Iranian Mathematical Olympiad

From $O$, we draw a perpendicular to $AM$ which intersects $AD$ and $BC$ at $P'$ and $X$ respectively. Since $O$ and $E$ are both on the perpendicular bisector of $MD$, then $OE \perp MD$. Since $E$ is the orthocenter of $\triangle OXM$ (this implies $XE \perp OM$, $ME \perp OX$, and $OE \perp XM$), and we have $OE \perp MD$, it implies $XM$ and $MD$ are the same line. Consequently $X$, $M$, $D$ are collinear. So we have:
$$XP' \cdot XO = XD \cdot XM$$
Also, $XD \cdot XM = XB \cdot XC$. Thus, $XP' \cdot XO = XB \cdot XC$. This implies $OP'BC$ is cyclic. So:
$$\begin{aligned} \angle P'BC &= 180^\circ - \angle P'OC \\ &= 180^\circ - (\angle P'OM + \angle MOC) \\ &= 180^\circ - (90^\circ - (\frac{B-C}{2}) + \angle A) \\ &= 90^\circ + (\frac{B-C}{2}) - \angle A \end{aligned}$$
$$\Rightarrow \angle P'BO = \angle P'BC - (90^\circ - \angle A) = \frac{B-C}{2}$$
$$\Rightarrow \angle P'BA = \angle OBA - \angle OBP' = (90^\circ - C) - \frac{B-C}{2} = (90^\circ - B) + \frac{B-C}{2} = \angle P'CA$$

Now we show that point $P$ on $AD$ is unique. Assume for contradiction that this is not the case, and for two distinct points $Y$ and $Z$ on $AD$ ($Y \neq Z$) we have $\angle YBA = \angle YCA$ and $\angle ZBA = \angle ZCA$. Then we have:
$$\frac{ZY}{YA} = \frac{BZ}{BA} \cdot \frac{\sin \angle ZBY}{\sin \angle YBA} = \frac{CZ}{CA} \cdot \frac{\sin \angle ZCY}{\sin \angle YCA}$$
Given $\angle YBA = \angle YCA$ and $\angle ZBA = \angle ZCA$. If we assume $\angle ZBY = \angle ZCY$, then:
$$\frac{BZ}{BA} = \frac{CZ}{CA}$$
$$\Leftrightarrow \frac{BZ}{BA} = \frac{CZ}{CA} \Leftrightarrow \frac{BZ}{CZ} = \frac{BA}{CA}$$
So $Z$, and similarly $Y$, lie on the Apollonian circle with respect to vertex $A$ and segment $BC$ (locus of points $K$ such that $KB/KC = AB/AC$). Since $Y$, $Z$ are on $AD$, and also on this Apollonian circle, this leads to a contradiction if $Y$, $Z \neq A$ and $AD$ is not that Apollonian circle. (Note that $Y \neq A$ and $Z \neq A$. Also, if one of these points lies on $BC$, it implies $AB = AC$, which contradicts the problem's assumption if $AB \neq AC$.)
Thus $P \equiv P'$ and the proof is complete. ■