Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:
Let ABCABC be an acute triangle with orthocenter HH, and let MM and NN denote the midpoints of ABAB and ACAC. Rays MHMH and NHNH intersect the circumcircle of ABCABC again at points XX and YY. Prove that the four points M,N,X,YM, N, X, Y lie on a circle.

Solution

Solution:
Let Ω\Omega be the circumcircle of ABC\triangle ABC. Let PP and QQ be the reflections of HH across MM and NN.
Figure 1
Since APB=AHB=180C\angle APB = \angle AHB = 180^\circ - \angle C, point PP lies on Ω\Omega; thus so does QQ. Moreover, MNPQMN \parallel PQ, so
XMN=XPQ=XYQ \angle XMN = \angle XPQ = \angle XYQ
and we are done.

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