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Statement

If $T=(A,B,\pi/3)$ and $\sqrt3\tan(A/4)\in\mathbb Q$, then $T$ has a tiling by $R=(A/2,B/2,2\pi/3)$, and every such tiling is nonsquare.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 31, pp. 17–18. Complete rewritten proof with the external existence input identified below. This result is needed for Theorem 3.

Proof

Put $\alpha=A/2$, $\beta=B/2$, and $\gamma=2\pi/3$. Then $\alpha+\beta=\pi/3$ and $T=(2\alpha,2\beta,\alpha+\beta)$. Proposition 10 gives $\sqrt3\sin\alpha,\cos\alpha\in\mathbb Q$. Laczkovich, Tilings of triangles (1995), Theorem 2.5, states that under these hypotheses the displayed $T$ can be tiled by $R$.

Normalize the tile and boundary sides to integers. Comparing the area formulas relative to the sides opposite $\gamma$ and $\pi/3$ gives

$$N=q^2\frac{\sin2\alpha\sin2\beta\sin(2\pi/3)} {\sin\alpha\sin\beta\sin(\pi/3)} =4q^2\cos\alpha\cos\beta$$

for a positive rational $q$. Put $t=\tan(\alpha/2)/\sqrt3\in\mathbb Q$, with $0<t<1/3$. Since $\cos\beta=(\cos\alpha+\sqrt3\sin\alpha)/2$, Proposition 10 transforms the count into

$$N=q^2\frac{2(1-3t^2)(1+6t-3t^2)}{(1+3t^2)^2}.$$

If $N$ were square, then $2(1-3t^2)(1+6t-3t^2)$ would be a rational square. Write $t=a/b$ with coprime positive integers $a,b$. Clearing the denominator $b^4$ would make

$$P=2(b^2-3a^2)(b^2+6ab-3a^2)$$

an integer square. Modulo $3$, this is $2b^4$. If $3\nmid b$, it is $2\pmod3$, impossible for a square. Thus $b=3c$, and $P=9Q$, where

$$Q=2(3c^2-a^2)(3c^2+6ac-a^2).$$

Then $Q$ must also be a square; modulo $3$ it equals $2a^4$. The same argument forces $3\mid a$, contradicting coprimality. Hence $N$ cannot be square.

Bears on. Problem 633.