11.5. In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
Answer: $30^{\circ}$.

Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.

First method. We use the "folding" method. Reflect triangle $P Q R$ symmetrically relative to line $P R$, and triangle $T S R$ - relative to line $T R$ (see Fig. 11.5a). From the equality (*) and the equality $R Q=R S$, it follows that the images of points $Q$ and $S$ are the same point $O$.

Note that triangle $T O P$ is equilateral. Moreover, $O R=O P=O T$. Therefore, $O$ is the center of the circumscribed circle of triangle $P R T$. Then $\angle P R T = 0.5 \angle P O T = 30^{\circ}$.

Second method. We will prove that $Q P T S$ is a parallelogram (see Fig. 11.5b). Indeed, using the equality of the angles at the bases of the isosceles triangles $P Q R$ and $R S T$ and the equality (*), we get: $\angle Q P T + \angle P T S = \angle Q P R + \angle R P T + \angle R T P + \angle S T R = \angle P R Q + \angle T R S + (180^{\circ} - \angle P R T) = 180^{\circ}$.

Thus, $P Q \| S T$ and $P Q = S T$ (by condition), so $Q P T S$ is a parallelogram. Then $Q S = P T$, hence triangle $Q R S$ is equilateral. Therefore, $\angle P R T = 0.5 \angle Q R S = 30^{\circ}$.

Grading criteria.

"+" A complete and justified solution is provided. 
"±" A generally correct solution is provided, containing minor gaps or inaccuracies (for example, the fact that the images of points $Q$ and $S$ coincide under the symmetries is used but not justified).
"Ғ" The correct answer is obtained based on the fact that $Q P T S$ is a parallelogram, but this is not proven.
"-" Only the answer is provided or the answer is obtained by considering a regular pentagon.
"-" An incorrect solution is provided or no solution is provided.