IMO shortlist 1985/2 solution

26th IMO 1985 shortlist

Problem 2

A convex quadrilateral ABCD is inscribed in a circle radius 1. Show that 0 < |AB + BC + CD + DA - AC - BD| < 2.

Solution

The triangle inequality gives: AB + BC > AC, BC + CD > BD, CD + DA > AC, DA + AB > BD. Adding and dividing by 2 gives AB + BC + CD + DA - AC - BC > 0, which gives the first part.

Let O be the center of the circle. Let angle AOB = 2a, angle BOC = 2b, angle COD = 2c, angle DOA = 2d. Then AB = 2 sin a, BC = 2 sin b, CD = 2 sin c, DA = 2 sin d, AC = 2 sin(a+b), BD = 2 sin(b+c). So it remains to show that sin a + sin b + sin c + sin d - sin(a+b) - sin(b+c) < 1. We have d = 180^o - (a+b+c), so this is equivalent to: sin a + sin b + sin c + sin(a+b+c) < sin(a+b) + sin(b+c) + 1. We claim that the stronger inequality sin a + sin b + sin c + sin(a+b+c) < sin(a+b) + sin(b+c) + sin(c+a) (*) holds.

We have sin b + sin c = sin( (b+c)/2 + (b-c)/2) + sin( (b+c)/2 - (b-c)/2) = 2 sin( (b+c)/2) cos( (b-c)/2), sin a + sin(a+b+c) = sin(a + (b+c)/2 - (b+c)/2) + sin(a + (b+c)/2 + (b+c)/2) = 2 sin(a + (b+c)/2) cos( (b+c)/2), sin(a+b) + sin(a+c) = sin(a + (b+c)/2 + (b-c)/2) + sin(a + (b+c)/2 - (b-c)/2) = 2 sin(a + (b+c)/2) cos( (b-c)/2), sin(b+c) = 2 sin( (b+c)/2) cos( (b+c)/2). So (*) is equivalent to: sin( (b+c)/2) cos( (b-c)/2) + sin(a + (b+c)/2) cos( (b+c)/2) < sin(a + (b+c)/2) cos( (b-c)/2) + sin( (b+c)/2) cos( (b+c)/2), and hence to: (sin(a + (b+c)/2) - sin( (b+c)/2) )(cos( (b-c)/2) - cos( (b+c)/2) > 0.

There is no loss of generality in taking a to be the smallest of a, b, c, d. So a + (b+c)/2 ≤ (a+d)/2 + (b+c)/2 <= 90^o. But sin x is an increasing function over 0, 90^o, so sin(a + (b+c)/2) - sin( (b+c)/2) > 0. Since b/2 and c/2 are both positive with sum less than 90^o, we have |(b-c)/2| < |(b+c)/2| and hence cos( (b-c)/2) - cos( (b+c)/2) > 0. So (*) is established.

Note that the inequalities are the best possible. If we take A, B, C, D almost coincident, then we can get AB + BC + CD + DA - AC - BD as close as we like to 0. If we take A, B, D almost coincident and AC a diameter, then we can get AB + BC + CD + DA - AC - BD as close as we like to 0 + 2 + 2 + 0 - 2 - 0 = 2.

26th IMO shortlist 1985