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length of AP depends only on BC=a and < BAC = \alpha

Source: Norwegian Mathematical Olympiad 2017 - Abel Competition p4

September 3, 2019
geometryangleCircumcenterorthocenter

Problem Statement

Let a>0a > 0 and 0<α<π0 < \alpha <\pi be given. Let ABCABC be a triangle with BC=aBC = a and ∠BAC=α\angle BAC = \alpha , and call the cicumcentre OO, and the orthocentre HH. The point PP lies on the ray from AA through OO. Let SS be the mirror image of PP through ACAC, and TT the mirror image of PP through ABAB. Assume that SATHSATH is cyclic. Show that the length APAP depends only on aa and α\alpha.
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