HMMT 十一月 2017 · 冲刺赛 · 第 31 题

HMMT November 2017 — Guts Round — Problem 31

[ 24 ] In unit square ABCD , points E, F, G are chosen on side BC, CD, DA respectively such that AE 404 is perpendicular to EF and EF is perpendicular to F G . Given that GA = , find all possible 1331 values of the length of BE .

解析

[ 24 ] In unit square ABCD , points E, F, G are chosen on side BC, CD, DA respectively such that AE 404 is perpendicular to EF and EF is perpendicular to F G . Given that GA = , find all possible values 1331 of the length of BE . Proposed by: Yuan Yao 9 Answer: 11 Let BE = x , then since triangles ABE, ECF, F DG are all similar, we have CE = 1 − x, CF = 2 2 2 x (1 − x ) , F D = 1 − x (1 − x ) , DG = x − x (1 − x ) , GA = 1 − x + x (1 − x ) = (1 − x )( x + 1), therefore it remains to solve the equation 404 2 (1 − x )( x + 1) = . 1331 p We first seek rational solutions x = for relatively prime positive integers p, q . Therefore we have q 2 2 ( q − p )( p + q ) 404 2 2 3 3 = . Since both q − p and p + q are relatively prime to q , we have q = 1331 ⇒ q = 11, 3 q 1331 2 2 so (11 − p )( p + 121) = 404 = 2 · 101, and it is not difficult to see that p = 9 is the only integral solution. We can therefore rewrite the original equation as 9 2 103 2 ( x − )( x − x + ) = 0 . 11 11 121 9 It is not difficult to check that the quadratic factor has no zeroes, therefore BE = x = is the only 11 solution.