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National Math Olympiad – First Round, 2015

Geometry Difficulty 4.6 AIME Prove it Slovenia

Let $\mathcal{K}$ be the circumcircle and $\mathcal{I}$ the incenter of the triangle $ABC$ . Denote by $D$ the midpoint of the arc $BC$ of the circle $\mathcal{K}$ which does not contain the point $A$ , and by $E$ the midpoint of the arc $CA$ of the circle $\mathcal{K}$ which does not contain the point $B$ . Let $F$ be the mirror image of the point $\mathcal{I}$ by reflection across the side $AB$ and suppose that $F$ lies on the circle $\mathcal{K}$ . What is the size of the angle $\angle DFE$ ?

Solution

Since the points $D$ and $E$ are the midpoints of the corresponding arcs of the circle $\mathcal{K}$ they lie on the angle bisectors of angles $\angle BAC$ and $\angle CBA$ . Due to symmetry we

have $\angle FAB = \angle BAI = \frac{1}{2}\angle BAC$ and thus $\angle FED = \angle FAD = \angle BAC$ . Similarly we get $\angle EDF = \angle CBA$ . It follows $\angle DFE = \angle ACB$ . Let's calculate the size of the angle $\angle ACB$ . The points $F, B, C$ , and $A$ are concyclic, therefore $\angle ACB = \pi - \angle BFA$ . Since $F$ is a mirror image of the point $I$ we have
$\begin{align*} \angle BFA &= \angle AIB = \pi - \frac{1}{2}\angle BAC - \frac{1}{2}\angle CBA = \pi - \frac{1}{2}(\angle BAC + \angle CBA) = \\ &= \pi - \frac{1}{2}(\pi - \angle ACB) = \frac{\pi}{2} + \frac{1}{2}\angle ACB. \end{align*}$

Therefore $\angle ACB = \pi - (\frac{\pi}{2} + \frac{1}{2}\angle ACB) = \frac{\pi}{2} - \frac{1}{2}\angle ACB$ which implies $\angle ACB = \frac{\pi}{3}$ and thus also $\angle DFE = \frac{\pi}{3}$ .

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