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Claim. Yan X Zhang, Tiling triangles with $2\pi/3$ angles, arXiv:2512.22696 (version 1 posted 27 December 2025, version 4 of 4 April 2026), proves (Theorem 4) that for a tile with sides $a\ge b$ and $c=\sqrt{a^2+ab+b^2}$, all three integers, and so with an angle of $2\pi/3$, the length $mab$ is equiconstructible for every integer $m\ge3\lceil(c^2-a-b)/(ab)\rceil$, and hence an equilateral triangle can be cut into $m^2ab$ congruent copies of the tile. So every such $m^2ab$ occurs as a value of $n$ for [[problems/discrete_geometry/E0634/_index|Problem 634]]. The integrality of all three sides is the standing assumption of the paper's Section 2.1, forced for the tiles in question by the rationality theorem of Beeson and Zhang; the construction cuts an equilateral triangle into three ideal trapezoids. Lemma 3 shows the family sharp for squarefree $a$ and $b$, and Conjecture 1 asserts that every equiconstructible length is a multiple of $ab$. The digest is on the [[../library/discrete_geometry/zhang_2025_tiling_triangles_angles/_index|source card]].

Covers. The values $m^2ab$ for every integer-sided tile $(a,b,c)$ with $c^2=a^2+ab+b^2$ and every $m\ge3\lceil(c^2-a-b)/(ab)\rceil$ occur. Nothing is claimed about other values of $n$; the paper's conjecture on which counts the $2\pi/3$ family admits is not a claim.

Depends on. No page of this wiki.

Standing. Claimed. The preprint has no journal record; the site's remarks credit the result, but the site labels the problem OPEN, so the credit is commentary, not acceptance. The site's remark states the result for arbitrary integers $a\ge b$, without the integrality of $c$, and misprints the third side as $\sqrt{a^2+b^2+2+ab}$. Harries's second manuscript (claim page) imports Theorem 4 and the row transfers of Propositions 8 to 11 and reports sharper thresholds for the same rows.