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area wanted, 3 angles 120^o (V Soros Olympiad 1998-99 Round 1 9.5)

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May 25, 2024
geometryareas

Problem Statement

In triangle ABCABC, ∠BAC=60o\angle BAC= 60^o. Point PP is taken inside the triangle so that angles ∠APB=∠BPC=∠CPA=120o\angle APB=\angle BPC= \angle CP A=120^o. It is known that AP=aAP = a. Find the area of triangle BPCBPC.
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