HMMT 二月 2013 · 冲刺赛 · 第 26 题

英文解析

26. [ 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 .
    First, note that the maximum possible value of a is , which occurs when ABC is equilateral. It 1
    remains to find a lower bound for a .3
    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 afterc b a b a cx xsome manipulation yields y < . Putting these conditions together yields x − 1 < , and afterx − 1 x − 1
    √
    3+ 5
    rearranging and solving a quadratic, we get x < .
    Guts Round 2
    We now use the condition a (1 + x + y ) = 1, and to find a lower bound for a , we need an upper boundx 1
    for 1 + x + y . We know that 1 + x + y < 1 + x + = x − 1 + + 3.
    x − 1 x − 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 1
    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 > = .
    3+ 54
    √ √
    √
    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 − apossible a , we can let b = c = . The triangle inequality and the altitude condition can both beverified algebraically.2
    √
    3 − 5 1
    We now conclude that the set of all possible a is < a ≤ .
    4 3