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HMMT 二月 2000 · 代数 · 第 9 题

HMMT February 2000 — Algebra — Problem 9

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. = = , a + b + d = 15. Find a + bd ad b . d 4 4 2
解析
  1. W e an view these onditions as a geometry diagram as seen b elo w. So, w e kno w that p e 3 3 3 3 2 2 = (sin e e = a b = d = f and w e kno w that e + f = 15 (sin e this is f 4 4 4 4 p p 2 2 2 2 a + b + d ). Also, note that a + bd ad b = ( a b )( d ) = ef . So, solving 2 2 2 2 2 2 for e and f , w e nd that e + f = 225, so 16 e + 16 f = 3600, so (4 e ) + (4 f ) = 3600, 2 2 2 2 2 2 2 so (3 f ) + (4 f ) = 3600, so f (3 + 4 ) = 3600, so 25 f = 3600, so f = 144 and f = 12. 3 Th us, e = 12 = 9. Therefore, ef = 9 12 = 108 . 4 e f a b d c 4 2