# Area of a triangle in a unit circle

What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

A bad first step is to put the center at the origin, one point at (1,0) , and one point at (sin x, cos x).

A start is the area of a triangle with included angle expression, $${a \times b \times \sin \theta} \over {2}$$

Assuming $\theta$ in radians. If theta is $\pi/2$ then we have a right triangle. Let a=b=1. Area expression is $$A=(\sin \theta) / 2$$ This is maximum for $\theta = \pi/2$.

Answer is maximum area for a right triangle.

• My guess would have been the origin, $(1,0)$, and $(0,1)$ which obviously has area of 1/2. Is the area of yours greater? – Shane Nov 5 '14 at 20:43

The area of a triangle with side lengths $a, b, c$ and angles $A, B, C$ can be expressed as $$\left|\triangle ABC\right| = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B.$$ Thus the area for a fixed $a = b = 1$ and variable central angle $C$ is maximized when $\sin C$ attains its maximum; i.e., $\angle C = \pi/2$ and $\sin C = 1$. Thus the triangle is right.
You have $$S=\frac{r^2}{2}\sin\theta=\frac{\sin\theta}{2},$$where $\theta$ is the angle at the center. The rest shouldn't be too hard.
Let the centre of the circle be $O$, let one of the vertices be $A$, and let the other vertex $X$ vary. The area is maximized when the height of the triangle with respect to the base $OB$ is maximal. This happens when $XO$ is perpendicular to $OB$.