Question

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

Answer 1

Please see below.

Explanation:

Let their be an isosceles triangle ABC inscribed in a circle as shown, in which equal sides `AC` and `BC` subtend an angle `x` at the center. It is apparent that side `AB` subtends an angle `360^0-x` at the center (as shown). Note that for equilateral triangles all these angles will be `(2pi)/3`.

As the area of the triangle portion subtended by an angle `x` is `R^2/2sinx`,

the complete area of triangle ABC is

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

= `R^2/2(2sinx-sin2x)`

= `R^2(sinx-sinxcosx)`

= `R^2sinx(1-cosx)`

For maximization we should have `(dA)/(dx)=0`

i.e. `R^2(cosx(1-cosx)+sinx xx sinx)=0`

or `cosx-cos^2x+1-cos^2x=0`

or `2cos^2x-cosx-1=0`

or `2cos^2x-2cosx+cosx-1=0`

or `2cosx(cosx-1)+1(cosx-1)=0`

or `(2cosx+1)(cosx-1)=0`

Hence `cosx=-1/2` or `cosx=1`

i.e. `x=(2pi)/3` or `x=0`

But for a triangle `x!=0`

hence `x=(2pi)/3`

and hence for maximum area triangle must be equilateral.