MathDB

2013.9

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

many points concyclic wanted, AB^2 + AC^2 = 2BC^2

Source: 2013 XVI All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p9

5/11/2021
Given a triangle PQRPQR, the inscribed circle ω\omega which touches the sides QR,RPQR, RP and PQPQ at points A,BA, B and CC, respectively, and AB2+AC2=2BC2AB^2 + AC^2 = 2BC^2. Prove that the point of intersection of the segments PA,QBPA, QB and RCRC, the center of the circle ω\omega, the point of intersection of the medians of the triangle ABCABC, the point AA and the midpoints of the segments ACAC and ABAB lie on one circle.
geometryConcyclicUkrainian TYM