HMMT 十一月 2019 · 冲刺赛 · 第 32 题

HMMT November 2019 — Guts Round — Problem 32

[17] A sequence of real numbers a , a , . . . , a with a = 0, a = 1, and a > 0 satisfies 0 1 9 0 1 2 a a a = a + a + a n +2 n n − 1 n +2 n n − 1 for all 1 ≤ n ≤ 7, but cannot be extended to a . In other words, no values of a ∈ R satisfy 10 10 a a a = a + a + a . 10 8 7 10 8 7 Compute the smallest possible value of a . 2

解析

[17] A sequence of real numbers a , a , . . . , a with a = 0, a = 1, and a > 0 satisfies 0 1 9 0 1 2 a a a = a + a + a n +2 n n − 1 n +2 n n − 1 for all 1 ≤ n ≤ 7, but cannot be extended to a . In other words, no values of a ∈ R satisfy 10 10 a a a = a + a + a . 10 8 7 10 8 7 Compute the smallest possible value of a . 2 Proposed by: Dylan Liu √ Answer: 2 − 1 a +1 − a +1 1 Say a = a . Then using the recursion equation, we have a = − 1, a = , a = , a = − , 2 3 4 5 6 a − 1 a +1 a 2 a a = − , and a = 1. 7 2 8 a − 1 Now we have a a a = a + a + a . No value of a can satisfy this equation iff a a = 1 and 10 8 7 10 8 7 10 8 7 2 a 2 a + a 6 = 0. Since a is 1, we want 1 = a = − , which gives a + 2 a − 1 = 0. The only positive 8 7 8 7 2 a − 1 √ root of this equation is 2 − 1. This problem can also be solved by a tangent substitution. Write a = tan α . The given condition n n becomes α + α + α = 0 . n +2 n n − 1 We are given α = 0 , α = π/ 4, and α ∈ (0 , π/ 2). Using this, we can recursively compute α , α , . . . 0 1 2 3 4 3 π in terms of α until we get to α = − 2 α . For a not to exist, we need α ≡ π/ 2 mod π . The 2 10 2 10 10 4 √ only possible value of α ∈ (0 , π/ 2) is α = π/ 8, which gives a = tan π/ 8 = 2 − 1. 2 2 2