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Slovenian Mathematical Olympiad, 2016

Geometry Difficulty 5.8 AIME, harder Prove it Slovenia

Let $ABC$ be a triangle with the angle $\angle BAC = 30^\circ$ , and let $P$ be an arbitrary point inside this triangle. Denote the mirror images of the point $P$ with respect to the sides $BC$ , $CA$ , and $AB$ by $P_A$ , $P_B$ , and $P_C$ , respectively. Let $P_A P_B P_C$ be an equilateral triangle. Prove that $\angle BPC = 90^\circ$ .

Solution

Let $D$ , $E$ and $F$ denote the intersections of the segments $PP_A$ , $PP_B$ and $PP_C$ with the segments $BC$ , $CA$ and $AB$ respectively. The points $D$ , $E$ and $F$ are also the midpoints of $PP_A$ , $PP_B$ and $PP_C$ , so the triangles $DEF$ and $P_A P_B P_C$ are similar. We conclude that $DEF$ is an equilateral triangle and $\angle EDF = 60^\circ$ .

From $\angle PEC = 90^\circ$ and $\angle CDP = 90^\circ$ it follows that the points $P$ , $D$ , $C$ and $E$ are concyclic, which implies $\angle DPC = \angle DEC$ . Similarly the points $B$ , $D$ , $P$ and $F$ are concyclic and $\angle BPD = \angle BFD$ . From here $\angle BPC = \angle BPD + \angle DPC = \angle BFD + \angle DEC = (180^\circ - \angle DFA) + (180^\circ - \angle AED) = 360^\circ - \angle DFA - \angle AED = \angle EDF + \angle FAE = 60^\circ + 30^\circ = 90^\circ$ .

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