Regular polygon with 2K sides/vertices; colours and rotations

Question

We have a regular polygon with $2K$ vertices that are coloured in $2K$ different colours(one vertex - one colour). There is another regular polygon with $2K$ vertices that are coloured in $2K$ different colours that overlaps the previous one (colours are the same, but vertex, that was coloured in the first polygon in red, for example, now has different color in that second polygon). Prove that we can rotate the second polygon such way, that there will be at least two places where the colour of a vertex of the first polygon will be the same as the colour of a vertex of the second polygon.

Answer 1

I will number the vertices rather than coloring them.

Let $n=2K$. Number the vertices of one polygon $0,1,2\dots,n-1$ in clockwise order. Number the vertices of the other polygon according to some permutation $\pi$ of $\{0,1,2,\dots,n-1\}$. If we rotate the second polygon $d$ spaces, and two pairs of corresponding vertices are colored the same, this mean that for some $k\neq m$, $$\pi(k+d)=k\\ \pi(m+d)=m$$ All operations are modulo $n$.

If the theorem is false, then for every $d,i,j$ if $\pi(i+d)=i$ and $i\neq j$ then $\pi(j+d)\neq j$. If we write $k=i+d,\ m=j+d$ we have $$\pi(k)-k=-d\neq \pi(m)-m$$ That is, $f:\mathbb Z_m\to\mathbb Z_n$ given by $$f(i)=\pi(i)-i$$ is an injection, hence a bijection.

This cannot be so if $n$ is even. $$\sum_{i\in\mathbb Z_n}\pi(i)-i=\sum_{i\in\mathbb Z_n}\pi(i)-\sum_{i\in\mathbb Z_n}i=0$$ because $\pi$ is a bijection so that $$\sum_{i\in\mathbb Z_n}\pi(i)=\sum_{i\in\mathbb Z_n}i$$

On the other hand, since $f$ is a bijection, $$\sum_{i\in\mathbb Z_n}\pi(i)-i=\sum_{i\in\mathbb Z_n}i=K$$ This is so because every element of $\mathbb Z_n$ other than $0$ and $K$ cancels with its additive inverse.

This contradiction proves the theorem.