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A triangle has rational side lengths and rational angles measured as degrees. Is such a triangle necessary equilateral?

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You might find this helpful: yaroslavvb.com/papers/olmsted-rational.pdf . It says that the only rational values $\sin(x)$ and $\cos(x)$ can take for rational (in degrees) $x$ are $0,\pm \frac{1}{2}, \pm 1$. –  Sebastian Feb 4 '11 at 11:06
    
Got it. Thanks! –  user6610 Feb 4 '11 at 11:39

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up vote 6 down vote accepted

As Sebastian pointed out, if the angle $\theta$ is rational measured in degrees, so if $\theta / \pi$ is rational, and if $\cos \theta$ is also rational, then $\theta \in \left\{ 0, \pi \pm \pi/2, \pm \pi/3, \pm/ 2\pi/3 \right\}$. The "modern" proof of this fact is that if $\theta / \pi$ is rational, $z=e^{i\theta}$ is an algebraic integer (it is a root of $1$), so $2 \cos \theta = z + \bar{z}$ also, and the only rationals which are algebraic integers are the integers, so $2 \cos \theta \in \mathbb{Z}$, and the result follows.

Back to your question, the law of cosines implies that the cosines of the three angles are rational, and from the previous lemma, the (undirected) angles can be $\pi/3$, $\pi/2$ or $2 \pi/3$ (I assume the triangle is not flat). The only possible case is if the three angles are $\pi/3$ (the sum has to be $\pi$), so yes, the triangle is equilateral.

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