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Statement

If $\alpha+\beta$ is $\pi/3$ or $2\pi/3$, the triangle with angles $(\alpha,\beta,\gamma)$ has rational side ratios exactly when

$$\sqrt3\sin\alpha\in\mathbb Q,\quad\cos\alpha\in\mathbb Q,$$

or, equivalently, when $\sqrt3\tan(\alpha/2)\in\mathbb Q$. Writing $t=\tan(\alpha/2)/\sqrt3$, these equivalent conditions give

$$\cos\alpha=\frac{1-3t^2}{1+3t^2},\qquad \sin\alpha=\frac{2\sqrt3t}{1+3t^2},\qquad t\in\mathbb Q.$$

If $\alpha+\beta=\pi/3$, then $0<t<1/3$.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 10, pp. 4–5. Complete rewritten proof.

Proof

Here $\sin\gamma=\sqrt3/2$, and $\sin\beta=(\sqrt3/2)\cos\alpha\pm(1/2)\sin\alpha$, with the minus sign for $\alpha+\beta=\pi/3$ and the plus sign for $2\pi/3$. Thus

$$\frac ac=\frac{2\sqrt3}{3}\sin\alpha,\qquad \frac bc=\cos\alpha\pm\frac{\sqrt3}{3}\sin\alpha.$$

Both ratios are rational exactly when the two stated trigonometric quantities are rational. If $t$ is rational, the half-angle formulas give the displayed parametrization, hence those quantities are rational. Conversely, $\sqrt3\tan(\alpha/2)=\sqrt3\sin\alpha/(1+\cos\alpha)$ is rational when they are; its denominator is nonzero since $0<\alpha<\pi$. Finally $0<\alpha<\pi/3$ implies $0<\tan(\alpha/2)<1/\sqrt3$, hence $0<t<1/3$.

Bears on. Problem 633.