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Statement

Let $T=(A,2A,\pi-3A)$ have incommensurable angles and $\sin(A/2)\in\mathbb Q$. Set

$$\alpha=A,\quad\beta=(\pi-3A)/2,\quad\gamma=(\pi+A)/2.$$

Then $T$ has a tiling by $R=(\alpha,\beta,\gamma)$, and every such tiling has a nonsquare number of tiles.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 27, pp. 14–15. Complete rewritten deduction from the exact external existence and counting inputs below.

Proof

Here $0<A<\pi/3$, so all three tile angles are positive and $3\alpha+2\beta=\pi$. The external existence result is Laczkovich, Tilings of triangles (1995), Theorem 2.4: if this angle relation holds and $\sin(\alpha/2)$ is rational, then $(\alpha,2\alpha,2\beta)$ can be tiled by congruent copies of $R$.

For any such tiling, put $s=2\sin(\alpha/2)=a/c$. Then $0<s<1$ and $s\in\mathbb Q$. Beeson, Triangle tiling: the case $3\alpha+2\beta=\pi$, arXiv:1206.2229v3, Lemmas 10–11, pp. 13–14, give the alternating tile coloring used by its Theorem 9, p. 55. That theorem supplies an integer $M$ (the signed difference of the two color classes) such that

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

Because $N>0$ and the fraction is positive and finite, $M\ne0$. If $N$ were square, multiplying by the rational square $((1-s)(2+s)/M)^2$ would make $(2-s^2)(3-s^2)$ a rational square. That contradicts Proposition 20.

Dependency boundary. This page does not reproduce the tiling existence construction, tile-coloring theorem, or its counting-equation proof.

Bears on. Problem 633.