2025 International Mathematical Olympiad China National Team Selection Test

Question

For a finite non-empty set A of real numbers, let (A) denote its maximum element, and define: P(A) = _B A |B| odd m(B), Q(A) = _ B A |B| even m(B), where m(B) is the median of finite non-empty set B: if B = b_1, b_2, , b_n (b_1 < b_2 < < b_n), then m(B) = 12(b_ n+12 + b_ n+12 ). Find the smallest real number c such that for any set A of 2025 distinct positive real numbers, P(A) - Q(A) c (A).

Accepted Answer

Without loss of generality, let the elements of A in increasing order be x_0, x_1, , x_2024. We first compute P(A). For each x_m, we count how many odd-sized subsets have x_m as their median. A subset B of odd size has x_m as its median if and only if |B x_0, , x_m-1| = |B x_m+1, , x_2024|. If this common value is d, the number of such subsets B is md2024-md. Thus, the total number of such B is _d md 2024-md = _d mm-d 2024-md = 2024m, where the last equality follows from Vandermonde's identity. Therefore, P = _m=0^2024 2024m x_m. Next, we compute Q(A). For each x_m, we count how many even-sized subsets have x_m as one of their two middle elements. The number of such subsets is aligned & _d ( md+1 2024-md + md 2024-md+1 ) &= _d mm-d-1 2024-md + _d mm-d 2024-md+1 &= 2024m-1 + 2024m+1. align…