HMMT 二月 2022 · ALGNT 赛 · 第 7 题
HMMT February 2022 — ALGNT Round — Problem 7
题目详情
- Let ( x , y ), ( x , y ), ( x , y ), ( x , y ), and ( x , y ) be the vertices of a regular pentagon centered at 1 1 2 2 3 3 4 4 5 5 (0 , 0). Compute the product of all positive integers k such that the equality k k k k k k k k k k x + x + x + x + x = y + y + y + y + y 1 2 3 4 5 1 2 3 4 5 must hold for all possible choices of the pentagon. b i 2
解析
- Let ( x , y ), ( x , y ), ( x , y ), ( x , y ), and ( x , y ) be the vertices of a regular pentagon centered at 1 1 2 2 3 3 4 4 5 5 (0 , 0). Compute the product of all positive integers k such that the equality k k k k k k k k k k x + x + x + x + x = y + y + y + y + y 1 2 3 4 5 1 2 3 4 5 must hold for all possible choices of the pentagon. Proposed by: Daniel Zhu Answer: 1152 Solution: Without loss of generality let the vertices of the pentagon lie on the unit circle. Then, if P 4 k f ( θ ) = cos( θ ) and g ( θ ) = f ( θ + 2 jπ/ 5), the condition becomes g ( θ ) = g ( π/ 2 − θ ), or g ( θ ) = j =0 g ( θ + π/ 2), since g is an odd function. iθ − iθ Write f ≍ g if f = cg for some nonzero constant c that we don’t care about. Since cos θ ≍ e + e , we find that X k iℓθ f ( θ ) ≍ e , k + ℓ 2 ℓ ∈ Z a where is defined to be zero if b is not an integer in the interval [0 , a ]. It is also true that b ( 4 iθ X 5 e 5 | ℓ iℓ ( θ +2 jπ/ 5) e = 0 else , j =0 so X k iℓθ g ( θ ) ≍ e . k + ℓ 2 ℓ ∈ 5 Z This is periodic with period π/ 2 if and only if all terms with ℓ not a multiple of 4 are equal to 0. However, we know that the nonzero terms are exactly the ℓ that (1) are multiples of 5, (2) are of the same parity as k , and (3) satisfy | ℓ | ≤ k . Hence, if k is even, the condition is satisfied if and only if k < 10 (else the ℓ = 10 term is nonzero), and if k is odd, the condition is satisfied if and only if k < 5 (else the ℓ = 5 term is nonzero). Our final answer is 1 · 2 · 3 · 4 · 6 · 8 = 1152. b i 2