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Source: original paper, printed p. 540, Theorem 8.

Statement and source correction

For every even integer $n\ge10$, there is a set of $N=3n/2$ points in $\mathbb R^{n+2}$ containing at least $N^3/15$ triangles with side lengths $1,1,\sqrt2$. This supplies the paper's sufficiently-large-parameter conclusion with an explicit admissible endpoint.

The source displays $z^2=1/n$ for its set $B$, but calls $B$ a circle and uses unit distance from all the points $e_i$. The required equation is $y^2+z^2=1/n$, as the following calculation shows. Both scans omit the $y^2$ term.

Full proof

Let

$$A=\{e_1,\ldots,e_n\}\subset\mathbb R^{n+2},\qquad B=\{(1/n,\ldots,1/n,y,z):y^2+z^2=1/n\}.$$

Distinct points of $A$ have distance $\sqrt2$. For $b\in B$ and every $i$,

$$|b-e_i|^2=(1-1/n)^2+(n-1)/n^2+y^2+z^2 =1-1/n+1/n=1.$$

Choose any $n/2$ distinct points of the circle $B$, and adjoin them to $A$. The two sets are disjoint and their union has $N=3n/2$ points. Each pair from $A$ and point from $B$ gives a distinct right unit triangle. Their number is

$$\binom n2\frac n2 =\frac2{27}\left(1-\frac1n\right)N^3 \ge\frac1{15}N^3,$$

where the last inequality is equivalent to $n\ge10$. Other triangles need not be counted.

For completeness, the source chooses the ratio by maximizing $\alpha/(1+\alpha)^3$ for $\alpha>0$. Its derivative is $(1-2\alpha)/(1+\alpha)^4$, so the maximum occurs at $\alpha=1/2$, precisely the choice above.