26th BMO 1990 Problem 4 A rectangle is inscribed in a triangle if its vertices all lie on the boundary of the triangle. Given a triangle T, let d be the shortest diagonal for any rectangle inscribed in T. Find the maximum value of d 2 /area T for all triangles T. Solution
: (4√3)/7. Suppose two vertices are on the side length a and that the corresponding altitude is length h. Let the side parallel to the altitude have length x. Then the other side has length (h - x)a/h. So the square of the diagonal is x 2 + a 2 (1 - x/h) 2 = (1 + a 2 /h 2 )(x - a 2 h/(a 2 +h 2 ) ) 2 + a 2 h 2 /(a 2 +h 2 ). Hence d 2 = a 2 h 2 /(a 2 +h 2 ), or rather that is true provided we have picked the right side. Suppose b is another side length with corresponding altitude k and. Let A = area T, then ah = bk = 2A. So we require a 2 + h 2 ≥ b 2 + k 2 or a 2 + 4A 2 /a 2 ≥ b 2 + 4A 2 /b 2 or (a 2 - b 2 )(1 - 4A 2 /(a 2 b 2 ) ) ≥ 0. But ab ≥ 2A (with equality iff the sides a and b are perpendicular, but the longest side is never perpendicular to another side). So we get the shortest diagonal by taking the longest side. The required ratio d 2 /A is thus 2ah/(a 2 + h 2 ), where a is the longest side. We may assume A = 1/2 (since the ratio is unaffected by the scale of the triangles). So we have to maximise 2/(a 2 + h 2 ) or equivalently to minimise a 2 + 1/a 2 . The area of an equilateral triangle with side 1 is (√3)/4  1, but a 2 + 1/a 2 is an increasing function of a for a > 1, so we want to minimise a subject to the constraint that A is fixed. That obviously occurs for an equilateral triangle. An equilateral triangle with side 1 has altitude (√3)/2, so d 2 /A has maximum value √3/(1 + 3/4) = (4√3)/7. 26th BMO 1990 © John Scholes jscholes@kalva.demon.co.uk 15 Aug 2002