IMO 1988 SL 23

Origin: SIN

Problem

Let Q be the center of the inscribed circle of a triangle ABC. Prove that for any point P, a(PA)2+b(PB)2+c(PC)2 = a(QA)2+b(QB)2+c(QC)2+(a+b+c)(QP)2, where a = BC, b = CA, and c = AB.

Solution

Denote by R the intersection point of lines AQ and BC. We know that BR : RC = c : b and AQ : QR = (b + c) : a. By applying Stewart’s theorem to ∆PBC and ∆PAR we obtain a \cdot AP 2 + b \cdot BP 2 + c \cdot CP 2 = aPA2 + (b + c)PR2 + (b + c)RB \cdot RC = (a + b + c)QP 2 + (b + c)RB \cdot RC + (a + b + c)QA \cdot QR. (1) On the other hand, putting P = Q into (1), we get that a \cdot AQ2 + b \cdot BQ2 + c \cdot CQ2 = (b + c)RB \cdot RC + (a + b + c)QA \cdot QR, and the required statement follows. Second solution. At vertices A, B, C place weights equal to a, b, c in some units respectively, so that Q is the center of gravity of the system. The left side of the equality to be proved is in fact the moment of inertia of the system about the axis through P and perpendicular to the plane ABC. On the other side, the right side expresses the same, due to the parallel axes theorem. Alternative approach. Analytical geometry. The fact that all the variable segments appear squared usually implies that this is a good approach. Assign coordinates A(xa, ya), B(xb, yb), C(xc, yc), and P(x, y), use that (a + b + c)Q = aA + bB + cC, and calculate. Alternatively, differentiate f(x, y) = a \cdot AP 2 + b \cdot BP 2 + c \cdot CP 2 −(a + b + c)QP 2 and show that it is constant.