HMMT 二月 2006 · 冲刺赛 · 第 15 题

英文解析

15. Let Y be as in problem 14. Find the maximum Z such that three circles of radius Zcan simultaneously fit inside an equilateral triangle of area Y without overlapping eachother.
    √
    Answer: 10 3 − 15
    √
    Solution: We first find that, in problem 15, each of the circles of radius Z is the
    ° ° °
    incircle of a 30 -60 -90 triangle formed by cutting the equilateral one in half. The
    √
    2 Y
    √
    equilateral triangle itself has sidelength , so the said inradius is
    √ √34
    √
    1 + 3 − 2 1 2 Y
    √
    Z = · · ,
    2 24
    so that 3
    √ √ √
    ( − 1 + 3) 4 − 2 3 2 3 − 32
    Z = √ Y = √ Y = Y.
    4 3 4 36
    Now we guess that X = 2 and see that, miraculously, everything works: in the problem
    14, say a crimson flower is placed first. Then there are 2 possibilities for C C , 4
    for C C , 2 for C C , and 2 for C C , giving a total of 10. Of course, the first flower can be of any of the three hues, so Y = 3 · 10 = 30. We compute Z andcheck X in a straightforward manner.
    √
    If X > 2, then Y > 30, and Z > 10 3 − 15, with the result that X ≤ 2, a contradiction.
    Assuming X < 2 results in a similar contradiction.
    √
    1+ an