# When Is a Triangle Like a Circle?

## This Week's Fiddler

To find the differential radius, we must find a value for $r$ in terms of $s$ such that $\frac{dA}{dr}=P$ for equilateral triangles.

For an equilateral triangle with side length $s$, its area $A=\frac{\sqrt{3}}{4}s^2$, and its perimeter $P=3s$.

So $3s=P=\frac{dA}{dr}=\frac{d}{dr}(\frac{\sqrt{3}}{4}s^2)=\frac{\sqrt{3}}{2}s\cdot\frac{ds}{dr}$, meaning $\frac{ds}{dr}=2\sqrt{3}$.

Taking the integral, we find that $s=(2\sqrt{3})r+C$. Since $P$ must scale with $s$ and $r$ must scale with $P$, we know that $s$ and $r$ should be directly proportional; thus, $C=0$.

We have that $s=(2\sqrt{3})r$, or $r=\frac{\sqrt{3}}{6}s$, giving us our answer.

Stepping back, this makes sense given the height of an equilateral triangle (the distance from one vertex to the midpoint of the opposite line segment) being $h=\frac{\sqrt{3}}{2}s$. This makes $r$ exactly $\frac{h}{3}$, the distance from the midpoint of any side to the center of the circle—an intuitive value for a "radius".

**Answer:** s * sqrt(3) / 6

## Extra Credit

Honestly, I have no evidence for this aside from that it works for the case of a square. A guess of sorts.

**Answer:** (a + b) / 2