# A triangle has one vertex at a circle's center and two vertices on the circle. Can the three enclosed regions have rational areas?

A triangle has one vertex at a circle's center and two vertices on the circle. Can the three enclosed regions have rational areas?

Let $$r=$$ radius of circle, $$\theta=$$ angle at vertex of triangle at center of circle.

Assume the three regions have rational areas. The area of the circle is rational, so $$r^2$$ is a rational multiple of $$1/\pi$$. Then (since the area of the triangle is rational) $$\sin\theta$$ is a rational multiple of $$\pi$$, and (since the area of the segment is rational) $$\theta$$ is a rational multiple of $$\pi$$.

So I think the question is equivalent to:

Can $$\theta$$ and $$\sin\theta$$ both be rational multiples of $$\pi$$? ($$0<\theta<\pi$$)

I thought about Niven's Theorem, but it doesn't seem to help.

(I suspect the answer is no.)

I think I can answer my own question. The sine of any rational multiple of $$\pi$$ is algebraic, as shown here, so it cannot be a rational multiple of $$\pi$$, so the answer to my question is no.