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Slovenia — National Math Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Slovenia

Let $O$ be the circumcentre of the acute triangle $ABC$ and denote the circumcircle by $\kappa$ . The bisector of the inner angle at $A$ meets $\kappa$ again at $D$ . The bisector of the inner angle at $B$ meets $\kappa$ again at $E$ . Let $I$ denote the incentre of the triangle $ABC$ . How much does the angle $\angle ACB$ measure if the points $D, E, O$ and $I$ lie on the same circle?

Solution

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

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