# Uniqueness of a polygon

Suppose I have two $n$-sided polygons A and B. Is there a non-trivial upper bound on the number of parameters (eg. area, perimeter, etc) of the two polygons, that need to be the same, for A and B to be identical? (For example, if we consider a simpler case to the problem when say A is constrained to lie inside B, then if area(A) = area(B), then A and B are identical. Hence area is the only non-trivial parameter in this case)

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I guess you are looking for a nice answer, but here is a stupid one. On the other hand I'm sure that there is no "nice answer".

A polygon can is uniquely determined by length of sides $\ell_i$ and angles $\alpha_i$. Thus we have to find a complete set of invariants for sequence $(\alpha_1,\ell_1,\alpha_2,\ell_2,\dots\alpha_n,\ell_n)$ which survive after even cyclic shifts and reversing order.

Then you prepare symmetric polynomials for your group. Say take all monomials of degree at most one in each $\alpha_i$ and $\ell_i$ and take its mean value it along the group. You obtain a big collection of polynomial expressions in $\alpha_i$ and $\ell_i$ which gives complete invariant (perimeter will be one of them).

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I liked the first version of your answer better --- somehow symmetric polynomials of complex coordinates of vertices seem more natural to me. –  t3suji Jan 21 '10 at 5:21
@t3suji. Well, it was not quite correct... –  Anton Petrunin Jan 21 '10 at 15:42

The moduli space of $n$-gons up to orientation preserving similarity can be identified with $\mathbb{C}P^{n-2}$, so I would say $2n-3$.

See On the moduli space of polygons in the Euclidean plane by Kapovich and Millson for more about this space.

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In what sense do two parameters suffice to pick out a triangle? Naively one needs three (say, the lengths of the sides). –  Michael Lugo Jan 21 '10 at 3:37
Oh, sorry, it's the moduli space up to orientation preserving similarity. So two angles suffice for a triangle. –  Richard Kent Jan 21 '10 at 4:03
Edited to fix this. –  Richard Kent Jan 21 '10 at 4:08