# How to show that a triangle of maximum area inscribed in a given circle is an equilateral triangle?

Jun 15, 2018

As the radius of the circle is not relevant we can assume we have a circle of unitary radius centered in the origin, whose equation is:

${x}^{2} + {y}^{2} = 1$

The triangle will then have three vertices $A , B , C$ lying on the circle. Because of the circular symmetry we can assume without loss of generality that the side $A B$ is parallel to the $y$ axis and intersecting the positive $x$ axis, so that the coordinates of $A$ and $B$ are:

$\left\{\begin{matrix}{x}_{A} = \cos \left(\alpha\right) \\ {y}_{A} = \sin \left(\alpha\right)\end{matrix}\right.$

$\left\{\begin{matrix}{x}_{B} = \cos \left(\alpha\right) \\ {y}_{A} = - \sin \left(\alpha\right)\end{matrix}\right.$

where $2 \alpha = \angle \left(A O B\right)$ so that the length of $A B$ is:

$\overline{A B} = 2 \sin \left(\alpha\right)$

Now, as a first step let's consider for any possible choice of $A B$ the triangle $A B C$ letting $C$ be the third vertex with coordinates:

$\left\{\begin{matrix}{x}_{C} = \cos \left(\theta\right) \\ {y}_{C} = \sin \left(\theta\right)\end{matrix}\right.$

where $\theta$ is the angle between the $x$ axis and $O C$ as we generally define in trigonometry.

The height of the triangle is going to be parallel to the $x$ axis.
As ${x}_{A}$ is positive by construction the height is:

$h = {x}_{A} - {x}_{C} = \cos \alpha - \cos \theta$

and the area of the triangle is:

$S = \frac{h \overline{A B}}{2} = \sin \alpha \left(\cos \alpha - \cos \theta\right)$

and as $1 - \le \cos \theta \le 1$ clearly the are is maximum when:

$\cos \theta = - 1$

$\theta = \pi$

that is when the triangle $A B C$ is isosceles and then the area is:

$S \left(\alpha\right) = \sin \alpha \left(\cos \alpha + 1\right)$

Maximize now this function with $\alpha \in \left(0 , \frac{\pi}{2}\right)$:

$\frac{\mathrm{dS}}{d \alpha} = \cos \alpha \left(\cos \alpha + 1\right) - {\sin}^{2} \alpha$

$\frac{\mathrm{dS}}{d \alpha} = {\cos}^{2} \alpha - {\sin}^{2} \alpha + \cos \alpha$

$\frac{\mathrm{dS}}{d \alpha} = {\cos}^{2} \alpha - \left(1 - {\cos}^{2} \alpha\right) + \cos \alpha$

$\frac{\mathrm{dS}}{d \alpha} = 2 {\cos}^{2} \alpha + \cos \alpha - 1$

The critical points are then:

$\cos \alpha = \frac{- 1 \pm \sqrt{1 + 8}}{2} = \frac{- 1 \pm 3}{2} = \left\{\begin{matrix}- 1 \\ \frac{1}{2}\end{matrix}\right.$

Excluding $\cos \alpha = - 1$ because by construction $\cos \alpha > 0$ we have:

$\alpha = \frac{\pi}{3}$

where:

$\frac{{d}^{2} S}{\mathrm{da} l p h {a}^{2}} = - 4 \sin \alpha \cos \alpha - \sin \alpha = - \sin \alpha \left(1 + 4 \cos \alpha\right)$

$\frac{{d}^{2} S}{\mathrm{da} l p h {a}^{2}} {|}_{\frac{\pi}{3}} = - \frac{\sqrt{3}}{2} \left(1 + 4 \left(\frac{1}{2}\right)\right) = - \frac{3 \sqrt{3}}{2} < 0$

so that in fact $\alpha = \frac{\pi}{3}$ is a local maximum.

Remembering that:

$\angle \left(A O B\right) = 2 \alpha = \frac{2 \pi}{3}$

as the angle $\angle \left(A C B\right) = \frac{1}{2} \angle \left(A O B\right)$ we have:

$\angle \left(A C B\right) = \frac{\pi}{3}$

and as the triangle is isosceles the angles at the base are each:

$\angle \left(B A C\right) = \angle \left(A B C\right) = \frac{1}{2} \left(\pi - \frac{\pi}{3}\right) = \frac{\pi}{3}$

which means the triangle is equilateral.