Japan Mathematical Olympiad

Let us denote by $(XYZ)$ the directed angle determined by the 3 points $X,Y,Z$. More precisely, we set $(XYZ)$ to be equal to $\alpha$ if the line $XY$ comes on top of the line $YZ$ when $XY$ is rotated around $Y$ counterclockwise by $\alpha$ degrees. When an integral multiple of $180^{\circ }$ is added to such an $\alpha$, the same condition on the line $XY$ will be satisfied, so, in the sequel we identify two values differing by an integral multiple of $180^{\circ }$, when we refer to directed angles. Let us prove the following simple lemma.

Lemma. If four distinct points $X,Y,Z,W$ lie on the circumference of a same circle, then $(XZY)=(XWY)$ holds.

Proof. The circumference of the circle are divided into 2 disjoint arcs by the points $X,Y$. If the 2 points $Z,W$ lie on a same arc given by this division, then by the well-known theorem on inscribed angles, we have $\angle XZY=\angle XWY$, and since these angles have the same direction, we have $(XZY)=(XWY)$ in this case. On the other hand, if $Z$ and $W$ lie on different arcs, from a well-known property of a quadrilateral inscribed in a circle, we can conclude that $\angle XZY=180^{\circ }-\angle XWY$. Since the direction of these angles are reversed in this situation, we have $(XZY)=180^{\circ }-(-(XWY))=(XWY)$ in this case also.

Let us denote by $M,N$ the midpoints of the sides $AB$ and $AC$ of the triangle $ABC$, respectively. From the fact that $OM\perp AM,ON\perp AN$ it follows that the four points $A,O,M,N$ lie on the circumference of a same circle, and hence by Lemma, we have $(MON)=(MAN)$. Since $A,O,P,Q$ also lie on the circumference of a same circle, we also have $(POQ)=(PAQ)$. Furthermore, since the three points $A,P,M$ lie on a same line and so do the three points $A,Q,N$, we have $(MAN)=(PAQ)$, and therefore, we conclude that $(MON)=(POQ)$, from which we obtain $(MOP)=(MON)-(PON)=(POQ)-(PON)=(NOQ)$.

If we denote by $B^{\prime }$ the point situated symmetrically to the point $B$ with respect to the point $P$, then we have ${\overrightarrow{AB}}^{\prime }=2\overrightarrow{MP}$ since $\overrightarrow{BA}=2\overrightarrow{BM}$ and ${\overrightarrow{BB}}^{\prime }=2\overrightarrow{BP}$ hold. Similarly, if we let $C^{\prime }$ be the point situated symmetrically to the point $C$ with respect to the point $Q$, then we have ${\overrightarrow{AC}}^{\prime }=2\overrightarrow{NQ}$.

Since $\overrightarrow{MP}\cdot (\overrightarrow{NQ})$ coincides with the vector obtained by rotating $\overrightarrow{OM}\cdot (\overrightarrow{ON})$, respectively counterclockwise by $90^{\circ }$ and multiplied by $\tan (MOP)(=\tan (NOQ))$, we can conclude that ${\overrightarrow{AB}}^{\prime }\cdot ({\overrightarrow{AC}}^{\prime })$ coincides with the vector obtained by rotating $\overrightarrow{OM}\cdot (\overrightarrow{ON})$, respectively counterclockwise by $90^{\circ }$ and multiplied by $2\tan (MOP)$. Therefore, ${\overrightarrow{B^{\prime }C}}^{\prime }$ coincides with the vector obtained by rotating $\overrightarrow{MN}$ counterclockwise by $90^{\circ }$ and multiplied by $2\tan (MOP)$. In particular, we have ${\overrightarrow{B^{\prime }C}}^{\prime }\perp \overrightarrow{BC}$, since $\overrightarrow{MN}\parallel \overrightarrow{MN}$.

Denote by $H$ the point of intersection of the lines $B^{\prime }{C}^{\prime }$ and $BC$, and let $P^{\prime }(Q^{\prime })$ be the foot of the perpendicular line drawn from $P(Q)$, respectively, to the line $BC$. Then, since $P^{\prime }(Q^{\prime })$, respectively, is the midpoint of the line segment $HB(HC$, respectively), we have $P^{\prime }{Q}^{\prime }=\frac{1}{2}BC=\frac{1}{2}PQ$.

If we let $\theta$ be the angle formed by the lines $PQ$ and $BC$ ($0^{\circ }\le \theta \le 90^{\circ }$ and $\theta =0$ if the two lines are parallel), then we see that $P^{\prime }{Q}^{\prime }=PQ\cos \theta$ holds so that we get $\cos \theta =\frac{1}{2}$ and therefore, $\theta =60^{\circ }$.