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.

(I thought about this question on and off for a few days before posting it here. Then, for some reason, almost immediately after posting the question here, without receiving any replies, I found the answer.)


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