An Angle from a Harmonic Condition

★ 7/10plane geometryLaw of SinestrigonometryTriangle Geometry

Problem

In △ABC\triangle ABC, ∠A=60∘\angle A=60^\circ, and ∠BAP=∠CAP\angle BAP=\angle CAP with PP inside the triangle. The extension of BPBP meets ACAC at QQ, and

1∣BP∣+1∣CP∣=1∣PQ∣.\frac 1{|BP|}+\frac 1{|CP|}=\frac 1{|PQ|}.

Find ∠BPC\angle BPC.

Answer

Solution

Difficulty7/10
Topicsplane geometry, Law of Sines, trigonometry, Triangle Geometry

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