IMO 1976 Problem A1 A plane convex quadrilateral has area 32, and the sum of two opposite sides and a diagonal is 16. Determine all possible lengths for the other diagonal.
At first sight, the length of the other diagonal appears unlikely to be significantly constrained. However, a little experimentation shows that it is hard to get such a low value as 16. This suggests that 16 may be the smallest possible value. If the diagonal which is part of the 16 has length x, then the area is the sum of the areas of two triangles base x, which is xy/2, where y is the sum of the altitudes of the two triangles. y must be at most (16 - x), with equality only if the two triangles are right-angled. But x(16 - x)/2 = (64 - (x - 8) 2 )/2 ≤ 32 with equality only iff x = 8. Thus the only way we can achieve the values given is with one diagonal length 8 and two sides perpendicular to this diagonal with lengths totalling 8. But in this case the other diagonal has length 8√2. s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 18th IMO 1976 © John Scholes jscholes@kalva.demon.co.uk 10 Oct 1998