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!

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.

