IMO 1988 SL 18

Origin: LUX

Problem

Consider two concentric circles of radii R and r (R > r) with center O. Fix P on the small circle and consider the variable chord PA of the small circle. Points B and C lie on the large circle; B, P, C are collinear and BC is perpendicular to AP. (a) For what value(s) of \angleOPA is the sum BC2 + CA2 + AB2 extremal? (b) What are the possible positions of the midpoints U of BA and V of AC as \angleOPA varies?

Solution

(i) Define \angleAPO = \varphi and S = AB2 + AC2 + BC2. We calculate PA = 2r cos \varphi and PB, PC =

R2 −r2 cos2 \varphi\pmr sin \varphi. We also have AB2 = PA2 + PB2, AC2 = PA2 + PC2 and BC = BP + PC. Combining all these we obtain S = AB2 + AC2 + BC2 = 2(PA2 + PB2 + PC2 + PB \cdot PC) = 2(4r2 cos2 \varphi + 2(R2 −r2 cos2 \varphi + r2 sin2 \varphi) + R2 −r2) = 6R2 + 2r2.

Hence it follows that S is constant; i.e., it does not depend on \varphi. (ii) Let B1 and C1 respectively be points such that APBB1 and APCC1 are rectangles. It is evident that B1 and C1 lie on the larger circle and that −−\to PU = 1 −−\to PB1 and −−\to PV = 1 −−\to PC1. It is evident that we can arrange for an arbitrary point on the larger circle to be B1 or C1. Hence, the locus of U and V is equal to the circle obtained when the larger circle is shrunk by a factor of 1/2 with respect to point P.