Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Spain

Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $E$ and $F$ be the intersections of the bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$ . If $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$ , then find the value of $\angle BAC$ .

Solution

Let us denote by $I$ the incenter of $\triangle abc$ . From the figure immediately follows $\angle IBC = \angle IFE$ and $\angle ICB = \angle IEF$ . So, $\triangle IBC \sim \triangle IFE$ . Since both have the same height (the radii of the incircle) because $BC$ and $EF$ are both tangent to $\gamma$ , then $\triangle IBC = \triangle IFE$ . Therefore, $IB = IF$ .

Now, if we denote $\angle ABC = 2\alpha$ and $\angle ACB = 2\beta$ , then $\angle BIF = \alpha + \beta$ . Now, from isosceles $\triangle IFB$ follows
$\angle IBF = \angle IFB = \angle CFB = \angle BAC = 180^\circ - 2(\alpha + \beta)$
Adding up the angles of $\triangle IFB$ yields
$(\alpha + \beta) + 180^\circ - 2(\alpha + \beta) + 180^\circ - 2(\alpha + \beta) = 180^\circ$
From which follows $\alpha + \beta = 60^\circ$ and $\angle BAC = 60^\circ$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.