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Easy Geometry equality

Source: EGMO 2018 Moldova TST

February 11, 2018
geometrycircumcircle

Problem Statement

Let ABCABC be a triangle with AB=cAB=c , BC=aBC=a and AC=bAC=b. If x,y∈R x,y\in\mathbb{R} satisfy 1x+1y+z=1a \frac{1}{x} +\frac{1}{y+z} = \frac{1}{a} , 1y+1x+z=1b \frac{1}{y} +\frac{1}{x+z} = \frac{1}{b} , 1z+1y+x=1c \frac{1}{z} +\frac{1}{y+x} = \frac{1}{c} . Prove that the following equality holds x(p−a)+y(p−b)+z(p−c)=3r2+12R∗r, x(p-a) + y(p-b) + z(p-c) = 3r^2 + 12R*r , Where pp is semi-perimeter, RR is the circumradius and rr is the inradius.
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