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HMMT 二月 2013 · 冲刺赛 · 第 26 题

HMMT February 2013 — Guts Round — Problem 26

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 17 ] Triangle ABC has perimeter 1. Its three altitudes form the side lengths of a triangle. Find the set of all possible values of min( AB, BC, CA ).
解析
  1. [ 17 ] Triangle ABC has perimeter 1. Its three altitudes form the side lengths of a triangle. Find the set of all possible values of min( AB, BC, CA ). √ 3 − 5 1 Answer: ( , ] Let a, b, c denote the side lengths BC, CA , and AB , respectively. Without 4 3 loss of generality, assume a ≤ b ≤ c ; we are looking for the possible range of a . 1 First, note that the maximum possible value of a is , which occurs when ABC is equilateral. It 3 remains to find a lower bound for a . Now rewrite c = xa and b = ya , where we have x ≥ y ≥ 1. Note that for a non-equilateral triangle, x > 1. The triangle inequality gives us a + b > c , or equivalently, y > x − 1. If we let K be the 2 K 2 K 2 K 1 1 1 area, the condition for the altitudes gives us + > , or equivalently, > − , which after c b a b a c x x some manipulation yields y < . Putting these conditions together yields x − 1 < , and after x − 1 x − 1 √ 3+ 5 rearranging and solving a quadratic, we get x < . 2 Guts Round We now use the condition a (1 + x + y ) = 1, and to find a lower bound for a , we need an upper bound x 1 for 1 + x + y . We know that 1 + x + y < 1 + x + = x − 1 + + 3. x − 1 x − 1 1 Now let f ( x ) = x − 1 + + 3. If 1 < x < 2, then 1 + x + y ≤ 1 + 2 x < 5. But for x ≥ 2, we see x − 1 that f ( x ) attains a minimum of 5 at x = 2 and continues to strictly increase after that point. Since ( ) √ √ √ 3+ 5 3+ 5 x < , we have f ( x ) < f = 3 + 5 > 5, so this is a better upper bound than the case for 2 2 ( ) √ 1 3 − 5 √ which 1 < x < 2. Therefore, a > = . 4 3+ 5 √ √ √ 3 − 5 1+ 5 For any a such that 5 − 2 ≥ a > , we can let b = a and c = 1 − a − b . For any other 4 2 1 − a possible a , we can let b = c = . The triangle inequality and the altitude condition can both be 2 verified algebraically. √ 3 − 5 1 We now conclude that the set of all possible a is < a ≤ . 4 3