Let's define almost mean of numbers $a_1, a_2, \ldots, a_n$ as $\frac{a_1 + a_2 + \ldots + a_n}{n+1}$. Oleksiy has positive real numbers $b_1, b_2, \ldots, b_{2023}$, not necessarily distinct. For each pair $(i, j)$ with $1 \leq i, j \leq 2023$, Oleksiy wrote on a board almost mean of numbers $b_i, b_{i+1}, \ldots, b_j$. Prove that there are at least $45$ distinct numbers on the board. Proposed by Anton Trygub
Problem
Source: Ukrainian Mathematical Olympiad 2024. Day 1, Problem 11.3
Tags: algebra, combinatorics, poset
R8kt
20.03.2024 12:51
Another cool problem by the master troll Anton. For every i from 1 to 2024, denote f(i) as the normal mean of the first i numbers, and g(i) as they’re almost mean. By Erdös there exist either an non-increasing or non-decreasing subsequence of length at least 45. WLOG it is non-increasing, and these numbers are a1, a2, … a45. Then g(a1), g(a2),… g(a45) are different.
BR1F1SZ
08.09.2024 03:02
R8kt wrote: Another cool problem by the master troll Anton. For every i from 1 to 2024, denote f(i) as the normal mean of the first i numbers, and g(i) as they’re almost mean. By Erdös there exist either an non-increasing or non-decreasing subsequence of length at least 45. WLOG it is non-increasing, and these numbers are a1, a2, … a45. Then g(a1), g(a2),… g(a45) are different. Hi! Could someone please explain to me what Erdös is? And how is it used in this problem? Are there any alternate solutions that don't use this? Thanks!
ohiorizzler1434
08.09.2024 06:17
blud that solution doesnt work
dgrozev
09.09.2024 16:45
@^^: Erdos - Szekeres theorem. @^: Of course it works, but it needs more attention exactly when $a_1,a_2,\dots, a_{45}$ is non-increasing. As written in #2, it works when this sequence is non-decreasing. In the former case we take the almost mean of the segments $(a_i, a_{45}), i=1,2,\dots,45$.