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Statement

An isosceles set is a set of points among any three of which at most two distances occur (§7.1, p. 46); see Theorem 7.2.2 for the chapter's conventions.

Theorem 7.2.5 (p. 48). Quoted, because the problem pages rest on its wording: "Let $X$ be an isosceles set in $R^d$, then $\operatorname{card}(X)\le\frac12(d+1)(d+2)$. Equality implies that $X$ is a two-distance set, or a spherical two-distance set together with its center."

In the corpus's words: every isosceles set in $\mathbb{R}^d$ has at most $\binom{d+2}{2}$ points, and a set attaining the bound is either a two-distance set or a two-distance set lying on a sphere with the center of that sphere adjoined. The theorem gives no lower bound and does not say for which $d$ the bound is attained.

Source. A. Blokhuis, Few-distance sets, CWI Tract 7, Centrum voor Wiskunde en Informatica, Amsterdam, 1984; Theorem 7.2.5 on printed p. 48, its proof on p. 49. The edition read is identified in the source digest.

Read depth. Claims checked: the statement was read clause by clause on the page image and its proof read in full and followed. Nothing here is independently reviewed.

Proof pointer

Induction on $d$ (p. 49). For $d=1$ an isosceles set has at most $3$ points. For $d=2$ the proof cites Kelly (the tract's reference [K], Amer. Math. Monthly 54 (1947), 227--229) for the maximum $6$, attained only by the regular pentagon with its center. For $d>2$: a two-distance set obeys the bound by Theorem 4.1.1; otherwise Theorem 7.2.2 gives a decomposition $(X_1,X_2)$. If $\dim(X_1)\ne0$, Lemma 7.2.1 gives $0<\dim(X_1)<d$ and $\dim(X_1)+\dim(X_2)\le d$, and the induction hypothesis applied to both parts gives a strict inequality. If $\dim(X_1)=0$, then $X_1$ is a single point and $X_2$ lies on a sphere centered at it; when $X_2$ is not a two-distance set it decomposes again and the first case applies, and otherwise $\operatorname{card}(X)\le1+\tfrac12d(d+3)=\tfrac12(d+1)(d+2)$. The bound $\tfrac12d(d+3)$ for a two-distance set on a sphere in $\mathbb{R}^d$ is used at this step without a reference on the page; it is the spherical bound of Delsarte, Goethals and Seidel, whose work the tract's introduction cites (p. 1). Equality thus forces a two-distance set or a centered spherical two-distance set.

Dependencies

Within the tract: Theorem 4.1.1 with $s=2$, Lemma 7.2.1 (p. 46) and Theorem 7.2.2. Outside it: Kelly's planar result for $d=2$ and the spherical two-distance bound $\tfrac12d(d+3)$.

Bears on

• Problem 503: an upper bound $\binom{d+2}{2}$ on the size of the largest $A\subseteq\mathbb{R}^d$ in which every three points determine an isosceles triangle, the quantity the problem asks for; the theorem does not determine that quantity.
• Problem 1088: since three points have pairwise distinct distances exactly when they do not form an isosceles triangle, the theorem gives $f_d(3)\le\binom{d+2}{2}+1$ in that problem's notation, as the problem's claim page for this tract records.