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.
I wouldn't go constructive on this one. Rather, I'd prove the following:
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 $A_{n,p}$).
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.)
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 $P$.)