MathDB

6

Part of 1980 Poland - Second Round

Problems(1)

a PA^2 + b PB^2 + c PC^2 is fixed if P lies on incircle of (ABC)

Source: Polish MO Recond Round 1980 p6

9/9/2024
Prove that if the point P P runs through a circle inscribed in the triangle ABC ABC , then the value of the expression a⋅PA2+b⋅PB2+c⋅PC2 a \cdot PA^2 + b \cdot PB^2 + c \cdot PC^2 is constant (a,b,c a, b, c are the lengths of the sides opposite the vertices A,B,C A, B, C , respectively).
geometryincirclefixedconstant