I am asked to proof the following : Using the pigeonhole principle, prove that among any 8 points on a circle of radius 1, there are at least two points whose distance is less than 1.
Just by using a drawing it seems logical, but i am having issues formally writing it, I am getting confused, i did a proof without using the pigeonhole saying :
Let's take the maximum angle in which we can separate the 8 points which is : $\frac{360}{8}=45$, convert 45 degrees to radian we get 0.7... we know that the segment linking any of those two points will be less than 0.7 so less than 1.
But i can not find a way to use pigeonhole. Thanks for your help
You may partition the circle into $7$ half-open arcs (each including one endpoint), each arc subtends $360^\circ/7$ at the centre of the circle. Within each partition, any two points have distance less than $1$ between them.
By the pigeonhole principle, at least $2$ of the $8$ points are in the same partition.