4th Swedish 1964 Problem 1 Find the side lengths of the triangle ABC with area S and ∠BAC = x such that the side BC is as short as possible.
BC = 2√(S tan(x/2)), AB = AC = √(S/(sin(x/2)cos(x/2)) ) Solution For given BC, the locus of A with ∠BAC = x is the arc of a circle. The area is BC times the height, so we maximise the area by taking AB = AC. Since we want BC as small as possible, we must have AB = AC. So suppose BC = 2k, height = h. Then S = hk, and h = k cot(x/2). So S = k 2 cot(x/2). Hence k = √(S tan(x/2)). Then the other sides have length k/sin(x/2). 4th Swedish 1964 © John Scholes jscholes@kalva.demon.co.uk 30 December 2003 Last corrected/updated 30 Dec 03