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!
 A: As you mentioned you get an octagon (not necessarily regular) with points on the circle. Each chord (side of the octagon) is shorter than the corresponding arclength it subtends. All the arclengths add to $2\pi r = 2\pi$, which is the circumference of the circle. The chords add up to the perimeter of the octagon. Therefore the perimeter of the octagon is less than the circumference of the circle.
$\frac{2\pi}{8} = 0.785$
Assume all the sides of the octagon have length $\ge \frac{2\pi}{8}$. Then the perimeter of the octagon will be $\ge 8(\frac{2\pi}{8}) = 2\pi$. But this is impossible because we know the perimeter of the octagon is less than the circumference of the circle.
So at least one of the sides of the octagon has length $< \frac{2\pi}{8} = 0.785 < \frac{9}{10}$
A: 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.
A: Below is a geometric proof by contradiction. The grey circle in the center has radius 1, while each of the red circles has radius 0.9 and its center marked with a red X. We begin by placing one point on the center circle's circumference, in this case, we start with the point on the right labeled 1. The next point, labelled 2, must be at a distance of at least 0.9 from point 1. To save space, we put it as close as possible, exactly 0.9 units from point 1. We do the same proceeding around the circle, putting each consecutive point exactly 0.9 units away from the previous one. By the time we get to point number 7, we find that there is no more room to place a point on the grey circle that is at least 0.9 units from both points 6 and 1. Therefore, it is not possible to put 8 points (or even 7 points) on a circle of radius 1, such that no two points are closer than 0.9 units apart.

