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Statement

Suppose a non-isosceles triangle $T$ is tiled by a nonsimilar triangle $R=(\alpha,\beta,\gamma)$. Up to the indicated labeling of $T$ and of the tile, it belongs to the following table.

The first two rows have $C=\pi/3$; the next two have $B=2A$; the last two have $C=A/2+B$ and $C=2A+B/2$, respectively.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Lemma 12 and Propositions 13–17, pp. 5–8. Complete rewritten proofs of the case deductions, conditional on Theorem 11. The four component propositions are proved here once, under their own labels.

Common setup and Lemma 12

By Theorem 11, the angles of $T$ and $R$ are incommensurable and $R$ has rational side ratios. In either group, $\alpha$ and $\beta$ are linearly independent over $\mathbb Q$: a rational ratio between them, together with their group's equation for $\pi$, would make both rational multiples of $\pi$.

The six patterns of Theorem 11 immediately give the four relations stated after the table. This proves Lemma 12. It remains to check that a triangle satisfying one of those relations cannot arise from an inappropriate pattern, and then to identify its rationality condition.

Proposition 14: $C=\pi/3$

In Group 1, one of $\alpha,2\alpha,2\beta,\beta,\alpha+\beta$ would equal $\pi/3$. The first four possibilities force commensurable angles. The last, together with $3\alpha+2\beta=\pi$, forces $\beta=0$. Therefore this group cannot occur.

In Group 2, no positive integer multiple of $\alpha$ or $\beta$ can be $\pi/3$, since this would again force commensurability. Neither $2\alpha+\beta$ nor $\alpha+2\beta$ can equal $\alpha+\beta$. Thus $C$ must be the angle $\alpha+\beta$, leaving only the first two rows. In the first row Proposition 10 applied to $R$ gives $\sqrt3\tan(A/4)\in\mathbb Q$, for either choice of $A$ among $2\alpha,2\beta$. In the second row the side ratios of $T$ are rational by the boundary observation in Theorem 11. Apply Proposition 10 to $T$, whose other two angles sum to $2\pi/3$, to obtain $\sqrt3\tan(A/2)\in\mathbb Q$ for either remaining angle.

Proposition 15: $B=2A$

In a fixed group, represent an angle $u\alpha+v\beta$ by its coefficient pair $(u,v)$. Linear independence implies that one angle is twice another only if their coefficient pairs have that same relation. Among the Group 1 patterns, the only such pair is $\alpha,2\alpha$ in $(\alpha,2\alpha,2\beta)$; in $(2\alpha,\beta,\alpha+\beta)$ there is none. Among the Group 2 patterns the only such pair is again $\alpha,2\alpha$, now in $(\alpha,2\alpha,3\beta)$. Thus $A=\alpha$. Apply Proposition 9 in Group 1 and Proposition 10 in Group 2. They give the fourth and third rows respectively. This coefficient check is the source's permutation exclusion written explicitly.

Proposition 16: $C=A/2+B$

Write $A=p\alpha+q\beta$ and $B=s\alpha+t\beta$, with nonnegative integer coefficients from the six patterns. Summing the angles gives

$$2\pi=(3p+4s)\alpha+(3q+4t)\beta.$$

In Group 1, independence implies $3p+4s=6$ and $3q+4t=4$. The only nonnegative integer solutions are $(p,s)=(2,0)$ and $(q,t)=(0,1)$. Hence $(A,B,C)=(2\alpha,\beta,\alpha+\beta)$, and Proposition 9 gives $\sin(A/4)\in\mathbb Q$. In Group 2 both right sides are $6$, forcing $s=t=0$ and hence $B=0$, impossible.

Proposition 17: $C=2A+B/2$

The same notation now gives

$$2\pi=(6p+3s)\alpha+(6q+3t)\beta.$$

In Group 1 this requires $6q+3t=4$, impossible. In Group 2 both coefficients equal $6$. Thus $(p,s)$ and $(q,t)$ are each either $(1,0)$ or $(0,2)$. Using the same choice twice makes $A$ or $B$ zero, so the choices must differ. They give $(A,B,C)=(\alpha,2\beta,2\alpha+\beta)$ or its version with $\alpha,\beta$ interchanged. Proposition 10 supplies $\sqrt3\tan(A/2)\in\mathbb Q$. All six table rows are proved.

Bears on. Problem 633.