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Statement

Let $T=(A,2A,\pi-3A)$ have incommensurable angles and $\sqrt3\tan(A/2)\in\mathbb Q$. Put $\alpha=A$, $\beta=\pi/3-A$, and $\gamma=2\pi/3$. Then $T$ has a tiling by $R=(\alpha,\beta,\gamma)$, and every such tiling is nonsquare.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 28, p. 15. Complete rewritten proof with external existence input Laczkovich (1995), Theorem 2.5.

Proof

The positive angles of $T$ imply $0<A<\pi/3$. Proposition 10 gives the rationality hypotheses $\sqrt3\sin\alpha,\cos\alpha\in\mathbb Q$ of Laczkovich's Theorem 2.5, whose relevant conclusion is that $(\alpha,2\alpha,3\beta)$ is tiled by congruent copies of $(\alpha,\beta,2\pi/3)$. This proves existence.

Normalize tile sides to integers. Boundary sides of $T$ are then integers. A triangle with angles $u,v,w$ and side $d$ opposite $w$ has area $d^2\sin u\sin v/(2\sin w)$. Applying this to $R$ with $w=\beta$ and to $T$ with $w=\pi-3\alpha$ (whose sine is $\sin3\alpha$) shows that, for some $q\in\mathbb Q_{>0}$,

$$N=q^2\frac{\sin2\alpha\sin\beta}{\sin3\alpha\sin\gamma} =q^2\frac{2\cos\alpha(\cos\alpha-\sin\alpha/\sqrt3)} {(2\cos\alpha-1)(2\cos\alpha+1)}.$$

We used $\sin3\alpha=\sin\alpha(2\cos\alpha-1)(2\cos\alpha+1)$ and $\sin\beta/\sin\gamma=\cos\alpha-\sin\alpha/\sqrt3$. Put $t=\tan(\alpha/2)/\sqrt3\in\mathbb Q$. By Proposition 10, $0<t<1/3$. Substituting its formulas for cosine and sine gives

$$N=q^2\frac{2(3t^2-1)}{3(3t+1)(t-1)}.$$

For example the cancellation uses $1-2t-3t^2=(1+t)(1-3t)$ and $1-9t^2=(1-3t)(1+3t)$; these factors are nonzero in the stated interval. The remaining rational factor is not a square by Proposition 21. Since $q\ne0$, neither is $N$.

Bears on. Problem 633.