above relation shows that triangles $RMP$ and $RCA$ are similar (side-side-angle), so $PM \parallel AC$.
Analogously, we have $\frac{RM}{RB} = \frac{MQ}{AB}$, and, since angles $\angle BAR$ and $\angle MQP$ are both acute, the side-side-angle case shows that triangles $RMQ$ and $RAB$ are also similar, so $QM \parallel AB$.

A third approach. Let $P'$ be the projection of $B$ on line $AI$ and $S$ the intersection point of lines $BP'$ and $AC$. Since $AP'$ is both bisector and altitude of triangle $ABS$, it follows that $P'$ is the midpoint of $BS$. Then $MP'$ is the midline of triangle $BCS$, so $MP' \parallel AC$. Also, if $Q'$ is the projection of $C$ on line $AI$, a similar reasoning leads to $MQ' \parallel AB$.
Next, we shall prove that $MP' = MQ' = MD$, which assure us that $P = P'$, $Q = Q'$ and leads to conclusion.

The quadrilateral $BIP'D$ is inscribed in the circle of diameter $AI$, so $\angle DP'Q' = \angle IBD = \frac{1}{2} \angle B$. Then
$$\angle DP'M = \angle DP'Q' + \angle Q'P'M = \frac{1}{2} \angle B + \frac{1}{2} \angle A$$
Since $\angle P'MD = \angle C$, considering the triangle $DP'M$ it results that
$$\begin{aligned} \angle P'DM &= 180^\circ - \angle P'MD - \angle DP'M = 180^\circ - \angle C - \frac{1}{2}(\angle A + \angle B) \\ &= \frac{1}{2}(\angle A + \angle B) = \angle DP'M, \end{aligned}$$
so $MP' = MD$. Similarly, one can prove that $MQ' = MD$.