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HMMT 十一月 2017 · GEN 赛 · 第 9 题

HMMT November 2017 — GEN Round — Problem 9

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Find the minimum possible value of √ √ √ 2 58 − 42 x + 149 − 140 1 − x where − 1 ≤ x ≤ 1.
解析
  1. Find the minimum possible value of √ √ √ 2 58 − 42 x + 149 − 140 1 − x where − 1 ≤ x ≤ 1. Proposed by: Serina Hu √ Answer: 109 √ 2 2 2 2 2 Substitute x = cos θ and 1 − x = sin θ , and notice that 58 = 3 + 7 , 42 = 2 · 3 · 7, 149 = 7 + 10 , and 140 = 2 · 7 · 10. Therefore the first term is an application of Law of Cosines on a triangle that has two sides 3 and 7 with an angle measuring θ between them to find the length of the third side; similarly, the second is for a triangle with two sides 7 and 10 that have an angle measuring 90 − θ between them. ”Gluing” these two triangles together along their sides of length 7 so that the merged triangles form a right angle, we see that the minimum length of the sum of their third sides occurs when the glued triangles form a right triangle. This right triangle has legs of length 3 and 10, so its √ hypotenuse has length 109.