HMMT 二月 2003 · 冲刺赛 · 第 12 题

HMMT February 2003 — Guts Round — Problem 12

[7] As shown in the figure, a circle of radius 1 has two equal circles whose diameters cover a chosen diameter of the larger circle. In each of these smaller circles we similarly draw three equal circles, then four in each of those, and so on. Compute the area of the region enclosed by a positive even number of circles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND 2 2 4 4

解析

As shown in the figure, a circle of radius 1 has two equal circles whose diameters cover a chosen diameter of the larger circle. In each of these smaller circles we similarly draw three equal circles, then four in each of those, and so on. Compute the area of the region enclosed by a positive even number of circles. Solution: π/e 2 At the n th step, we have n ! circles of radius 1 /n ! each, for a total area of n ! · π/ ( n !) = π/n !. The desired area is obtained by adding the areas of the circles at step 2, then subtracting those at step 3, then adding those at step 4, then subtracting those at step 5, and so forth. Thus, the answer is ( ) π π π 1 1 1 1 − 1 − + − · · · = π − + − + · · · = πe . 2! 3! 4! 0! 1! 2! 3! 2 2 4 4