# Does the perimeter of a polygon necessarily decrease if more edges are added to it, with the constraint of constant area?

A circle has the lowest perimeter for a 2D shape of a given area. To my understanding, it can also be approximated by a polygon of infinite sides. So, if I take an n-sided polygon and gradually add edges to it, keeping my area constant, will the perimeter also gradually decrease(Since I am approaching a circle)?

Thanks!

• This is certainly true if your $n$-gon is regular. But I'm not at all certain if you allow any $n$-sided polygon, in particular if it's not convex. – Semiclassical Aug 26 '16 at 15:50
• What if I add the constraint that the polygon has to be convex? – Arun Aug 26 '16 at 16:06
• Interesting question. I suspect taht convexity, without necessarly regularity, could suffice... – guestDiego Aug 26 '16 at 16:13
• I have to make an essential precisation to the previous comment: I meant that you keep increasing the number of sides of the convex polygon by adding new vertices WITHOUT removing the old ones (and then scaling down to conserve the area). Otherwise the statement can be easily contradicted (think again to the flattening of the shape of the polygon) – guestDiego Aug 26 '16 at 16:29
• Some implicit assertions in the question are not true. Particularly, every polygon is "approximated by a polygon of infinitely many sides" (think, e.g., of a triangle approximated by "increasingly flat" arcs of circles approximated by $n$-gons with large $n$). – Andrew D. Hwang Aug 26 '16 at 16:31

Convexity is not sufficient. Take a unit square with area $1$ and perimeter $4$. Replace one side with an isosceles triangle with legs of length $100$. The area is now about $51$ and the perimeter is $203$. Scaling down linearly by $\sqrt {51}$ to make unit area leaves the perimeter $\frac {203}{\sqrt {51}}\approx 28.42$

• Essentially, this example shows why "number of sides" is a terrible way to control a polygon: one can have any number of microscopic sides that inflate the side count but contribute virtually nothing to perimeter nor to area. – Erick Wong Aug 26 '16 at 16:47
• Thanks, this is a good example. To stretch this further, does my assumption hold if I include the constraints of both convexity and regularity? – Arun Aug 26 '16 at 17:05
• Yes, and regular implies convex. In that case guestDiego's answer applies. The polygon then inscribes in a circle and you can draw radii from the center to each vertex and the center of each side, making twice as many right triangles as there are vertices. You can then evaluate the perimeter and area. This is how Archimedes approximated $\pi$ – Ross Millikan Aug 26 '16 at 19:38
• Maybe better to take triangle with legs of length 6? Then perimeter will be precisely 7.5. – Somnium Aug 30 '16 at 17:03

Depends on the details of how you add edges. You could get polygons looking like this.

• It's not clear to me how one deduces a counterexample from this... – Semiclassical Aug 26 '16 at 15:54
• @Semiclassical: compare it to a regular polygon of half the sides and the same area. It will look very much like a circle that is halfway between the inner points and the outer points. It will have much smaller perimeter. – Ross Millikan Aug 26 '16 at 16:25
• Hmm, thats interesting. It seems like my question was far too open-ended. Does your answer change if I add the constraint that the polygon should be convex? – Arun Aug 26 '16 at 16:30
• @Arun: No, it takes more than convexity. See my answer. – Ross Millikan Aug 26 '16 at 16:44

I guess that you are supposing that you are dealing with convex regular polygons. If you don't make explicit this hypothesis, the claim is false. In fact, by adding sides and keeping constant area you can obtain polygons with perimeters tending to infinity. You simply need to make the shape flatter and flatter. On the other side, if you consider a sequence of convex regular polygons with the same area and an increasing number of sides, the claim is true. Of course it need to be proved carefully. You need to prove that for any convex $n$-regular polygon inscribed in a unit circle, whose perimeter is $P_n$ and area $A_n$, the ratio $A_n/P_n^2$ is increasing with $n$.