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$n$ vertices from a regular polygon with $2n$ sides are chosen and colored red. The other $n$ vertices are colored blue. Afterwards, the $\binom{n}{2}$ lengths of the segments formed with all pairs of red vertices are ordered in a non-decreasing sequence, and the same procedure is done with the $\binom{n}{2}$ lengths of the segments formed with all pairs of blue vertices. Prove that both sequences are identical.

I have no idea what to do. Any suggestion please?

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  • $\begingroup$ not sure if this helps, but I think it is sufficient to prove that there exists an axis of symmetry in the coloured polygon $\endgroup$ – iamwhoiam Nov 23 '17 at 15:23
  • $\begingroup$ There doesn't have to be an axis of symmetry. You could have RRRBRRBBBB, for example. $\endgroup$ – Ross Millikan Nov 23 '17 at 15:52
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We will refer to the decagon with vertices colored $RRRBRRBBBB$ as an example in what follows. First consider the sides of the decagon, which are the shortest length. Go around the decagon and count how many sides have ends with different colors, calling the number $k$, which must be even. In our example there are four. Then there are $n$ red vertices grouped into $\frac k2$ groups. The number of red-red edges is $n-\frac k2$, as is the number of blue-blue edges. Here $k=4$ and the number of red-red or blue-blue edges is three.

That argument works for any length that results from a vertex spacing that is coprime to $2n$, so for our example segments that result from vertices three apart. If you draw the segments you get a ten pointed star that is just another order to look at the vertices. For a spacing of two vertices you get two pentagons. In our example one is $RRRBB$ and one is $RBRBB$. Again if you look at the number of red-blue edges, here $6$, you have $\frac 62$ groups of red vertices and $n-\frac k2=2$ red-red or blue-blue edges.

This extends to the diameters. There are $k$ red-blue diameters, then $n-\frac k2$ red-red or blue-blue diameters. Here $k=6$ and there are two red-red and two blue-blue diameters.

Added: a much simpler approach. Consider all the segments of a given length other than the diameter. There are $2n$ of them. $2n$ of the ends are red and $2n$ are blue. Let there be $k$ that have one red end and one blue end. There are $2n-k$ red ends unaccounted for, so there must be $\frac 12(2n-k)$ segments with both ends red and the same number with both ends blue. For the diameters there are $n$ of them, with $n$ red ends and $n$ blue ends. Again, let $k$ of them have one red and one blue end. There must be $\frac 12(n-k)$ with red at both ends and the same number with blue at both ends.

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