MathDB

XI.6

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

AQ, BN,CT equal and concurrent, centroids, AK^2-BC^2=BK^2-AC^ =CK^2-AB^2

Source: XI All-Ukrainian Tournament of Young Mathematicians, Qualifying p6

5/27/2021
Prove that there exists a point KK in the plane of △ABC\vartriangle ABC such that AK2−BC2=BK2−AC2=CK2−AB2.AK^2 - BC^2 = BK^2 - AC^2 = CK^2 - AB^2. Let Q,N,TQ, N, T be the points of intersection of the medians of the triangles BKC,CKA,AKBBKC, CKA, AKB, respectively. Prove that the segments AQ,BNAQ, BN and CTCT are equal and have a common point.
geometryconcurrentconcurrencyequal segmentsCentroidUkrainian TYM