hii

 

Now since the circle is the cirumcircle of the triangle we know the following results ..

 

well a = 2rsinA etc

 

Also area of triangle is 1/2. ab sinC

 

                         =  1/2. (2r sinA)(2r sinB)sinC

                         =  2.r² sinA.sinB.sinC

 

so now to maximize the area we need to maximize sinA.sinB.sinC 

 

Now sinA.sinB.sinC = sinA.sinB.sin(A+B)

 

To maximize we differentiate w.r.t A and put it zero ..

 

so sinB.cos(A+B) + cosB.sin(A+B) = 0

 

or sin(A+2B) = 0

 

or A + 2B = 180

 

but we know A + B + C = 180

 

so,  B = C

 

By symetry we get A = B = C = 60

 

and hence the triangle is equilateral ..

 

cheers

 

Another approach could be as follows:

let ABC be a triangle inscribed in the circle with center O and radius r.

If the area of this triangle is maximum, then vertex C should be at a maximum distance from the base AB i.e. CD must be perpendicular to AB.

Hence, ABC is an isosceles triangle.
If  BCD = , where D is the mid-point of BC, then BOD = 2

so, AB = 2 BD

= 2r sin 2

CD = CO + OD = r + r cos 2

If S be the area of the triangle ABC, then

S = (1/2) AB x CD

= (1/2) x 2r sin 2 (r + r cos 2)

ds/d = r²[sin2 (-2 sin2) + (1 + cos2)(2 cos2)]

= 2r²[cos²2 - sin²2 + cos2] = 2r²(cos4 + cos2)

For maximum and mimimum

ds/d = 0

or, cos4 + cos2 = 0

or, 2 cos3 cos = 0

so, Either cos3 = 0 or, cos =0

If cos = 0 or 3 = /2 or = /6

(d²s/d²) = -ve, for = /6

ACB = 2 = 2(/6) = /3 = ABC = BAC

so ABC is an equilateral triangle.