HMMT 二月 2006 · 冲刺赛 · 第 15 题

HMMT February 2006 — Guts Round — Problem 15

[7] Let Y be as in problem 14. Find the maximum Z such that three circles of radius Z can simultaneously fit inside an equilateral triangle of area Y without overlapping each other. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND √ 1+ a n

解析

Let Y be as in problem 14. Find the maximum Z such that three circles of radius Z can simultaneously fit inside an equilateral triangle of area Y without overlapping each other. √ 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 4 3 √ √ √ 1 + 3 − 2 1 2 Y √ Z = · · , 4 2 2 3 so that √ √ √ 2 ( − 1 + 3) 4 − 2 3 2 3 − 3 Z = √ Y = √ Y = Y. 6 4 3 4 3 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 and check 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+ a n