On Isosceles Sets in the 4-Dimensional Euclidean Space

Abstract

A subset 𝑋 in the 𝑘-dimensional Euclidean space ℝ^𝑘 that contains 𝑛 points (elements) is called an 𝑛-point isosceles set if every triplet of points selected from them forms an isosceles triangle. In this paper, we show that there exist exactly two 11-point isosceles sets in ℝ⁴ up to isomorphisms and that the maximum cardinality of isosceles sets in ℝ⁴ is 11.

1. Introduction

Let ℝ^𝑘 be the 𝑘-dimensional Euclidean space, let 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑘) and 𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑘) be in ℝ^𝑘, and 𝑑(𝑥,𝑦)=∑^𝑘_𝑖=1(𝑥_𝑖−𝑦_𝑖)².

For a finite set 𝑋⊂ℝ^𝑘, let𝐴(𝑋)=𝑑𝑥,𝑦∣𝑥,𝑦∈𝑋,𝑥≠𝑦. (1.1) If |𝐴(𝑋)|=𝑠, we call 𝑋 an 𝑠-distance set.

Two subsets in ℝ^𝑘 are said to be isomorphic if there exists similar transformation from one to the other.

We have the following interesting problems on 𝑠-distance sets. (1)What is the cardinality of points when the number of 𝑠-distance sets in ℝ^𝑘 is finite up to isomorphisms? (2)What is the maximum cardinality of 𝑠-distance sets in ℝ^𝑘? (3)Can we say something about the ratios of distances in an 𝑠-distance set?

As regards question (1), Einhorn and Schoenberg [1] showed that the number of 2-distance sets in ℝ^𝑘 is finite if cardinalities are more than or equal to 𝑘+2.

For question (2) with 𝑠=2 and 𝑘≤8, Erdös and Kelly [2], Croft [3], and Lisoněk [4] gave the maximum cardinalities. Their results are summarized in Table 1 (see [4, 5]).

Table 1 

The maximum cardinality of 2-distance sets.

As regards question (3), Larman et al. [6] showed that if |𝑋|>2𝑘+3, the ratio of 2 distances in any 2-distance set 𝑋 is given by √𝛼−1∶√𝛼, where 𝛼is an integer 𝛼satisfying 𝛼≤1/2+𝑘/2.

Bannai et al. [7] and Blokhuis [8] proved that the cardinality of an 𝑠-distance set in ℝ^𝑘 is bounded above by 𝑠𝑘+𝑠. For the case 𝑠=3 and 𝑘=2, Shinohara [9] gave the answers to questions (1) and (2) by classifying 3-distance sets in ℝ². He proved that there are finitely many 3-distance sets when cardinalities are more than or equal to 5. He also proved that the maximum cardinality of 3-distance sets is 7. The complete classification of 3-distance sets in ℝ² was also given. Recently Shinohara [10] showed uniqueness of maximum 3-distance sets in ℝ³.

In this paper, we deal with isosceles sets which are defined in the following.

We call a set in ℝ^𝑘 with 𝑛 points an 𝑛-point isosceles set if every triplet of points selected from them forms an isosceles triangle.

Here three collinear points will be interpreted as forming an isosceles triangle if and only if one of them is the mid-point of the other pair.

We remark that all 𝑛-point 2-distance sets are 𝑛-point isosceles sets.

In this paper, we consider classification and the maximum cardinality of isosceles sets in ℝ⁴. The following theorem and corollary are the main results.

Theorem 1.1. There exist exactly two 11-point isosceles sets in ℝ⁴ up to isomorphisms. They are 𝑋 and 𝑌, which will be explicitly defined in the following section.

Corollary 1.2. There is no 12-point isosceles set in ℝ⁴. Therefore the maximum cardinality of isosceles sets in ℝ⁴ is 11.

We prove them by expanding the method by Croft [3] into ℝ⁴.

2. Known Results and Example of Isosceles Sets

The following are the known facts about isosceles sets so far. (i)Ten-point isosceles sets in ℝ⁴ exist infinitely many up to isomorphisms. For example, {(cos(2𝑗/5)𝜋,sin(2𝑗/5)𝜋,0,0)∣0≤𝑗≤4}∪{𝑐(0,0,cos(2𝑘/5)𝜋,sin(2𝑘/5)𝜋)∣0≤𝑘≤4} is a 10-point isosceles set for any positive real number 𝑐. It is nonisomorphic to {(cos(2𝑗/5)𝜋,sin(2𝑗/5)𝜋,0,0)∣0≤𝑗≤4}∪{𝑐^′(0,0,cos(2𝑘/5)𝜋,sin(2𝑘/5)𝜋)∣0≤𝑘≤4} for any positive real number 𝑐^′ satisfying 𝑐^′≠𝑐. (ii)No 9-point isosceles set in ℝ³ exists (Croft [3]). (iii)There exists a unique 8-point isosceles set in ℝ³ up to isomorphisms. It is in Figure 1 (Kido [11]). (iv)Seven-point isosceles sets in ℝ³ exist infinitely many up to isomorphisms. (v)No 7-point isosceles set in ℝ² exists (Erdös and Golomb [12], Erdös and Kelly [2]). (vi)There exists a unique 6-point isosceles set in ℝ² up to isomorphisms. It consists of five points of a regular pentagon and its center (Erdös and Golomb [12], Erdös and Kelly [2]). (vii)There exist exactly three 5-point isosceles sets in ℝ² up to isomorphisms. They are four points of a square and its center, five points of a regular pentagon, and four points of a regular pentagon and its center (Fishburn [13], Erdös and Golomb [12]). (viii)Four-point isosceles sets in ℝ² exist infinitely many up to isomorphisms.

Figure 1 

A unique 8-point isosceles set in ℝ3 (from Kido [11]).

Now we define two examples 𝑋 and 𝑌 which are mentioned in Theorem.

2.1. Example of 11-Point Isosceles Sets in ℝ⁴

Let 𝑒_𝑖, 1≤𝑖≤4 be the canonical basis of ℝ⁴. Then 11-point sets 𝑋 and 𝑌 in ℝ⁴ defined as follows are isosceles sets:𝑋=𝑋^′∪𝑢₀, (2.1) where𝑋^′=𝑒_𝑖+𝑒_𝑗∣1≤𝑖<𝑗≤4∪−𝑒_𝑘+𝑢∣1≤𝑘≤4, (2.2) and 𝑢₀=((5+√5)/10,(5+√5)/10,(5+√5)/10,(5+√5)/10) and 𝑢=((3+√5)/4,(3+√5)/4,(3+√5)/4,(3+√5)/4):𝑌=cos2𝑗5𝜋,sin2𝑗5∪𝜋,0,0∣0≤𝑗≤40,0,cos2𝑘5𝜋,sin2𝑘5𝜋∣0≤𝑘≤4∪{(0,0,0,0)}. (2.3)

Remark 2.1. In above 𝑋^′ is known as a unique 10-point 2-distance set (see Lisoněk [4]). It is constructed by the Petersen graph (Figure 2) and it is on a 3-dimensional sphere whose center is 𝑢₀. Also we can easily see that 𝑋^′ and 𝑋 contain a square and that 𝑌 contains a regular pentagon.

Figure 2 

The Petersen graph.

3. Notation and Some Isosceles Set Configurations

We introduce the following notation (see [3]): apex: a point of a set of three or more points equidistant from all the others.

Let 𝒫={𝑃₁,…,𝑃_𝑛} be an 𝑛-point isosceles set. We define the vertex-number 𝑉(𝑃_𝑖) of a point 𝑃_𝑖∈𝒫 by the number of distinct isosceles triangles of which 𝑃_𝑖 is an apex. It is easy to see that𝑉𝑃₁+⋯+𝑉𝑃_𝑛≥𝑛3. (3.1) Especially let 𝛼be the number of regular triangles in 𝒫:𝑉𝑃₁+⋯+𝑉𝑃_𝑛𝑛3=2𝛼+. (3.2)

We further say that a point 𝑃_𝑖∈𝒫 is of type (𝑟,𝑠,…,𝑢) if the lines joining it to the remaining points in 𝒫 are constituted; thus: 𝑟 of length 𝑎, 𝑠 of length 𝑏,…,𝑢 of length 𝑙, where 𝑎,𝑏,…,𝑙 are no two of them equal. Setting 𝑟≥𝑠≥⋯≥𝑢, 𝑟+𝑠+⋯+𝑢=𝑛−1 clearly holds. Moreover if 𝑃_𝑖 is of type (𝑟,𝑠,…,𝑢), then𝑉𝑃_𝑖=𝑟2+𝑠2𝑢2+⋯+. (3.3)

Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in ℝ⁴, and suppose that 𝑃₁ has the largest vertex-number. Then the type of 𝑃₁ satisfies one of the following cases (A)–(H): Case (A): (10), Case (B): (9,1), (8,2), (8,1,1), Case (C): (7,3), (7,2,1), Case (D): (6,4), (6,3,1), (5,5), (5,4,1), Case (E): (7,1,1,1), Case (F): (6,2,2), Case (G): (6,2,1,1), Case (H): (6,1,1,1,1).

Proof. Since 𝑉(𝑃₁)+⋯+𝑉(𝑃₁₁)≥₃¹¹=165 by (3.1), we have 𝑉(𝑃₁)≥15. Let (𝑟,𝑠,…,𝑢) be the type of 𝑃₁. Then we have 𝑟2+𝑠2𝑢2+⋯+≥15, (3.4) and we have 𝑟+𝑠+⋯+𝑢=10. (3.5) In order to satisfy (3.4) and (3.5), (𝑟,𝑠,…,𝑢) must be one in the list of the lemma.

Throughout this paper, we refer to the condition (𝑋) as “four points in a set lie on a circle.”

We first show the following lemma.

Lemma 3.2. If an 11-point isosceles set in ℝ⁴ exists, then the condition (𝑋) is true for it.

In Sections 4–11, we prove Lemma 3.2 case by case according to eight cases (A)–(H) of types of 𝑃₁ given in Lemma 3.1. In Sections 12 and 13, we deal with 11-point isosceles sets satisfying the condition (𝑋). In Section 14, we complete the proofs of Theorem 1.1 and Corollary 1.2.

The following propositions are useful for us to prove Lemma 3.2 and Theorem 1.1. We can prove Propositions 3.3 and 3.4 using a similar method to Lemma 3.1.

Proposition 3.3. In a 10-point isosceles set in ℝ⁴, let 𝑃 be a point that has the largest vertex-number. Let (𝑟,𝑠,…,𝑢) be the type of 𝑃. If 𝑟<6, then it must be (5,4), (5,3,1), (5,2,2), or (4,4,1).

Proposition 3.4. In a 6-point isosceles set in ℝ³, let 𝑃 be a point that has the largest vertex-number. Then the type of 𝑃 is one of (5), (4,1), and (3,2).

Proposition 3.5. Let an 𝑛-point isosceles set in ℝ⁴ be constituted thus: 𝑃₁, which is the center of a 3-dimensional sphere 𝑆; upon 𝑆 lie 𝑃₂,𝑃₃,𝑃₄,…, being at least 3 and less than or equal to 𝑛−2 points; and at least one 𝑃, say 𝑃_𝑛, does not lie on 𝑆. Then those points of the set that lie on 𝑆 lie on one of two disjoint 2-dimensional spheres.

Proof. We may assume that the equation of 𝑆 is 𝑥²+𝑦²+𝑧²+𝑤²=1 and 𝑃_𝑛=(𝑘,0,0,0), where 𝑘>0 and 𝑘≠1. Then 𝑃₁=(0,0,0,0). For a point 𝑃_𝑖=(𝑥_𝑖,𝑦_𝑖,𝑧_𝑖,𝑤_𝑖) on 𝑆, we consider Δ𝑃₁𝑃_𝑖𝑃_𝑛.
When 𝑃₁𝑃_𝑖=𝑃_𝑖𝑃_𝑛=1 holds, we have 𝑥²_𝑖+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1 and (𝑥_𝑖−𝑘)²+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1. Then 𝑥_𝑖=𝑘/2 and 𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1−𝑘²/4.
On the other hand, when 𝑃₁𝑃_𝑛=𝑃_𝑖𝑃_𝑛=𝑘 holds, we have 𝑥²_𝑖+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1 and (𝑥_𝑖−𝑘)²+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=𝑘². Then 𝑥_𝑖=1/2𝑘 and 𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1−1/4𝑘².
Combining them, it holds that 𝑃_𝑖 is on one of two disjoint 2-dimensional spheres.

Proposition 3.6. If three points, 𝑃₁, 𝑃₂, 𝑃₃, say, in an 𝑛-point isosceles set in ℝ⁴ are collinear in this order, then the other points of the set all lie on a 2-dimensional sphere.

Proof. We may assume that 𝑃₁=(−1,0,0,0),𝑃₂=(0,0,0,0), and 𝑃₃=(1,0,0,0). We consider the position of 𝑃_𝑖=(𝑥_𝑖,𝑦_𝑖,𝑧_𝑖,𝑤_𝑖) for 𝑖=4,…,𝑛. By a similar method used in the proof of Kelly [2] or Lemma 6 in Croft [3], in a plane, 𝑃₁,𝑃₂,𝑃₃, and 𝑃_𝑖 must satisfy Figure 3.
Hence 𝑥²_𝑖+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1, (𝑥_𝑖+1)²+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=2, and (𝑥_𝑖−1)²+𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=2 hold. Then we have 𝑥_𝑖=0 and 𝑦²_𝑖+𝑧²_𝑖+𝑤²_𝑖=1.
Therefore the other points lie on a 2-dimensional sphere.

Figure 3 

Corollary 3.7. For 𝑛≥11, there is no 𝑛-point isosceles set in ℝ⁴ which has three collinear points.

Proof. We may show that this corollary holds for 𝑛=11. By Proposition 3.6, the other eight points must lie on a 2-dimensional sphere. So they form an 8-point isosceles set in the 2-dimensional sphere (⊂ℝ³). We know that there exists a unique 8-point isosceles set in ℝ³ and it is in Figure 1. But looking at the figure, we see that eight points in it do not lie on a 2-dimensional sphere. Hence there is no 8-point isosceles set in the 2-dimensional sphere.
Therefore there is no 11-point isosceles set in ℝ⁴ which has three collinear points.

Proposition 3.8. Let 𝒫={𝑃₁,…,𝑃₆} be a 6-point isosceles set in a 2-dimensional sphere 𝑆. Then four points in 𝒫 lie on a circle; the condition (𝑋) holds.

Proof. Let 𝑃₁ be a point that has the largest vertex-number in 𝒫. By Proposition 3.4, the type of 𝑃₁ is one of (5), (4,1), and (3,2). If the type of 𝑃₁ is (5) or (4,1), then at least four points among 𝑃₂,…,𝑃₆ are on the intersection of 𝑆 and the sphere whose center is 𝑃₁. So at least four points are on a circle, the condition (𝑋) holds.
Thus we suppose that 𝑃₁ is of type (3,2) with corresponding distances 𝑟₁ and 𝑟₂. For 𝑖=1,2, let 𝑆_𝑖 be the sphere centered at 𝑃₁ with radius 𝑟_𝑖. Let 𝑈₁=𝒫∩𝑆₁={𝑃₂,𝑃₃,𝑃₄} and 𝑈₂=𝒫∩𝑆₂={𝑃₅,𝑃₆}.
Now 𝒫 is a 2- or an 𝑠-distance set (𝑠≥3). We suppose that it is a 2-distance set. We know that there exist exactly six 6-point 2-distance sets in ℝ³. These six figures are in Figure 4. Two figures contain all points of a square, and the others contain four points of a regular pentagon. All points of a square and four points of a regular pentagon are both on a circle. Therefore the condition (𝑋) holds.
On the other hand, we suppose that 𝒫 is an 𝑠-distance set (𝑠≥3). So there exists a pair of points in 𝒫 whose distance is 𝑐, that is, distinct from 𝑟₁ and 𝑟₂. Since 𝑃₁𝑃_𝑖=𝑟₁ or 𝑟₂(𝑖=2,…,6), 𝑐 is the distance apart of a pair of points in {𝑃₂,…,𝑃₆}. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑈₁ and 𝑃_𝑗∈𝑈₂, then Δ𝑃₁𝑃_𝑖𝑃_𝑗 is scalene with sides 𝑟₁,𝑟₂,𝑐. Thus the following condition holds:𝑃_𝑖𝑃_𝑗=𝑟₁or𝑟₂forany𝑃_𝑖∈𝑈₁,𝑃_𝑗∈𝑈₂. (3.6) Because 𝑐 is the distance apart of a pair of points in 𝑈₁ or 𝑈₂, at least one of 𝑃₂𝑃₃, 𝑃₂𝑃₄, 𝑃₃𝑃₄, and 𝑃₅𝑃₆ is 𝑐.
We suppose that 𝑃₅𝑃₆=𝑐. Let 𝑃_𝑖∈𝑈₁ and consider Δ𝑃_𝑖𝑃₅𝑃₆. Since 𝑃_𝑖𝑃₅ and 𝑃_𝑖𝑃₆ are of length 𝑟₁ or 𝑟₂ by (3.6), we have 𝑃_𝑖𝑃₅=𝑃_𝑖𝑃₆. Thus three points 𝑃₂, 𝑃₃, and 𝑃₄ are on the plane perpendicularly bisecting 𝑃₅𝑃₆, the sphere 𝑆₁, and the sphere 𝑆. But the plane and the two spheres intersect at exactly two points. This is a contradiction. Therefore 𝑃₅𝑃₆≠𝑐, without loss of generality we may assume 𝑃₂𝑃₃=𝑐.
Next we suppose that 𝑃₂𝑃₃=𝑐 and 𝑃₂𝑃₄=𝑑 (𝑑≠𝑟₁, 𝑑≠𝑟₂, but we can admit 𝑐=𝑑). Let 𝑃_𝑗∈𝑈₂ and consider Δ𝑃₂𝑃₃𝑃_𝑗. Because 𝑃₂𝑃_𝑗 and 𝑃₃𝑃_𝑗 are of length 𝑟₁ or 𝑟₂ by (3.6), we have 𝑃₂𝑃_𝑗=𝑃₃𝑃_𝑗. When we consider Δ𝑃₂𝑃₄𝑃_𝑗 similarly, we have 𝑃₂𝑃_𝑗=𝑃₄𝑃_𝑗. Thus 𝑃₆ and 𝑃₇ are on the plane perpendicularly bisecting 𝑃₂𝑃₃, the plane perpendicularly bisecting 𝑃₂𝑃₄, the sphere 𝑆₂, and the sphere 𝑆. Since the segment 𝑃₂𝑃₃ and the segment 𝑃₂𝑃₄ are not mutually parallel, the two planes and the two spheres have no intersection. Hence 𝑃₂𝑃₃=𝑐 and 𝑃₂𝑃₄=𝑑 do not hold. Similarly we can show that 𝑃₂𝑃₃=𝑐 and 𝑃₃𝑃₄=𝑑 do not hold.
So in 𝒫, there is exactly one pair 𝑃₂𝑃₃ whose distance is distinct from 𝑟₁ and 𝑟₂. When we consider Δ𝑃₂𝑃₃𝑃_𝑘 for 𝑘=4,5,6, 𝑃₂𝑃_𝑘=𝑃₃𝑃_𝑘 holds by the configuration hypothesis. And we have 𝑃₁𝑃₂=𝑃₁𝑃₃. Then four points 𝑃₁, 𝑃₄, 𝑃₅, and 𝑃₆ on the plane perpendicularly bisecting 𝑃₂𝑃₃ and the sphere 𝑆. The intersection of them is a circle, the condition (𝑋) holds.

Figure 4 

All 6-point 2-distance sets in ℝ3 (from Einhorn and Schoenberg [14]).

4. Case (A) in Lemma 3.1

We consider the case (A) in Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which 𝑃₁ is of type (6.1). Let 𝑆 be the sphere centered at 𝑃₁ and 𝑉=𝒫∩𝑆={𝑃₂,…,𝑃₁₁}.

We notice that 𝑉 is a 10-point isosceles set. Let 𝑃₂ be a point that has the largest vertex-number in 𝑉. Let (𝑟,𝑠,…,𝑢) be the type of 𝑃₂ in 𝑉, the type of 𝑃₂ is 𝑟≥6, (5,4), (5,3,1), (5,2,2), or (4,4,1) by Proposition 3.3.

Proposition 4.1. Let (𝑟,𝑠,…,𝑢) be the type of 𝑃₂ in 𝑉. If the type of 𝑃₂ satisfies 𝑟≥6, then the condition (𝑋) holds.

Proof. If the type of 𝑃₂ in 𝑉 satisfies 𝑟≥6, then at least six points among 𝑃₃,…,𝑃₁₁ are on the intersection of S and the sphere whose center is 𝑃₂. So they are on a 2-dimensional sphere. By Proposition 3.8, the condition (𝑋) holds.

Proposition 4.2. If the type of 𝑃₂ is (5,4) in 𝑉, then the condition (𝑋) holds.

Proof. We suppose that 𝑃₂ is of type (5,4) in 𝑉. We see that five points in 𝑉 are distributed on a 2-dimensional sphere which is the intersection of 𝑆 and the sphere whose center is 𝑃₂ and another four points in 𝑉 are distributed on another 2-dimensional sphere. These two spheres are disjoint.
We will call them 𝑆₁ (on which 𝑃₃,…,𝑃₇ are) and 𝑆₂ (on which 𝑃₈,…,𝑃₁₁ are). Let 𝑉₁=𝑉∩𝑆₁={𝑃₃,…,𝑃₇} and 𝑉₂=𝑉∩𝑆₂={𝑃₈,…,𝑃₁₁}. For 𝑃_𝑖∈𝑉₁, let 𝑃₂𝑃_𝑖=𝑎 and for 𝑃_𝑗∈𝑉₂, let 𝑃₂𝑃_𝑗=𝑏.
If 𝑉 is a 2-distance set, then the types of ten points in 𝑉 must be all (6,3) by looking at the Petersen graph (Figure 2). But 𝑃₂ is of type (5,4), 𝑉 is not a 2-distance set. Hence 𝑉 is an 𝑠-distance set (𝑠≥3), there exists a pair of points in 𝑃₂,…,𝑃₁₁ whose distance is 𝑐, that is, distinct from 𝑎 and 𝑏. Because 𝑃₂𝑃_𝑘=𝑎 or 𝑏(𝑘=3,…,11), 𝑐 is the distance between a pair of distinct points in {𝑃₃,…,𝑃₁₁}. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑉₁ and 𝑃_𝑗∈𝑉₂, then Δ𝑃₂𝑃_𝑖𝑃_𝑗 is scalene with sides 𝑎,𝑏,𝑐. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑉₁,𝑃_𝑗∈𝑉₂. (4.1) So 𝑐 is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁. Without loss of generality we may assume 𝑃₃𝑃₄=𝑐. For 𝑃_𝑗∈𝑉₂ we consider Δ𝑃₃𝑃₄𝑃_𝑗. Since 𝑃₃𝑃_𝑗 and 𝑃₄𝑃_𝑗 are of length 𝑎 or 𝑏 by (4.1), 𝑃₃𝑃_𝑗=𝑃₄𝑃_𝑗 holds. Thus four points 𝑃₈,…,𝑃₁₁ are on the hyperplane perpendicularly bisecting 𝑃₃𝑃₄, the sphere 𝑆, and the sphere 𝑆₂. The intersection of them is a circle. Therefore the condition (𝑋) holds.
We can repeat the similar discussion when we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₂.

Next we consider that the type of 𝑃₂ is (5,3,1) or (5,2,2) in V. We see that five points in 𝑉 are distributed on a 2-dimensional sphere which is the intersection of 𝑆 and the sphere whose center is 𝑃₂ and another two or three points in 𝑉 are distributed on another 2-dimensional sphere. These two spheres are disjoint.

We will call them 𝑆₁ (on which 𝑃₃,…,𝑃₇ are) and 𝑆₂ (on which 𝑃₈ and 𝑃₉ are). Let 𝑉₁=𝑉∩𝑆₁={𝑃₃,…,𝑃₇} and 𝑉₂=𝑉∩𝑆₂={𝑃₈,𝑃₉}. For 𝑃_𝑖∈𝑉₁, let 𝑃₂𝑃_𝑖=𝑎 and for 𝑃_𝑗∈𝑉₂, let 𝑃₂𝑃_𝑗=𝑏. Moreover let 𝑃₂𝑃₁₁=𝑐.

Proposition 4.3. Let 𝑋₁={𝑃₂,…,𝑃₉}. If 𝑋₁ is an 𝑠-distance set (𝑠≥3), then the condition (𝑋) holds.

Proof. Because we suppose that 𝑋₁ is an 𝑠-distance set (𝑠≥3), there exists a pair of points 𝑃₂,…,𝑃₉ whose distance is 𝑑, that is, distinct from 𝑎 and 𝑏 (but we can admit 𝑐=𝑑).
Since 𝑃₂𝑃_𝑖=𝑎 or 𝑏(𝑖=3,…,9), 𝑑 is the distance apart of a pair of points in {𝑃₃,…,𝑃₉}. If 𝑃_𝑖𝑃_𝑗=𝑑 holds for some 𝑃_𝑖∈𝑉₁ and 𝑃_𝑗∈𝑉₂, then Δ𝑃₂𝑃_𝑖𝑃_𝑗 is scalene with sides 𝑎,𝑏,𝑑. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑉₁,𝑃_𝑗∈𝑉₂. (4.2) So 𝑑 is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose that 𝑑 is the distance between a pair of distinct points on 𝑆₂, that is, 𝑃₈𝑃₉=𝑑. In this case, if we repeat the similar discussion as Proposition 4.2, then the condition (𝑋) holds. Hence we suppose that 𝑑 is the distance between a pair of distinct points on 𝑆₁. For 𝑃₃,…,𝑃₇ on 𝑆₁, we consider 5-point graphs in Table 2. Edges in a graph represent the distance, that is, distinct from 𝑎 and 𝑏. We regard the others, (i.e., transparent edges) as the distances 𝑎 and 𝑏. Here we need not consider the graph which has no edge, because we suppose that there is at least one pair whose distance is distinct from 𝑎 and 𝑏. We remark that 33 graphs in Table 2 and the graph which has no edge are all 5-point graphs.
We can classify 33 graphs into the following:
(i)a 4-point subgraph is “connected”; graphs satisfying it are (5,3,1), (5,3,3), (5,4,1), (5,4,2), (5,4,3), (5,4,5), (5,4,6), and (5,𝑎,∗) for 5≤𝑎≤10 (∗ is arbitrary);(ii)another four graphs whose a 3-point subgraph is

and no edge between them and the other two points; they are (5,2,1), (5,3,2), (5,3,4), and (5,4,4);(iii)(5,2,2);(iv)(5,1,1).
We observe each case. In the case (i), we may assume that the 4-point subgraph with 𝑃₃,…,𝑃₆ is connected. Without loss of generality we may assume 𝑃₃𝑃₄=𝑑. For 𝑖=8,9, consider Δ𝑃₃𝑃₄𝑃_𝑖. Then we have 𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖 by (4.2). Since the 4-point subgraph with 𝑃₃,…,𝑃₆ is connected, we have 𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖=𝑃₅𝑃_𝑖=𝑃₆𝑃_𝑖 by the similar discussion. Moreover we have 𝑃₃𝑃_𝑗=𝑃₄𝑃_𝑗=𝑃₅𝑃_𝑗=𝑃₆𝑃_𝑗 for 𝑗=1,2 by the assumption. Then four points 𝑃₁, 𝑃₂, 𝑃₈, 𝑃₉ are equidistant from 𝑃₃,…,𝑃₆ on the 2-dimensional sphere 𝑆₁. If 𝑃₃,…,𝑃₆ are on a plane, then they are on a circle; the condition (𝑋) holds. On the other hand, if 𝑃₃,…,𝑃₆ are not on a plane, then 𝑃₁, 𝑃₂, 𝑃₈, and 𝑃₉ are on a line. We cannot take four points on a line. This is a contradiction.
In the case (ii), we may assume that 𝑃₃𝑃₄=𝑑 and 𝑃₃𝑃₅=𝑒 (we can admit 𝑑=𝑒). For 𝑖=8,9, consider Δ𝑃₃𝑃₄𝑃_𝑖 and Δ𝑃₃𝑃₅𝑃_𝑖. Then we have 𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖 and 𝑃₃𝑃_𝑖=𝑃₅𝑃_𝑖 by (4.2). In this case, 𝑃₃𝑃_𝑗, 𝑃₄𝑃_𝑗, and 𝑃₅𝑃_𝑗 are 𝑎 or 𝑏 for 𝑗=6,7. When we consider Δ𝑃₃𝑃₄𝑃_𝑗 and Δ𝑃₃𝑃₅𝑃_𝑗, 𝑃₃𝑃_𝑗=𝑃₄𝑃_𝑗 and 𝑃₃𝑃_𝑗=𝑃₅𝑃_𝑗 hold. By the assumption we have 𝑃₃𝑃_𝑘=𝑃₄𝑃_𝑘 and 𝑃₃𝑃_𝑘=𝑃₅𝑃_𝑘 for 𝑘=1,2. Then six points 𝑃₁, 𝑃₂, 𝑃₆, 𝑃₇, 𝑃₈, 𝑃₉ are equidistant from 𝑃₃, 𝑃₄, and 𝑃₅. Hence they are in the 2-dimensional Euclidean space, that is, {𝑃₁,𝑃₂,𝑃₆,𝑃₇,𝑃₈,𝑃₉} is a 6-point isosceles set in ℝ². We know that there exists a unique 6-point isosceles set in ℝ² up to isomorphisms. It consists of five points of a regular pentagon and its center. So four points in {𝑃₁,𝑃₂,𝑃₆,𝑃₇,𝑃₈,𝑃₉} lie on a circle; the condition (𝑋) holds.
In the case (iii), we may assume that 𝑃₃𝑃₄=𝑑 and 𝑃₅𝑃₆=𝑒 (we can admit 𝑑=𝑒). For 𝑖=8,9, consider Δ𝑃₃𝑃₄𝑃_𝑖 and Δ𝑃₅𝑃₆𝑃_𝑖. Then we have 𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖 and 𝑃₅𝑃_𝑖=𝑃₆𝑃_𝑖 by (4.2). In this case, 𝑃₃𝑃₇, 𝑃₄𝑃₇, 𝑃₅𝑃₇, and 𝑃₆𝑃₇ are 𝑎 or 𝑏. When we consider Δ𝑃₃𝑃₄𝑃₇ and Δ𝑃₅𝑃₆𝑃₇, 𝑃₃𝑃₇=𝑃₄𝑃₇ and 𝑃₅𝑃₇=𝑃₆𝑃₇ hold. By the assumption we have 𝑃₃𝑃_𝑗=𝑃₄𝑃_𝑗 and 𝑃₅𝑃_𝑗=𝑃₆𝑃_𝑗 for 𝑗=1,2. Then five points 𝑃₁,𝑃₂,𝑃₇,𝑃₈,𝑃₉ are on the hyperplane perpendicularly bisecting 𝑃₃𝑃₄ and the hyperplane perpendicularly bisecting 𝑃₅𝑃₆. For the intersection of them, there are two cases:
(𝛼) since two hyperplanes are same, the intersection is a 3-dimensional Euclidean space.(𝛽)a 2-dimensional Euclidean space.
In the case (𝛼), since 𝑃₃,…,𝑃₆ are on the 2-dimensional sphere 𝑆₁, the segment 𝑃₃𝑃₄ and the segment 𝑃₅𝑃₆ are mutually parallel. Then there is a plane that contains 𝑃₃,…,𝑃₆. So they are on a circle; the condition (𝑋) holds.
In the case (𝛽), {𝑃₁,𝑃₂,𝑃₇,𝑃₈,𝑃₉} is a 5-point isosceles set in ℝ². We know that there exist exactly three 5-point isosceles sets in ℝ² up to isomorphisms. They are four points of a square and its center, five points of a regular pentagon, and four points of a regular pentagon and its center. So four points in {𝑃₁,𝑃₂,𝑃₇,𝑃₈,𝑃₉} lie on a circle; the condition (𝑋) holds.
In the case (iv), we may assume that 𝑃₃𝑃₄=𝑑. Then we see that there is exactly one pair 𝑃₃𝑃₄ whose distance is distinct from 𝑎 and 𝑏 in 𝑋₁. When we consider Δ𝑃₃𝑃₄𝑃_𝑖 for 𝑖=2,5,…,9, 𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖 holds by the configuration hypothesis. Thus six points 𝑃₂,𝑃₅,𝑃₆,𝑃₇,𝑃₈, and 𝑃₉ are on the hyperplane perpendicularly bisecting 𝑃₃𝑃₄. This hyperplane is a 3-dimensional Euclidean space. Since 𝐴({𝑃₂,𝑃₅,𝑃₆,𝑃₇,𝑃₈,𝑃₉})={𝑎,𝑏}, this is a 2-distance set in ℝ³. We know that there exist exactly six 6-point 2-distance sets in ℝ³. Any set contains four points lying on a circle. Hence the condition (𝑋) holds.

Table 2 

5-point graphs.

Proposition 4.4. Similarly let 𝑋₁={𝑃₂,…,𝑃₉}. If 𝑋₁ is a 2-distance set, then the condition (𝑋) holds.

Proof. We consider the sum of all vertex-numbers in 𝒫. Because 𝑃₂ has the largest vertex-number in 𝑉, 𝑉(𝑃₁)+⋯+𝑉(𝑃₁₁)≤₂¹⁰+10×{⁵₂+³₂+¹₂}=175. Let 𝛼be the number of regular triangles in 𝒫. Then 2𝛼+₃¹¹≤175 holds by (3.2). Thus 𝛼≤5.
Let 𝑉₁={𝑃₃,…,𝑃₇}. We notice that 𝑉₁ on 𝑆₁ is a 2-distance set in ℝ³. We consider 5-point graphs in Table 2 again. Edges in a graph represent the distance 𝑏. We regard the others, (i.e., transparent edges) as the distance 𝑎. Here we need not consider the graph which has no edge, because there is no 5-point 1-distance set in ℝ³. Similarly we need not consider the complete graph (5,10,1).
If 𝑃_𝑖𝑃_𝑗=𝑎 for 𝑖,𝑗∈{3,…,7} (𝑖≠𝑗), then Δ𝑃₂𝑃_𝑖𝑃_𝑗 is a regular triangle. Since 𝛼≤5, there are at most five pairs in 𝑉₁ whose distances are 𝑎. The number of pairs in 𝑉₁ is ⁵₂=10. Thus there are at least five pairs in 𝑉₁ whose distances are 𝑏.
Hence we have only to consider the 19 graphs between (5,5,1) and (5,9,1) in Table 2. In any graph, a 4-point subgraph is “connected”. We may assume that their four points are 𝑃₃,…,𝑃₆ and that there is an edge between 𝑃₃ and 𝑃₄, that is, 𝑃₃𝑃₄=𝑏. We consider Δ𝑃₂𝑃₃𝑃₁₁ and Δ𝑃₂𝑃₄𝑃₁₁. Since 𝑃₂𝑃₃=𝑃₂𝑃₄=𝑎 and 𝑃₂𝑃₁₁=𝑐, 𝑃₃𝑃₁₁ and 𝑃₄𝑃₁₁ are 𝑎 or 𝑐. Then we consider Δ𝑃₃𝑃₄𝑃₁₁, we have 𝑃₃𝑃₁₁=𝑃₄𝑃₁₁. Because the 4-point subgraph with 𝑃₃,…,𝑃₆ is connected, we have 𝑃₃𝑃₁₁=𝑃₄𝑃₁₁=𝑃₅𝑃₁₁=𝑃₆𝑃₁₁ by the similar discussion. Moreover we have 𝑃₃𝑃_𝑘=𝑃₄𝑃_𝑘=𝑃₅𝑃_𝑘=𝑃₆𝑃_𝑘 for 𝑘=1,2 by the assumption. Thus three points 𝑃₁, 𝑃₂, 𝑃₁₁ are equidistant from 𝑃₃,…,𝑃₆ on the 2-dimensional sphere 𝑆₁. If 𝑃₃,…,𝑃₆ are on a plane, then they are on a circle; the condition (𝑋) holds. On the other hand, if 𝑃₃,…,𝑃₆ are not on a plane, then 𝑃₁, 𝑃₂, and 𝑃₁₁ are on a line. By Corollary 3.7, this is a contradiction.
Therefore if 𝑋₁ is a 2-distance set, then the condition (𝑋) holds.

Combining Propositions 4.3 and 4.4, we have the following proposition.

Proposition 4.5. If the type of 𝑃₂ is (5,3,1) or (5,2,2) in 𝑉, then the condition (𝑋) holds.

The last case is what the type of 𝑃₂ is (4,4,1) in 𝑉. We see that four points in 𝑉 are distributed on a 2-dimensional sphere which is the intersection of 𝑆 and the sphere whose center is 𝑃₂ and another four points in 𝑉 are distributed on another 2-dimensional sphere. These two spheres are disjoint.

We will call these two spheres 𝑆₁ (on which 𝑃₃,…,𝑃₆ are) and 𝑆₂ (on which 𝑃₇,…,𝑃₁₀ are). Let 𝑉₁=𝑉∩𝑆₁={𝑃₃,…,𝑃₆} and 𝑉₂=𝑉∩𝑆₂={𝑃₇,…,𝑃₁₀}. For 𝑃_𝑖∈𝑉₁, let 𝑃₂𝑃_𝑖=𝑎 and for 𝑃_𝑗∈𝑉₂, let 𝑃₂𝑃_𝑗=𝑏. Moreover let 𝑃₂𝑃₁₁=𝑐. Because 𝑉(𝑃₂)=12, 𝑉(𝑃_𝑘)=12 for 𝑘=3,…,11. Thus the type of 𝑃_𝑘 is (4,4,1) in 𝑉 for any 𝑘. (Since 𝑉(𝑃_𝑘)=12, the type of 𝑃_𝑘 may be (5,2,2) in 𝑉. In this case, if we apply Proposition 4.5, then the condition (𝑋) holds.)

Proposition 4.6. If the type of 𝑃₂ is (4,4,1) in 𝑉, then 𝑃₁₁ is equidistant from four points on one of the 2-dimensional spheres 𝑆₁ and 𝑆₂.

Proof. Since the type of 𝑃₁₁ is (4,4,1) in 𝑉 and 𝑃₂𝑃₁₁=𝑐, the distance 𝑐 corresponds to 1 or 4 of type (4,4,1). If 𝑐 corresponds to 1 of type (4,4,1), then 𝑃_𝑖𝑃₁₁≠𝑐 for 𝑖=3,…,10. Considering Δ𝑃₂𝑃_𝑖𝑃₁₁, we have 𝑃₃𝑃₁₁=𝑃₄𝑃₁₁=𝑃₅𝑃₁₁=𝑃₆𝑃₁₁=𝑎 and 𝑃₇𝑃₁₁=𝑃₈𝑃₁₁=𝑃₉𝑃₁₁=𝑃₁₀𝑃₁₁=𝑏. Thus this proposition holds.
On the other hand, if 𝑐 corresponds to 4 of type (4,4,1), then for 𝑗=3,…,10, there are exactly three points such that 𝑃_𝑗𝑃₁₁=𝑐. We may assume that 𝑃₃𝑃₁₁=𝑐. We have three means to select the other two points.
(i)𝑃₄𝑃₁₁=𝑃₅𝑃₁₁=𝑐. (Both points are on 𝑆₁.)(ii)𝑃₄𝑃₁₁=𝑃₇𝑃₁₁=𝑐. (One is on 𝑆₁ and the other is on 𝑆₂.)(iii)𝑃₇𝑃₁₁=𝑃₈𝑃₁₁=𝑐. (Both points are on 𝑆₂.)
In the case (i), considering Δ𝑃₂𝑃_𝑘𝑃₁₁ for 𝑘=6,…,10, we have 𝑃₆𝑃₁₁=𝑎 and 𝑃₇𝑃₁₁=𝑃₈𝑃₁₁=𝑃₉𝑃₁₁=𝑃₁₀𝑃₁₁=𝑏. Thus this proposition holds for 𝑆₂. In the case (ii), considering Δ𝑃₂𝑃𝑃₁₁ for =5,6,8,9,10, we have 𝑃₅𝑃₁₁=𝑃₆𝑃₁₁=𝑎 and 𝑃₈𝑃₁₁=𝑃₉𝑃₁₁=𝑃₁₀𝑃₁₁=𝑏. Then the type of 𝑃₁₁ is (4,3,2), not (4,4,1). This is a contradiction. In the case (iii), considering Δ𝑃₂𝑃_𝑚𝑃₁₁ for 𝑚=4,5,6,9,10, we have 𝑃₄𝑃₁₁=𝑃₅𝑃₁₁=𝑃₆𝑃₁₁=𝑎 and 𝑃₉𝑃₁₁=𝑃₁₀𝑃₁₁=𝑏. Then the type of 𝑃₁₁ is (4,3,2), not (4,4,1). This is a contradiction.
Therefore 𝑃₁₁ is equidistant from four points on one of the 2-dimensional spheres 𝑆₁ and 𝑆₂.

Proposition 4.7. If the type of 𝑃₂ is (4,4,1) in 𝑉, then the condition (𝑋) holds.

Proof. By Proposition 4.6, 𝑃₁₁ is equidistant from four points on one of the 2-dimensional spheres 𝑆₁ and 𝑆₂. We may assume that it is 𝑆₁. Moreover we have 𝑃_𝑖𝑃₃=𝑃_𝑖𝑃₄=𝑃_𝑖𝑃₅=𝑃_𝑖𝑃₆ for 𝑖=1,2 by the assumption. Thus three points 𝑃₁,𝑃₂,𝑃₁₁ are equidistant from 𝑃₃,…,𝑃₆ on the 2-dimensional sphere 𝑆₁. If 𝑃₃,…,𝑃₆ are on a plane, then they are on a circle; the condition (𝑋) holds. On the other hand, if 𝑃₃,…,𝑃₆ are not on a plane, then 𝑃₁, 𝑃₂, and 𝑃₁₁ are on a line. By Corollary 3.7, this is a contradiction.
Therefore if the type of 𝑃₂ is (4,4,1) in 𝑉, then the condition (𝑋) holds.

Summing up the results of Propositions 4.1, 4.2, 4.5, and 4.7, we have the following.

Lemma 4.8. For any 11-point isosceles set in ℝ⁴ in which 𝑃₁ is of type (6.1), the condition (𝑋) holds.

5. Case (B) in Lemma 3.1

We consider the case (B) in Lemma 3.1. We see that at least eight points in an 11-point isosceles set are distributed on a 3-dimensioal sphere, and at least one point does not lie on the sphere.

Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ satisfies the case (B). Let 𝑆 be the sphere centered at 𝑃₁ with radius 𝑎 and 𝑉=𝒫∩𝑆={𝑃₂,…,𝑃₉}. Let 𝑃₁₁ be the point which is not on 𝑆 and 𝑃₁𝑃₁₁=𝑏.

Lemma 5.1. The condition (𝑋) holds for any 11-point isosceles set in which the type of 𝑃₁ satisfies the case (B) in Lemma 3.1.

Proof. By Proposition 3.5, eight points 𝑃₂,…,𝑃₉ are on one of two disjoint 2-dimensional spheres 𝑆₁ and 𝑆₂, where 𝑃_𝑖 on 𝑆₁ satisfies 𝑃_𝑖𝑃₁₁=𝑎 and 𝑃_𝑗 on 𝑆₂ satisfies 𝑃_𝑗𝑃₁₁=𝑏 (consider Δ𝑃₁𝑃_𝑘𝑃₁₁ for 𝑘=2,…,9).
If more than or equal to six points lie on one sphere, then the condition (𝑋) holds by Proposition 3.8. So we consider the following cases.
(i)Five points lie on one sphere; the other three points lie on the other sphere.(ii)Four points lie on one sphere; the other four points lie on the other sphere.
We consider the case (i). We may suppose that 𝑃₂,…,𝑃₆ are on 𝑆₁ and 𝑃₇, 𝑃₈, 𝑃₉ are on 𝑆₂. Let 𝑉₁=𝑉∩𝑆₁={𝑃₂,…,𝑃₆} and 𝑉₂=𝑉∩𝑆₂={𝑃₇,𝑃₈,𝑃₉}.
Here the 10-point set {𝑃₁,…,𝑃₉,𝑃₁₁} is not a 2-distance set, because 𝑃₁ is of type (8,1) in it, not of type (6,3) in the Petersen graph. Hence it is an 𝑠-distance set (𝑠≥3), there exists a pair of points in {𝑃₂,…,𝑃₉} whose distance is 𝑐, that is, distinct from 𝑎 and 𝑏. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑉₁ and 𝑃_𝑗∈𝑉₂, then Δ𝑃_𝑖𝑃_𝑗𝑃₁₁ is scalene. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑉₁,𝑃_𝑗∈𝑉₂. (5.1) So 𝑐 is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₂. Without loss of generality we may assume 𝑃₇𝑃₈=𝑐. For 𝑃_𝑖∈𝑉₁ we consider Δ𝑃_𝑖𝑃₇𝑃₈. Since 𝑃_𝑖𝑃₇ and 𝑃_𝑖𝑃₈ are of length 𝑎 or 𝑏 by (5.1), we have 𝑃_𝑖𝑃₇=𝑃_𝑖𝑃₈. Thus five points 𝑃₂,…,𝑃₆ are on the hyperplane perpendicularly bisecting 𝑃₇𝑃₈ and the 2-dimensional sphere 𝑆₁. The intersection of them is a circle. Hence the condition (𝑋) holds.
On the other hand, we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁. Without loss of generality we may assume 𝑃₂𝑃₃=𝑐.
Next we suppose that there exist more than or equal to two pairs of points on 𝑆₁ whose distances are distinct from 𝑎 and 𝑏. One is 𝑃₂𝑃₃=𝑐. We have two cases as the second pair whose distance is distinct from 𝑎 and 𝑏.
Case 1. 𝑃₂𝑃₃=𝑐 and 𝑃₄𝑃₅=𝑑 (𝑑≠𝑎, 𝑑≠𝑏, but we can admit 𝑐=𝑑).
Let 𝑃_𝑗∈𝑉₂ and consider Δ𝑃₂𝑃₃𝑃_𝑗. Because 𝑃₂𝑃_𝑗 and 𝑃₃𝑃_𝑗 are of length 𝑎 or 𝑏 by (5.1), we have 𝑃₂𝑃_𝑗=𝑃₃𝑃_𝑗. When we consider Δ𝑃₄𝑃₅𝑃_𝑗 similarly, we have 𝑃₄𝑃_𝑗=𝑃₅𝑃_𝑗. Thus three points 𝑃₇,𝑃₈ and 𝑃₉ are on the hyperplane perpendicularly bisecting 𝑃₂𝑃₃, the hyperplane perpendicularly bisecting 𝑃₄𝑃₅, and the 2-dimensional sphere 𝑆₂. For the intersection of them, there are two cases:
(𝛼) because two hyperplanes are same, the intersection is a circle.(𝛽)two points.
In the case (𝛼), because 𝑃₂,…,𝑃₅ are on the 2-dimensional sphere 𝑆₁, the segment 𝑃₂𝑃₃ and the segment 𝑃₄𝑃₅ are mutually parallel. Then there is a plane that contains 𝑃₂,…,𝑃₅. So they are on a circle; the condition (𝑋) holds.
In the case (𝛽), we cannot put one of 𝑃₇, 𝑃₈, and 𝑃₉. This is a contradiction.
Case 2. 𝑃₂𝑃₃=𝑐 and 𝑃₂𝑃₄=𝑑 (𝑑≠𝑎, 𝑑≠𝑏, but we can admit 𝑐=𝑑).
We can repeat the same discussion. (But the case (α) does not exist, only the case (β) exists.)
Hence we suppose that there is exactly one pair 𝑃₂𝑃₃ which is distinct from 𝑎 and 𝑏 in 𝑉₁. When we consider Δ𝑃₂𝑃₃𝑃_𝑘 for 𝑘=1,4,…,9,11, 𝑃₂𝑃_𝑘=𝑃₃𝑃_𝑘 holds by the configuration hypothesis. Thus eight points 𝑃₁,𝑃₄,…,𝑃₉ and 𝑃₁₁ are on the hyperplane perpendicularly bisecting 𝑃₂𝑃₃. This hyperplane is a 3-dimensional Euclidean space. Moreover {𝑃₁,𝑃₄,…,𝑃₉,𝑃₁₁} is a 2-distance set with the distances 𝑎 and 𝑏. But we know that there exists no 𝑛-point 2-distance set in ℝ³ for 𝑛≥7. This is a contradiction.
We consider the case (ii). If we repeat this discussion similarly, then we can see that the condition (𝑋) holds.

6. Case (C) in Lemma 3.1

We consider the case (C) in Lemma 3.1. We see that seven points in an 11-point isosceles set are distributed on a 3-dimensioal sphere, another at least two points are distributed on another 3-dimensioal sphere, where these are concentric spheres. The center of the spheres is in it.

Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ satisfies the case (C). 𝑃₁ will denote the common center of the two spheres, which we will call 𝑆₁ (on which 𝑃₂,…,𝑃₈ are), 𝑆₂ (on which 𝑃₉ and 𝑃₁₀ are), radii 𝑎, 𝑏, respectively.

Lemma 6.1. The condition (𝑋) holds for any 11-point isosceles set 𝒫 in which the type of 𝑃₁ satisfies the case (C) in Lemma 3.1.

Proof. The 10-point set {𝑃₁,…,𝑃₁₀} is not a 2-distance set, because 𝑃₁ is of type (7,2) in it, not of type (6,3) in the Petersen graph. Hence it is an 𝑠-distance set (𝑠≥3), there exists a pair of points in {𝑃₂,…,𝑃₁₀} whose distance is 𝑐, that is, distinct from 𝑎 and 𝑏. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂, then Δ𝑃₁𝑃_𝑖𝑃_𝑗 is scalene. Thus 𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑆₁,𝑃_𝑗∈𝑆₂. (6.1) So 𝑐 is the distance between a pair of distinct points on the same 3-dimensional sphere.
We suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₂, that is, 𝑃₉𝑃₁₀=𝑐. For 𝑃_𝑖∈𝑆₁ we consider Δ𝑃_𝑖𝑃₉𝑃₁₀. Since 𝑃_𝑖𝑃₉ and 𝑃_𝑖𝑃₁₀ are of length 𝑎 or 𝑏 by (6.1), we have 𝑃_𝑖𝑃₉=𝑃_𝑖𝑃₁₀. Thus seven points 𝑃₂,…,𝑃₈ are on the hyperplane perpendicularly bisecting 𝑃₉𝑃₁₀ and the 3-dimensional sphere 𝑆₁. The intersection of them is a 2-dimensional sphere. By Proposition 3.8, the condition (𝑋) holds.
Thus we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁. By Proposition 3.5, seven points 𝑃₂,…,𝑃₈ are on one of two disjoint 2-dimensional spheres 𝑆₁₁ and 𝑆₁₂, where 𝑃_𝑖 on 𝑆₁₁ satisfies 𝑃_𝑖𝑃₉=𝑎 and 𝑃_𝑗 on 𝑆₁₂ satisfies 𝑃_𝑗𝑃₉=𝑏 (consider Δ𝑃₁𝑃_𝑘𝑃₉ for 𝑘=2,…,8).
If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑆₁₁ and 𝑃_𝑗∈𝑆₁₂, then Δ𝑃_𝑖𝑃_𝑗𝑃₉ is scalene. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑆₁₁,𝑃_𝑗∈𝑆₁₂. (6.2) So 𝑐 is the distance between a pair of distinct points on the same 2-dimensional sphere.
If more than or equal to six points lie on one sphere, then the condition (𝑋) holds by Proposition 3.8. So we consider the following cases:
(i)Five points lie on one sphere; the other two points lie on the other sphere.(ii)Four points lie on one sphere; the other three points lie on the other sphere.
We consider the case (i). We may suppose that 𝑃₂,…,𝑃₆ are on 𝑆₁₁ and 𝑃₇, 𝑃₈ are on 𝑆₁₂.
We suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁₂, that is, 𝑃₇𝑃₈=𝑐. We consider Δ𝑃_𝑖𝑃₇𝑃₈ for 𝑃_𝑖∈𝑆₁₁. By (6.2), we have 𝑃_𝑖𝑃₇=𝑃_𝑖𝑃₈. Thus five points 𝑃₂,…,𝑃₆ are on the hyperplane perpendicularly bisecting 𝑃₇𝑃₈ and the 2-dimensional sphere 𝑆₁₁. The intersection of them is a circle. Hence the condition (𝑋) holds.
On the other hand, we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁₁. Without loss of generality we may assume 𝑃₂𝑃₃=𝑐.
Next we suppose that there exist more than or equal to two pairs of points on 𝑆₁₁ whose distances are distinct from 𝑎 and 𝑏. One is 𝑃₂𝑃₃=𝑐. We have two cases as the second pair whose distance is distinct from 𝑎 and 𝑏.
Case 1. 𝑃₂𝑃₃=𝑐 and 𝑃₄𝑃₅=𝑑 (𝑑≠𝑎, 𝑑≠𝑏, but we can admit 𝑐=𝑑).
Let 𝑃_𝑗∈𝑆₁₂∪𝑆₂ and consider Δ𝑃₂𝑃₃𝑃_𝑗. By (6.1) and (6.2), we have 𝑃₂𝑃_𝑗=𝑃₃𝑃_𝑗. When we consider Δ𝑃₄𝑃₅𝑃_𝑗 similarly, we have 𝑃₄𝑃_𝑗=𝑃₅𝑃_𝑗. For 𝑃₁ we have 𝑃₁𝑃₂=𝑃₁𝑃₃ and 𝑃₁𝑃₄=𝑃₁𝑃₅. Thus five point 𝑃₁, 𝑃₇, 𝑃₈, 𝑃₉, and 𝑃₁₀ are on the hyperplane perpendicularly bisecting 𝑃₂𝑃₃ and the hyperplane perpendicularly bisecting 𝑃₄𝑃₅. For the intersection of them, there are two cases:
(α) since two hyperplanes are same, the intersection is a 3-dimensional Euclidean space;(β)a 2-dimensional Euclidean space.
In the case (α), since 𝑃₂,…,𝑃₅ are on the 2-dimensional sphere 𝑆₁₁, the segment 𝑃₂𝑃₃ and the segment 𝑃₄𝑃₅ are mutually parallel. Then there is a plane that contains 𝑃₂,…,𝑃₅. So they are on a circle; the condition (𝑋) holds.
In the case (β), {𝑃₁,𝑃₇,𝑃₈,𝑃₉,𝑃₁₀} is a 5-point isosceles set in ℝ². We know that there exist three 5-point isosceles sets in ℝ² up to isomorphisms. They are four points of a square and its center, five points of a regular pentagon, and four points of a regular pentagon and its center. So four points in {𝑃₁,𝑃₇,𝑃₈,𝑃₉,𝑃₁₀} lie on a circle; the condition (𝑋) holds.
Case 2. 𝑃₂𝑃₃=𝑐 and 𝑃₂𝑃₄=𝑑 (𝑑≠𝑎, 𝑑≠𝑏, but we can admit 𝑐=𝑑).
We can repeat the same discussion. (But the case (α) does not exist, only the case (β) exists.)
Hence we suppose that there is exactly one pair 𝑃₂𝑃₃ whose distance is distinct from 𝑎 and 𝑏 in {𝑃₁,…,𝑃₁₀}. When we consider Δ𝑃₂𝑃₃𝑃_𝑘 for 𝑘=1,4,…,10, 𝑃₂𝑃_𝑘=𝑃₃𝑃_𝑘 holds by the configuration hypothesis. Thus eight points 𝑃₁, 𝑃₄,…, 𝑃₉, and 𝑃₁₀ are on the hyperplane perpendicularly bisecting 𝑃₂𝑃₃. This hyperplane is a 3-dimensional Euclidean space. Moreover {𝑃₁,𝑃₄,…,𝑃₁₀} is a 2-distance set with the distances 𝑎 and 𝑏. But there exists no 𝑛-point 2-distance set in ℝ³ for 𝑛≥7. This is a contradiction.
We consider the case (ii). If we repeat this discussion similarly, then we see that the condition (𝑋) holds.

7. Case (D) in Lemma 3.1

Lemma 7.1. The condition (𝑋) holds for any 11-point isosceles set in which the type of 𝑃₁ satisfies the case (D) in Lemma 3.1.

Proof. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set. When the type of 𝑃₁ is (6,4) or (6,3,1), 𝑃₁ will denote the common center of the two spheres, which we will call 𝑆₁ (on which 𝑃₂,…,𝑃₇ are), 𝑆₂ (on which 𝑃₈, 𝑃₉, and 𝑃₁₀ are).
Let 𝒫^′={𝑃₁,…,𝑃₁₀}. 𝒫^′ can be the 2-distance set 𝑋^′ mentioned in Section 2. Since 𝑋^′ contains a square, the condition (𝑋) holds.
Hence we may suppose that 𝒫^′ is an 𝑠-distance set (𝑠≥3). In this case, we can show that the condition (𝑋) holds by repeating the similar discussion as the proof of Lemma 6.1.
When the type of 𝑃₁ is (5,5) or (5,4,1), we can show that the condition (𝑋) holds by repeating the similar discussion as the proof of Lemma 6.1.

8. Case (E) in Lemma 3.1

We consider the case (E) in Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ is (7,1,1,1). We may assume that 𝑃₁𝑃₂=𝑃₁𝑃₃=⋯=𝑃₁𝑃₈=𝑎, 𝑃₁𝑃₉=𝑏, 𝑃₁𝑃₁₀=𝑐, and 𝑃₁𝑃₁₁=𝑑. Let 𝑋₁={𝑃₁,…,𝑃₈,𝑃₉}, 𝑋₂={𝑃₁,…,𝑃₈,𝑃₁₀}, and 𝑋₃={𝑃₁,…,𝑃₈,𝑃₁₁}.

Proposition 8.1. For 𝑋₁,…,𝑋₃ above, if 2-distance sets exist, then the number of them is at most one.

Proof. We suppose that 𝑋₁ and 𝑋₂ are 2-distance sets. We may prove that this leads a contradiction.
We have 𝐴(𝑋₁)={𝑎,𝑏} and 𝐴(𝑋₂)={𝑎,𝑐} by the hypothesis above. For 𝑖,𝑗=2,…,8(𝑖≠𝑗), 𝑃_𝑖𝑃_𝑗 must be 𝑎, 𝑏, or 𝑐.
If 𝑃_𝑖𝑃_𝑗=𝑏, then 𝐴(𝑋₂)≠{𝑎,𝑐} for 𝑋₂. If 𝑃_𝑖𝑃_𝑗=𝑐, then 𝐴(𝑋₁)≠{𝑎,𝑏} for 𝑋₁. So we have 𝑃_𝑖𝑃_𝑗=𝑎. Hence 𝐴({𝑃₁,…,𝑃₈})={𝑎}, {𝑃₁,…,𝑃₈} is a 1-distance set. But there is no 8-point 1-distance set in ℝ⁴. This is a contradiction.
Therefore the number of 2-distance sets is at most one.

Lemma 8.2. The condition (𝑋) holds for any 11-point isosceles set 𝒫 in which the type of P₁ is (7,1,1,1).

Proof. By Proposition 8.1, at least two sets of 𝑋₁,…,𝑋₃ are 𝑠-distance sets (𝑠≥3). We may suppose 𝑋₁ and 𝑋₂ are 𝑠-distance sets. Espesially we notice that 𝑋₁ is an 𝑠-distance set. Thus there is a distance apart of a pair of points in 𝑋₁ which is distinct from 𝑎 and 𝑏. This is one of 𝑐,𝑑, and 𝑒, where 𝑒 is distinct from 𝑎,𝑏,𝑐, and 𝑑. We may assume that it is 𝑐.
Let 𝑆 be the sphere centered at 𝑃₁ with radius 𝑎 and 𝑉=𝑋₁∩𝑆={𝑃₂,…,𝑃₈}. By Proposition 3.5, seven points 𝑃₂,…,𝑃₈ are on one of two disjoint 2-dimensional spheres 𝑆₁ and 𝑆₂, where 𝑃_𝑖 on 𝑆₁ satisfies 𝑃_𝑖𝑃₉=𝑎 and 𝑃_𝑗 on 𝑆₂ satisfies 𝑃_𝑗𝑃₉=𝑏 (consider Δ𝑃₁𝑃_𝑘𝑃₉ for 𝑘=2,…,8).
We remark that there is the distance 𝑐 in 𝑉. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂, then Δ𝑃_𝑖𝑃_𝑗𝑃₉ is scalene. Thus 𝑃_𝑖𝑃_𝑗=𝑎 or 𝑏 for any 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂. So 𝑐 is the distance between a pair of distinct points on the same 2-dimensional sphere.
If more than or equal to six points lie on one sphere, then the condition (𝑋) holds by Proposition 3.8. So we consider the following cases:
(I)Five points lie on one sphere; the other two points lie on the other sphere.(II)Four points lie on one sphere; the other three points lie on the other sphere.
We consider the case (I). We may suppose that 𝑃₂,…,𝑃₆ are on 𝑆₁ and 𝑃₇,𝑃₈ are on 𝑆₂.
If 𝑐 is the distance between a pair of distinct points on 𝑆₂, then we can show that the condition (𝑋) holds by the similar discussion as the proof of Lemma 5.1.
Thus we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁. Without loss of generality we may assume 𝑃₂𝑃₃=𝑐. For 𝑃₂,…,𝑃₆ on 𝑆₁, we consider 5-point graphs in Table 2 again. Edges in a graph represent the distance, that is, distinct from 𝑎 and 𝑏. We regard the others, (i.e., transparent edges) as the distances 𝑎 and 𝑏. Here we need not consider the graph which has no edge, because we suppose that there is at least one pair whose distance is distinct from 𝑎 and 𝑏.
We observe the cases (i)–(iv) in the proof of Proposition 4.3 similarly. In any case, we can show that the condition (𝑋) holds.
In the case (II), we can apply the similar discussion as the proof of Lemma 5.1. If we apply it, then we see that the condition (𝑋) holds.

9. Case (F) in Lemma 3.1

We consider the case (F) in Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ is (6,2,2). We may assume that 𝑃₁𝑃₂=𝑃₁𝑃₃=⋯=𝑃₁𝑃₇=𝑎, 𝑃₁𝑃₈=𝑃₁𝑃₉=𝑏, and 𝑃₁𝑃₁₀=𝑃₁𝑃₁₁=𝑐. Let 𝑋₁={𝑃₁,…,𝑃₇,𝑃₈,𝑃₉} and 𝑋₂={𝑃₁,…,𝑃₇,𝑃₁₀,𝑃₁₁}.

Proposition 9.1. For 𝑋₁ and 𝑋₂ above, at least one of them is an 𝑠-distance set (𝑠≥3).

Proof. We can show that the condition (𝑋) holds by repeating the similar discussion as the proof of Proposition 8.1.

Lemma 9.2. The condition (𝑋) holds for any 11-point isosceles set 𝒫 in which the type of 𝑃₁ is (6,2,2).

Proof. By Proposition 9.1, at least one of 𝑋₁ and 𝑋₂ is an 𝑠-distance set (𝑠≥3). We may suppose 𝑋₁ is an 𝑠-distance set.
𝑃₁ will denote the common center of the two spheres, which we will call 𝑆₁ (on which 𝑃₂,…,𝑃₇ are), 𝑆₂ (on which 𝑃₈ and 𝑃₉ are), radii 𝑎, 𝑏, respectively.
There is a distance apart of a pair of points in {𝑃₂,…,𝑃₉} which is distinct from 𝑎 and 𝑏. This is 𝑐 or 𝑑, where 𝑑 is distinct from 𝑎,𝑏, and 𝑐. We may assume that it is 𝑐. If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂, then Δ𝑃₁𝑃_𝑖𝑃_𝑗 is scalene. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑆₁,𝑃_𝑗∈𝑆₂. (9.1) So 𝑐 is the distance between a pair of distinct points on the same 3-dimensional sphere.
If 𝑐 is the distance between a pair of distinct points on 𝑆₂, then we can show that the condition (𝑋) holds by the similar discussion as the proof of Lemma 6.1.
Thus we suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁. By Proposition 3.5, six points 𝑃₂,…,𝑃₇ are on one of two disjoint 2-dimensional spheres 𝑆₁₁ and 𝑆₁₂, where 𝑃_𝑖 on 𝑆₁₁ satisfies 𝑃_𝑖𝑃₉=𝑎 and 𝑃_𝑗 on 𝑆₁₂ satisfies 𝑃_𝑗𝑃₉=𝑏 (consider Δ𝑃₁𝑃_𝑘𝑃₉ for 𝑘=2,…,7).
If 𝑃_𝑖𝑃_𝑗=𝑐 holds for some 𝑃_𝑖∈𝑆₁₁ and 𝑃_𝑗∈𝑆₁₂, then Δ𝑃_𝑖𝑃_𝑗𝑃₉ is scalene. Thus𝑃_𝑖𝑃_𝑗=𝑎or𝑏forany𝑃_𝑖∈𝑆₁₁,𝑃_𝑗∈𝑆₁₂. (9.2) So 𝑐 is the distance between a pair of distinct points on the same 2-dimensional sphere.
If six points lie on one sphere, then the condition (𝑋) holds by Proposition 3.8. So we consider the following cases:
(I)Five points lie on one sphere; the other one point lies on the other sphere.(II)Four points lie on one sphere; the other two points lie on the other sphere.(III)Three points lie on one sphere, the other three points lie on the other sphere.
As for the case (I), we can apply the similar discussion as the proof of the case (I) of Lemma 8.2. Thus the condition (𝑋) holds in the case (I).
In the case (II), we can apply the similar discussion as the proof of Lemma 6.1 in Case (C). If we apply it, then we see that the condition (𝑋) holds.
We consider the case (III). We suppose that 𝑃₂,…,𝑃₄ are on 𝑆₁₁ and 𝑃₅,…,𝑃₇ are on 𝑆₁₂.
We may suppose that 𝑐 is the distance between a pair of distinct points on 𝑆₁₂. Without loss of generality we may assume 𝑃₆𝑃₇=𝑐. Next we suppose that there exist more than or equal to two pairs of points on 𝑆₁₂ whose distances are distinct from 𝑎 and 𝑏. One is 𝑃₆𝑃₇=𝑐. Without loss of generality the second is 𝑃₅𝑃₇=𝑒 (𝑒≠𝑎, 𝑒≠𝑏, but we can admit 𝑐=𝑒).
Let 𝑃_𝑖∈𝑆₁₁∪𝑆₂ and consider Δ𝑃₆𝑃₇𝑃_𝑖. By (9.1) and (9.2), we have 𝑃₆𝑃_𝑖=𝑃₇𝑃_𝑖. When we consider Δ𝑃₅𝑃₇𝑃_𝑖 similarly, we have 𝑃₅𝑃_𝑖=𝑃₇𝑃_𝑖. For 𝑃₁ we have 𝑃₁𝑃₆=𝑃₁𝑃₇ and 𝑃₁𝑃₅=𝑃₁𝑃₇. Thus six points 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄, 𝑃₈, and 𝑃₉ are on the hyperplane perpendicularly bisecting 𝑃₆𝑃₇ and the hyperplane perpendicularly bisecting 𝑃₅𝑃₇. The intersection of them is a 2-dimensional Euclidean space. Then {𝑃₁,𝑃₂,𝑃₃,𝑃₄,𝑃₈,𝑃₉} is a 6-point isosceles set in ℝ². There exist a unique 6-point isosceles set in ℝ² up to isomorphisms and it contains four points on a circle. Thus the condition (𝑋) holds.
Hence we suppose that there is exactly one pair 𝑃₆𝑃₇ whose distance is distinct from 𝑎 and 𝑏 on 𝑆₁₂. If we repeat the similar discussion above, then there is also at most one pair whose distance is distinct from 𝑎 and 𝑏 on 𝑆₁₁. Without loss of generality this is 𝑃₂𝑃₃.
When we consider Δ𝑃₆𝑃₇𝑃_𝑘 for 𝑘=1,…,5,8,9, 𝑃₆𝑃_𝑘=𝑃₇𝑃_𝑘 holds by (9.1), (9.2), and the configuration hypothesis. Thus seven points 𝑃₁,…,𝑃₅,𝑃₈, and 𝑃₉ are on the hyperplane perpendicularly bisecting 𝑃₆𝑃₇. This hyperplane is a 3-dimensional Euclidean space. Particularly {𝑃₁,𝑃₃,𝑃₄,𝑃₅,𝑃₈,𝑃₉} is a 6-point 2-distance set in ℝ³ with distances 𝑎 and 𝑏. There exist exactly six 6-point 2-distance sets in ℝ³. Any set contains four points lying on a circle. Hence the condition (𝑋) holds.
Therefore if the type of 𝑃₁ is (6,2,2), then the condition (𝑋) holds.

10. Case (G) in Lemma 3.1

We consider the case (G) in Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ is (6,2,1,1). We may assume that 𝑃₁𝑃₂=𝑃₁𝑃₃=⋯=𝑃₁𝑃₇=𝑎, 𝑃₁𝑃₈=𝑃₁𝑃₉=𝑏, 𝑃₁𝑃₁₀=𝑐, and 𝑃₁𝑃₁₁=𝑑. Let 𝑋₁={𝑃₁,…,𝑃₇,𝑃₈,𝑃₉}, 𝑋₂={𝑃₁,…,𝑃₇,𝑃₁₀}, and 𝑋₃={𝑃₁,…,𝑃₇,𝑃₁₁}.

Proposition 10.1. For 𝑋₁,…,𝑋₃ above, if 2-distance sets exist, then the number of them is at most one.

Proof. We can show this proposition by repeating the similar discussion as Proposition 8.1.

Lemma 10.2. The condition (𝑋) holds for any 11-point isosceles set 𝒫 in which the type of 𝑃₁ is (6,2,1,1).

Proof. By Proposition 10.1, at least two sets of 𝑋₁,…,𝑋₃ are 𝑠-distance sets (𝑠≥3). If 𝑋₁ is an 𝑠-distance set, then we can show that the condition (𝑋) holds by repeating the similar discussion as Lemma 9.2. Hence we may assume that 𝑋₁ is a 2-distance set and that 𝑋₂ and 𝑋₃ are 𝑠-distance sets. Since |𝐴({𝑃₂,…,𝑃₇})|≥2 and 𝑋₁ is a 2-distance set with distances 𝑎 and 𝑏, it holds that 𝐴𝑃₂,…,𝑃₇={𝑎,𝑏}. (10.1) Thus 𝑏 is the third distance in 𝑋₂ and 𝑋₃.
Let 𝑆 be the sphere centered at 𝑃₁ with radius 𝑎. By Proposition 3.5, six points 𝑃₂,…,𝑃₇ are on one of two disjoint 2-dimensional spheres 𝑆₁ and 𝑆₂, where 𝑃_𝑖 on 𝑆₁ satisfies 𝑃_𝑖𝑃₁₀=𝑎 and 𝑃_𝑗 on 𝑆₂ satisfies 𝑃_𝑗𝑃₁₀=𝑐 (consider Δ𝑃₁𝑃_𝑘𝑃₁₀ for 𝑘=2,…,7).
We remark that there is the distance 𝑏 in {𝑃₂,…,𝑃₇}. If 𝑃_𝑖𝑃_𝑗=𝑏 holds for some 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂, then Δ𝑃_𝑖𝑃_𝑗𝑃₁₀ is scalene. Thus 𝑃_𝑖𝑃_𝑗=𝑎 or 𝑐 for any 𝑃_𝑖∈𝑆₁ and 𝑃_𝑗∈𝑆₂. Combining this and (10.1), the following condition holds:𝑃_𝑖𝑃_𝑗=𝑎forany𝑃_𝑖∈𝑆₁,𝑃_𝑗∈𝑆₂. (10.2) So 𝑏 is the distance between a pair of distinct points on the same 2-dimensional sphere.
If six points lie on one sphere, then the condition (𝑋) holds by Proposition 3.8. So we consider the following cases.
(I)Five points lie on one sphere; the other one point lies on the other sphere.(II)Four points lie on one sphere; the other two points lie on the other sphere.(III)Three points lie on one sphere, the other three points lie on the other sphere.
As for the case (I), we can apply the similar discussion as the proof of the case (I) of Lemma 8.2. Thus the condition (𝑋) holds in the case (I).
In the case (II), we can apply the similar discussion as the proof of Lemma 6.1. If we apply it, then we see that the condition (𝑋) holds.
We consider the case (III). We suppose that 𝑃₂,…,𝑃₄ are on 𝑆₁ and 𝑃₅,…,𝑃₇ are on 𝑆₂.
We may suppose that 𝑏 is the distance between a pair of distinct points on 𝑆₂. Without loss of generality we may assume 𝑃₆𝑃₇=𝑏. Next we suppose that there exist more than or equal to two pairs of points on 𝑆₂ whose distances are 𝑏. In this assumption, we can apply the similar discussion as the proof of the case (III) of Lemma 9.2. If we apply it, then we see that the condition (𝑋) holds.
Hence we suppose that there is exactly one pair 𝑃₆𝑃₇ whose distance is 𝑏 on 𝑆₂. If we repeat the similar discussion above, then there is also at most one pair whose distance is 𝑏 on 𝑆₁. Without loss of generality this is 𝑃₂𝑃₃.
When we consider Δ𝑃₆𝑃₇𝑃_𝑘 for 𝑘=2,…,5, 𝑃₆𝑃_𝑘=𝑃₇𝑃_𝑘 holds by (10.2) and the configuration hypothesis. Thus 𝑃₂,…,𝑃₅ are on the hyperplane perpendicularly bisecting 𝑃₆𝑃₇ and on 𝑆. The intersection of them is a 2-dimensional sphere. By (10.1), (10.2), and the assumption, 𝑃₂𝑃_𝑖=𝑃₃𝑃_𝑖=𝑃₄𝑃_𝑖=𝑃₅𝑃_𝑖=𝑎 for 𝑖=1,6,7. Thus 𝑃₁,𝑃₆, and 𝑃₇ are equidistant from 𝑃₂,…,𝑃₅ on a 2-dimensional sphere. If 𝑃₂,…,𝑃₅ are on a plane, then they are on a circle; the condition (𝑋) holds. On the other hand, if 𝑃₂,…,𝑃₅ are not on a plane, then 𝑃₁,𝑃₆, and 𝑃₇ are on a line. By Corollary 3.7, this is a contradiction.
Therefore if the type of 𝑃₁ is (6,2,1,1), then the condition (𝑋) holds.

11. Case (H) in Lemma 3.1

We consider the case (H) in Lemma 3.1. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set in which the type of 𝑃₁ is (6,1,1,1,1). We may assume that 𝑃₁𝑃₂=𝑃₁𝑃₃=⋯=𝑃₁𝑃₇=𝑎, 𝑃₁𝑃₈=𝑏, 𝑃₁𝑃₉=𝑐, 𝑃₁𝑃₁₀=𝑑, and 𝑃₁𝑃₁₁=𝑒.

We consider the sum of all vertex-numbers in 𝒫. Since 𝑃₁ has the largest vertex-number in 𝒫, 𝑉(𝑃₁)+⋯+𝑉(𝑃₁₁)≤11×{⁶₂+¹₂+¹₂+¹₂+¹₂}=165. On the other hand, 𝑉(𝑃₁)+⋯+𝑉(𝑃₁₁)≥165 by (3.1). Thus 𝑉(𝑃₁)+⋯+𝑉(𝑃₁₁)=165. Let 𝛼be the number of regular triangles in 𝒫. Then 𝛼=0 holds by (3.2). Moreover 𝑉(𝑃_𝑖)=15 holds for any 𝑃_𝑖∈𝒫; the type of 𝑃_𝑖 is (6,1,1,1,1).

Lemma 11.1. There is no 11-point isosceles set in which the type of 𝑃₁ is (6,1,1,1,1).

Proof. We notice that the type of 𝑃₂ is (6,1,1,1,1). So the distance 𝑎 corresponds to 6 or 1 of type (6,1,1,1,1). If 𝑎 corresponds to 6, then at least one of 𝑃₂𝑃₃,…,𝑃₂𝑃₇ is 𝑎. We may suppose that 𝑃₂𝑃₃=𝑎. Then Δ𝑃₁𝑃₂𝑃₃ is a regular triangle with the distance 𝑎. This contradicts 𝛼=0. Thus 𝑎 corresponds to 1. Then 𝑃₂𝑃₈=𝑏, 𝑃₂𝑃₉=𝑐, 𝑃₂𝑃₁₀=𝑑, and 𝑃₂𝑃₁₁=𝑒 hold by considering Δ𝑃₁𝑃₂𝑃_𝑖 for 𝑖=8,…,11. This means that one of 𝑏,𝑐,𝑑, and 𝑒 corresponds to 6 of type (6,1,1,1,1). We may assume that this is 𝑏. Then 𝑃₂𝑃₃=⋯=𝑃₂𝑃₈=𝑏.
Next we notice that the type of 𝑃₃ is (6,1,1,1,1). We see that 𝑎 corresponds to 1 of type (6,1,1,1,1) by repeating the discussion for 𝑃₂. Thus 𝑃₃𝑃₈=𝑏, 𝑃₃𝑃₉=𝑐, 𝑃₃𝑃₁₀=𝑑, and 𝑃₃𝑃₁₁=𝑒 hold by considering Δ𝑃₁𝑃₃𝑃_𝑖 for 𝑖=8,…,11, 𝑏 corresponds to 6 of type (6,1,1,1,1). Then 𝑃₂𝑃₃=𝑃₃𝑃₄=⋯=𝑃₃𝑃₈=𝑏. But Δ𝑃₂𝑃₃𝑃₄ is a regular triangle with the distance 𝑏. This contradicts 𝛼=0.
Therefore there is no 11-point isosceles set in which the type of 𝑃₁ is (6,1,1,1,1).

Therefore combining Lemmas 3.1, 4.8, 5.1, 6.1, 7.1, 8.2, 9.2, 10.2, and 11.1, we have Lemma 3.2.

By Lemma 3.2, at least four points, say 𝑃₁,…,𝑃₄, in 𝒫 lie on a circle. We keep to this notation of suffixes in what follows. Lemma 11.2 can be proved by the same method given in the proof of Lemma 18 in Croft [3].

Lemma 11.2. 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄ are either all the vertices of a square, or four of the vertices of a regular pentagon.

From now on, we observe two cases in Lemma 11.2 respectively.

12. Observation of 11-Point Isosceles Sets in ℝ⁴ Containing Four Points of a Regular Pentagon

Proposition 12.1. Suppose an 𝑛-point isosceles set 𝒫={𝑃₁,…,𝑃_𝑛} contains four vertices of a regular pentagon, 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄ (in order, with the “gap” between 𝑃₄ and 𝑃₁). We may suppose that 𝑃₁=((−1−√5)/4,(10+2√5)/4,0,0), 𝑃₂=(−1/2,0,0,0), 𝑃₃=(1/2,0,0,0), 𝑃₄=((1+√5)/4,(10+2√5)/4,0,0). (The mid-point of 𝑃₂𝑃₃ is the origin. Each side of this regular pentagon is 1.)
Then the only other possible coordinates for the remaining points are as follows:
(i)(0,(3+√5)/2(10+2√5),𝑧,𝑤), where 𝑧 and 𝑤 are arbitrary,(ii)𝑇=(0,(5+3√5)/2(10+2√5),0,0), the remaining vertex of the pentagon, or(iii)(0,(1−√5)/2(10+2√5),𝑧,𝑤), where 𝑧 and 𝑤 satisfy 𝑧²+𝑤²=((10+2√5)/2√5)².

Proof. We expand the proof of Lemma 22 in Croft [3] into ℝ⁴, then we obtain this proposition.

Proposition 12.2. Let 𝑄 be a point satisfying (iii) in the previous proposition. Then no 𝑛-point isosceles set can contain 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄, 𝑇, and 𝑄.

Proof. It holds that 𝑄𝑇=(10+2√5)/2. Then Δ𝑃₁𝑄𝑇 is scalene with 1, (1+√5)/2, (10+2√5)/2. This is contrary to the configuration hypothesis.
Therefore no 𝑛-point isosceles set can contain 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄, 𝑇, and 𝑄.

We observe the detail for 𝑛=11 in Proposition 12.1. The space which satisfies the case (i) in Proposition 12.1 is a plane and that satisfying the case (iii) in Proposition 12.1 is a circle. The maximum cardinality of isosceles sets in ℝ² is 6 and we see that that on a circle is 5. We consider them and Proposition 12.2. If an 11-point isosceles set exists, then it satisfies one row of Table 3.

Table 3 

The distribution of the remaining points.

Proposition 12.3. Any 11-point isosceles set in ℝ⁴ satisfying ⟨1⟩ in Table 3 is isomorphic to 𝑌 in Theorem 1.1.

Proof. Any 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} in ℝ⁴ satisfying ⟨1⟩ in Table 3 contains all the vertices of a regular pentagon. And the other six points are in a 2-dimensional Euclidean space. Then they are all the vertices of a regular pentagon and its center. Hence we can fix 𝑃₁,𝑃₂,𝑃₃,𝑃₄,𝑃₅=𝑇, and 𝑃₆=(0,(3+√5)/2(10+2√5),0,0), we consider the configuration of the other five points which form a regular pentagon in the 2-dimensional Euclidean space 𝑥=0, 𝑦=(3+√5)/2(10+2√5).
Let 𝑃_𝑖=(0,(3+√5)/2(10+2√5),𝑧,𝑤) for 𝑖∈{7,…,11}. We consider Δ𝑃₅𝑃₆𝑃_𝑖, we have (𝑃₅𝑃₆)²=((1+√5)/(10+2√5))², (𝑃₆𝑃_𝑖)²=𝑧²+𝑤²(>0), and (𝑃₅𝑃_𝑖)²=(𝑃₅𝑃₆)²+(𝑃₆𝑃_𝑖)². Since 𝑃₅𝑃_𝑖>𝑃₅𝑃₆ and 𝑃₅𝑃_𝑖>𝑃₆𝑃_𝑖, 𝑃₅𝑃₆=𝑃₆𝑃_𝑖 holds by the configuration hypothesis. Thus 𝑃₇,…,𝑃₁₁ which form a regular pentagon are on the circle satisfying 𝑧²+𝑤²=((1+√5)/(10+2√5))². This 11-point isosceles set 𝒫 is isomorphic to 𝑌 in Theorem 1.1.

Next we observe ⟨5⟩ and ⟨6⟩ in Table 3. For any 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} in ℝ⁴ satisfying ⟨5⟩ or ⟨6⟩ in Table 3, the other seven points 𝑃₅,…,𝑃₁₁ are in the 3-dimensional Euclidean space 𝑥=0, and four points in {𝑃₅,…,𝑃₁₁} are on a circle. We may assume that they are 𝑃₅,…,𝑃₈. Then they are all the vertices of a square, or four points of a regular pentagon. Moreover we may assume that 𝑃₁₀ and 𝑃₁₁ are in the 2-dimensional Euclidean space 𝑥=0, 𝑦=(3+√5)/2(10+2√5).

The following two propositions in ℝ³ are useful for us. We quote them from Kido [11] (or Croft [3]).

Proposition 12.4. Let four points 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄ of an 𝑛-point isosceles set in ℝ³ form a square. We may suppose that 𝑃₁=(−1/2,−1/2,0), 𝑃₂=(1/2,−1/2,0), 𝑃₃=(1/2,1/2,0), 𝑃₄=(−1/2,1/2,0). And let the center (0,0,0) be 𝑂, and let the plane that contains the square be Π.
Then the only other possible situations for the remaining points are:
(i)on the vertical line 𝐿 through 𝑂, or(ii)at some of 𝑄₁,…,𝑄₈, where 𝑄₁=10,−2,√32,𝑄₂=12√,0,32,𝑄₃=10,2,√32,𝑄₄=−12√,0,32,𝑄₅=10,−2√,−32,𝑄₆=12√,0,−32,𝑄₇=10,2√,−32,𝑄₈=−12√,0,−32. (12.1) (The square 𝑄₁, 𝑄₂, 𝑄₃, 𝑄₄, and 𝑄₅, 𝑄₆, 𝑄₇, 𝑄₈ both have sides of length √2/2.)

Proposition 12.5. Suppose an 𝑛-point isosceles set in ℝ³ contains four vertices of a regular pentagon, 𝑃₁, 𝑃₂, 𝑃₃, 𝑃₄ (in order, with the “gap” between 𝑃₄ and 𝑃₁), lying in a horizontal plane. Then the only other possible situations for the remaining points are:(i)at 𝑇 the remaining vertex of the pentagon; or(ii)at two points 𝑄₁, 𝑄₂, which are the only points 𝑄 such that Δ𝑄𝑃₄𝑃₁ and Δ𝑄𝑃₂𝑃₃ are both equilaternal; or(iii)on the vertical line 𝐿 through the center of the pentagon.

Proposition 12.6. Any 11-point isosceles set in ℝ⁴ satisfies neither ⟨5⟩ nor ⟨6⟩ in Table 3.

Proof. We consider when 𝑃₅,…,𝑃₈ form a square. If each side of the square 𝑃₁𝑃₂𝑃₃𝑃₄ in Proposition 12.4 is (10+2√5)/2√5×√2=(5+√5)/√5, then we can change them into 𝑃₅,…,𝑃₈. Now 𝑃₁₀ and 𝑃₁₁ are in 𝑥=0,𝑦=(3+√5)/2(10+2√5). By Proposition 12.4, we see that there is exactly one point (0,(3+√5)/2(10+2√5),0,0) in 𝑥=0, 𝑦=(3+√5)/2(10+2√5). Thus we cannot put one of 𝑃₁₀ and 𝑃₁₁. This is a contradiction.
On the other hand, we consider when 𝑃₅,…,𝑃₈ form four points of a regular pentagon. Here 𝑃₁₀ and 𝑃₁₁ are in 𝑥=0,𝑦=(3+√5)/2(10+2√5). By Proposition 12.5, we see that there is exactly one point (0,(3+√5)/2(10+2√5),0,0) in 𝑥=0,𝑦=(3+√5)/2(10+2√5). Thus we cannot put one of 𝑃₁₀ and 𝑃₁₁. This is a contradiction.
Therefore any 11-point isosceles set in ℝ⁴ satisfies neither ⟨5⟩ nor ⟨6⟩ in Table 3.

The last cases are ⟨2⟩-⟨4⟩ in Table 3. For any 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} in ℝ⁴ satisfying one of ⟨2⟩-⟨4⟩ in Table 3, we may assume that 𝑃₅ satisfies (iii) in Proposition 12.1 and that 𝑃₈,…,𝑃₁₁ satisfy (i) in Proposition 12.1. We may suppose that 𝑃₅=(0,(1−√5)/2(10+2√5),0,(10+2√5)/2√5) because of symmetry. We remark that 𝑃₅,…,𝑃₁₁ are in the 3-dimensional Euclidean space 𝑥=0.

Proposition 12.7. Suppose that an 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} in ℝ⁴ satisfying one of ⟨2⟩-⟨4⟩ in Table 3, then the possible situations for 𝑃₈,…,𝑃₁₁ are the following:(I)on the line 𝐿 which satisfies 𝑤=(10+2√5)/4√5,(II)𝑅₁=(0,(3+√5)/2(10+2√5),0,(1/20)(−510+2√5+50+10√5), or(III)𝑅₂=(0,(3+√5)/210+2√5,0,(1/20)(510+2√5+50+10√5).

Proof. For 𝑖=8,…,11, let 𝑃_𝑖=(0,(3+√5)/210+2√5,𝑧,𝑤). We consider Δ𝑃₂𝑃₅𝑃_𝑖. Because 𝑃₂𝑃₅=1, one of the following (a-1)–(a-3) must hold to satisfy the configuration hypothesis:(a-1)𝑧²+𝑤²=(5−√5)/10 when 𝑃₂𝑃_𝑖=1,(a-2)((1+√5)/(10+2√5))²+𝑧²+(𝑤−(10+2√5)/2√5)²=1 when 𝑃₅𝑃_𝑖=1,(a-3)𝑤=(10+2√5)/4√5 when 𝑃₂𝑃_𝑖=𝑃₅𝑃_𝑖.
On the other hand, we consider Δ𝑃₁𝑃₅𝑃_𝑖. Since 𝑃₁𝑃₅=(1+√5)/2, one of the following (b-1)–(b-3) must hold to satisfy the configuration hypothesis:
(b-1)𝑧²+𝑤²=(5+2√5)/5 when 𝑃₁𝑃_𝑖=(1+√5)/2,(b-2)((1+√5)/(10+2√5))²+𝑧²+(𝑤−(10+2√5)/2√5)²=(3+√5)/2 when 𝑃₅𝑃_𝑖=(1+√5)/2,(b-3)𝑤=(10+2√5)/4√5 when 𝑃₁𝑃_𝑖=𝑃₅𝑃_𝑖.
Hence combining one of (a-1)–(a-3) and one of (b-1)–(b-3), we see that the possible situations for 𝑃_𝑖 must be in the list of the proposition.

Proposition 12.8. Any 11-point isosceles set in ℝ⁴ cannot satisfy one of ⟨2⟩-⟨4⟩ in Table 3.

Proof. The previous proposition implies that 𝑃₈,…,𝑃₁₁ satisfy one of the following conditions:(i)four points on 𝐿,(ii)three points on 𝐿 and the other is one of 𝑅₁ and 𝑅₂, (iii)two points on 𝐿 and the others are 𝑅₁ and 𝑅₂.
In the case (i), we cannot take four points on a line. This is a contradiction. In the case (ii), three collinear points are contained. By Corollary 3.7, this is a contradiction. In the case (iii), Δ𝑃₅𝑅₁𝑅₂ is scalene with 1, (1+√5)/2, 1+((1+√5)/2)². This is contrary to the configuration hypothesis.
Therefore any 11-point isosceles set in ℝ⁴ cannot satisfy one of ⟨2⟩-⟨4⟩ in Table 3.

Thus we have the following lemma.

Lemma 12.9. There exists a unique 11-point isosceles set in ℝ⁴ containing four vertices of a regular pentagon. This is 𝑌 in Theorem 1.1.

13. Observation of 11-Point Isosceles Sets in ℝ⁴ Containing a Square

Proposition 13.1. Let 𝑃₁,𝑃₂,𝑃₃,𝑃₄ in an 𝑛-point isosceles set 𝒫={𝑃₁,…,𝑃_𝑛} form a square. We may suppose that 𝑃₁=(−1/2,−1/2,0,0),𝑃₂=(1/2,−1/2,0,0),𝑃₃=(1/2,1/2,0,0),𝑃₄=(−1/2,1/2,0,0).
Then the only other possible coordinates for the remaining points are
(i)(0,0,𝑧,𝑤), where 𝑧 and 𝑤 are arbitrary, or(ii)one of (0,−1/2,𝑧,𝑤), (1/2,0,𝑧,𝑤), (0,1/2,𝑧,𝑤), and (−1/2,0,𝑧,𝑤), where 𝑧 and 𝑤 satisfy 𝑧²+𝑤²=3/4.

Proof. We expand the proof of Lemma 19 in Croft [3] into ℝ⁴, then we obtain this proposition.

We observe the detail for 𝑛=11 in Proposition 13.1. The space which satisfies the case (i) in Proposition 13.1 is a plane. The maximum cardinality of isosceles sets in ℝ² is 6. Hence if an 11-point isosceles set exists, then it satisfies one row of Table 4.

Table 4 

The distribution of the remaining points.

We observe ⟨1⟩-⟨3⟩ in Table 4. We see that another point 𝑃_𝑖 of an 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} which satisfies (ii) in Proposition 13.1 is on one of four circles.

Let 𝑆₁ be 𝑥=0, 𝑦=−1/2, 𝑧²+𝑤²=3/4,𝑆₂ be 𝑥=1/2, 𝑦=0, 𝑧²+𝑤²=3/4, 𝑆₃ be 𝑥=0, 𝑦=1/2, 𝑧²+𝑤²=3/4, and 𝑆₄ be 𝑥=−1/2, 𝑦=0, 𝑧²+𝑤²=3/4. We remark that 𝑆₁ and 𝑆₃ are the subsets of the 3-dimensional Euclidean space 𝑥=0, and 𝑆₂ and 𝑆₄ are the subsets of the 3-dimensional Euclidean space 𝑦=0.

When 𝒫 satisfies one of ⟨1⟩-⟨3⟩ in Table 4, the remaining at least five points are distributed on some of 𝑆₁,…,𝑆₄. If they are distributed on one circle, then they form a regular pentagon. By Lemma 12.9, such any 11-point isosceles set is isomorphic to 𝑌 in Theorem 1.1.

Hence we may suppose that they are distributed on more than or equal to two circles. We may assume that we choose 𝑆₁ as the first circle because of symmetry. Now we separate the choice of the second circle into two cases whether 𝑆_𝑖 is the subset of the 3-dimensional Euclidean space 𝑥=0 or not for 𝑖=2,3,4. So one is 𝑆₃, the other is 𝑆₂ or 𝑆₄.

Proposition 13.2. One considers the first case above. One fixes a point 𝑃_𝑖 on 𝑆₁. Then the possible situations for the points on 𝑆₃ are at most three. Moreover the distance between a pair of distinct points from these three points must be 1 or 2√6/3.

Proof. We may assume that 𝑃_𝑖=(0,−1/2,√3/2,0) because 𝑆₁ and 𝑆₃ are on the 3-dimensional Euclidean space 𝑥=0 and we have only to investigate the relation between the points on 𝑆₁ and those on 𝑆₃. Let 𝑃_𝑗=(0,1/2,𝑧,𝑤) on 𝑆₃, where 𝑧²+𝑤²=3/4. We consider Δ𝑃₁𝑃_𝑖𝑃_𝑗. Since 𝑃₁𝑃_𝑖=1 and 𝑃₁𝑃_𝑗=√2, we have 𝑃_𝑖𝑃_𝑗=1 or √2. When 𝑃_𝑖𝑃_𝑗=1, 𝑃_𝑗 is (0,1/2,√3/2,0). When 𝑃_𝑖𝑃_𝑗=√2, 𝑃_𝑗 is (0,1/2,√3/6,√6/3) or (0,1/2,√3/6,−√6/3).
Therefore the possible situations for 𝑃_𝑗 are at most three. Moreover we see easily that the distance between a pair of distinct points from these three points must be 1 or 2√6/3.

We consider the other case. We may suppose that the choice of the second circle is 𝑆₂ because of symmetry.

Proposition 13.3. One considers that the choice of the second circle is 𝑆₂. One fixes a point 𝑃_𝑖=(0,−1/2,𝑧_𝑖,𝑤_𝑖) on 𝑆₁, where 𝑧²_𝑖+𝑤²_𝑖=3/4. Then the possible situations for the points on 𝑆₂ are at most two. And the distance between the two points must be one of √3, √15/3, (√5+1)/2, and (√5−1)/2.

Proof. Let 𝑃_𝑗=(1/2,0,𝑧,𝑤) on 𝑆₂, where 𝑧²+𝑤²=3/4. We consider Δ𝑃₁𝑃_𝑖𝑃_𝑗. Since 𝑃₁𝑃_𝑖=1 and 𝑃₁𝑃_𝑗=√2, we have 𝑃_𝑖𝑃_𝑗=1 or √2. Then (𝑧−𝑧_𝑖)²+(𝑤−𝑤_𝑖)²=1/2 or (𝑧−𝑧_𝑖)²+(𝑤−𝑤_𝑖)²=3/2 holds. From them, we have 𝑧_𝑖𝑧+𝑤_𝑖𝑤−1/2=0 or 𝑧_𝑖𝑧+𝑤_𝑖𝑤=0. And 𝑧 and 𝑤 satisfy 𝑧²+𝑤²=3/4. Hence the possible situations for 𝑃_𝑗 are at most four.
Since (1/2,0,0,0) is on 𝑧_𝑖𝑧+𝑤_𝑖𝑤=0, the distance between (1/2,0,0,0) and 𝑧_𝑖𝑧+𝑤_𝑖𝑤−1/2=0 is |−1/2|/𝑧²_𝑖+𝑤²_𝑖=1/2/√3/2=√3/3 in spite of the way to fix 𝑃_𝑖. So let 𝑄₁,𝑄₂,𝑄₃,𝑄₄ be the four possible points for 𝑃_𝑗, the distances between 𝑄-points are in Figure 5 in spite of the way to fix 𝑃_𝑖. Looking at Figure 5, all triangles that we choose from 𝑄-points are scalene.
Therefore if we fix a point 𝑃_𝑖 on 𝑆₁, then the possible situations for the points on 𝑆₂ are at most two. Moreover we see easily that the distance between the two points must be one of √3, √15/3, (√5+1)/2, and (√5−1)/2 by Figure 5.

Figure 5 

For the supposition of Proposition 13.3, moreover we suppose that there is a point on 𝑆₄, too. Then the distance between two points on 𝑆₂ must be 1 or 2√6/3 by an analogue of Proposition 13.2, we take at most one point on 𝑆₂. Thus we see that we cannot take five points satisfying (ii) in Proposition 13.1, we have the following proposition.

Proposition 13.4. Any 11-point isosceles set in ℝ⁴ cannot satisfy one of ⟨1⟩-⟨3⟩ in Table 4.

Next we observe ⟨5⟩-⟨7⟩ in Table 4. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set. We suppose that 𝑃₅ is on some of 𝑆₁,…,𝑆₄ and that 𝑃₈,…,𝑃₁₁ are on the plane 𝑥=𝑦=0. We may assume that 𝑃₅ is on 𝑆₁ because of symmetry. So 𝑃₅ is one of (0,−1/2,𝑎,3/4−𝑎²) and (0,−1/2,𝑎,−3/4−𝑎²), where −√3/2≤𝑎≤√3/2. We can choose that 𝑃₅=(0,−1/2,𝑎,3/4−𝑎²). (For the latter we can repeat the similar discussion.)

Proposition 13.5. If an 11-point isosceles set 𝒫={𝑃₁,…,𝑃₁₁} contains a square satisfying one of ⟨5⟩-⟨7⟩ in Table 4, then the possible situations for 𝑃₈,…,𝑃₁₁ are as follows:(I)on the line 𝐿 which satisfies 𝑎𝑧+𝑤3/4−𝑎²=1/4, or(II)at some of 𝑅₁,…,𝑅₄, where 𝑅₁=0,0,−2𝑎−53−4𝑎²6,2√5𝑎−√3−4𝑎²6,𝑅₂=0,0,−2𝑎+53−4𝑎²6,−2√5𝑎−√3−4𝑎²6,𝑅₃=√0,0,2𝑎−3−4𝑎²2,2𝑎+√3−4𝑎²2,𝑅₄=√0,0,2𝑎+3−4𝑎²2,−2𝑎+√3−4𝑎²2. (13.1)

Proof. For 𝑖=8,…,11, let 𝑃_𝑖=(0,0,𝑧,𝑤). We consider Δ𝑃₁𝑃₅𝑃_𝑖. Because 𝑃₁𝑃₅=1, one of the following (a-1)–(a-3) must hold to satisfy the configuration hypothesis:(a-1)𝑧²+𝑤²=1/2 when 𝑃₁𝑃_𝑖=1,(a-2)(𝑧−𝑎)²+(𝑤−3/4−𝑎²)²=3/4 when 𝑃₅𝑃_𝑖=1,(a-3)𝑎𝑧+𝑤3/4−𝑎²=1/4 when 𝑃₁𝑃_𝑖=𝑃₅𝑃_𝑖.
On the other hand, we consider Δ𝑃₃𝑃₅𝑃_𝑖. Since 𝑃₃𝑃₅=√2, one of the following (b-1)–(b-3) must hold to satisfy the configuration hypothesis:
(b-1)𝑧²+𝑤²=3/2 when 𝑃₃𝑃_𝑖=√2,(b-2)(𝑧−𝑎)²+(𝑤−3/4−𝑎²)²=7/4 when 𝑃₅𝑃_𝑖=√2,(b-3)𝑎𝑧+𝑤3/4−𝑎²=1/4 when 𝑃₃𝑃_𝑖=𝑃₅𝑃_𝑖.
Hence conbining one of (a-1)–(a-3) and one of (b-1)–(b-3), we see that the possible situations for 𝑃_𝑖 must be in the list of the proposition.

Proposition 13.6. Any 11-point isoseles set in ℝ⁴ cannot satisfy one of ⟨5⟩-⟨7⟩ in Table 4.

Proof. By Proposition 13.5, four points 𝑃₈,…,𝑃₁₁ are in the list of the proposition. We observe (II) in Proposition 13.5. Because 𝑅₁𝑅₂=√15/3, 𝑅₁𝑅₃=𝑅₂𝑅₄=(5−√5)/2, 𝑅₁𝑅₄=𝑅₂𝑅₃=(5+√5)/2, and 𝑅₃𝑅₄=√3, any triangle selected from 𝑅-points is scalene. So we choose at most two 𝑅-points. On the other hand, we observe (I) in Proposition 13.5. Since 𝐿 is a line, we choose at most three points on 𝐿. By Corollary 3.7, we cannot choose three points. So we choose at most two points on 𝐿. Hence we must choose two 𝑅-points and two points on 𝐿 for 𝑃₈,…,𝑃₁₁. Let 𝑃₈, 𝑃₉ be two 𝑅-points and 𝑃₁₀, 𝑃₁₁ be two points on 𝐿. For each choice of two 𝑅-points, we see that the possible situations for 𝑃₁₀ and 𝑃₁₁ are five by considering Δ𝑃₈𝑃₉𝑃₁₀ and Δ𝑃₈𝑃₉𝑃₁₁ and the calculations.
The number of the choices of 𝑃₈ and 𝑃₉ is ⁴₂=6 and the number of the choices of 𝑃₁₀ and 𝑃₁₁ is ⁵₂=10. Thus the number of the choices of 𝑃₈,…,𝑃₁₁ is 6×10=60. We have only to check 60 cases whether 𝑃₁,…,𝑃₅,𝑃₈,…,𝑃₁₁ form an isosceles set or not. But for all cases we see that they contain a scalene by the calculations.
Therefore any 11-point isoseles set cannot satisfy one of ⟨5⟩-⟨7⟩ in Table 4.

Finally we observe ⟨4⟩ in Table 4. Let 𝒫={𝑃₁,…,𝑃₁₁} be an 11-point isosceles set. Four points 𝑃₅,…,𝑃₈ lie on some of 𝑆₁,…,𝑆₄. We may assume that one point 𝑃₅=(0,−1/2,√3/2,0) because of symmetry. 𝑃₉,…,𝑃₁₁ are in the plane 𝑥=𝑦=0.

Proposition 13.7. If there exists an 11-point isosceles set in ℝ⁴ satisfying ⟨4⟩ in Table 4, then it is isomorphic to 𝑋 or 𝑌 in Theorem 1.1.

Proof. If 𝑃₅,…,𝑃₈ are distributed on 𝑆₁, then they are all points of a square or four points of a regular pentagon.
We consider when 𝑃₅,…,𝑃₈ form a square. If each side of the square 𝑃₁𝑃₂𝑃₃𝑃₄ in Proposition 12.4 is √3/2×√2=√6/2, then we can change them into 𝑃₅,…,𝑃₈ in the 3-dimensional Euclidean space 𝑥=0. Now the other points 𝑃₉, 𝑃₁₀, 𝑃₁₁ are in the 2-dimensional Euclidean space 𝑥=𝑦=0. By Proposition 12.4, we see that there is exactly one point (0,0,0,0) in 𝑥=𝑦=0. Thus we cannot take three points in 𝑥=𝑦=0. This is a contradiction. Hence 𝑃₅,…,𝑃₈ do not form a square.
If 𝑃₅,…,𝑃₈ form four points of a regular pentagon, then such any 11-point isosceles set is isomorphic to 𝑌 in Theorem 1.1 by Lemma 12.9.
Hence they are distributed on more than or equal to two circles. By the proof of Proposition 13.2, the number of the possible points on 𝑆₃ for 𝑃₆, 𝑃₇, 𝑃₈ is at most three. They are 𝑈₁=(0,1/2,√3/2,0), 𝑈₂=(0,1/2,√3/6,√6/3), and 𝑈₃=(0,1/2,√3/6,−√6/3). By the proof of Proposition 13.3, the number of the possible points on 𝑆₂ for 𝑃₆, 𝑃₇, 𝑃₈ is at most four. They are 𝑈₄=(1/2,0,√3/3,√15/6), 𝑈₅=(1/2,0,√3/3,−√15/6), 𝑈₆=(1/2,0,0,√3/2), and 𝑈₇=(1/2,0,0,−√3/2). Similarly the number of the possible points on 𝑆₄ for 𝑃₆, 𝑃₇, 𝑃₈ is at most four by the proof of Proposition 13.3. They are 𝑈₈=(−1/2,0,√3/3,√15/6), 𝑈₉=(−1/2,0,√3/3,−√15/6), 𝑈₁₀=(−1/2,0,0,√3/2), and 𝑈₁₁=(−1/2,0,0,−√3/2).
If we apply Proposition 13.5 to 𝑃₅, then 𝑃₉, 𝑃₁₀, and 𝑃₁₁ in the plane 𝑥=𝑦=0 must satisfy one of the following situations:(I)√onthelinewhichsatisfies𝑧=36,()√IIatsomeof0,0,−36,√156,√0,0,−36√,−156,√0,0,32,√32,√0,0,32√,−32. (13.2)We apply Proposition 13.5 and its analogue to 𝑈-points, we have only to check whether there exist three points in 𝑥=𝑦=0 which satisfy (13.2) or not. When we take 𝑈₁, 𝑈₄, and 𝑈₈, there exist three points (0,0, √3/6, √15/30), (0,0, −√3/6, √15/6), and (0,0, √3/6, −√15/6) satisfying (13.2). Then {𝑃₁,…,𝑃₅,𝑈₁,𝑈₄,𝑈₈, (0,0, √3/6, √15/30), (0,0, −√3/6, √15/6), (0,0, √3/6, −√15/6)} is an 11-point isosceles set which is isomorphic to 𝑋 in Theorem 1.1. Similarly when we take 𝑈₁, 𝑈₅, and 𝑈₉, there exist three points (0,0,√3/6,−√15/30), (0,0,−√3/6,−√15/6), and (0,0,√3/6,√15/6) satisfying (13.2). Then {𝑃₁,…,𝑃₅,𝑈₁,𝑈₅,𝑈₉,(0,0,√3/6,−√15/30),(0,0,−√3/6,−√15/6),(0,0,√3/6,√15/6)} is an 11-point isosceles set which is isomorphic to 𝑋 in Theorem 1.1, too. In the other cases, three points satisfying (13.2) do not exist.
On the other hand, if we apply the proofs of Propositions 13.2 and 13.3 to 𝑈-points, then there are some possible points on 𝑆₁ except for 𝑃₅. We apply Proposition 13.5 and its analogue to them. But three points satisfying (13.2) do not exist. Hence we cannot take points on 𝑆₁ except for 𝑃₅.
Therefore if there exists an 11-point isosceles set in ℝ⁴ satisfying ⟨4⟩ in Table 4, then it is isomorphic to 𝑋 or 𝑌 in Theorem 1.1.

We remark that 𝑌 in Theorem 1.1 does not contain a square. Thus we have the following lemma.

Lemma 13.8. There exists a unique 11-point isosceles set in ℝ⁴ containing a square. This is 𝑋 in Theorem 1.1.

14. Completion of the Proofs of Theorem 1.1 and Corollary 1.2

First, Lemma 3.1 holds if an 11-point isosceles set exists. In any case of Lemma 3.1, if there exists an 11-point isosceles set, then the condition (𝑋) holds by Lemmas 4.8, 5.1, 6.1, 7.1, 8.2, 9.2, 10.2, and 11.1.

When the condition (𝑋) holds, four points that lie on a circle are either all the vertices of a square, or four of the vertices of a regular pentagon by Lemma 11.2. If they are four of the vertices of a regular pentagon, then Lemma 12.9 implies that there exists a unique 11-point isosceles set 𝑌. On the other hand, if they are all the vertices of a square, then there exists a unique 11-point isosceles set 𝑋 by Lemma 13.8.

Therefore there are exactly two 11-point isosceles sets 𝑋 and 𝑌 in ℝ⁴ up to isomorphisms. Moreover we see that there is no 12-point isosceles set in ℝ⁴ by the calculation, and he maximum cardinality of isosceles sets in ℝ⁴ is 11.