Let triangle ABC satisfy AB < AC < BC. Let and be the incenter and the incircle of triangle , respectively. Let be a point on line , different from , such that the line through and parallel to is tangent to . Similarly, let be a point on line , different from , such that the line through and parallel to is tangent to . Line intersects the circumcircle of triangle ABC at P ≠ A. Let and be the midpoints of and , respectively.
Prove that ∠KIL + ∠YPX = 180°.
Proposed by Dominik Burek, Poland
Solution
Notice that such a point and are unique (For , let be the intersection of the tangent with . Then since is inscribed inside we have , by Thales where . Solving this equation, we have unique hence is unique. ).
A construction of such points and is easy : take the symmetric of about . Draw the tangents of to and their intersections and with . The parallelism is obvious since we have a rhombus.
Notice that (the south pole) belong to the bisector of .
Now the following lemma : are concyclic (as well as ). Sketch of the proof : Angle chase with and .
Remark with Thales theorem we have . Finally , as desired.