22nd Swedish 1982 Problem 4 ABC is a triangle with AB = 33, AC = 21 and BC = m, an integer. There are points D, E on the sides AB, AC respectively such that AD = DE = EC = n, an integer. Find m.
30 Solution Cosine rule applied to ADE gives n 2 = n 2 + (21-n) 2 - 2n(21-n) cos A. Cosine rule applied to ABC gives m 2 = 21 2 + 33 2 - 2·21·33 cos A. Hence cos A = (21-n)/2n and = (1530-m 2 )/1386. So (2223-m 2 )n = 14553 = 3 3 7 2 11. But n < 21, so n = 3, 7, 9 or 11. Checking each in turn, n = 3 gives m 2 = 2223 - 4851, which is impossible; n = 7 gives m 2 = 2223 - 2079 = 144, so m = 12; n = 9 gives m 2 = 2223 - 1617 = 606, which is not a square; n = 11 gives m 2 = 2223 - 1323 = 900, giving m = 30. But if m = 12, then the triangle is degenerate (33 = 21 + 12), so the only possible solution is n = 11, m = 30. Thanks to Suat Namli 22nd Swedish 1982 © John Scholes jscholes@kalva.demon.co.uk 27 February 2004 Last corrected/updated 27 Feb 04