39th Swedish 1999 Problem 4 An equilateral triangle side x has its vertices on the sides of a square side 1. What are the possible values of x?
any value in [1,√6 - √2] Solution If two vertices of the triangle are on the same side of the square, then the third vertex would be a height < 1 above the side and hence in the interior of the square. Contradiction. So there must be one vertex of the triangle on each of three sides of the square. So two vertices A, B of the triangle are on opposite sides of the square. Hence AB ≥ 1. Start with AB parallel to a side of the square, then gradually increase ∠BAD. AB increases and ∠CAX decreases, so AX increases. We can adjust the position of A to get C on the top side until AC cos CAX becomes too large. Thus we can get any value of x from 1 to the value in the right-hand of the top three diagrams. That value is easily found by Pythagoras, we have √(x 2 -1) + x/√2 = 1, so x = √6 - √2 = about 1.035. 39th Swedish 1999 © John Scholes jscholes@kalva.demon.co.uk 16 January 2004 Last corrected/updated 16 Jan 04