HMMT 二月 2013 · 冲刺赛 · 第 10 题
HMMT February 2013 — Guts Round — Problem 10
题目详情
- [ 6 ] Wesyu is a farmer, and she’s building a cao (a relative of the cow) pasture. She starts with a ◦ ◦ triangle A A A where angle A is 90 , angle A is 60 , and A A is 1. She then extends the pasture. 0 1 2 0 1 0 1 1 First, she extends A A to A such that A A = A A and the new pasture is triangle A A A . 2 0 3 3 0 2 0 1 2 3 2 1 Next, she extends A A to A such that A A = A A . She continues, each time extending A A 3 1 4 4 1 3 1 n n − 2 6 1 to A such that A A = A A . What is the smallest K such that her pasture never n +1 n +1 n − 2 n n n − 2 2 − 2 exceeds an area of K ?
解析
- [ 6 ] Wesyu is a farmer, and she’s building a cao (a relative of the cow) pasture. She starts with a ◦ ◦ triangle A A A where angle A is 90 , angle A is 60 , and A A is 1. She then extends the pasture. 0 1 2 0 1 0 1 1 First, she extends A A to A such that A A = A A and the new pasture is triangle A A A . 2 0 3 3 0 2 0 1 2 3 2 1 Next, she extends A A to A such that A A = A A . She continues, each time extending A A 3 1 4 4 1 3 1 n n − 2 6 1 to A such that A A = A A . What is the smallest K such that her pasture never n +1 n +1 n − 2 n n n − 2 2 − 2 exceeds an area of K ? √ Answer: 3 First, note that for any i , after performing the operation on triangle A A A , i i +1 i +2 the resulting pasture is triangle A A A . Let K be the area of triangle A A A . From i +1 i +2 i +3 i i i +1 i +2 Guts Round 1 1 A A = A A and A A = A A + A A , we have A A = (1+ ) A A . n +1 n − 2 n n n − 2 n n +1 n n − 2 n − 2 n +1 n n +1 n n n − 2 2 − 2 2 − 2 We also know that the area of a triangle is half the product of its base and height, so if we let the base of 1 triangle A A A be A A , its area is K = hA A . The area of triangle A A A n − 2 n − 1 n n n − 2 n − 2 n n − 2 n − 1 n n +1 2 1 is K = hA A . The h ’s are equal because the distance from A to the base does not change. n − 1 n n +1 n − 1 2 n K A A K K K K n − 1 n n +1 1 2 − 1 3 7 3 7 1 2 2 1 We now have = = 1 + = . Therefore, = , = = · = , n n K A A 2 − 2 2 − 2 K 2 K K K 6 2 4 n − 2 n n − 2 0 0 1 0 n +1 K K K K 15 7 15 2 − 1 3 3 2 n = = · = . We see the pattern = , which can be easily proven by induction. n K K K 14 4 8 K 2 0 2 0 0 K n As n approaches infinity, grows arbitrarily close to 2, so the smallest K such that the pasture never K 0 √ exceeds an area of K is 2 K = 3. 0