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Statement

Suppose $T=(A,B,\pi/3)$ has incommensurable angles, $A<B$, and $\sqrt3\tan(A/2)\in\mathbb Q$. Then $T$ admits a tiling by the triangle $R=(\alpha,\beta,\gamma)$ with

$$\alpha=A,\qquad\beta=\pi/3-A,\qquad\gamma=2\pi/3.$$

Every tiling by this shape of tile uses a nonsquare number of tiles.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 26, p. 14, with Lemma 25, pp. 13–14. Complete rewritten proof using the external existence theorem stated below.

Proof

Since $A+B=2\pi/3$ and $A<B$, we have $0<A<\pi/3$, so $R$ is a nondegenerate triangle. Proposition 10 gives $\sqrt3\sin\alpha,\cos\alpha\in\mathbb Q$. The external existence input is Laczkovich, Tilings of triangles (1995), Theorem 2.5: for $\alpha+\beta=\pi/3$ with these rationality conditions, the triangle $(\alpha,\alpha+2\beta,\alpha+\beta)$ can be tiled by congruent $(\alpha,\beta,2\pi/3)$ triangles. This is $T$.

Normalize the tile sides $a,b,c$ to positive integers. The cosine rule gives $c^2=a^2+ab+b^2$. Lemma 25 is the identity

$$\sin(\alpha+2\beta)=\sin\alpha+\sin\beta.$$

Indeed $\alpha=\pi/3-\beta$, so subtracting the addition formulas for $\sin(\pi/3+\beta)$ and $\sin(\pi/3-\beta)$ gives $\sin\beta$. This proves that lemma without relying on its optional geometric diagram.

The sine rule now shows that the sides of $T$ opposite $(\alpha,\alpha+2\beta,\alpha+\beta)$ are proportional to $(a,a+b,c)$. Write them as $(am,(a+b)m,cm)$. Boundary sides are integers, so $m>0$ is rational. Division of areas gives

$$N=\frac{am(a+b)m\sin(\pi/3)}{ab\sin(2\pi/3)} =\frac{a+b}{b}m^2.$$

If $N$ were square, $b(a+b)$ would be a rational square, hence an integer square. Apply Proposition 19 with the roles of $a,b$ interchanged: this is incompatible with $c^2=a^2+ab+b^2$.

Dependencies. Proposition 10, Proposition 19, and the cited external existence theorem. The proof of Laczkovich's theorem is not reproduced here.

Bears on. Problem 633.