1st Eötvös 1894 Problem 3 A triangle has sides length a, a + d, a + 2d and area S. Find its sides and angles in terms of d and S. Give numerical answers for d = 1, S = 6.
It is slightly more convenient to use b = a+d. Then by Heron S 2 = (3b 2 /4)(b 2 /4 - d 2 ). This is a quadratic in b 2 . We need a positive solution, so b 2 = 2(d 2 + √(d 4 + 4S 2 /3) ). That gives the sides a, b, c = b+d. Now sin A = 2S/bc, and sin B = 2S/ac. Since c is the longest side, A and B must be acute. Finally, C = 180 o - A - B. For d = 1, S = 6, we find b = 4. Hence a = 3, c = 5, A = sin -1 (3/5). We recognize the triangle, so C = 90 o , B = 90 o - A. 1st Eötvös 1894 © John Scholes jscholes@kalva.demon.co.uk 11 Oct 2003 Last corrected/updated 11 Oct 03