2n points on a circle in two different colors. Prove that pairwise distances of same-color points are the same There are $2n$ points on a circle. The distance (defined by shortest distance you would take to walk from one point to another along the circle) between adjacent points are the same. $n$ points are black and $n$ points are white.
Now we compute the pairwise distances between all the black points and pairwise distances between all the white points. Prove they have the same collection ( with multiplicities) of pairwise distances.
It looks like there has to be a simple trick to map from one group of points to the other through some reflection principle. But I haven't figured out a way...
 A: consider the set of all pairwise distance of black, if distance $j$ is not in it, then $x \mapsto x+j$ is a bijection between black and white.
If we take multiplicity into consideration, the multiplicity of black $j$ is number of points that itself is black and maps to black, it is the same as # of points that is white and maps to white.
A: (Fill in any gaps yourself.)
Step 1: Consider any ring comprising of B's and W's (with possibly an unequal number of them).

Show that the number of $BW$ (in that order) is equal to the number of $WB$.


Show that the number of $BB$ is equal to the number of $B$ minus the number of $BW$.

Step 2: Given the setup, fix $d$. Then, construct a ring of $B$ and $W$ by taking point 1, going around distance $d$ till we loop back.
If there are leftover points (when $\gcd(d,2n)\ne 1$), then take another starting point to form multiple loops.
Step 3: For the (possibly multiple) loops corresponding to $d$, each individual loop has the same number of BW and WB. The total across all these loops have $n$ $B$'s and $n$ $W$'s.

Show that the total number of $BB$ and $WW$ are equal.

Hence conclude that in the setup, the number of distances of length $d$ are equal.
Hence, the multi-set of distances are equal.
