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Korea Second Round 2012 Problem 1

Source: Korea Second Round 2012 Problem 1

August 19, 2012
geometrycircumcircletrigonometryrectangletrapezoidgeometry proposed

Problem Statement

Let ABC ABC be an obtuse triangle with ∠A>90∘ \angle A > 90^{\circ} . Let circle O O be the circumcircle of ABC ABC . D D is a point lying on segment AB AB such that AD=AC AD = AC . Let AK AK be the diameter of circle O O . Two lines AK AK and CD CD meet at L L . A circle passing through D,K,L D, K, L meets with circle O O at P(≠K) P ( \ne K ) . Given that AK=2,∠BCD=∠BAP=10∘ AK = 2, \angle BCD = \angle BAP = 10^{\circ} , prove that DP=sin⁡(∠A2) DP = \sin ( \frac{ \angle A}{2} ).
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