Which type of triangle has the largest circularity value, $\sqrt{\frac{4\pi\cdot\text{area}}{\text{perimeter}^2}}$?

Question

Equation 7 of "Particle Shape Factors and Their Use in Image Analysis–Part 1: Theory" (PDF) defines circularity as

$$\text{circularity} = \sqrt{\frac{4\pi\cdot\text{area}}{\text{perimeter}^2}}$$

Which type of triangle has the largest circularity value, and how does one prove it?

Answer 1

Since the circularity measure of similar triangle is the same, we may normalize the triangles to have a fixed perimeter, say $1$, and then the problem is simply to find the triangle of maximum area and given perimeter.

It is well-known that the answer is an equilateral triangle. This is easy to show using Heron's formula and Lagrange multipliers, as is done here.

Answer 2

In other words you want to find the triangle type that has the biggest ratio between area and perimeter squared.

First, let us show that it has to be isosceles triangle. Suppose the opposite, that the triangle $ABC$ is optimal. Draw line through point $C$ parallel with $AB$ and find the mirror point $A'$ of $A$ with respect to that line. Line $BA'$ meets the parallel line at point $C_1$.

Obviously, triangles $ABC$ and $ABC_1$ have the same area but triangle $AMC_1$ has smaller perimeter because:

$$AC_1+BC_1=A'C_1+C_1B<BC+CA'=BC+AC$$

So triangle $ABC$ is not optimal and we should consider only triangles that are isosceles.

Supose that the perimeter $P=2s$ is fixed, base of the thriangle is equal to $a$ and leg to $b$. Let us maximize the area.

$$b=s-\frac a2$$

$$h=\sqrt{b^2-\frac{a^2}{4}}$$

$$h=\sqrt{(s-\frac a2)^2-\frac{a^2}{4}}$$

$$h=\sqrt{s^2-as}$$

$$A=\frac{ah}2=\frac12\sqrt{a^2s^2-a^3s}$$

So the area is maximized when the function

$$f(a)=a^2s^2-a^3s$$

...has maximm value. By solving equation:

$$f'(a)=0$$

...we get:

$$2as^2-3a^2s=0\implies a=\frac{2s}{3} \implies a=b=\frac P3$$

So the most "circular" triangle is actually an equilateral triangle.

Answer 3

Let's denote the circularity by $Z$. $$Z=(some\ constant)*\frac{(s(s-a)(s-b)(s-c))^{1/4}}{s},$$ since area$=(s(s-a)(s-b)(s-c))^{1/2}$ and perimeter$=2s$.

Note that $Z$ depends on $a,b$ and $c$. So, for $Z$ to be maximum, it's partial derivatives with respect to $a,b$ and $c$ will be $0$. It'll be a little long calculation, but at last, it'll give you three equations:- $$-\frac{s}{s-a}+\frac{s}{s-b}+\frac{s}{s-c}=3\ \ \ \ \ \ \ \ (1)$$ $$\frac{s}{s-a}-\frac{s}{s-b}+\frac{s}{s-c}=3\ \ \ \ \ \ \ \ (2)$$ $$\frac{s}{s-a}+\frac{s}{s-b}-\frac{s}{s-c}=3\ \ \ \ \ \ \ \ (3)$$ Add those to get $$\frac{s}{s-a}+\frac{s}{s-b}+\frac{s}{s-c}=9\ \ \ \ \ \ \ \ (4)$$ Now, by AM-HM inequality, $$\frac{(s-a)+(s-b)+(s-c)}{3}\geq \frac{3}{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}\ \ \ \ \ \ \ \ (5)$$ You can rearrange $(5)$ to get $(4)$. Note that equality holds iff all those terms are equal $\implies s-a=s-b=s-c\implies a=b=c$