This week’s Fiddler is about a generalized notion of “radius”.

For a circle with radius $r$, its area is $\pi r^2$ and its circumference is $2\pi r$. If you take the derivative of the area formula with respect to $r$, you get the circumference formula! Let’s define the term “differential radius.” The differential radius $r$ of a shape with area $A$ and perimeter $P$ (both functions of $r$) has the property that $dA/dr = P$. (Note that $A$ always scales with $r^2$ and $P$ always scales with $r$.)

For example, consider a square with side length $s$. Its differential radius is $r = s/2$. The square’s area is $s^2$, or $4r^2$, and its perimeter is $4s$, or $8r$. Sure enough, $dA/dr = d(4r^2)/dr = 8r = P$. What is the differential radius of an equilateral triangle with side length s?

*Extra credit:*

What is the differential radius of a rectangle with sides of length $a$ and $b$?

My solution:

[Show Solution]

In general, if $A$ scales with $r^2$ and $P$ scales with $r$ and $P$ is the derivative of $A$, then it must be the case that:

\[

A = \alpha r^2\qquad\text{and}\qquad P = 2\alpha r

\]for some constant of proportionality $\alpha$. If we have a certain shape and we know formulas for its area and perimeter, then we can solve for $\alpha$ and $r$ in terms of $A$ and $P$ and we obtain:

\[

\alpha = \frac{P^2}{4A}\qquad\text{and}\qquad r = \frac{2A}{P}

\]

We can verify this formula for a square of sidelength $s$. Here, $P=4s$ and $A=s^2$. Substituting into the above formulas, we obtain $\alpha = 4$ and $r = \tfrac{s}{2}$, just as the problem stated.

In general, we can solve the case for a regular polygon with $n$ sides of length $s$. The perimeter is $P=ns$ and the area is $A = \frac{ns^2}{4\tan(\pi/n)}$. Therefore, we obtain:

\[

\alpha = n\tan\bigl(\tfrac{\pi}{n}\bigr),\qquad\text{and}\qquad r = \frac{s}{2}\cot\bigl(\tfrac{\pi}{n}\bigr)

\]So, in particular, for $n=3$, we get $r=\frac{\sqrt{3}}{6}s$. For regular polygons, the differential radius is the apothem (the line joining the center of the polygon to the midpoint of one of its sides. It is also the “inradius” (radius of the inscribed circle). Note that based on the general formula for $r$, we have:

\[

A = \frac{P\cdot r}{2}

\]In the case of the regular polygon, this means the area is one half of the perimeter times the differential radius (apothem). This makes sense, since we can unfold a regular polygon and compute its area using the length of the bases (the perimeter) and the height (apothem). See below for a visual demonstration:

And here is an animated version:

### Extra credit

For the case of a rectangle of side lengths $a$ and $b$, the perimeter is $P=2(a+b)$ and the area is $A=ab$. Applying the formula above, we obtain:

\[

\alpha = \frac{(a+b)^2}{ab}\qquad\text{and}\qquad r = \frac{ab}{a+b}

\]The differential radius is half of the harmonic mean of $a$ and $b$. It also satisfies the nice formula:

\[

\frac{1}{r} = \frac{1}{a} + \frac{1}{b}

\]which is incidentally how you add resistors in parallel. We can construct the differential radius geometrically by drawing a line connecting opposite corners, and then another line at 45 degrees to the other corner and marking the intersection. The result is illustrated in the image below:

The triangles on the right are what you get when you “unfold” the rectangle on the left. The area of the rectangle, $ab$, is simply the base of the triangles (perimeter of rectangle), times the height ($r$, or the differential radius), times one half.

This works because triangles AFP and PGC are similar. It follows that:

\begin{align*}

\frac{AF}{FP} &= \frac{AD-DF}{FP} = \frac{b-r}{r},\quad\text{and}\\[2mm]

\frac{PG}{GC} &= \frac{PG}{DC-DG} = \frac{r}{a-r}

\end{align*}Equating them, we obtain:

\[

\frac{r}{a-r}=\frac{b-r}{r}

\]Solving this equation for $r$, we obtain $r = \frac{ab}{a+b}$, as required.

Here is an animation of the rectangle unfolding into four triangles with equal height and whose bases sum to the perimeter of the rectangle!

Thanks for the great general solution!

I’m not sure that r = s*sqrt(3)/2 is correct for a triangle, though.

I came up with r = s*sqrt(3)/6,

and if you plug n= 3 into r = (s/2)*cot(pi/n), I think you get

r = (s/2)*cot(pi/3) = (s/2)*sqrt(3)/3) = s*sqrt(3)/6

Also, sqrt(3)/2 > 1, and it seems that the apothem for an equilateral triangle should be < 1.

Thanks for catching that! I meant to write $\frac{1}{2\sqrt{3}}$ but wrote $\frac{\sqrt{3}}{2}$ by accident. I fixed it now.