The quadrilateral Q contains a circle which touches each side. It has side lengths a, b, c, d and area √(abcd). Prove it is cyclic.
Solution
It helps a lot to know the following result about the area of quadrilaterals: the area of a quadrilateral side a, b, c, d is maximised by making it cyclic, in which case its area is √( (s - a)(s - b)(s - c)(s - d) ) (*), where 2s = a + b + c + d.
If we assume this result, then the problem is fairly easy. Assume that the sides are in the order a, b, c, d, so that the side length a is opposite that length c. If Q contains a circle C which touches each side, then a + c = b + d. [The distances from a vertex or the quadrilateral to the two points of contact of its two sides are equal. So for some w, x, y, z we have a = w + x, b = x + y, c = y + z, d = z + w and hence a + c = b + d.]
But this means that s - a = (- a + c + b + d)/2 = (-a + c + a + c)/2 = c. Similarly, s - b = d, s - c = a and s - d = b, so the maximum possible area of the quadrilateral is √(abcd), with equality iff it is cyclic. But we are given that the area is √(abcd), so it must be cyclic.
To establish the result about quadrilaterals, drop the assumption that a + c = b + d, and let the angle between a and b be θ and the opposite angle be φ. Let the area be A. Then 2A = ab sin θ + cd sin φ, so 16A2 = 4(a2b2 sin2θ + c2d2 sin2φ ) + 8abcd sin θ sin φ (*).
By the cosine rule we have a2 + b2 - 2ab cos θ = c2 + d2 - 2cd cos φ, so (a2 + b2 - c2 - d2)2 = 4(a2b2 cos2θ + c2d2 cos2φ ) - 8abcd cos θ cos φ. Adding to (*) gives 16A2 = 4(a2b2 + c2d2) - (a2 + b2 - c2 - d2)2 - 8abcd cos(θ + φ) ≤ 4(a2b2 + c2d2) - (a2 + b2 - c2 - d2)2 + 8abcd (**), with equality iff cos(θ + φ) = -1, in other words, iff the quadrilateral is cyclic.
But we can easily check that (-a + b + c + d)(a - b + c + d)(a + b - c + d)(a + b + c - d) = 4(a2b2 + c2d2) - (a2 + b2 - c2 - d2)2 + 8abcd.
Comment. I am not sure whether one is expected to prove the generalised Heron's formula. I suspect it was considered to be fairly well-known bookwork in 1970, so maybe not. On the other hand, I do not think I knew it when I was an undergraduate (Cambridge 1968-71). However, the problem is not so much proving it - it is fairly easy to prove - but seeing how to do the question if one is not familar with it. Unless I am missing something, it looks relatively hard to do any other way.
© John Scholes
jscholes@kalva.demon.co.uk
25 Aug 2001