IMO 1989 SL 13

Origin: ICE

Problem

The quadrilateral ABCD has the following properties: (i) AB = AD + BC; (ii) there is a point P inside it at a distance x from the side CD such that AP = x + AD and BP = x + BC. Show that \sqrtx \geq \sqrt AD + \sqrt BC .

Solution

Let us construct the circles \sigma1 with center A and radius R1 = AD, \sigma2 with center B and radius R2 = BC, and \sigma3 with center P and radius x. The points C and D lie on \sigma2 and \sigma1 respectively, and CD is tangent to \sigma3. From this it is plain that the greatest value of x occurs when CD is also tangent to \sigma1 and \sigma2. We shall show that in this case the required inequality is really an equality, i.e., that \sqrtx = \sqrt AD + \sqrt BC . Then the inequality will immediately follow. Denote the point of tangency of CD with \sigma3 by M. By the Pythagorean theorem we have CD =

(R1 + R2)2 −(R1 −R2)2 = 2\sqrtR1R2. On the

other hand, CD = CM + MD = 2\sqrtR2x + 2\sqrtR1x. Hence, we obtain \sqrtx = \sqrtR1 + \sqrtR2 .