BMO 1966 Problem 5 A spanner has an enclosed hole which is a regular hexagon side 1. For what values of s can it turn a square nut side s? Solution
: 3 - √3 >= s > √(3/2). Note that 3 - √3 = 1.27, √(3/2) = 1.22, so the range is fairly narrow. The square must be small enough to fit into the hole and big enough to stick. The sticking condition is easy. Inscribe a circle in the hexagon. If the square will fit into the circle, then it obviously does not stick. So we require the diagonal of the square to be > √3, or s > √(3/2). On the other hand if the diagonal exceeds √3, then it obviously does stick because the parallel opposite sides of the hexagon are only a distance √3 apart. It seems fairly clear that the maximum square has sides parallel to a pair of opposite sides and each vertex a distance x from on endpoint of those sides. For the square to have equal sides, we then need √3 (1 - x) = 1 + x, or x = 2 - √3, and hence s = 3 - √3. But how do we prove it? 2nd BMO 1966 © John Scholes jscholes@kalva.demon.co.uk 30 Aug 2000