IMO 1967 LL HUN22

Origin: HUN

Problem

The distance between the centers of the circles k1 and k2 with radii r is equal to r. Points A and B are on the circle k1, symmetric with respect to the line connecting the centers of the circles. Point P is an arbitrary point on k2. Prove that PA2 + PB2 \geq2r2. When does equality hold?

Solution

Let O1 and O2 be the centers of circles k1 and k2 and let C be the midpoint of the segment AB. Using the well-known relation for elements of a triangle, we obtain PA2 + PB2 = 2PC2 + 2CA2 \geq2O1C2 + 2CA2 = 2O1A2 = 2r2. Equality holds if P coincides with O1 or if A and B coincide with O2.