4

I've found the following interesting exercise.

Let $z_1,...,z_n \in B(0,1) = \{ z \in \mathbb{C} \mid | z | \leq 1 \}$ be $n$ complex numbers in the unit sphere. Show that there exists a $w \in B(0,1)$ for which is $$\sum_{i=1}^n | z_i - w | \geq n.$$

I tried to proove it by induction, but it didn't seem to work since I don't know how to choose $w$ with respect to $z_1,...,z_n$.

Lukas
  • 129

3 Answers3

1

Hint: $\,$ let $\,g = \frac{1}{n} \sum z_i \,\in\, B = B(0,1)\,$ be the centroid of points $\,z_i\,$, then by the triangle inequality $\,\sum | z_i - w | \geq \left| \sum (z_i - w)\right| = \left| \sum z_i - n w\right| = n \left|g - w\right|\,$, and the problem reduces to showing that for any point $\,g \in B\,$ there exists a point $\,w \in B\,$ such that $\,|g-w| \ge 1\,$.

dxiv
  • 76,497
1

You can also show that for every $w$, $w'$ opposite points on the sphere, we have $\sum_k |z_k - w| \ge n$ or $\sum_k |z_k - w'| \ge n$. Indeed their sum is $$\sum_k (|z_k - w| + |z_k - w'|) \ge \sum_k |w-w'| = 2n $$

The points $z_k$ can be anywhere in the given space.

Note: I've seen this problem proposed quite a while ago by Titu Andreescu.

orangeskid
  • 53,909
0

Technical nit: this problem is about numbers in the closed unit disc not the 1-sphere.

As is often the case this can be interpreted as a result about polynomials.
$p(w):=\prod_{k=1}^n(w-z_k)$ and suppose for contradiction that a satisfying $w$ doesn't exist
$\implies\big \vert p(w)\big \vert \leq\Big(\frac{1}{n}\sum_{k=1}^n | z_k - w|\Big)^n\lt 1$
for all $w\in \overline B(0,1)$ by $\text{GM}\leq\text{AM}$.. Then $w^n-p(w)$ is a polynomial with degree $\lt n$ but has exactly $n$ solutions in the unit disc per Rouche, and this is a contradiction.

user8675309
  • 10,034