PUMaC 2022 · 团队赛 · 第 1 题

PUMaC 2022 — Team Round — Problem 1

Have b, c ∈ R satisfy b ∈ (0 , 1) and c > 0, then let A, B denote the points of intersection of 2 the line y = bx + c with y = | x | , and let O denote the origin of R . Let f ( b, c ) denote the area ∞ P 1 2 of triangle △ OAB . Let k = , and for n ≥ 1 let k = k . If the sum f ( k , k ) 0 n n n − 1 n − 1 2022 n =1 p can be written as for relatively prime positive integers p, q , find the remainder when p + q is q divided by 1000.

解析

Have b, c ∈ R satisfy b ∈ (0 , 1) and c > 0, then let A, B denote the points of intersection of 2 the line y = bx + c with y = | x | , and let O denote the origin of R . Let f ( b, c ) denote the area ∞ P 1 2 of triangle △ OAB . Let k = , and for n ≥ 1 let k = k . If the sum f ( k , k ) 0 n n n − 1 n − 1 2022 n =1 p can be written as for relatively prime positive integers p, q , find the remainder when p + q is q divided by 1000. Proposed by Sunay Joshi Answer: 484 c c Note that the points A, B have x -coordinates < 0 and > 0. Thus the area of the right − 1 − b 1 − b √ √ 2 1 c c c triangle △ OAB equals f ( b, c ) = · 2 · 2 = . As a result, the desired sum equals 2 2 1+ b 1 − b 1 − b ∞ n n P 2 2 2 k k k . We claim that this sum equals . To see this, expand the term as a n +1 2 n +1 2 2 1 − k 1 − k 1 − k n =1 ∞ P n +1 n j 2 +2 geometric series to find k . The exponents of this series contain all positive integers j =0 n n +1 n n +1 m ≡ 2 (mod 2 ). Since the set of positive integers m such that m ≡ 2 (mod 2 ) for ∞ P 2 2 ℓ k some n ≥ 1 is exactly the set of even positive integers, our sum reduces to k = , 2 1 − k ℓ =1 1 1 as claimed. Plugging in k = , we find a sum of . Thus p + q = 4088484 and our 2022 4088483 remainder is 484.