Smallest area of polygon with $n$ sides all of length $1$

Question

Given an odd number $n$, consider all non-self-intersecting polygons with $n$ sides, all of length $1$. What is the infimum of their areas? We can approach $\sqrt 3/4$ by approximating an equilateral triangle of side $1$, like this:

Can we do better?

Answer 1

One cannot do better. I'm coming short of a proof, but following is the skeleton of a proof.

Idea: recursively chop down the polygon into polygons having all unit sides except one side, of length $x \leq 1$ showing that the minimum area of the parts is some function of $x$. It results in a series of small geometric problems.

Definitions

For $0 \leq x \leq 1$, let an “odd-$x$-poly” be a non-self-intersecting polygon with an odd number of sides, where all sides except one have unit length. The other side has length $x$.

For $0 \leq x \leq 1$, let an “even-$x$-poly” be a non-self-intersecting polygon with an even number of sides, where all sides except one have unit length. The other side has length $x$.

Induction

By induction we will construct the proofs for

$O_n$ : The area of an $n$-sided odd-$x$-poly is at least ${x \over 2} \sqrt {1 - {x ^2 \over 4}}$

$E_n$ : The area of an $n$-sided even-$x$-poly is at least ${1 - x \over 2} \sqrt{1 - {{(1-x)^2} \over 4}}$

$O_3$ is proven trivially by considering the formula for the area of the isosceles triangle.

$E_4$ is proven in [...] (figure 2).

Let’s consider an $n$-sided odd-$x$-poly $p$.

There’s three cases:

1. No vertices or edges from p in the triangle created by ag and two unit sides. Figure 4.

   In this case, this area without vertices is already large enough to satisfy $O_n$.

2. A vertice is present, say, $d$. Figure 1

   In this case, the area of $p$ is either:

   • the area of a $l$-sided even-$y$-poly ($abcd$) plus the area of another $m$-sided even-$z$-poly ($defg$), where $y+z > x$, plus the area of $adg$. Figure 1. $O_n$ is true by [...] ( Some proof missing using geometry and $E_{i}, i<n$. )

   • the area of a $l$-sided odd-$y$-poly ($abc$) plus the area of another $m$-sided odd-$z$-poly ($cde$), where $y+z > x$, plus the area of $ace$. Figure 3. $O_n$ is true by [...] ( Some proof missing using geometry and $O_{i}, i<n$. )

In both cases, it should be possible to prove $O_n$, from $O_x$ and $E_y$, with $x<n, y<n$.

3. An edge goes through the triangle created by ag and two unit sides. (Figure 5). Given how $c, d$ must be at distance of at least 1 of $a, b$, a construction showing $O_n$ is possible, given $O_i$ and $O_j$, with $i<n, j<n$.

Let’s consider an $n$-sided even-$x$-poly $p$. Similar construction as above, missing, switching odd and even sub-polygons.

Very interesting problem, but I've used the time I had to put towards it. Sorry for the missing parts.

Finally, once the missing bits in the proof are added, one can observe that the polygon described in the problem is a even-$x$-poly with $x=1$. Thus we find the specified minimal area.

Answer 2

No, one cannot do better.

Consider a polygon $P_0$ with $n$ odd, edges or length 1. We can repeatedly construct $P_l$ with the same number of unit edges, so that $area(P_{l+1}) \leq area (P_l)$. Until $area(P_{n-1 \over 2})$ is $\sqrt(3) \over 4$.

As such, $area(P_0) \geq {\sqrt(3) \over 4}$. The construction follows.

From $P_l$, take 5 consecutive corners, label them 1,2,3,4,5. Observe that by merging (moving arbitrarily close) corners 2 and 4, we create a sliver $(2,3,4)$, we reduce the area covered by $(1,2,4,5)$ and we keep constant the rest of the area. This gives us $P_{l+1}$. Moving forward, we consider 2 and 4 as a single corner, and disregard 5.

By repeating this process, we always end up with an equilateral triangle and a number of slivers of area almost 0. And we know the original polygon had larger area.

Two missings item in this reasonning that we could elaborate on:

• Showing that overlapping is not an issue. We might need to generalize to a definition of area that counts doubly the overlapping area, so that 2 and 4 can always be "merged". By the end of the process, though, any overlapping is gone. We just need a definition of area that decreases as we simplify the polygon.
• Showing that merging corners 2 and 4 always reduce the area. It's pretty evident, but I didn't write up the math for it.