Olympiad Mathematics Again

New Zealand, Camp Selection 2018, problem 4

June 16, 2019

Problem. Let P be a point inside triangle ABC such that CPA=90 and CBP=CAP. Let X and Y be the midpoints of AB and AC respectively. Prove that PXY=90.

Origin: New Zealand, Camp Selection 2018, problem 4.

Solution. Let's consider a homothety with center A and scale factor 2. This homothety maps XB, YC, PP, where P is the point on the ray AP such that AP=PP.

Notice that since CPA=90, P lies on the circle with diameter AC and center Y. Thus, AY=YP. These two segments, AY and YP, map to the segments AC and CP respectively under the homothety considered above. Therefore, AC=CP. This, in turn, implies that PAC=APC. When we combine this with the condition CBP=CAP, we obtain PPC=CBP. This implies that the the quadrilateral CPBP is inscribed (cyclic). Therefore, CBP=CPP=90. Finally, notice that under the homothety considered above the angle PXY maps to the angle PBC, thus PXY=PBC=90, and we are done.