5. In an equilateral triangle $ABC$ with side length $a$, points $M, N, P, Q$ are positioned as shown in the figure. It is known that $MA + AN = PC + CQ = a$. Find the measure of angle $NOQ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_09b0e7336bad4773d41eg-3.jpg?height=397&width=462&top_left_y=2114&top_left_x=837)
Answer: $60^{\circ}$

Solution. According to the problem, $A N=a-A M$, hence $A N=M C$. Similarly, $A P=Q C$. From these equalities and the equality $\angle A=\angle C=60^{\circ}$, it follows that $\triangle A N P=\triangle C M Q$. Therefore, $\angle A N P=\angle Q M C, \angle A P N=\angle M Q C$. By the theorem on the sum of angles in a triangle, $\angle A N P+\angle A P N=120^{\circ}$, so $\angle O M P+\angle O P M=120^{\circ}$, and thus $\angle M O P=60^{\circ}$. Angles $M O P$ and $N O Q$ are vertical, so $\angle N O Q=60^{\circ}$.

Grading criteria.

- Any complete and correct solution - 7 points.
- It is proven that triangles $A N P$ and $Q C M$ are equal, but there is no further progress or it is incorrect - 2 points.
- Only the answer is provided - 0 points.