What is the largest possible area of a triangle in which one of the vertices is the center

What is the largest possible area of a triangle in which one of the vertices is the center of a circle of radius 2, and the other two vertices lie on this circle?

Let the angle between the sides of the triangle AO and OC be equal to a.

ΔAO is isosceles, because AO = AC = 2 (AO and OC go out from the center of the circle, the radius of which is r = 2).

OS = AO = r = 2;

ОМ = r * cos (a / 2);

MS = r * sin (a / 2).

Area S of triangle AOC:

S = (AC / 2) * OM = MC * OM = r * cos (a / 2) * r * sin (a / 2) = r ^ 2 * cos (a / 2) * sin (a / 2) =

= r ^ 2/2 * sin a = (2 ^ 2/2) * sin a = 2 sin a.

The largest sin a = 1 at a = 90 °. The largest value of the area S = 2 * 1 = 2.

Answer: The largest possible area of a triangle is 2.