Maths Olympiad Prep

Library / /30 of 397

Algebra Difficulty 4.8 AIME Prove it Taiwan

Given a positive even integer n>2n > 2, suppose the real numbers x1,x2,…,xnx_1, x_2, \dots, x_n satisfy
∑1≤i<j≤n(xi−xj)2=1. \sum_{1 \le i < j \le n} (x_i - x_j)^2 = 1.
Find the maximum value of ∑i=1n(xi−xi+1)2\sum_{i=1}^{n} (x_i - x_{i+1})^2, where xn+1=x1x_{n+1} = x_1.

Solution

The maximum value is 4n\frac{4}{n}.
First, one can check that this value is attained when x1=x3=⋯=xn−1,x2=⋯=xnx_1 = x_3 = \cdots = x_{n-1}, x_2 = \cdots = x_n.
Hence it suffices to prove that (x1−x2)2+⋯+(xn−1−xn)2+(xn−x1)2≤4n(x_1 - x_2)^2 + \cdots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \le \frac{4}{n}.
We may simultaneously add or subtract a constant to all the xix_i so that x1+⋯+xn=0x_1 + \cdots + x_n = 0; then the original condition becomes equivalent to n(x12+⋯+xn2)=1n(x_1^2 + \cdots + x_n^2) = 1.
Since for any real numbers a,ba, b we have
(a−b)2≤2a2+2b2, (a-b)^2 \le 2a^2 + 2b^2,
it follows that
(x1−x2)2+⋯+(xn−1−xn)2+(xn−x1)2≤4(x12+⋯+xn2) (x_1 - x_2)^2 + \cdots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \le 4(x_1^2 + \cdots + x_n^2)
which completes the proof.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from the original; metadata (topic, difficulty) added by this project.