42nd Balkan Mathematical Olympiad

Let $M$ be the midpoint of $BC$ and let $D$ be the reflection of $C$ in $P$. From the length condition in the statement, we have $DA=DP-AP=DP-(PC-AB)=AB$. This together with $PM$ being the midline in $△CDB$ implies $\angle MPC=\angle BDA=\frac{\angle BAC}{2}=\angle MBA$. So $ABMP$ is cyclic. Denote its circumcircle by $\omega$.

Next, let $R$ denote the intersection of $AB$ and $PM$. We have $\angle RMB=\angle PMB=180^{\circ }-\angle BAP=180^{\circ }-2\angle MBR$, so $MR=MB$. Since $MB=MC$ as well, $M$ is the circumcentre of $△BRC$ and $\angle BRC=90^{\circ }$. (*)



Let $\Gamma$ denote the circumcircle of $△CBD$, and let $W$ be the centre of $\Gamma$. Let $N$ be the midpoint of arc $\text{overarc}CBD$ of $\Gamma$. We showed that $AB$ bisects $\angle CBD$, so $AB\perp BN$. Also, since $P$ is the midpoint of $CD$, $\angle NPA=90^{\circ }$ which shows that $N$ lies on $\omega$. Thus, $NB$ is the radical axis of $\Gamma$ and $\omega$, which yields $NB\perp OW$. We conclude $AB\parallel OW$, and since $OQ\parallel AB$, we have that $Q$ lies on line $OW$.

Let $Q^{\prime }$ be the intersection of $AM$ and $OW$. We will show that $B,Q^{\prime }$, and $P$ are collinear which, by the previous result, is sufficient to show $Q^{\prime }=Q$ and finish the problem.

Note that $\angle WCB=90^{\circ }-\angle BDA=90^{\circ }-\angle CBR=\angle RCB$, so $W$ lies on line $CR$. Let $A^{\prime }$ be the intersection of $MW$ (which is the perpendicular bisector of $BC$) and $AB$.

Applying Menelaus' theorem to $△AMC$ and points $Q^{\prime },B$, and $P$ we have $$Q^{\prime },B,P\text{ collinear}⟺-1=\frac{A{Q}^{\prime }}{Q^{\prime }M}\cdot \frac{MB}{BC}\cdot \frac{CP}{PA}=\frac{A{Q}^{\prime }}{Q^{\prime }M}\cdot \frac{-1}{2}\cdot \frac{CP}{PA}$$ From $Q^{\prime }W\parallel A{A}^{\prime }$, we get $\frac{A{Q}^{\prime }}{Q^{\prime }M}=\frac{A^{\prime }W}{WM}$. Thus, defining $F$ as the midpoint of $A^{\prime }W$, we can rewrite the condition as $$Q^{\prime },B,P\text{ collinear}⟺\frac{AP}{PC}=\frac{A^{\prime }W/2}{WM}=\frac{FW}{WM}.(†)$$ Since $F$ is the circumcenter of the right-angled $△A^{\prime }RW$, we have $$\angle FR{A}^{\prime }+\angle BRM=\angle R{A}^{\prime }F+\angle MBR=\angle B{A}^{\prime }M+\angle MB{A}^{\prime }=90^{\circ }⟹\angle MRF=90^{\circ }.$$

This also shows $$\angle RFM=\angle RFW=2\angle R{A}^{\prime }F=2(90^{\circ }-\angle MBA)=\angle CAR.$$ Combining this with $\angle ARC=90^{\circ }$ yields $△ARC\sim △FRM$ and, since $\angle MRW=\angle PRC$, $W$ and $P$ are corresponding points in these triangles from which (\dagger) follows.