PUMaC 2013 · 几何（B 组） · 第 8 题

PUMaC 2013 — Geometry (Division B) — Problem 8

[ 8 ] Triangle A B C is an equilateral triangle with sidelength 1. For each n > 1, we construct 1 1 1 triangle A B C from A B C according to the following rule: A , B , C are points n n n n − 1 n − 1 n − 1 n n n on segments A B , B C , C A respectively, and satisfy the following: n − 1 n − 1 n − 1 n − 1 n − 1 n − 1 A A B B C C 1 n − 1 n n − 1 n n − 1 n = = = A B B C C A n − 1 n n − 1 n n − 1 n n − 1 So for example, A B C is formed by taking the midpoints of the sides of A B C . Now, we 2 2 2 1 1 1 | A B C | 5 5 5 m can write = where m, n are relatively prime positive integers. Find m + n . (For a | A B C | n 1 1 1 triangle 4 ABC, | ABC | denotes its area.) 1

解析

[ 8 ] Triangle A B C is an equilateral triangle with sidelength 1. For each n > 1, we construct 1 1 1 triangle A B C from A B C according to the following rule: A , B , C are points n n n n 1 n 1 n 1 n n n on segments A B , B C , C A respectively, and satisfy the following: n 1 n 1 n 1 n 1 n 1 n 1 A A B B C C 1 n 1 n n 1 n n 1 n = = = A B B C C A n 1 n n 1 n n 1 n n 1 So for example, A B C is formed by taking the midpoints of the sides of A B C . Now, we 2 2 2 1 1 1 | A B C | 5 5 5 m can write = where m, n are relatively prime positive integers. Find m + n . (For a | A B C | n 1 1 1 triangle 4 ABC, | ABC | denotes its area.) Solution It will su ce to compute | A B C | / | A B C | for the first few terms. It is n n n n 1 n 1 n 1 immediate that this ratio is proportional to the square of | A B | / | A B | . Noting that n n n 1 n 1 2 based on the construction A B C is equilateral for all n we have by the law of cosines n n n 2 2 2 2 | A B | ( n 1) | A B | ( n 1) | A B | n 1 n 1 n 1 n 1 n 1 n 1 2 | A B | = + 2 n n 2 2 2 n n n ✓ ◆ 2 2 | A B | 1 + ( n 1) ( n 1) n n = 2 | A B | n n 1 n 1 ✓ ◆ 2 | A B | 1 2 2 = | A B | 4 1 1 ✓ ◆ 2 | A B | 1 3 3 = | A B | 3 2 2 ✓ ◆ 2 | A B | 7 4 4 = | A B | 16 3 3 ✓ ◆ 2 | A B | 13 5 5 = | A B | 25 4 4 4 Y | A B C | | A B C | 5 5 5 i +1 i +1 i +1 ) = | A B C | | A B C | 1 1 1 i i i i =1 ✓ ◆ 4 2 Y | A B | i +1 i +1 = | A B | i i i =1 91 = 4800 Giving a final answer of 4891 . 3