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The set of all polygons with $m$ sides and perimeter $1$ has an element with maximal area.

I read this fact in a book, and the reference was in German. Does anyone here know?

I know how to prove that the maximal area polygonal must be regular.

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  • $\begingroup$ If the maximal area polygon must be regular, then the regular polygon in the set of all polygons with m sides and perimeter 1 is an element with maximal area. $\endgroup$ – Jonny Apr 14 '15 at 2:04
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By translation invariance, you can assume one of the vertices is $(0,0)$. Then the rest are in the closed unit disk $D$. Encoding the polygon as a finite sequence of vertices, you can identify it with a point in $D^m$. Clearly, $D^m$ is a compact set. It remains to show that the area of polygon is a continuous function of the sequence of vertices $P_i$. This can be done, for example, by writing the area in terms of 2D cross-products: $$ A = \frac12 \sum_i ({P_{i+1}P_i} \times OP_i) \tag{1} $$


By 2D cross product of two vectors $\langle x_1,y_1\rangle$ and $\langle x_2,y_2\rangle$ I mean the determinant $$\begin{vmatrix} x_1& y_1 \\ x_2 & y_2 \end{vmatrix}$$

It gives the signed area of the parallelogram spanned by the vectors; division by $2$ makes it the area of triangle $OP_iP_{i+1}$ in the formula (1). The area being signed takes care of the possible wiggly shapes of the polygon. One way to prove (1) is to express the area using Green's formula, as in anorton's blog post.

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  • $\begingroup$ This makes sense, but what is a 2D cross product? $\endgroup$ – Lorenzo Najt Apr 14 '15 at 18:13
  • $\begingroup$ I added an explanation. $\endgroup$ – user147263 Apr 14 '15 at 18:19

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