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

Question

Noah Atkinson · Accepted Answer

Please see below.

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.

Caroline Aucoin · Answer

undefined To prove that the largest isosceles triangle that can be drawn in a circle is an equilateral triangle, we can use the following steps:

1. Let's assume we have a circle with center O and radius r. We want to inscribe an isosceles triangle in this circle.

2. Consider an isosceles triangle ABC inscribed in the circle, with AB = AC. Let M be the midpoint of BC.

3. Since ABC is isosceles, AM is a perpendicular bisector of BC. Therefore, AM is the altitude of triangle ABC.

4. Let's denote the length of AM as h and the length of BC as 2x.

5. By the Pythagorean theorem in triangle OAM, we have: $OA^2 = h^2 + r^2$.

6. By the Pythagorean theorem in triangle OBM, we have: $OB^2 = x^2 + r^2$.

7. Since OA = OB (both are radii of the circle), we can equate the two equations: $h^2 + r^2 = x^2 + r^2$.

8. Simplifying, we get: $h^2 = x^2$.

9. This implies that the altitude of the isosceles triangle is equal to half the base, which means the triangle is equilateral.

10. Therefore, the largest isosceles triangle that can be drawn in a circle is an equilateral triangle.