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AP^2+PX^2=BP^2+PY^2=CP^2+PZ^2, projections , circumcenter, incenter

Source: 2021 IMOC qualification problem, G1

December 30, 2021
geometryCircumcenterincenterprojectionscircumcircle

Problem Statement

Let OO be the circumcenter and II be the incenter of △\vartriangle, PP is the reflection from II through OO, the foot of perpendicular from PP to BC,CA,ABBC,CA,AB is X,Y,ZX,Y,Z, respectively. Prove that AP2+PX2=BP2+PY2=CP2+PZ2AP^2+PX^2=BP^2+PY^2=CP^2+PZ^2.
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