Selection Examinations for the IMO

Let us write $\angle ACB=\gamma$. Then $\angle CBA=2\gamma$ and $\angle BAC=\pi -3\gamma$. If we draw a figure where $\angle CBA=2\angle ACB$, we notice that $I{A}_1{C}_1$ is a right triangle. Let us prove this. Since $AB{A}_1{B}_1$ is a cyclic quadrilateral we have $\angle C_1{A}_{1}I=\angle B_1{A}_{1}A=\angle B_{1}BA=\frac{\angle BCA}{2}=\gamma$. The angle $\angle A_{1}IC$ is equal to $\pi -\angle CIA$. Since $\angle CIA=\pi -\angle CAI-\angle ACI$, we get $$\angle A_{1}IC=\angle CAI+\angle ACI=\frac{\pi -3\gamma }{2}+\frac{\gamma }{2}=\frac{\pi }{2}-\gamma .$$ So, $\angle A_{1}IC+\angle C_1{A}_{1}I=\frac{\pi }{2}$ and $\angle I{C}_1{A}_1=\frac{\pi }{2}$. Thales' theorem states that the circumcentre of the triangle $I{A}_1{C}_1$ (i.e. the point S) is the midpoint of the segment $I{A}_1$. The central angle is twice the inscribed angle, so $\angle C_{1}SI=2\angle C_1{A}_{1}I=2\frac{\angle CBA}{2}=2\gamma$. The quadrilateral $AB{A}_1{C}_1$ is cyclic, so $\angle C{A}_{1}A=\angle CBA=2\gamma$. Hence, $C{A}_1$ is parallel to DS. But D is the midpoint of the segment CE and S is the midpoint of the segment $I{A}_1$, so DS is the midsegment of the quadrilateral $SI{A}_1{C}_1$. Since DS is parallel to $C{A}_1$ it must also be parallel to EI. So, $\angle EIA=\angle DSA=2\gamma$.

For the inner angles of the triangle ABI we have $\angle BAI=\angle BAC=\frac{\pi -3\gamma }{2}$, $\angle IBA=\angle CBA=\frac{\pi -3\gamma }{2}$. Since the sum of the inner angles in an arbitrary triangle is equal to $\pi$ we get $\frac{\pi -3\gamma }{2}+\gamma +\pi -2\gamma =\pi$ or $\gamma =\frac{\pi }{5}$.

The angles of the triangle ABC measure $\angle BAC=\angle CBA=\frac{2\pi }{5}$ and $\angle ACB=\frac{\pi }{5}$.