# Prove that the largest isosceles triangle that can be drawn in a circle, is an equilateral triangle?

Mar 19, 2017

#### Explanation:

Let their be an isosceles triangle ABC inscribed in a circle as shown, in which equal sides $A C$ and $B C$ subtend an angle $x$ at the center. It is apparent that side $A B$ subtends an angle ${360}^{0} - x$ at the center (as shown). Note that for equilateral triangles all these angles will be $\frac{2 \pi}{3}$.

As the area of the triangle portion subtended by an angle $x$ is ${R}^{2} / 2 \sin x$,

the complete area of triangle ABC is

A=R^2/2(sinx+sinx+sin(360-2x)

= ${R}^{2} / 2 \left(2 \sin x - \sin 2 x\right)$

= ${R}^{2} \left(\sin x - \sin x \cos x\right)$

= ${R}^{2} \sin x \left(1 - \cos x\right)$

For maximization we should have $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

i.e. ${R}^{2} \left(\cos x \left(1 - \cos x\right) + \sin x \times \sin x\right) = 0$

or $\cos x - {\cos}^{2} x + 1 - {\cos}^{2} x = 0$

or $2 {\cos}^{2} x - \cos x - 1 = 0$

or $2 {\cos}^{2} x - 2 \cos x + \cos x - 1 = 0$

or $2 \cos x \left(\cos x - 1\right) + 1 \left(\cos x - 1\right) = 0$

or $\left(2 \cos x + 1\right) \left(\cos x - 1\right) = 0$

Hence $\cos x = - \frac{1}{2}$ or $\cos x = 1$

i.e. $x = \frac{2 \pi}{3}$ or $x = 0$

But for a triangle $x \ne 0$

hence $x = \frac{2 \pi}{3}$

and hence for maximum area triangle must be equilateral.