BMO Shortlist, 2019

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

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 $\triangle ADC \equiv \triangle ADM(SAS) \Rightarrow DC = DM(= DE)$ and $m(\angle AMD) = m(\angle ACD) = 45^\circ$. It follows that the triangle $\triangle DME$ is isosceles with $m(\angle DME) = m(\angle DEM) = 45^\circ$. In the triangle $\triangle BDE$ 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 $m(\angle FCP') = 15^\circ = m(\angle FCE)$, and therefore $P' \in CE$ and $P' = P$, which means that $P$ is the midpoint of the segment $AD$.

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 $\triangle DPE$ and $\triangle APE$ we get
$$\begin{aligned} \frac{DP}{\sin 30^\circ} &= \frac{PE}{\sin 15^\circ}, & \frac{AP}{\sin 105^\circ} &= \frac{PE}{\sin 30^\circ} & \Rightarrow \frac{DP}{\sin 30^\circ} \cdot \frac{\sin 105^\circ}{AP} &= \frac{PE}{\sin 15^\circ} \cdot \frac{\sin 30^\circ}{PE}, \\ \frac{DP}{AP} &= \frac{\sin 30^\circ}{\sin 105^\circ \cdot \sin 15^\circ} &= \frac{1}{4 \cdot \sin 105^\circ \cdot \sin 15^\circ} &= \frac{1}{2 \cdot (\cos 90^\circ - \cos 120^\circ)} = \frac{1}{2 \cdot \frac{1}{2}} = 1. \end{aligned}$$
Thus, $QP = AP$. $\square$