Proving 2 chords are the same length in a circle divided into $n$ equal arcs

Question

Let us have a circle that is divided in to $n$ equal arcs by $n$ points on the circumference. There are $\dfrac{n}{2}$ chords joining pairs of points. For what values of $n$ would it be necessary for there to be at least 2 chords with the same length, and how would you prove this?

Answer 1

The number of chords rather is ${n\choose 2}$ instead.

Your question then comes down to $${n\choose 2}=\frac{n!}{(n-2)!\ 2!}=\frac{n\ (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.

But then, even so from $n>3$ on there are chords $(i,\ j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i,\ (i+1)\mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.

--- rk

Answer 2

For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?

This is necessary for all odd $n$.

For any even $n$ there are exactly $\frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $\lfloor\frac{n}{2}\rfloor$ distinct chord lengths. If the selection $\frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.