The answers to a recent question established that it is possible to construct families of polygons all with the same area and perimeter. Some comments on some of the answers inspired this very specific question:

Prove that for any n-sided polygon P, and any integer m greater than n, there is an m-sided polygon with the same area and perimeter as P.


  • I define a polygon as not having two successive edges collinear, so you can't just insert a vertex to the middle of an edge.
  • I do not care if the polygons in question are convex or not. So it needs to work if P is not convex, but it does not need to produce convex polygons.
  • I would like a proper written proof, rather than just a description of how one might construct a proof.
  • 1
    $\begingroup$ One rather simple approach (inspired by one of your approaches in there) takes care of a lot of cases: Pick one of the corners of $P$, cut it off, and attach it somewhere on one of the other sides of $P$. This has the same area and perimeter, but $n+3$ sides all together. Iterating this, we conclude that the case of $m=n$ mod 3 is correct. $\endgroup$ Commented Aug 5, 2014 at 11:06
  • $\begingroup$ ...actually, doesn't your cut-flip-join method offer a direct proof? Each step takes you to an $n+1$-gon. Also, note that one doesn't have to 'cut' along a line connecting two vertices: almost every way of cutting off and flipping a particular corner will do. So one just needs to take a corner, cut-flip it, then repeat this with the new corner, etc. $\endgroup$ Commented Aug 5, 2014 at 11:17
  • $\begingroup$ @Semiclassical If you take off a small triangular corner and turn it over you can fix it at one of the new corners to add just two sides. By adjusting the size of the triangle you cut, you can bring that down to one extra side. This does not guarantee convexity. $\endgroup$ Commented Aug 5, 2014 at 11:18
  • $\begingroup$ @MarkBennet: I think it'll be convex as long as the corner one picks is an obtuse angle. And the only figure which doesn't have to have an obtuse corner is a triangle, I think. $\endgroup$ Commented Aug 5, 2014 at 11:21
  • $\begingroup$ @Semiclassical - you have to take a little care, I think, because the neighbouring angle might be too obtuse - but then you cut that corner off instead. $\endgroup$ Commented Aug 5, 2014 at 11:27

1 Answer 1


I wouldn't go constructive on this one. Rather, I'd prove the following:

  1. Show that among all polygons with $n$ sides and a fixed perimeter $p$, the regular $n$-gon has the largest area (we'll call it $A_{n,p}$).

  2. Show that for any area $A < A_{n,p}$, there exists an $n$-gon with perimeter $p$ and area $A$. (Imagine folding up an $n$-gon, being careful not to leave 2 adjacent sides collinear, to generate any area between $0$ and the max.)

  3. Show that for $n>0$, $A_{n,p} < A_{n+1,p}$.

Your theorem follows from these.

(Start with the regular $m$-gon with perimeter the same as your $n$-gon, $P$. It has area greater then $P$, because it's area exceeds the area of the regular $n$-gon, which is the max area for $n$-gons. The regular $m$-gon can then be "squished" down to an $m$-gon with the same area as $P$.)

  • $\begingroup$ I would prefer an actual proof rather than a description of how one might construct one. I'll edit my original question to reflect this. $\endgroup$ Commented Aug 6, 2014 at 8:06
  • $\begingroup$ Also, I am not 100% convinced the squishing process really does cover all areas between 0 and the maximum possible area, especially for polygons with an odd number of edges. $\endgroup$ Commented Aug 12, 2014 at 16:21

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