Consider 8 points on a circle of radius 1. Show that at least two points have a distance less than $\frac{9}{10}$ from each other. Consider 8 points on a circle of radius 1. Show that at least two points have a
distance less than $\frac{9}{10}$ from each other.
What I know so far:
I think I start off with 8 points on the circumference which creates an octagon with side lengths $\frac{1}{2}$. Hence all points have at most a maximum distance of $\frac{1}{2}$.
However I have no idea what to do next. Any help would be great!

We use the following claim:

Claim 1: Let $d_0,\ldots, d_m$ be $m$ real numbers. Then if $\sum_{i=0}^m d_i=K$ then there exists an $i$ such that $d_i \le \frac{K}{m+1}$

So now we use Claim 1. Write the points $x_0,\ldots, x_7$, in the order around the circle. Let us write as $d'(x_i,x_j)$ the length of the shortest walk between $x_i$ and $x_j$ on the circle.
The circumference of the circle is $2\pi$, so $\sum_{i=0}^7 d'(x_i,x_{i+1}) = 2\pi$ the circumference of the circle. By Claim 1,  there exists an $i$ such that $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} <.8 < \frac{9}{10}$.
As the euclidian distance between $x_i$ and $x_{i+1}$ is shorter than the length of the shortest segement on the circle between $x_i$ and $x_{i+1}$, if the inequality $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} < \frac{9}{10}$ is satisfied then the euclidena distance between $x_i$ and $x_{i+1}$, which is less than $d'(x_i,x_{i+1})$, must also be less than 9/10.
