Maths Olympiad Prep

Track / Stage 5 / 282 of 400 #882 of 1964

Problem 882

AIME late
Algebra Difficulty 5.7 Prove it

Let x1,x2,,xn,y1,y2,,ynx_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{n} be complex numbers with modulus equal to 1. Let
zi=xyi+yxixiyi(i=1,2,,n), z_{i}=x y_{i}+y x_{i}-x_{i} y_{i}(i=1,2, \cdots, n),

where, x=1ni=1nxi,y=1ni=1nyix=\frac{1}{n} \sum_{i=1}^{n} x_{i}, y=\frac{1}{n} \sum_{i=1}^{n} y_{i}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Prove: i=1nzin\sum_{i=1}^{n}\left|z_{i}\right| \leqslant n.
Example 4 is not difficult, only of medium difficulty level in the national competition, but some students made mistakes by incorrectly using homogeneity. Since the inequality is homogeneous with respect to a1,a2,,ana_{1}, a_{2}, \cdots, a_{n} and b1,b2,,bnb_{1}, b_{2}, \cdots, b_{n}, therefore, without loss of generality, we can assume i=1nai=1\sum_{i=1}^{n} a_{i}=1 and i=1nbi=1\sum_{i=1}^{n} b_{i}=1, and under these conditions, the problem was solved. In fact, the key in this problem is the case where i=1nai=i=1nbi=0\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=0.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.