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

1 Answer
Mar 19, 2017

Answer:

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#.

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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.