If an $n$-gon $P$ is isospectral to a regular $n$-gon $Q$, what could we say about the shape of the $P$. Otherwise, what could we say about $Q$? In fact, some hints or simply some ideas would be appreciated.


**Clarification :** I talk about the spectrum of the Laplacian on the interior of the polygon, acting on the space of functions vanishing on the boundary.



A $n$-gon is isospectral to a regular $n$-gon (Isospectral $\implies$ isometry ?)
Rowlett is hosting *The Sound of Symmetry* [here](http://www.math.chalmers.se/~rowlett/monthly.pdf). The proof of Theorem 4 is exactly as Noam Elkies suggests: Via the Dirichlet heat trace's asymptotic expansion, both area and perimeter are determined by the spectrum, and so for any $n$-gon $\Omega$ the isoperimetric ratio $|\Omega|/|\partial\Omega|^2$ is determined by the spectrum. The content of the proof is that this ratio is globally maximized among $n$-gons at the regular $n$-gon. This determines the regular $n$-gon and implies that any $n$-gon isospectral to the regular $n$-gon is in fact isometric to it. Unless there's a point in the proof where you're confused, I think this settles the question.


(I have a suspicion that the third coefficient of the heat trace in convex polygons 
$$ \frac{1}{24}\sum\_{\mbox{angles }\alpha\_i} \bigg(\frac{\pi}{\alpha\_i} - \frac{\alpha\_i}{\pi}\bigg) $$
is also extremal at the regular $n$-gon. This may be another way to show a version of the desired result for convex $n$-gons.)


