International Journal of Combinatorics

Volume 2010 (2010), Article ID 803210, 30 pages

http://dx.doi.org/10.1155/2010/803210

## On Isosceles Sets in the 4-Dimensional Euclidean Space

Faculty of Mathematics, Kyushu University, 744 Motooka, Nishi-ku, Fukuoka 819-0395, Japan

Received 22 July 2010; Accepted 4 November 2010

Academic Editor: Gerard Jennhwa Chang

Copyright © 2010 Hiroaki Kido. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

A subset in the -dimensional Euclidean space that contains points (elements) is called an -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 up to isomorphisms and that the maximum cardinality of isosceles sets in is 11.

#### 1. Introduction

Let be the -dimensional Euclidean space, let and be in , and .

For a finite set , let
If , we call an *-distance set*.

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

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

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

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

As regards question (3), Larman et al. [6] showed that if , the ratio of 2 distances in any 2-distance set is given by , where is an integer satisfying .

Bannai et al. [7] and Blokhuis [8] proved that the cardinality of an -distance set in is bounded above by . For the case and , Shinohara [9] gave the answers to questions (1) and (2) by classifying 3-distance sets in . 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 was also given. Recently Shinohara [10] showed uniqueness of maximum 3-distance sets in .

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

We call a set in with points an *-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 -point 2-distance sets are -point isosceles sets.

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

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

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

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

#### 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 exist infinitely many up to isomorphisms. For example, is a 10-point isosceles set for any positive real number . It is nonisomorphic to for any positive real number satisfying . (ii)No 9-point isosceles set in exists (Croft [3]). (iii)There exists a unique 8-point isosceles set in up to isomorphisms. It is in Figure 1 (Kido [11]). (iv)Seven-point isosceles sets in exist infinitely many up to isomorphisms. (v)No 7-point isosceles set in exists (Erdös and Golomb [12], Erdös and Kelly [2]). (vi)There exists a unique 6-point isosceles set in 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 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 exist infinitely many up to isomorphisms.

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

##### 2.1. Example of 11-Point Isosceles Sets in

Let , be the canonical basis of . Then 11-point sets and in defined as follows are isosceles sets: where and and :

*Remark 2.1. *In above 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 . Also we can easily see that and contain a square and that contains a regular pentagon.

#### 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 be an -point isosceles set. We define the *vertex-number * of a point by the number of distinct isosceles triangles of which is an apex. It is easy to see that
Especially let be the number of regular triangles in :

We further say that a point is of *type * if the lines joining it to the remaining points in are constituted; thus: of length , of length of length , where are no two of them equal. Setting , clearly holds. Moreover if is of type , then

Lemma 3.1. *Let be an 11-point isosceles set in , and suppose that has the largest vertex-number. Then the type of 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 by (3.1), we have . Let be the type of . Then we have
and we have
In order to satisfy (3.4) and (3.5), must be one in the list of the lemma.

Throughout this paper, we refer to the *condition * 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 exists, then the condition 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 given in Lemma 3.1. In Sections 12 and 13, we deal with 11-point isosceles sets satisfying the condition . 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 , let be a point that has the largest vertex-number. Let be the type of . If , 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 , let be a point that has the largest vertex-number. Then the type of is one of (5), (4,1), and (3,2).*

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

*Proof. *We may assume that the equation of is and , where and . Then . For a point on , we consider .

When holds, we have and . Then and .

On the other hand, when holds, we have and . Then and .

Combining them, it holds that is on one of two disjoint 2-dimensional spheres.

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

*Proof. *We may assume that , and . We consider the position of for . By a similar method used in the proof of Kelly [2] or Lemma 6 in Croft [3], in a plane, , and must satisfy Figure 3.

Hence , , and hold. Then we have and .

Therefore the other points lie on a 2-dimensional sphere.

Corollary 3.7. *For , there is no -point isosceles set in which has three collinear points.*

*Proof. *We may show that this corollary holds for . 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 (). We know that there exists a unique 8-point isosceles set in 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 which has three collinear points.

Proposition 3.8. *Let be a 6-point isosceles set in a 2-dimensional sphere . Then four points in lie on a circle; the condition holds.*

*Proof. *Let be a point that has the largest vertex-number in . By Proposition 3.4, the type of is one of (5), (4,1), and (3,2). If the type of is (5) or (4,1), then at least four points among are on the intersection of and the sphere whose center is . So at least four points are on a circle, the condition holds.

Thus we suppose that is of type (3,2) with corresponding distances and . For , let be the sphere centered at with radius . Let and .

Now is a 2- or an -distance set (). We suppose that it is a 2-distance set. We know that there exist exactly six 6-point 2-distance sets in . 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 holds.

On the other hand, we suppose that is an -distance set (). So there exists a pair of points in whose distance is , that is, distinct from and . Since or , is the distance apart of a pair of points in . If holds for some and , then is scalene with sides . Thus the following condition holds:
Because is the distance apart of a pair of points in or , at least one of , , , and is .

We suppose that . Let and consider . Since and are of length or by (3.6), we have . Thus three points , , and are on the plane perpendicularly bisecting , the sphere , and the sphere . But the plane and the two spheres intersect at exactly two points. This is a contradiction. Therefore , without loss of generality we may assume .

Next we suppose that and (, , but we can admit ). Let and consider . Because and are of length or by (3.6), we have . When we consider similarly, we have . Thus and are on the plane perpendicularly bisecting , the plane perpendicularly bisecting , the sphere , and the sphere . Since the segment and the segment are not mutually parallel, the two planes and the two spheres have no intersection. Hence and do not hold. Similarly we can show that and do not hold.

So in , there is exactly one pair whose distance is distinct from and . When we consider for , holds by the configuration hypothesis. And we have . Then four points , , , and on the plane perpendicularly bisecting and the sphere . The intersection of them is a circle, the condition holds.

#### 4. Case (A) in Lemma 3.1

We consider the case (A) in Lemma 3.1. Let be an 11-point isosceles set in which is of type (6.1). Let be the sphere centered at and .

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

Proposition 4.1. *Let be the type of in . If the type of satisfies , then the condition holds.*

*Proof. *If the type of in satisfies , then at least six points among are on the intersection of and the sphere whose center is . So they are on a 2-dimensional sphere. By Proposition 3.8, the condition holds.

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

*Proof. *We suppose that is of type (5,4) in . We see that five points in are distributed on a 2-dimensional sphere which is the intersection of and the sphere whose center is and another four points in are distributed on another 2-dimensional sphere. These two spheres are disjoint.

We will call them (on which are) and (on which are). Let and . For , let and for , let .

If is a 2-distance set, then the types of ten points in must be all (6,3) by looking at the Petersen graph (Figure 2). But is of type (5,4), is not a 2-distance set. Hence is an -distance set (), there exists a pair of points in whose distance is , that is, distinct from and . Because or , is the distance between a pair of distinct points in . If holds for some and , then is scalene with sides . Thus
So is the distance between a pair of distinct points on the same 2-dimensional sphere.

We suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume . For we consider . Since and are of length or by (4.1), holds. Thus four points are on the hyperplane perpendicularly bisecting , the sphere , and the sphere . The intersection of them is a circle. Therefore the condition holds.

We can repeat the similar discussion when we suppose that is the distance between a pair of distinct points on .

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

We will call them (on which are) and (on which and are). Let and . For , let and for , let . Moreover let .

Proposition 4.3. *Let . If is an -distance set , then the condition holds.*

*Proof. *Because we suppose that is an -distance set , there exists a pair of points whose distance is , that is, distinct from and (but we can admit ).

Since or , is the distance apart of a pair of points in . If holds for some and , then is scalene with sides . Thus
So is the distance between a pair of distinct points on the same 2-dimensional sphere.

We suppose that is the distance between a pair of distinct points on , that is, . In this case, if we repeat the similar discussion as Proposition 4.2, then the condition holds. Hence we suppose that is the distance between a pair of distinct points on . For on , we consider 5-point graphs in Table 2. Edges in a graph represent the distance, that is, distinct from and . We regard the others, (i.e., transparent edges) as the distances and . 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 and . 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,,) for ( is arbitrary);(ii)another four graphs whose a 3-point subgraph is

We observe each case. In the case (i), we may assume that the 4-point subgraph with is connected. Without loss of generality we may assume . For , consider . Then we have by (4.2). Since the 4-point subgraph with is connected, we have by the similar discussion. Moreover we have for by the assumption. Then four points , , , are equidistant from on the 2-dimensional sphere . If are on a plane, then they are on a circle; the condition holds. On the other hand, if are not on a plane, then , , , and are on a line. We cannot take four points on a line. This is a contradiction.

In the case (ii), we may assume that and (we can admit ). For , consider and . Then we have and by (4.2). In this case, , , and are or for . When we consider and , and hold. By the assumption we have and for . Then six points , , , , , are equidistant from , , and . Hence they are in the 2-dimensional Euclidean space, that is, is a 6-point isosceles set in . We know that there exists a unique 6-point isosceles set in up to isomorphisms. It consists of five points of a regular pentagon and its center. So four points in lie on a circle; the condition holds.

In the case (iii), we may assume that and (we can admit ). For , consider and . Then we have and by (4.2). In this case, , , , and are or . When we consider and , and hold. By the assumption we have and for . Then five points are on the hyperplane perpendicularly bisecting and the hyperplane perpendicularly bisecting . For the intersection of them, there are two cases:

() since two hyperplanes are same, the intersection is a 3-dimensional Euclidean space.()a 2-dimensional Euclidean space.

In the case (), since are on the 2-dimensional sphere , the segment and the segment are mutually parallel. Then there is a plane that contains . So they are on a circle; the condition holds.

In the case (), is a 5-point isosceles set in . We know that there exist exactly three 5-point isosceles sets in 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 lie on a circle; the condition holds.

In the case (iv), we may assume that . Then we see that there is exactly one pair whose distance is distinct from and in . When we consider for , holds by the configuration hypothesis. Thus six points , and are on the hyperplane perpendicularly bisecting . This hyperplane is a 3-dimensional Euclidean space. Since , this is a 2-distance set in . We know that there exist exactly six 6-point 2-distance sets in . Any set contains four points lying on a circle. Hence the condition holds.

Proposition 4.4. *Similarly let . If is a 2-distance set, then the condition holds.*

*Proof. *We consider the sum of all vertex-numbers in . Because has the largest vertex-number in , . Let be the number of regular triangles in . Then holds by (3.2). Thus .

Let . We notice that on is a 2-distance set in . We consider 5-point graphs in Table 2 again. Edges in a graph represent the distance . We regard the others, (i.e., transparent edges) as the distance . Here we need not consider the graph which has no edge, because there is no 5-point 1-distance set in . Similarly we need not consider the complete graph (5,10,1).

If for (), then is a regular triangle. Since , there are at most five pairs in whose distances are . The number of pairs in is . Thus there are at least five pairs in whose distances are .

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 and that there is an edge between and , that is, . We consider and . Since and , and are or . Then we consider , we have . Because the 4-point subgraph with is connected, we have by the similar discussion. Moreover we have for by the assumption. Thus three points , , are equidistant from on the 2-dimensional sphere . If are on a plane, then they are on a circle; the condition () holds. On the other hand, if are not on a plane, then , , and are on a line. By Corollary 3.7, this is a contradiction.

Therefore if is a 2-distance set, then the condition () holds.

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

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

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

We will call these two spheres (on which are) and (on which are). Let and . For , let and for , let . Moreover let . Because , for . Thus the type of is (4,4,1) in for any . (Since , the type of may be (5,2,2) in . In this case, if we apply Proposition 4.5, then the condition () holds.)

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

*Proof. *Since the type of is (4,4,1) in and , the distance corresponds to 1 or 4 of type (4,4,1). If corresponds to 1 of type (4,4,1), then for . Considering , we have and . Thus this proposition holds.

On the other hand, if corresponds to 4 of type (4,4,1), then for , there are exactly three points such that . We may assume that . We have three means to select the other two points.

(i). (Both points are on .)(ii). (One is on and the other is on .)(iii). (Both points are on .)

In the case (i), considering for , we have and . Thus this proposition holds for . In the case (ii), considering for , we have and . Then the type of is (4,3,2), not (4,4,1). This is a contradiction. In the case (iii), considering for , we have and . Then the type of is (4,3,2), not (4,4,1). This is a contradiction.

Therefore is equidistant from four points on one of the 2-dimensional spheres and .

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

*Proof. *By Proposition 4.6, is equidistant from four points on one of the 2-dimensional spheres and . We may assume that it is . Moreover we have for by the assumption. Thus three points are equidistant from on the 2-dimensional sphere . If are on a plane, then they are on a circle; the condition holds. On the other hand, if are not on a plane, then , , and are on a line. By Corollary 3.7, this is a contradiction.

Therefore if the type of is (4,4,1) in , then the condition () 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 in which is of type (6.1), the condition 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 be an 11-point isosceles set in which the type of satisfies the case (B). Let be the sphere centered at with radius and . Let be the point which is not on and .

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

*Proof. *By Proposition 3.5, eight points are on one of two disjoint 2-dimensional spheres and , where on satisfies and on satisfies (consider for ).

If more than or equal to six points lie on one sphere, then the condition () 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 are on and , , are on . Let and .

Here the 10-point set is not a 2-distance set, because is of type (8,1) in it, not of type (6,3) in the Petersen graph. Hence it is an -distance set (), there exists a pair of points in whose distance is , that is, distinct from and . If holds for some and , then is scalene. Thus
So is the distance between a pair of distinct points on the same 2-dimensional sphere.

We suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume . For we consider . Since and are of length or by (5.1), we have . Thus five points are on the hyperplane perpendicularly bisecting and the 2-dimensional sphere . The intersection of them is a circle. Hence the condition () holds.

On the other hand, we suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume .

Next we suppose that there exist more than or equal to two pairs of points on whose distances are distinct from and . One is . We have two cases as the second pair whose distance is distinct from and .*Case 1. * and (, , but we can admit ).

Let and consider . Because and are of length or by (5.1), we have . When we consider similarly, we have . Thus three points and are on the hyperplane perpendicularly bisecting , the hyperplane perpendicularly bisecting , and the 2-dimensional sphere . For the intersection of them, there are two cases:

() because two hyperplanes are same, the intersection is a circle.()two points.

In the case (), because are on the 2-dimensional sphere , the segment and the segment are mutually parallel. Then there is a plane that contains . So they are on a circle; the condition () holds.

In the case (), we cannot put one of , , and . This is a contradiction.*Case 2. * and (, , but we can admit ).

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 which is distinct from and in . When we consider for , holds by the configuration hypothesis. Thus eight points and are on the hyperplane perpendicularly bisecting . This hyperplane is a 3-dimensional Euclidean space. Moreover is a 2-distance set with the distances and . But we know that there exists no -point 2-distance set in for . This is a contradiction.

We consider the case (ii). If we repeat this discussion similarly, then we can see that the condition 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 be an 11-point isosceles set in which the type of satisfies the case (C). will denote the common center of the two spheres, which we will call (on which are), (on which and are), radii , , respectively.

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

*Proof. *The 10-point set is not a 2-distance set, because is of type (7,2) in it, not of type (6,3) in the Petersen graph. Hence it is an -distance set (), there exists a pair of points in whose distance is , that is, distinct from and . If holds for some and , then is scalene. Thus
So is the distance between a pair of distinct points on the same 3-dimensional sphere.

We suppose that is the distance between a pair of distinct points on , that is, . For we consider . Since and are of length or by (6.1), we have . Thus seven points are on the hyperplane perpendicularly bisecting and the 3-dimensional sphere . The intersection of them is a 2-dimensional sphere. By Proposition 3.8, the condition holds.

Thus we suppose that is the distance between a pair of distinct points on . By Proposition 3.5, seven points are on one of two disjoint 2-dimensional spheres and , where on satisfies and on satisfies (consider for ).

If holds for some and , then is scalene. Thus
So 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 () 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 are on and , are on .

We suppose that is the distance between a pair of distinct points on , that is, . We consider for . By (6.2), we have . Thus five points are on the hyperplane perpendicularly bisecting and the 2-dimensional sphere . The intersection of them is a circle. Hence the condition () holds.

On the other hand, we suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume .

Next we suppose that there exist more than or equal to two pairs of points on whose distances are distinct from and . One is . We have two cases as the second pair whose distance is distinct from and .*Case 1. * and (, , but we can admit ).

Let and consider . By (6.1) and (6.2), we have . When we consider similarly, we have . For we have and . Thus five point , , , , and are on the hyperplane perpendicularly bisecting and the hyperplane perpendicularly bisecting . 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 are on the 2-dimensional sphere , the segment and the segment are mutually parallel. Then there is a plane that contains . So they are on a circle; the condition holds.

In the case (*β*), is a 5-point isosceles set in . We know that there exist three 5-point isosceles sets in 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 lie on a circle; the condition holds.*Case 2. * and (, , but we can admit ).

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 whose distance is distinct from and in . When we consider for , holds by the configuration hypothesis. Thus eight points , ,, , and are on the hyperplane perpendicularly bisecting . This hyperplane is a 3-dimensional Euclidean space. Moreover is a 2-distance set with the distances and . But there exists no -point 2-distance set in for . This is a contradiction.

We consider the case (ii). If we repeat this discussion similarly, then we see that the condition () holds.

#### 7. Case (D) in Lemma 3.1

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

*Proof. *Let be an 11-point isosceles set. When the type of is (6,4) or (6,3,1), will denote the common center of the two spheres, which we will call (on which are), (on which , , and are).

Let . can be the 2-distance set mentioned in Section 2. Since contains a square, the condition holds.

Hence we may suppose that is an -distance set (). In this case, we can show that the condition holds by repeating the similar discussion as the proof of Lemma 6.1.

When the type of is (5,5) or (5,4,1), we can show that the condition 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 be an 11-point isosceles set in which the type of is (7,1,1,1). We may assume that , , , and . Let , , and .

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

*Proof. *We suppose that and are 2-distance sets. We may prove that this leads a contradiction.

We have and by the hypothesis above. For , must be , , or .

If , then for . If , then for . So we have . Hence , is a 1-distance set. But there is no 8-point 1-distance set in . This is a contradiction.

Therefore the number of 2-distance sets is at most one.

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

*Proof. *By Proposition 8.1, at least two sets of are -distance sets (). We may suppose and are -distance sets. Espesially we notice that is an -distance set. Thus there is a distance apart of a pair of points in which is distinct from and . This is one of , and , where is distinct from , and . We may assume that it is .

Let be the sphere centered at with radius and . By Proposition 3.5, seven points are on one of two disjoint 2-dimensional spheres and , where on satisfies and on satisfies (consider for ).

We remark that there is the distance in . If holds for some and , then is scalene. Thus or for any and . So 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 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 are on and are on .

If is the distance between a pair of distinct points on , then we can show that the condition holds by the similar discussion as the proof of Lemma 5.1.

Thus we suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume . For on , we consider 5-point graphs in Table 2 again. Edges in a graph represent the distance, that is, distinct from and . We regard the others, (i.e., transparent edges) as the distances and . 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 and .

We observe the cases (i)–(iv) in the proof of Proposition 4.3 similarly. In any case, we can show that the condition () 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 () holds.

#### 9. Case (F) in Lemma 3.1

We consider the case (F) in Lemma 3.1. Let be an 11-point isosceles set in which the type of is (6,2,2). We may assume that , , and . Let and .

Proposition 9.1. *For and above, at least one of them is an -distance set .*

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

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

*Proof. *By Proposition 9.1, at least one of and is an -distance set (). We may suppose is an -distance set.

will denote the common center of the two spheres, which we will call (on which are), (on which and are), radii , , respectively.

There is a distance apart of a pair of points in which is distinct from and . This is or , where is distinct from , and . We may assume that it is . If holds for some and , then is scalene. Thus
So is the distance between a pair of distinct points on the same 3-dimensional sphere.

If is the distance between a pair of distinct points on , then we can show that the condition () holds by the similar discussion as the proof of Lemma 6.1.

Thus we suppose that is the distance between a pair of distinct points on . By Proposition 3.5, six points are on one of two disjoint 2-dimensional spheres and , where on satisfies and on satisfies (consider for ).

If holds for some and , then is scalene. Thus
So 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 () 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 () 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 () holds.

We consider the case (III). We suppose that are on and are on .

We may suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume . Next we suppose that there exist more than or equal to two pairs of points on whose distances are distinct from and . One is . Without loss of generality the second is (, , but we can admit ).

Let and consider . By (9.1) and (9.2), we have . When we consider similarly, we have . For we have and . Thus six points , , , , , and are on the hyperplane perpendicularly bisecting and the hyperplane perpendicularly bisecting . The intersection of them is a 2-dimensional Euclidean space. Then is a 6-point isosceles set in . There exist a unique 6-point isosceles set in up to isomorphisms and it contains four points on a circle. Thus the condition holds.

Hence we suppose that there is exactly one pair whose distance is distinct from and on . If we repeat the similar discussion above, then there is also at most one pair whose distance is distinct from and on . Without loss of generality this is .

When we consider for , holds by (9.1), (9.2), and the configuration hypothesis. Thus seven points , and are on the hyperplane perpendicularly bisecting . This hyperplane is a 3-dimensional Euclidean space. Particularly is a 6-point 2-distance set in with distances and . There exist exactly six 6-point 2-distance sets in . Any set contains four points lying on a circle. Hence the condition holds.

Therefore if the type of is (6,2,2), then the condition holds.

#### 10. Case (G) in Lemma 3.1

We consider the case (G) in Lemma 3.1. Let be an 11-point isosceles set in which the type of is (6,2,1,1). We may assume that , , , and . Let , , and .

Proposition 10.1. *For 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 () holds for any 11-point isosceles set in which the type of is (6,2,1,1).*

*Proof. *By Proposition 10.1, at least two sets of are -distance sets (). If is an -distance set, then we can show that the condition () holds by repeating the similar discussion as Lemma 9.2. Hence we may assume that is a 2-distance set and that and are -distance sets. Since and is a 2-distance set with distances and , it holds that
Thus is the third distance in and .

Let be the sphere centered at with radius . By Proposition 3.5, six points are on one of two disjoint 2-dimensional spheres and , where on satisfies and on satisfies (consider for ).

We remark that there is the distance in . If holds for some and , then is scalene. Thus or for any and . Combining this and (10.1), the following condition holds:
So 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 () 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 () 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 () holds.

We consider the case (III). We suppose that are on and are on .

We may suppose that is the distance between a pair of distinct points on . Without loss of generality we may assume . Next we suppose that there exist more than or equal to two pairs of points on whose distances are . 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 () holds.

Hence we suppose that there is exactly one pair whose distance is on . If we repeat the similar discussion above, then there is also at most one pair whose distance is on . Without loss of generality this is .

When we consider for , holds by (10.2) and the configuration hypothesis. Thus are on the hyperplane perpendicularly bisecting and on . The intersection of them is a 2-dimensional sphere. By (10.1), (10.2), and the assumption, for . Thus , and are equidistant from on a 2-dimensional sphere. If are on a plane, then they are on a circle; the condition () holds. On the other hand, if are not on a plane, then , and are on a line. By Corollary 3.7, this is a contradiction.

Therefore if the type of is (6,2,1,1), then the condition () holds.

#### 11. Case (H) in Lemma 3.1

We consider the case (H) in Lemma 3.1. Let be an 11-point isosceles set in which the type of is (6,1,1,1,1). We may assume that , , , , and .

We consider the sum of all vertex-numbers in . Since has the largest vertex-number in , . On the other hand, by (3.1). Thus . Let be the number of regular triangles in . Then holds by (3.2). Moreover holds for any ; the type of is (6,1,1,1,1).

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

*Proof. *We notice that the type of is (6,1,1,1,1). So the distance corresponds to 6 or 1 of type (6,1,1,1,1). If corresponds to 6, then at least one of is . We may suppose that . Then is a regular triangle with the distance . This contradicts . Thus corresponds to 1. Then , , , and hold by considering for . This means that one of , and corresponds to 6 of type (6,1,1,1,1). We may assume that this is . Then .

Next we notice that the type of is (6,1,1,1,1). We see that corresponds to 1 of type (6,1,1,1,1) by repeating the discussion for . Thus , , , and hold by considering for , corresponds to 6 of type (6,1,1,1,1). Then . But is a regular triangle with the distance . This contradicts .

Therefore there is no 11-point isosceles set in which the type of 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 , in 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. *, , , 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 Containing Four Points of a Regular Pentagon

Proposition 12.1. *Suppose an -point isosceles set contains four vertices of a regular pentagon, , , , (in order, with the “gap” between and ). We may suppose that , , , . (The mid-point of 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)*, where and are arbitrary,*(ii)*, the remaining vertex of the pentagon, or*(iii)*, where and satisfy .*

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

Proposition 12.2. *Let be a point satisfying (iii) in the previous proposition. Then no -point isosceles set can contain , , , , , and .*

*Proof. *It holds that . Then is scalene with 1, , . This is contrary to the configuration hypothesis.

Therefore no -point isosceles set can contain , , , , , and .

We observe the detail for