Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

Explanation

Let ABC be a triangle inscribed in a given circle with centre O and radius r.
For maximum area, the vertex A should be at a maximum distance from the base BC.
Therefore, A must lie on the diameter, perpendicular to BC. Thus,
`AD bot BC`.

So, triangle ABC must be isosceles.
Let `angleCAD = theta`
Now, `BC = 2CD = 2OC sin 2 theta = 2r sin 2 theta`
and, `AD = (OA + OD) = (r + r cos 2 theta)`
Let A be the area of the triangle
The, `A = (1)/(2) BC xx AD = r^(2) sin 2 theta (1 + r cos 2 theta)`
`:. (dA)/(d theta) = r^(2) [sin 2 theta (-2 sin 2 theta) + (1 + cos 2 theta) .2 cos 2 theta]`
`= r^(2) [2(cos^(2) 2 theta - sin^(2) 2 theta) + 2 cos 2 theta] = 2r^(2) [cos 4 theta + cos 2 theta]`.
And, `(d^(2)A)/(d theta^(2)) = 2r^(2) [-4 sin 4 theta - 2 sin 2 theta] = -4r^(2) (2 sin 4 theta + sin 2 theta)`
Now, `(dA)/(d theta) = 0 rArr cos 4 theta + cos 2 theta 0`
`rArr cos 4 theta = - cos 2 theta = cos (pi - 2 theta)` ltbtgt `rArr 4 theta = pi - 2 theta rArr theta = (pi)/(6)`
And, `[(d^(2)A)/(d theta^(2))]_(theta = (pi//6)) = -4r^(2) (2"sin"(2pi)/(3) + "sin"(Pi)/(3)) = - 6r^(2) sqrt3 lt 0`
`:. theta = (pi)/(6)` is a point of maximum.
So, in this case, each angle of the triangle is `(pi//3)`
Hence, ABC is an equilateral triangle.