@joe-shmo *said*

It works for an equilateral triangle as well:

Let the side length be S

P = 3S

A = 1/2* ( 3/16 ) ^( ½ ) * S^2

P = A → S * ( ( 3/16 ) ^( ½ ) * S - 3 ) = 0

S= 0, 4*( 3 )^( ½ )

let the radius of the inscribed circle be "r"

Then;

r / ( S / 2) = (1/2 ) / ( ( 3/4)^( ½ )

r =2

Actually...the general proof is pretty simple:

For any regular n-gon with side length S

P = n*S

The Area is can be divided into n triangles of base S and a vertex terminating at the centroid :

A = n * 1/2* S* h

The height of each triangle is the perpendicular bisector of the side length S, as such it is exactly the radius of the inscribed circle "r"

A = n * 1/2 * S * r

And we have constraints that:

P = A

Thus,

n*S = n * 1/2 * S * r

r = 2 ( the radius of the inscribed circle )

for all regular n-gons where their perimeters is equal to the areas.

Nice problem!