Benelux Mathematical Olympiad competition problems

Solution:

We claim $|Q R| = |B C|$, which will clearly imply that quantity $|Q R|$ is independent from the position of $P$ inside triangle $\triangle A B C$ (and independent from the position of $A$).
This equality will follow from the equality between triangles $\triangle B P C$ and $\triangle R B Q$. This in turn will be shown by means of three equalities (two sides and an angle): $|B P| = |R B|$, $|P C| = |B Q|$ and $\angle B P C = \angle R B Q$.

a. $|B P| = |R B|$
Obvious since $R$ is the reflection of $P$ in a line going through $B$.

b. $|P C| = |B Q|$
Let $U$ be the fourth vertex of parallelogram $B P C U$. Then $U$ is on line $P Q$ and $\angle B U P = \angle U P C = \alpha$. If $Q$ is on the same $\operatorname{arc} P B$ as $A$, then $\angle B Q P = \alpha$, and $\triangle Q P U$ is isosceles; hence, $|B Q| = |B U| = |P C|$. On the other way, if $Q$ is on the other $\operatorname{arc} P B$, then $\angle B Q P$ and $\alpha$ are supplementary, hence $B Q U = \alpha$, and again $\triangle Q P U$ is isosceles; the same conclusion follows.

c. $\angle B P C = \angle R B Q$
Define $T$ to be the midpoint of $P R$. Then line $B T$, tangent to circle $\Gamma$ in $B$, splits $\angle R B Q$ into two parts, $\angle R B T$ and $\angle T B Q$.
We first show that $\angle R B T = \alpha$. Indeed, by symmetry, $\angle R B T = \angle P B T$ and, since $B T$ is tangent to $\Gamma$, we have that $\angle P B T = \angle P A B$ (because they both intercept the same $\operatorname{arc} \widehat{P B}$ on circle $\Gamma$), from which our claim follows.
We then show that $\angle T B Q = \angle B P M$. Indeed, since $\angle T B Q$ and $\angle B P Q$ intercept opposite arcs on circle $\Gamma$, they are supplementary and we have $\angle T B Q = \pi - \angle B P Q = \angle B P M$. We finally conclude that
$$\angle R B Q = \angle R B T + \angle T B Q = \alpha + \angle B P M = \angle M P C + \angle B P M = \angle B P C$$
We have thus shown $\triangle B P C = \triangle R B Q$, which completes the proof.

Alternative 1 for (b). The law of sines in triangle $\triangle B Q M$ gives
$$\frac{|B M|}{\sin \angle B Q M} = \frac{|B Q|}{\sin \angle B M Q}.$$
Since $Q$ belongs to circle $\Gamma$, we have either $\angle B Q P = \angle B A P = \alpha$, hence $\angle B Q M = \angle M P C$, or these angles are supplementary; in both cases they have equal sines. We also have that $\angle B M Q$ and $\angle C M P$ are supplementary, hence have equal sines. Using these facts along with $|B M| = |M C|$ transforms (2) into
$$\frac{|M C|}{\sin \angle M P C} = \frac{|B Q|}{\sin \angle C M P}$$
from which the law of sines in triangle $\triangle C P M$ implies that $|B Q| = |P C|$.

Alternative 2 for (b). Let $S$ be the second intersection of line $C P$ with circle $\Gamma$. Then, $\angle B S P = \alpha$, so $B S$ and $M P$ are parallel; since $M$ is the midpoint of segment $B C$, $P$ is the midpoint of $S C$. If $Q$ is on the same $\operatorname{arc} P B$ as $A$, then the quadrilateral $Q P B S$ is an isosceles trapezoid, and $|Q B| = |S P| = |P C|$. If $Q$ is on the other arc $P B$, then the quadrilateral $P Q B S$ is an isosceles trapezoid, and again $|Q B| = |S P| = |P C|$.