# Answer: $90^{\circ}$.

## Solution:

Since $M C=B N$ and $A C=B C$, then $A M=A C-M C=B C-B N=N C$.

From the fact that $A M=N C, A B=A C$ and $\angle B A M=\angle C A N=60^{\circ}$, it follows that triangles $B A M$ and $A C N$ are congruent, so $\angle N A C=\angle M B A=\alpha$.

From triangle $A B M: \angle A M B$ (and it is also $\angle A M Q)=180^{\circ}-60^{\circ}-\alpha=120^{\circ}-\alpha$.

And from triangle $A Q M: \angle A Q M=180^{\circ}-\angle A M Q-\alpha=180^{\circ}-\left(120^{\circ}-\alpha\right)-\alpha=60^{\circ}$.

Then $\angle M Q N=120^{\circ}$, so $\angle M Q N+\angle M C N=180^{\circ}-$ from which it follows that quadrilateral $M Q N C$ is cyclic.

But then $\angle N Q C=\angle N M C$.

In triangle $N M C$, side $N C$ is half the length of side $M C$, and the angle between them is $60^{\circ}$, so this triangle is a right triangle with angles $90^{\circ}, 60^{\circ}, 30^{\circ}$. From this cyclic property: $\angle M Q C=\angle M N C=90^{\circ}$, then $\angle C Q B$ is also a right angle.

## Criteria:

If it is proven that $\angle M Q N=120^{\circ}$ or $\angle A Q M=60^{\circ}$ - award 2 points.

If this is not proven but used, award no more than 3 points for the solution.

Similarly for the equality $\angle M Q C=90^{\circ}$ (proven - award 2 points, not proven - award no more than 3 points for the solution).
3. Triangle $A B C$ is equilateral. On side $A C$, point $M$ is marked, and on side $B C$, point $N$, such that $M C=B N=2 A M$. Segments $M B$ and $A N$ intersect at point $Q$. Find the angle $C Q B$.

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