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Bulgarian Winter Tournament

Geometry Difficulty 4.6 AIME Prove it Bulgaria

Let $ABC$ be a triangle, satisfying $2AC = AB + BC$ . If $O$ and $I$ are its circumcenter and incenter, show that $\angle OIB = 90^\circ$ .

(Konstantin Delchev)

Solution

Let $D = BI \cap (ABC)$ . We apply Ptolemy's theorem
$AB \cdot DC + BC \cdot AD = AC \cdot BD$
which implies $BD = 2DA$ . From $AD = DI = DC$ it yields $I$ is midpoint of $BD$ so $\angle OIB = 90^\circ$ .
$\square$

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