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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Setting (p. 16). Tdunit(n)T_d^{\mathrm{unit}}(n) is the largest number of unit equilateral triangles spanned by nn points of Rd\mathbb R^d; 1P\mathbb 1_P is 11 when PP holds and 00 otherwise.

Proposition 25 (pp. 16--17). Let r≥3r\ge3 be a fixed integer. For every sufficiently large nn,

T2runit(n)=∑1≤i<j<k≤rninjnk+∑i∈[r](ni−1ni∉4Z)(n−ni),T_{2r}^{\mathrm{unit}}(n)=\sum_{1\le i<j<k\le r}n_in_jn_k +\sum_{i\in[r]}(n_i-\mathbb 1_{n_i\notin4\mathbb Z})(n-n_i),

with (n1,…,nr)(n_1,\ldots,n_r) the splitting of nn chosen in Theorem 3.

The expression is that of Theorem 3 without the terms ni−pi3+1pi>8(pi−8)\frac{n_i-p_i}3+\mathbb 1_{p_i>8}(p_i-8), which count the triangles with all three vertices on one circle, whose side differs from the others'. For r=3r=3 it is n3/27+O(n2)n^3/27+O(n^2), since the parts differ from n/3n/3 by at most 22.

Proof pointer

P. 17, a sketch only: the paper says the upper bound follows by an argument analogous to that of Theorems 3 and 5 with the triangles on one circle left uncounted, and the lower bound from the Lenz construction of § 2.2 without those triangles. In that construction, two points on different circles of radius 11 are at distance 2\sqrt2, so the counted triangles have side 2\sqrt2; scaling the circles to radius 1/21/\sqrt2 makes them unit triangles with the same count. No detailed proof is printed.

Read depth

Claims checked: the definition and Proposition 25 were read clause by clause on the page images of pp. 16--17 (arXiv version 4), with the sketch that follows. The paper prints no full proof, so none was checked. Nothing here is independently reviewed.

Dependencies

Theorem 3, Theorem 5 and, through them, Theorem 7.

Source. F. C. Clemen, A. Dumitrescu and D. Liu, The number of regular simplices in higher dimensions, arXiv:2507.19841 (2025), read in version 4 (28 July 2026); see the source card.

Bears on

  • Problem 755: the case r=3r=3 is the problem's count, triangles of side 11 among nn points of R6\mathbb R^6, and gives n3/27+O(n2)n^3/27+O(n^2) for it for large nn; the paper states it with a proof sketch only. The problem's bound also follows from Theorem 2, which counts triangles of all sizes.