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Statement

Construction (unnumbered, pp. 301–302). For $1 \le i \le m$ let $u_i^2 + v_i^2 = 1$, and put

$$X_i = (u_i, v_i, 0, 0, 0, 0),\qquad Y_i = (0, 0, u_i, v_i, 0, 0),\qquad Z_i = (0, 0, 0, 0, u_i, v_i).$$

Each triangle $X_iY_jZ_k$ is equilateral, and all $m^3$ of them are congruent. The paper counts them as coming "from only 3m points", so the $m$ points $(u_i,v_i)$ are taken distinct, though the print does not say so. The paper concludes that $f_6^s(n)$, $f_6^e(n)$ and $f_6^c(n)$ are all greater than $(n^3/27) - cn^2$. It says the construction also appeared in Erdős, On sets of distances of $n$ points in Euclidean space (1960), and in Erdős and Purdy, Some extremal problems in geometry (J. Combin. Theory 10 (1971)).

Side length. The print calls these triangles equilateral "with side one" [sic] (p. 302). With $u_i^2+v_i^2=1$, a point of one circle and a point of another are at distance $\sqrt{u_i^2+v_i^2+u_j^2+v_j^2}=\sqrt2$, so the printed triangles have side $\sqrt2$. Taking $u_i^2+v_i^2=\tfrac12$ instead, three circles of radius $1/\sqrt2$ about the origin, gives side one; the count is unchanged.

Notation. $f_6^e(n)$, $f_6^c(n)$ and $f_6^s(n)$ are the largest possible numbers of equilateral, of pairwise congruent and of pairwise similar triangles among $n$ distinct points of $E_6$ (pp. 291–292); $c$ is a positive constant (p. 291).

Source. P. Erdős and G. B. Purdy, Some extremal problems in geometry, III, Proceedings of the Sixth Southeastern Conference on Combinatorics, Graph Theory and Computing (Boca Raton, 1975), Congress. Numer. XIV, Utilitas Math., Winnipeg, 1975, pp. 291–308. The edition read is identified on the source card. The construction is in Section 3 on pp. 301–302; the question about it is in Section 6 on p. 307.

Read depth. Claims checked: the construction and its conclusion were read clause by clause on the printed pages.

Question posed, p. 307

The paper's Conclusion asks whether the inequality $f_6^e(n) \ge \frac{n^3}{27} - cn^2$ is best possible, and says it would be interesting even to show $f_6^e(n) \le (\frac16 - \epsilon)n^3$ for some $\epsilon > 0$. Here $f_6^e$ counts equilateral triangles of every size.

Proof pointer

Points on different circles lie in orthogonal coordinate planes, so their distance depends only on the two radii; with three equal radii every triangle with one vertex on each circle is equilateral of the same side. With $m=[n/3]$ this gives $m^3 \ge n^3/27 - cn^2$ triangles.

Bears on

• Problem 755: the problem asks whether $n$ points of $\mathbb R^6$ span at most $(\frac1{27}+o(1))n^3$ unit equilateral triangles. Rescaled to side one, this construction spans at least $n^3/27 - cn^2$ unit equilateral triangles, so the constant $\frac1{27}$ cannot be lowered. The paper proves no upper bound in six dimensions; its question on p. 307, whether $f_6^e(n) \ge \frac{n^3}{27} - cn^2$ is best possible, concerns equilateral triangles of every size.