Japan Junior Mathematical Olympiad

Let $B^{\prime }$ be the point symmetric to $B$ with respect to $P$, and $C^{\prime }$ be the point symmetric to $C$ with respect to $Q$. Since $B$ is the mid-point of the line segment $P{P}^{\prime }$ and $M$ is the mid-point of the line segment $BC$ we see that $PM//B{C}^{\prime }$, and similarly, we have $QM//C^{\prime }B$. Thus, it suffices to show that $B^{\prime }C\perp C^{\prime }B$ holds.

From $AP=BP=B^{\prime }P$, we see that $P$ is the orthocenter of the triangle $AB{B}^{\prime }$, and therefore, $B{B}^{\prime }$ is a diameter of the circum-circle of the triangle $AB{B}^{\prime }$. Thus, we have $\angle BA{B}^{\prime }=90^{\circ }$, from which we get $\angle CA{B}^{\prime }=\angle BA{B}^{\prime }-\angle BAC=90^{\circ }-30^{\circ }=60^{\circ }$. We also have $\angle CX{B}^{\prime }=\angle XBC+\angle XCB=60^{\circ }$. Consequently, we see that the 4 points $A,X,C,B^{\prime }$ lie on the circumference of a same circle, and this implies, in particular that $\angle XAC=\angle X{B}^{\prime }C$.

Similarly, we get the fact that $Q$ is the orthocenter of the triangle $A{C}^{\prime }B$, from which we obtain $\angle C^{\prime }AB=\angle C^{\prime }XB=60^{\circ }$ and $\angle XAB=\angle X{C}^{\prime }B$. From these, it follows that we also have $\angle C^{\prime }A{B}^{\prime }=\angle C^{\prime }AB+\angle BAC+\angle CA{B}^{\prime }=60^{\circ }+30^{\circ }+60^{\circ }=150^{\circ }$, $\angle C^{\prime }X{B}^{\prime }=\angle BXC=180^{\circ }-60^{\circ }=120^{\circ }$ and $\angle X{C}^{\prime }A+\angle X{B}^{\prime }A=360^{\circ }-(\angle C^{\prime }A{B}^{\prime }+\angle C^{\prime }X{B}^{\prime })=360^{\circ }-(150^{\circ }+120^{\circ })=90^{\circ }$.

Furthermore, we get $$\begin{array}{rl}\angle B{C}^{\prime }A+\angle C{B}^{\prime }A& =(\angle B{C}^{\prime }X+\angle X{C}^{\prime }A)+(\angle X{B}^{\prime }C+\angle X{B}^{\prime }A)\\ & =(\angle XAB+\angle XAC)+(\angle X{C}^{\prime }A+\angle X{B}^{\prime }A)\\ & =30^{\circ }+90^{\circ }=120^{\circ }.\end{array}$$

If we let $D$ be the point of intersection of the lines $B^{\prime }C$ and $C^{\prime }B$, then we get, putting together the facts obtained above $$\begin{array}{rl}\angle C^{\prime }D{B}^{\prime }& =360^{\circ }-(\angle D{C}^{\prime }A+\angle C^{\prime }A{B}^{\prime }+\angle A{B}^{\prime }D)\\ & =360^{\circ }-(150^{\circ }+\angle B{C}^{\prime }A+\angle C{B}^{\prime }A)\\ & =360^{\circ }-(150^{\circ }+120^{\circ })=90^{\circ },\end{array}$$ which implies that $B^{\prime }C\perp C^{\prime }B$.