Let ABCD be a convex quadrilateral such that the sides AB, AD, BC satisfy AB = AD + BC. There exists a point P inside the quadrilateral at a distance h from the line CD such that AP = h + AD and BP = h + BC. Show that:
1/√h ≥ 1/√AD + 1/√BC.
Solution
by Gerhard Wöginger, Technical University, Graz
Let CA be the circle center A, radius AD, and CB the circle center B, radius BC. The circles touch on AB. Let CP the the circle center P, radius h. CP touches CA and CB and CD. Let t be the common tangent to CA and CB whose two points of contact are on the same side of AB as C and D. Then CP is confined inside the curvilinear triangle whose sides are segments of t, CA and CB. Evidently h attains its maximum value, for given lengths AB, AD, BC, when CP touches t, in which case D must be the point at which t touches CA, and C the point at which it touches CB. Suppose E is the point at which t touches CP.
Angles ADC and BCD are right angles, so CD2 = AB2 - (AD - BC)2 = 4 AD BC. Similarly, DE2 = 4 h AD, and CE2 = 4 h BC. But CD = DE + CE, so 1/√h = 1/√AD + 1/√BC. This gives the maximum value of h, so in general we have the inequality stated.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Sep 2002