Let their be an isosceles triangle ABC inscribed in a circle as shown, in which equal sides ACAC and BCBC subtend an angle xx at the center. It is apparent that side ABAB subtends an angle 360^0-x3600−x at the center (as shown). Note that for equilateral triangles all these angles will be (2pi)/32π3.
![enter image source here]()
As the area of the triangle portion subtended by an angle xx is R^2/2sinxR22sinx,
the complete area of triangle ABC is
A=R^2/2(sinx+sinx+sin(360-2x)A=R22(sinx+sinx+sin(360−2x)
= R^2/2(2sinx-sin2x)R22(2sinx−sin2x)
= R^2(sinx-sinxcosx)R2(sinx−sinxcosx)
= R^2sinx(1-cosx)R2sinx(1−cosx)
For maximization we should have (dA)/(dx)=0dAdx=0
i.e. R^2(cosx(1-cosx)+sinx xx sinx)=0R2(cosx(1−cosx)+sinx×sinx)=0
or cosx-cos^2x+1-cos^2x=0cosx−cos2x+1−cos2x=0
or 2cos^2x-cosx-1=02cos2x−cosx−1=0
or 2cos^2x-2cosx+cosx-1=02cos2x−2cosx+cosx−1=0
or 2cosx(cosx-1)+1(cosx-1)=02cosx(cosx−1)+1(cosx−1)=0
or (2cosx+1)(cosx-1)=0(2cosx+1)(cosx−1)=0
Hence cosx=-1/2cosx=−12 or cosx=1cosx=1
i.e. x=(2pi)/3x=2π3 or x=0x=0
But for a triangle x!=0x≠0
hence x=(2pi)/3x=2π3
and hence for maximum area triangle must be equilateral.
