When Is a Triangle Like a Circle?

For a circle, the derivative of the area with respect to the radius equals the circumference: $\frac{d}{dr}(\pi r^2)=2\pi r$. Define the differential radius of any shape as the length $r$ for which $\frac{dA}{dr}=P$, where $A$ is the area and $P$ the perimeter as the shape is scaled. What is the differential radius of an equilateral triangle of side $s$?

The Fiddler, Zach Wissner-Gross, May 17, 2024(original post)

Solution

Scale any shape by a factor $t$. Its area grows as $A=A_0t^2$ and its perimeter as $P=P_0t$, while the differential radius, being a length, scales linearly, $r=\lambda t$. Then $$\frac{dA}{dr}=\frac{dA/dt}{dr/dt}=\frac{2A_0t}{\lambda},$$ and setting this equal to $P=P_0t$ forces $\lambda=2A_0/P_0$, so at the shape’s actual size ($t=1$), $$r=\frac{2A}{P}.$$ This is the area over the semiperimeter, which for any shape with an inscribed circle is exactly the inradius. For the equilateral triangle, $A=\tfrac{\sqrt3}{4}s^2$ and $P=3s$, giving $r=\tfrac{\sqrt3 s^2/2}{3s}$: $$\boxed{\ \frac{s}{2\sqrt3}=\frac{s\sqrt3}{6}\ }\approx0.2887\,s,$$ precisely the triangle’s inradius. A triangle behaves like a circle of its own inscribed circle.

The computation

Encode the definition rather than the answer: set the area and perimeter as functions of the scale $t$, form $\frac{dA}{dr}$ with $r=\lambda t$, and solve $\frac{dA}{dr}=P$ for $\lambda$. For the equilateral triangle it returns $\tfrac{\sqrt3}{6}s$.

import sympy as sp
t, s, a, b, lam = sp.symbols('t s a b lambda', positive=True)
def diff_radius(A_of_t, P_of_t):
    r = lam * t                                    # the differential radius scales linearly
    dA_dr = sp.diff(A_of_t, t) / sp.diff(r, t)
    lam_sol = sp.solve(sp.Eq(dA_dr, P_of_t), lam)[0]
    return sp.simplify((lam_sol * t).subs(t, 1))   # evaluate at actual size
print(diff_radius(sp.sqrt(3)/4 * (s*t)**2, 3*s*t))  # sqrt(3)*s/6

Extra Credit

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

Solution

The same $r=2A/P$ holds. With $A=ab$ and $P=2(a+b)$, $$r=\frac{2A}{P}=\frac{2ab}{2(a+b)}= \boxed{\ \frac{ab}{a+b}\ },$$ half the harmonic mean of the sides. For a square ($a=b$) it is $a/2$, the inradius again; for a long thin rectangle it tends to half the short side, the largest disc the rectangle can hold.

The computation

The same diff_radius routine, now with the rectangle’s area and perimeter, returns $\tfrac{ab}{a+b}$.

print(diff_radius(a*b * t**2, 2*(a+b) * t))         # a*b/(a + b)