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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.

Notes:

  • 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.
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DavidButlerUofA
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    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. – Semiclassical Aug 05 '14 at 11:06
  • ...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. – Semiclassical Aug 05 '14 at 11:17
  • @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. – Mark Bennet Aug 05 '14 at 11:18
  • @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. – Semiclassical Aug 05 '14 at 11:21
  • @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. – Mark Bennet Aug 05 '14 at 11:27
  • We need restrictions, else we can insert vertices along the edges. (The standard definition of polygon allows that.) – André Nicolas Aug 05 '14 at 15:53
  • @Semiclassical My method requires there to be a line segment fully inside the polygon such that it's possible to rejoin the new part without self-intersections. That might be an issue! – DavidButlerUofA Aug 05 '14 at 16:39
  • @MarkBennet I don't see how you can "bring it down to one extra side" every time. The case I am worried about is a star-shaped polygon. – DavidButlerUofA Aug 05 '14 at 16:42
  • @DavidButlerUofA I see what you mean. I think once you have a convex polygon which will do, you can, I think, maintain convexity, it is "intuitively obvious" that you can turn out the inner corners of the star, and then flatten the result to get an equal area - which means to say it is easier to write than to prove. – Mark Bennet Aug 05 '14 at 16:48
  • Here's an interesting thought: I feel pretty confident in being able to offer a 'proof' (possibly with some gaps of rigor) in the case presented here. On the other hand, here's something I definitely don't know if it's true or false: Given an $n$-sided polygon, can I always reduce the number of sides? (I suspect the answer is no, but I don't have a definite sense either way.) – Semiclassical Aug 06 '14 at 22:13
  • @Semiclassical It is not always possible to find a shape with less sides that has the same area and perimeter. For example consider a square with perimeter $4s$ and area $s^2$. The triangle with this perimeter with the biggest area is an equilateral triangle with area $\frac{4\sqrt3 s^2}{9}$, which is less than $s^2$. So there is no triangle with the same area and perimeter as the square. Indeed, there will be plenty of other quadrilaterals with perimeter $4s$ with no matching triangle -- all the ones with area more than $\frac{4\sqrt3 s^2}{9}$. – DavidButlerUofA Aug 10 '14 at 19:58

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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 $P$.)

Zimul8r
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  • 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. – DavidButlerUofA Aug 06 '14 at 08:06
  • 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. – DavidButlerUofA Aug 12 '14 at 16:21