Let $C=\{ z \in \mathbb{C} : |z|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, z_{240} \in C$ (not necessarily different) be $240$ complex numbers, satisfying the following two conditions: (1) For any open arc $\Gamma$ of length $\pi$ on $C$, there are at most $200$ of $j ~(1 \le j \le 240)$ such that $z_j \in \Gamma$. (2) For any open arc $\gamma$ of length $\pi/3$ on $C$, there are at most $120$ of $j ~(1 \le j \le 240)$ such that $z_j \in \gamma$. Find the maximum of $|z_1+z_2+\ldots+z_{240}|$.
Problem
Source: 2022 China TST, Test 1, P5
Tags: inequalities, complex numbers
28.03.2022 08:26
05.04.2022 05:25
Just notice that $z_1+z_2+...+z_{240}=\int_0^{2\pi}F(\theta)e^{i\theta}d\theta$,where F($\theta$) indicates the number of $z_i$s in the arc $(\theta-\frac{\pi}{6},\theta+\frac{\pi}{6})$ This intergal equals $\frac{1}{6} \int_0^{2\pi}(F(\theta)e^{i\theta}+F(\theta+\frac{\pi}{3})e^{i(\theta+\frac{\pi}{3})}+...F(\theta+\frac{5\pi}{3})e^{i(\theta+\frac{5\pi}{3})})d\theta$ WLOG $z_1+z_2+...+z_{240}$ belongs to R+. Ignoring the finite breaking point,all we have to notice is when there are 40 $i$s,40 $-i$s,80 $1$s,40 $w$s,40 $w^5$s($w=e^{\frac{i\pi}{3}}$),for all $\theta$,$Re((F(\theta)e^{i\theta}+F(\theta+\frac{\pi}{3})e^{i(\theta+\frac{\pi}{3})}+...F(\theta+\frac{5\pi}{3})e^{i(\theta+\frac{5\pi}{3})}))$are maximized. Thus the answer is $80+40\sqrt{3}$
05.04.2022 05:52
Masterpieces
19.04.2022 11:12
1、the problem can be reduced to 6 complex numbers, satisfying the following two conditions: (3)For any open arc $\Gamma$ of length $\pi$ on $C$,there are at most 5 complex numbers such that they $\in \Gamma$. (4)For any open arc $\gamma$ of length $\pi/3$ on $C$,there are at most 3 complex numbers such that they $\in \gamma$. 2、Not too hard to guess the maximum is $sqrt(3)+2$, when they are 1 -1 i i $exp(i*pi/3)$ $exp(i*2pi/3)$ ,just notice that they should as close as possible at the unit circle, thus one is 1 and one is -1 should be a good choice, since condition (3) restrict they can not be more close. 3、Next step is the key: when you want to prove the length of the sum of 6 complex numbers(=Y) $<=$ $sqrt(3)+2$, we try to prove the length of the component of Y in any direction is equal or less than $\sqrt(3)+2$. To accomplish this , just discuss how many complex numbers in a half unit circle, which is easy to do so.(at first 4, then 5, then 3, then 2, then 1, then 0). the answer is $80+40\sqrt(3)$, Done.