National Math Olympiad

The triangle $ABC$ is acute, so the points $I$ and $O$ lie on the same side of the line $ED$. The condition that the points $D$, $E$, $I$ and $O$ lie on the same circle therefore implies that $\angle DOE=\angle DIE$.

Let us denote the angles of the triangle by $\alpha$, $\beta$ and $\gamma$ and let us express the angles $\angle DOE$ and $\angle DIE$ in these terms.

We have $\angle DOE=\angle EOC+\angle COD$. Since the central angle is always twice the inscribed angle, we get $\angle COE=2\angle CBE$. The line $EB$ is the bisector of the angle $CBA$, so $\angle CBE=\frac{\beta }{2}$. Hence, $\angle COE=2\angle CBE=\beta$.

Similarly, we have $\angle DOC=2\angle DAC=2\cdot \frac{\alpha }{2}=\alpha$ and so $\angle DOE=\alpha +\beta$.

Also, $\angle DIE=\angle AIB=\pi -\angle BAI-\angle IBA=\pi -\frac{\alpha }{2}-\frac{\beta }{2}$.

The identity $\angle DOE=\angle DIE$ implies that $\pi =\frac{3}{2}(\alpha +\beta )$ or $\alpha +\beta =\frac{2\pi }{3}$.

We conclude that $\gamma =\pi -\alpha -\beta =\frac{\pi }{3}$.