BMO 2019 Shortlist

Let $P^{\prime }$ be the midpoint of the segment $AD$. We will prove that $P^{\prime }=P$. Let $F\in AC$ such that $DF\perp AC$. The triangle $CDF$ is isosceles with $FD=FC$ and the triangle $D{P}^{\prime }F$ is equilateral as $m(\angle ADF)=60^{\circ }$. Thus, the triangle $FC{P}^{\prime }$ is isosceles ($F{P}^{\prime }=FC$) and $m(\angle FC{P}^{\prime })=m(\angle F{P}^{\prime }C)=15^{\circ }$.

Figure 2: G2

We prove now that $m(\angle FCE)=15^{\circ }$. Let $M$ be the point on $[AB$ such that the triangle $ACM$ is equilateral. As $△ADC\equiv △ADM(SAS)\Rightarrow DC=DM(=DE)$ and $m(\angle AMD)=m(\angle ACD)=45^{\circ }$. It follows that the triangle $△DME$ is isosceles with $m(\angle DME)=m(\angle DEM)=45^{\circ }$. In the triangle $△BDE$ we have $m(\angle BDE)=60^{\circ }$ and thus $m(\angle CDE)=120^{\circ }$. As the triangle $DCE$ is isoscel with $m(\angle DCE)=m(\angle DEC)=30^{\circ }$. Finally $m(\angle ACE)=m(\angle ACB)-m(\angle BCE)=45^{\circ }-30^{\circ }=15^{\circ }$. Thus $m(\angle FC{P}^{\prime })=15^{\circ }=m(\angle FCE)$, and therefore $P^{\prime }\in CE$ and $P^{\prime }=P$, which means that $P$ is the midpoint of the segment $AD$.

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Alternative solution.

In the way as above we prove that $m(\angle BCE)=15^{\circ }$. So the quadrilateral $ACDE$ is inscribed in a circle. Now, applying the sine rules to $△DPE$ and $△APE$ we get $$\begin{array}{rlrlrl}\frac{DP}{\sin 30^{\circ }}& =\frac{PE}{\sin 15^{\circ }},& \frac{AP}{\sin 105^{\circ }}& =\frac{PE}{\sin 30^{\circ }}& \Rightarrow \frac{DP}{\sin 30^{\circ }}\cdot \frac{\sin 105^{\circ }}{AP}& =\frac{PE}{\sin 15^{\circ }}\cdot \frac{\sin 30^{\circ }}{PE},\\ \frac{DP}{AP}& =\frac{\sin 30^{\circ }}{\sin 105^{\circ }\cdot \sin 15^{\circ }}& =\frac{1}{4\cdot \sin 105^{\circ }\cdot \sin 15^{\circ }}& =\frac{1}{2\cdot (\cos 90^{\circ }-\cos 120^{\circ })}=\frac{1}{2\cdot \frac{1}{2}}=1.& & \end{array}$$ Thus, $QP=AP$. $\square$