Zhang 2025 tiling triangles angles

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Yan X Zhang, Tiling Triangles with $2π/3$ Angles. arXiv:2512.22696 (2025). The arXiv record (https://arxiv.org/abs/2512.22696, read 2026-10-02) names the Creative Commons Attribution 4.0 license. The held PDF is version 4 (arXiv:2512.22696v4, 4 April 2026), and the labels and pages cited here are that version's; an earlier version posed as a conjecture the rationality result that version 4 cites as Theorem 1 (footnote 4, p. 3).

The paper attacks the incommensurable-angles case of when a triangle T tiles into N congruent copies of a triangle R, the fine-grained form of Erdős' prize problem asking which N occur at all. Reptiling and commensurable-angle tiles are already understood (families N = k^2, h^2+k^2, 3k^2, 6k^2), so Zhang treats the tile R with an angle gamma = 2 pi/3 (the tile angle that occurs most often in Figure 1, Beeson's table of the incommensurable-angle cases, which carries the content of Laczkovich's 1995 Theorem 4.1: six of its cases have it) plus the related pi/3 case, assuming that the sides (a,b,c), with c^2 = a^2+ab+b^2, are integers; Theorem 1, cited from recent joint work of Beeson and Zhang, shows that this assumption loses nothing. The main tool is the ideal trapezoid; Theorem 4 shows that with M = 3 ceil((c^2-a-b)/(ab)), every m >= M makes mab equiconstructible, hence there is an m^2 ab tiling, obtained by cutting an equilateral triangle of side (r+s+t)ab into three tileable ideal trapezoids. Lemma 3 shows that for squarefree a,b any equiconstructible X must be a multiple of ab, the sense in which the paper calls Theorem 4's consequence "sharp" (p. 5); Theorem 4 and Lemma 3 then leave open only the lengths mab with m < M. Conjecture 1 asserts divisibility by ab in general (smallest interesting case (5,16,19), where it asks for divisibility by 16); the paper says confirming it would resolve its motivating problem and states the expected answer as Conjecture 2, that the possible N are exactly the m^2 ab with m >= M. Section 5 carries the constructions over to the other four of the six gamma = 2 pi/3 cases of Figure 1; the introduction calls this the first known construction for three of the six, while the abstract says only two of the six had a known construction. For problem 634, this supplies new admissible values of N and a conjectural complete answer for the 2 pi/3 family.

Source: https://arxiv.org/abs/2512.22696.

Bears on. #634

Results to transcribe.

• Theorem 4: For a tile (a,b,c) with a, b and c all integers and M = 3 ceil((c^2-a-b)/(ab)), for every integer m >= M the length mab is equiconstructible by the tile (a,b,c), giving a tiling of an equilateral triangle into m^2 ab congruent copies.
• Lemma 3: If a and b are squarefree, every equiconstructible X equals mab for some integer m; this is the sense in which the paper calls Theorem 4's consequence "sharp" (p. 5).
• Conjecture 1: All equiconstructible X are divisible by ab (smallest interesting case: tile (5,16,19), where it asks for divisibility by 16); the paper says confirming it would resolve its motivating problem, whose expected answer it states as Conjecture 2 (the possible N are exactly m^2 ab, m >= M).
• Theorem 1 (Beeson and Zhang, cited as [5, Theorem 1.2]): If a triangle T is tiled by a tile R that is not similar to T, is not a right triangle and has incommensurable angles, then R has commensurable sides; so assuming integer sides loses nothing.
• Section 5: Propositions 8 to 11 carry the equilateral and (2 alpha, 2 beta, alpha + beta) constructions over to the other four gamma = 2 pi/3 cases of Figure 1, so all six cases get families of tilings (Table 1, p. 13, lists them for the tile (3,5,7)); the introduction calls this the first known construction for three of the six, while the abstract says only two of the six had a known construction.