Is there a pattern in figures whose perimeter is the same as their area? Here's what I've found so far.


*

*Circumference of a circle with identical circumference and area: $4\pi$

*Side length of a triangle with identical perimeter and area: $4\sqrt{3}$


And so on...


*

*Square: $4$

*Pentagon: $4\sqrt{5 - 2\sqrt{5}}$

*Hexagon: $\frac{4\sqrt{3}}{3}$

*Heptagon: $\frac{4\cos{\frac{3\pi}{14}}}{1 + \sin{\frac{3\pi}{14}}}$

*Octagon: $4(1+ \sqrt{2})$


Is there a pattern here with the 4s?
 A: We can look at a regular $n$-gon of "radius" $r$, i.e., the convex hull of $r$ times the $n$-th roots of unity.
Connecting the vertices of the polygon to the origin gives you $n$ isosceles triangles of area $\frac{r^2}{2}\sin \frac{2\pi}{n}$. The total area of the $n$-gon is thus
$$A = \frac{nr^2}{2} \sin \frac{2\pi}{n}.$$
To calculate the length $l$ of the third side of these triangles, we can bisect the side to get two right triangles, and then
$$l = 2r\sin \frac{\pi}{n}$$
and the total perimeter is
$$P = 2r n \sin \frac{\pi}{n}.$$
Now we can solve for $r$:
$$\begin{align*}
\frac{nr^2}{2}\sin\frac{2\pi}{n} &= 2rn\sin \frac{\pi}{n}\\
r \sin \frac{2\pi}{n} &= 4\sin \frac{\pi}{n}\\
2r \sin \frac{\pi}{n} \cos \frac{\pi}{n} &= 4\sin \frac{\pi}{n}\\
r &= 2\sec \frac{\pi}{n}.
\end{align*}$$
Plugging this radius into the formula for perimeter gives you
$$P = 4n\tan \frac{\pi}{n}$$
and as you observed, this formula has a natural factor of 4 in it. And indeed,
$$\lim_{n\to\infty} 4n\tan\frac{\pi}{n} = 4\pi.$$
