IMO 1967 LL CZS8

Origin: CZS

Problem

ABCD is a parallelogram; AB = a, AD = 1, \alpha is the size of \angleDAB, and the three angles of the triangle ABD are acute. Prove that the four circles KA, KB, KC, KD, each of radius 1, whose centers are the vertices A, B, C, D, cover the parallelogram if and only if a \leq cos \alpha + \sqrt 3 sin \alpha.

Solution

The circles KA, KB, KC, KD cover the parallelogram if and only if for every point X inside the parallelogram, the length of one of the segments XA, XB, XC, XD does not exceed 1. Let O and r be the center and radius of the circumcircle of \triangleABD. For every point X inside \triangleABD, it holds that XA \leqr or XB \leqr or XD \leqr. Similarly, for X inside \triangleBCD, XB \leqr or XC \leqr or XD \leqr. Hence KA, KB, KC, KD cover the parallelogram if and only if r \leq1, which is equivalent to \angleABD \geq30◦. However, this last is exactly equivalent to a = AB = 2r sin \angleADB \leq2 sin(\alpha + 30◦) = \sqrt 3 sin \alpha + cos \alpha.