26th Hellenic Mathematical Olympiad

Since $BA=B\Gamma =BE$, point $B$ is the circumcenter of the triangle $A\Gamma E$. The angle $A\hat{E}\Gamma$ is inscribed to the circle $C(B,BA)$ and so: $$A\hat{E}\Gamma =\frac{1}{2}A\hat{B}\Gamma =30^{\circ }.$$

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Alternative solution.

From $BA=BE$ and $BA=B\Gamma$, we have $B\Gamma =BE$, and the triangle $B\Gamma E$ is isosceles. We draw the altitude from $B$, let $BZ$, $Z\in \Gamma E$.



Then $BZ$ is also median and bisector of the angle $EB\Gamma$ of the triangle $B\Gamma E$. Let $BZ$ meet $AE$ at $K$. Then the triangles $BK\Gamma$ and $BKE$ are equal because they have:



$$B\Gamma =BE,BK\text{ common side and }KB\Gamma =KBE.$$ Therefore we have: $B\Gamma K=B\hat{E}K$ and since $$B\hat{E}K=B\hat{A}K\text{ it follows that }B\Gamma K=B\hat{A}K.$$ Thus quadrilateral $ABK\Gamma$ is inscribable, and hence: $$ZKE=BKA=B\Gamma A=60^{\circ }$$ Finally we get: $$A\hat{E}\Gamma =K\hat{E}Z=90^{\circ }-EKZ=90^{\circ }-60^{\circ }=30^{\circ }.$$