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Statement

Suppose $T=(A,B,C)$ has incommensurable angles, $C=2A+B/2$, and $\sqrt3\tan(A/2)\in\mathbb Q$. Set $\alpha=A$, $\beta=B/2$, and $\gamma=2\pi/3$. There exists a tiling of $T$ by $R=(\alpha,\beta,\gamma)$, and every such tiling is nonsquare.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 30, pp. 16–17. Complete rewritten proof with external existence input Laczkovich (1995), Theorem 2.5.

Proof

Summing the angles of $T$ gives $\alpha+\beta=\pi/3$ and $T=(\alpha,2\beta,2\alpha+\beta)$. By Proposition 10, $\sqrt3\sin\alpha,\cos\alpha$ are rational, so Laczkovich's Theorem 2.5 gives a tiling of this triangle by $R$.

For any such tiling, normalize the sides of $R$, and hence the boundary sides of $T$, to integers. The area formula used in Proposition 28 gives, for some positive rational $q$,

$$N=q^2\frac{\sin\gamma\sin2\beta} {\sin\beta\sin(\pi/3+\alpha)} =q^2\frac{\sqrt3\cos\beta}{\sin(\pi/3+\alpha)}.$$

Here $2\alpha+\beta=\pi/3+\alpha$. Set $t=\tan(\alpha/2)/\sqrt3\in\mathbb Q$, so $0<t<1/3$ by Proposition 10. Using $\beta=\pi/3-\alpha$ and its parametrization,

$$\frac{\sqrt3\cos\beta}{\sin(\pi/3+\alpha)} =\frac{1+6t-3t^2}{1+2t-3t^2} =\frac{3t^2-6t-1}{(t-1)(3t+1)}.$$

Proposition 22 excludes square values of this factor throughout the interval. Multiplication by $q^2$ cannot change that conclusion.

Dependencies. Proposition 10, Proposition 22, and the cited external tiling-existence theorem.

Bears on. Problem 633.