# On Isosceles Sets in the 4-Dimensional Euclidean Space

Authors: Hiroaki Kido
Journal: International Journal of Combinatorics
 (2010)

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## Abstract

A subsetX in the k-dimensional Euclidean space ℝk 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.


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## Body

## 1. Introduction

Letℝk be the k-dimensional Euclidean space, let x=(x1,x2,…,xk) and y=(y1,y2,…,yk) be in ℝk, and d(x,y)=∑i=1k(xi-yi)2.For a finite setX⊂ℝk, let(1.1)A(X)={d(x,y)∣x,y∈X,x≠y}.
If |A(X)|=s, we call X an s-distance set.Two subsets inℝk are said to be isomorphic if there exists similar transformation from one to the other.We have the following interesting problems ons-distance sets.  (1)
What is the cardinality of points when the number ofs-distance sets in ℝk is finite up to isomorphisms?(2)
What is the maximum cardinality ofs-distance sets in ℝk?(3)
Can we say something about the ratios of distances in ans-distance set?As regards question (1), Einhorn and Schoenberg [1] showed that the number of 2-distance sets in ℝk is finite if cardinalities are more than or equal to k+2.For question (2) withs=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.
k(k+22)The maximum cardinality of 2-distance setsThe number of 2-distance sets giving the maximum cardinality133126513106641510152116162827173629184545≥1As 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 ℝk is bounded above by (k+ss). 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 inℝk 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 alln-point 2-distance sets are n-point isosceles sets.In this paper, we consider classification and the maximum cardinality of isosceles sets inℝ4. The following theorem and corollary are the main results.Theorem 1.1.
There exist exactly two 11-point isosceles sets inℝ4 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 inℝ4. 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 inℝ4 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 inℝ3 exists (Croft [3]).(iii)
There exists a unique 8-point isosceles set inℝ3 up to isomorphisms. It is in Figure 1 (Kido [11]).(iv)
Seven-point isosceles sets inℝ3 exist infinitely many up to isomorphisms.(v)
No 7-point isosceles set inℝ2 exists (Erdös and Golomb [12], Erdös and Kelly [2]).(vi)
There exists a unique 6-point isosceles set inℝ2 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 inℝ2 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 inℝ2 exist infinitely many up to isomorphisms.Figure 1
A unique 8-point isosceles set inℝ3 (from Kido [11]).Now we define two examplesX and Y which are mentioned in Theorem.


### 2.1. Example of 11-Point Isosceles Sets inℝ4

Letei, 1≤i≤4 be the canonical basis of ℝ4. Then 11-point sets X and Y in ℝ4 defined as follows are isosceles sets:(2.1)X=X′∪{u0},
where(2.2)X′={ei+ej∣1≤i<j≤4}∪{-ek+u∣1≤k≤4},
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):(2.3)Y={(cos⁡2j5π,sin2j5π,0,0)∣0≤j≤4}∪{(0,0,cos⁡2k5π,sin2k5π)∣0≤k≤4}∪{(0,0,0,0)}.Remark 2.1.
In aboveX′ 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.


## 2.1. Example of 11-Point Isosceles Sets inℝ4

Letei, 1≤i≤4 be the canonical basis of ℝ4. Then 11-point sets X and Y in ℝ4 defined as follows are isosceles sets:(2.1)X=X′∪{u0},
where(2.2)X′={ei+ej∣1≤i<j≤4}∪{-ek+u∣1≤k≤4},
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):(2.3)Y={(cos⁡2j5π,sin2j5π,0,0)∣0≤j≤4}∪{(0,0,cos⁡2k5π,sin2k5π)∣0≤k≤4}∪{(0,0,0,0)}.Remark 2.1.
In aboveX′ 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.Let𝒫={P1,…,Pn} be an n-point isosceles set. We define the vertex-numberV(Pi) of a point Pi∈𝒫 by the number of distinct isosceles triangles of which Pi is an apex. It is easy to see that(3.1)V(P1)+⋯+V(Pn)≥(n3).
Especially let αbe the number of regular triangles in 𝒫:(3.2)V(P1)+⋯+V(Pn)=2α+(n3).We further say that a pointPi∈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 Pi is of type (r,s,…,u), then(3.3)V(Pi)=(r2)+(s2)+⋯+(u2).Lemma 3.1.
Let𝒫={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.
SinceV(P1)+⋯+V(P11)≥(113)=165 by (3.1), we have V(P1)≥15. Let (r,s,…,u) be the type of P1. Then we have
(3.4)(r2)+(s2)+⋯+(u2)≥15,
and we have
(3.5)r+s+⋯+u=10.
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 thecondition(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 inℝ4 exists, then the condition (X) is true for it.In Sections4–11, 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 Lemma3.2 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 inℝ4, 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 inℝ3, 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 ann-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 Pn, 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 ofS is x2+y2+z2+w2=1 and Pn=(k,0,0,0), where k>0 and k≠1. Then P1=(0,0,0,0). For a point Pi=(xi,yi,zi,wi) on S, we consider ΔP1PiPn.
WhenP1Pi=PiPn=1 holds, we have xi2+yi2+zi2+wi2=1 and (xi-k)2+yi2+zi2+wi2=1. Then xi=k/2 and yi2+zi2+wi2=1-k2/4.
On the other hand, whenP1Pn=PiPn=k holds, we have xi2+yi2+zi2+wi2=1 and (xi-k)2+yi2+zi2+wi2=k2. Then xi=1/2k and yi2+zi2+wi2=1-1/4k2.
Combining them, it holds thatPi 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 thatP1=(-1,0,0,0),P2=(0,0,0,0), and P3=(1,0,0,0). We consider the position of Pi=(xi,yi,zi,wi) 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 Pi must satisfy Figure 3.
Hencexi2+yi2+zi2+wi2=1, (xi+1)2+yi2+zi2+wi2=2, and (xi-1)2+yi2+zi2+wi2=2 hold. Then we have xi=0 and yi2+zi2+wi2=1.
Therefore the other points lie on a 2-dimensional sphere.Figure 3Corollary 3.7.
Forn≥11, there is no n-point isosceles set in ℝ4 which has three collinear points.Proof.
We may show that this corollary holds forn=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 inℝ4 which has three collinear points.Proposition 3.8.
Let𝒫={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.
LetP1 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 thatP1 is of type (3,2) with corresponding distances r1 and r2. For i=1,2, let Si be the sphere centered at P1 with radius ri. Let U1=𝒫∩S1={P2,P3,P4} and U2=𝒫∩S2={P5,P6}.
Now𝒫 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 that𝒫 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 P1Pi=r1 or r2(i=2,…,6), c is the distance apart of a pair of points in {P2,…,P6}. If PiPj=c holds for some Pi∈U1 and Pj∈U2, then ΔP1PiPj is scalene with sides r1,r2,c. Thus the following condition holds:(3.6)PiPj=r1orr2foranyPi∈U1,Pj∈U2.
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 thatP5P6=c. Let Pi∈U1 and consider ΔPiP5P6. Since PiP5 and PiP6 are of length r1 or r2 by (3.6), we have PiP5=PiP6. 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 thatP2P3=c and P2P4=d (d≠r1, d≠r2, but we can admit c=d). Let Pj∈U2 and consider ΔP2P3Pj. Because P2Pj and P3Pj are of length r1 or r2 by (3.6), we have P2Pj=P3Pj. When we consider ΔP2P4Pj similarly, we have P2Pj=P4Pj. 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 in𝒫, there is exactly one pair P2P3 whose distance is distinct from r1 and r2. When we consider ΔP2P3Pk for k=4,5,6, P2Pk=P3Pk 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 inℝ3 (from Einhorn and Schoenberg [14]).


## 4. Case (A) in Lemma3.1

We consider the case (A) in Lemma3.1. 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 thatV 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 ofP2 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 ofP2 is (5,4) in V, then the condition (X) holds.Proof.
We suppose thatP2 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 themS1 (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 Pi∈V1, let P2Pi=a and for Pj∈V2, let P2Pj=b.
IfV 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 P2Pk=a or b(k=3,…,11), c is the distance between a pair of distinct points in {P3,…,P11}. If PiPj=c holds for some Pi∈V1 and Pj∈V2, then ΔP2PiPj is scalene with sides a,b,c. Thus(4.1)PiPj=aorbforanyPi∈V1,Pj∈V2.
So c is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose thatc is the distance between a pair of distinct points on S1. Without loss of generality we may assume P3P4=c. For Pj∈V2 we consider ΔP3P4Pj. Since P3Pj and P4Pj are of length a or b by (4.1), P3Pj=P4Pj 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 thatc is the distance between a pair of distinct points on S2.Next we consider that the type ofP2 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 themS1 (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 Pi∈V1, let P2Pi=a and for Pj∈V2, let P2Pj=b. Moreover let P2P11=c.Proposition 4.3.
LetX1={P2,…,P9}. If X1 is an s-distance set (s≥3), then the condition (X) holds.Proof.
Because we suppose thatX1 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).
SinceP2Pi=a or b(i=3,…,9), d is the distance apart of a pair of points in {P3,…,P9}. If PiPj=d holds for some Pi∈V1 and Pj∈V2, then ΔP2PiPj is scalene with sides a,b,d. Thus(4.2)PiPj=aorbforanyPi∈V1,Pj∈V2.
So d is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose thatd 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 withP3,…,P6 is connected. Without loss of generality we may assume P3P4=d. For i=8,9, consider ΔP3P4Pi. Then we have P3Pi=P4Pi by (4.2). Since the 4-point subgraph with P3,…,P6 is connected, we have P3Pi=P4Pi=P5Pi=P6Pi by the similar discussion. Moreover we have P3Pj=P4Pj=P5Pj=P6Pj 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 thatP3P4=d and P3P5=e (we can admit d=e). For i=8,9, consider ΔP3P4Pi and ΔP3P5Pi. Then we have P3Pi=P4Pi and P3Pi=P5Pi by (4.2). In this case, P3Pj, P4Pj, and P5Pj are a or b for j=6,7. When we consider ΔP3P4Pj and ΔP3P5Pj, P3Pj=P4Pj and P3Pj=P5Pj hold. By the assumption we have P3Pk=P4Pk and P3Pk=P5Pk 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 thatP3P4=d and P5P6=e (we can admit d=e). For i=8,9, consider ΔP3P4Pi and ΔP5P6Pi. Then we have P3Pi=P4Pi and P5Pi=P6Pi 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 P3Pj=P4Pj and P5Pj=P6Pj 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 thatP3P4=d. Then we see that there is exactly one pair P3P4 whose distance is distinct from a and b in X1. When we consider ΔP3P4Pi for i=2,5,…,9, P3Pi=P4Pi 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 letX1={P2,…,P9}. If X1 is a 2-distance set, then the condition (X) holds.Proof.
We consider the sum of all vertex-numbers in𝒫. Because P2 has the largest vertex-number in V, V(P1)+⋯+V(P11)≤(102)+10×{(52)+(32)+(12)}=175. Let αbe the number of regular triangles in 𝒫. Then 2α+(113)≤175 holds by (3.2). Thus α≤5.
LetV1={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).
IfPiPj=a for i,j∈{3,…,7} (i≠j), then ΔP2PiPj 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 Table2. 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 P3Pk=P4Pk=P5Pk=P6Pk 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 ifX1 is a 2-distance set, then the condition (X) holds.Combining Propositions4.3 and 4.4, we have the following proposition.Proposition 4.5.
If the type ofP2 is (5,3,1) or (5,2,2) in V, then the condition (X) holds.The last case is what the type ofP2 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 spheresS1 (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 Pi∈V1, let P2Pi=a and for Pj∈V2, let P2Pj=b. Moreover let P2P11=c. Because V(P2)=12, V(Pk)=12 for k=3,…,11. Thus the type of Pk is (4,4,1) in V for any k. (Since V(Pk)=12, the type of Pk 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 ofP2 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 ofP11 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 PiP11≠c for i=3,…,10. Considering ΔP2PiP11, we have P3P11=P4P11=P5P11=P6P11=a and P7P11=P8P11=P9P11=P10P11=b. Thus this proposition holds.
On the other hand, ifc corresponds to 4 of type (4,4,1), then for j=3,…,10, there are exactly three points such that PjP11=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), consideringΔP2PkP11 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 ΔP2PłP11 for ł=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 ΔP2PmP11 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.
ThereforeP11 is equidistant from four points on one of the 2-dimensional spheres S1 and S2.Proposition 4.7.
If the type ofP2 is (4,4,1) in V, then the condition (X) holds.Proof.
By Proposition4.6, 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 PiP3=PiP4=PiP5=PiP6 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 ofP2 is (4,4,1) in V, then the condition (X) holds.Summing up the results of Propositions4.1, 4.2, 4.5, and 4.7, we have the following.Lemma 4.8.
For any 11-point isosceles set inℝ4 in which P1 is of type (6.1), the condition (X) holds.


## 5. Case (B) in Lemma3.1

We consider the case (B) in Lemma3.1. 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.Let𝒫={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 Proposition3.5, eight points P2,…,P9 are on one of two disjoint 2-dimensional spheres S1 and S2, where Pi on S1 satisfies PiP11=a and Pj on S2 satisfies PjP11=b (consider ΔP1PkP11 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 thatP2,…,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 PiPj=c holds for some Pi∈V1 and Pj∈V2, then ΔPiPjP11 is scalene. Thus(5.1)PiPj=aorbforanyPi∈V1,Pj∈V2.
So c is the distance between a pair of distinct points on the same 2-dimensional sphere.
We suppose thatc is the distance between a pair of distinct points on S2. Without loss of generality we may assume P7P8=c. For Pi∈V1 we consider ΔPiP7P8. Since PiP7 and PiP8 are of length a or b by (5.1), we have PiP7=PiP8. 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 thatc 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 onS1 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).
LetPj∈V2 and consider ΔP2P3Pj. Because P2Pj and P3Pj are of length a or b by (5.1), we have P2Pj=P3Pj. When we consider ΔP4P5Pj similarly, we have P4Pj=P5Pj. 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 pairP2P3 which is distinct from a and b in V1. When we consider ΔP2P3Pk for k=1,4,…,9,11, P2Pk=P3Pk 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 Lemma3.1

We consider the case (C) in Lemma3.1. 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.Let𝒫={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 PiPj=c holds for some Pi∈S1 and Pj∈S2, then ΔP1PiPj is scalene. Thus
(6.1)PiPj=aorbforanyPi∈S1,Pj∈S2.
So c is the distance between a pair of distinct points on the same 3-dimensional sphere.
We suppose thatc is the distance between a pair of distinct points on S2, that is, P9P10=c. For Pi∈S1 we consider ΔPiP9P10. Since PiP9 and PiP10 are of length a or b by (6.1), we have PiP9=PiP10. 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 thatc 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 Pi on S11 satisfies PiP9=a and Pj on S12 satisfies PjP9=b (consider ΔP1PkP9 for k=2,…,8).
IfPiPj=c holds for some Pi∈S11 and Pj∈S12, then ΔPiPjP9 is scalene. Thus(6.2)PiPj=aorbforanyPi∈S11,Pj∈S12.
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 thatP2,…,P6 are on S11 and P7, P8 are on S12.
We suppose thatc is the distance between a pair of distinct points on S12, that is, P7P8=c. We consider ΔPiP7P8 for Pi∈S11. By (6.2), we have PiP7=PiP8. 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 thatc 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 onS11 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).
LetPj∈S12∪S2 and consider ΔP2P3Pj. By (6.1) and (6.2), we have P2Pj=P3Pj. When we consider ΔP4P5Pj similarly, we have P4Pj=P5Pj. 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 pairP2P3 whose distance is distinct from a and b in {P1,…,P10}. When we consider ΔP2P3Pk for k=1,4,…,10, P2Pk=P3Pk 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 Lemma3.1

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.
Let𝒫={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).
Let𝒫′={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 that𝒫′ 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 ofP1 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 Lemma3.1

We consider the case (E) in Lemma3.1. 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.
ForX1,…,X3 above, if 2-distance sets exist, then the number of them is at most one.Proof.
We suppose thatX1 and X2 are 2-distance sets. We may prove that this leads a contradiction.
We haveA(X1)={a,b} and A(X2)={a,c} by the hypothesis above. For i,j=2,…,8(i≠j), PiPj must be a, b, or c.
IfPiPj=b, then A(X2)≠{a,c} for X2. If PiPj=c, then A(X1)≠{a,b} for X1. So we have PiPj=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 Proposition8.1, 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.
LetS 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 Pi on S1 satisfies PiP9=a and Pj on S2 satisfies PjP9=b (consider ΔP1PkP9 for k=2,…,8).
We remark that there is the distancec in V. If PiPj=c holds for some Pi∈S1 and Pj∈S2, then ΔPiPjP9 is scalene. Thus PiPj=a or b for any Pi∈S1 and Pj∈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 thatP2,…,P6 are on S1 and P7,P8 are on S2.
Ifc 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 thatc 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 Proposition4.3 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 Lemma5.1. If we apply it, then we see that the condition (X) holds.


## 9. Case (F) in Lemma3.1

We consider the case (F) in Lemma3.1. 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.
ForX1 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 Proposition9.1, 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 PiPj=c holds for some Pi∈S1 and Pj∈S2, then ΔP1PiPj is scalene. Thus(9.1)PiPj=aorbforanyPi∈S1,Pj∈S2.
So c is the distance between a pair of distinct points on the same 3-dimensional sphere.
Ifc 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 thatc 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 Pi on S11 satisfies PiP9=a and Pj on S12 satisfies PjP9=b (consider ΔP1PkP9 for k=2,…,7).
IfPiPj=c holds for some Pi∈S11 and Pj∈S12, then ΔPiPjP9 is scalene. Thus(9.2)PiPj=aorbforanyPi∈S11,Pj∈S12.
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 Lemma8.2. Thus the condition (X) holds in the case (I).
In the case (II), we can apply the similar discussion as the proof of Lemma6.1 in Case (C). If we apply it, then we see that the condition (X) holds.
We consider the case (III). We suppose thatP2,…,P4 are on S11 and P5,…,P7 are on S12.
We may suppose thatc 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).
LetPi∈S11∪S2 and consider ΔP6P7Pi. By (9.1) and (9.2), we have P6Pi=P7Pi. When we consider ΔP5P7Pi similarly, we have P5Pi=P7Pi. 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 pairP6P7 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 considerΔP6P7Pk for k=1,…,5,8,9, P6Pk=P7Pk 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 ofP1 is (6,2,2), then the condition (X) holds.


## 10. Case (G) in Lemma3.1

We consider the case (G) in Lemma3.1. 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.
ForX1,…,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 Proposition8.1.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 Proposition10.1, 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
(10.1)A({P2,…,P7})={a,b}.
Thus b is the third distance in X2 and X3.
LetS 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 Pi on S1 satisfies PiP10=a and Pj on S2 satisfies PjP10=c (consider ΔP1PkP10 for k=2,…,7).
We remark that there is the distanceb in {P2,…,P7}. If PiPj=b holds for some Pi∈S1 and Pj∈S2, then ΔPiPjP10 is scalene. Thus PiPj=a or c for any Pi∈S1 and Pj∈S2. Combining this and (10.1), the following condition holds:(10.2)PiPj=aforanyPi∈S1,Pj∈S2.
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 Lemma8.2. Thus the condition (X) holds in the case (I).
In the case (II), we can apply the similar discussion as the proof of Lemma6.1. If we apply it, then we see that the condition (X) holds.
We consider the case (III). We suppose thatP2,…,P4 are on S1 and P5,…,P7 are on S2.
We may suppose thatb 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 pairP6P7 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 considerΔP6P7Pk for k=2,…,5, P6Pk=P7Pk 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, P2Pi=P3Pi=P4Pi=P5Pi=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 ofP1 is (6,2,1,1), then the condition (X) holds.


## 11. Case (H) in Lemma3.1

We consider the case (H) in Lemma3.1. 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 in𝒫. 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(Pi)=15 holds for any Pi∈𝒫; the type of Pi is (6,1,1,1,1).Lemma 11.1.
There is no 11-point isosceles set in which the type ofP1 is (6,1,1,1,1).Proof.
We notice that the type ofP2 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 ΔP1P2Pi 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 ofP3 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 ΔP1P3Pi 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 ofP1 is (6,1,1,1,1).Therefore combining Lemmas3.1, 4.8, 5.1, 6.1, 7.1, 8.2, 9.2, 10.2, and 11.1, we have Lemma 3.2.By Lemma3.2, 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 Lemma11.2 respectively.


## 12. Observation of 11-Point Isosceles Sets inℝ4 Containing Four Points of a Regular Pentagon

Proposition 12.1.
Suppose ann-point isosceles set 𝒫={P1,…,Pn} 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+25)/4,0,0), P2=(-1/2,0,0,0), P3=(1/2,0,0,0), P4=((1+5)/4,(10+25)/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+25),z,w), where z and w are arbitrary,(ii)
T=(0,(5+35)/2(10+25),0,0), the remaining vertex of the pentagon, or(iii)
(0,(1-5)/2(10+25),z,w), where z and w satisfy z2+w2=((10+25)/25)2.Proof.
We expand the proof of Lemma 22 in Croft [3] into ℝ4, then we obtain this proposition.Proposition 12.2.
LetQ 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 thatQT=(10+25)/2. Then ΔP1QT is scalene with 1, (1+5)/2, (10+25)/2. This is contrary to the configuration hypothesis.
Therefore non-point isosceles set can contain P1, P2, P3, P4, T, and Q.We observe the detail forn=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.
The number of points satisfying (i)(ii)The number of points satisfying (iii)〈1〉6T0〈2〉61〈3〉52〈4〉43〈5〉34〈6〉25Proposition 12.3.
Any 11-point isosceles set inℝ4 satisfying 〈1〉 in Table 3 is isomorphic to Y in Theorem 1.1.Proof.
Any 11-point isosceles set𝒫={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+25),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+25).
LetPi=(0,(3+5)/2(10+25),z,w) for i∈{7,…,11}. We consider ΔP5P6Pi, we have (P5P6)2=((1+5)/(10+25))2, (P6Pi)2=z2+w2(>0), and (P5Pi)2=(P5P6)2+(P6Pi)2. Since P5Pi>P5P6 and P5Pi>P6Pi, P5P6=P6Pi holds by the configuration hypothesis. Thus P7,…,P11 which form a regular pentagon are on the circle satisfying z2+w2=((1+5)/(10+25))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+25).The following two propositions inℝ3 are useful for us. We quote them from Kido [11] (or Croft [3]).Proposition 12.4.
Let four pointsP1, 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 lineL through O, or(ii)
at some ofQ1,…,Q8, where
(12.1)Q1=(0,-12,32),Q2=(12,0,32),Q3=(0,12,32),Q4=(-12,0,32),Q5=(0,-12,-32),Q6=(12,0,-32),Q7=(0,12,-32),Q8=(-12,0,-32).
(The square Q1, Q2, Q3, Q4, and Q5, Q6, Q7, Q8 both have sides of length 2/2.)Proposition 12.5.
Suppose ann-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)
atT the remaining vertex of the pentagon; or(ii)
at two pointsQ1, Q2, which are the only points Q such that ΔQP4P1 and ΔQP2P3 are both equilaternal; or(iii)
on the vertical lineL through the center of the pentagon.Proposition 12.6.
Any 11-point isosceles set inℝ4 satisfies neither 〈5〉 nor 〈6〉 in Table 3.Proof.
We consider whenP5,…,P8 form a square. If each side of the square P1P2P3P4 in Proposition 12.4 is (10+25)/25×2=(5+5)/5, then we can change them into P5,…,P8. Now P10 and P11 are in x=0,y=(3+5)/2(10+25). By Proposition 12.4, we see that there is exactly one point (0,(3+5)/2(10+25),0,0) in x=0, y=(3+5)/2(10+25). Thus we cannot put one of P10 and P11. This is a contradiction.
On the other hand, we consider whenP5,…,P8 form four points of a regular pentagon. Here P10 and P11 are in x=0,y=(3+5)/2(10+25). By Proposition 12.5, we see that there is exactly one point (0,(3+5)/2(10+25),0,0) in x=0,y=(3+5)/2(10+25). Thus we cannot put one of P10 and P11. This is a contradiction.
Therefore any 11-point isosceles set inℝ4 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+25),0,(10+25)/25) 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 set𝒫={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 lineL which satisfies w=(10+25)/45,(II)
R1=(0,(3+5)/2(10+25),0,(1/20)(-510+25+50+105), or(III)
R2=(0,(3+5)/210+25,0,(1/20)(510+25+50+105).Proof.
Fori=8,…,11, let Pi=(0,(3+5)/210+25,z,w). We consider ΔP2P5Pi. 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 P2Pi=1,(a-2)
((1+5)/(10+25))2+z2+(w-(10+25)/25)2=1 when P5Pi=1,(a-3)
w=(10+25)/45 when P2Pi=P5Pi.
On the other hand, we considerΔP1P5Pi. 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+25)/5 when P1Pi=(1+5)/2,(b-2)
((1+5)/(10+25))2+z2+(w-(10+25)/25)2=(3+5)/2 when P5Pi=(1+5)/2,(b-3)
w=(10+25)/45 when P1Pi=P5Pi.
Hence combining one of (a-1)–(a-3) and one of (b-1)–(b-3), we see that the possible situations forPi must be in the list of the proposition.Proposition 12.8.
Any 11-point isosceles set inℝ4 cannot satisfy one of 〈2〉–〈4〉 in Table 3.Proof.
The previous proposition implies thatP8,…,P11 satisfy one of the following conditions:(i)
four points onL,(ii)
three points onL and the other is one of R1 and R2,(iii)
two points onL 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 Corollary3.7, 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 inℝ4 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 inℝ4 containing four vertices of a regular pentagon. This is Y in Theorem 1.1.


## 13. Observation of 11-Point Isosceles Sets inℝ4 Containing a Square

Proposition 13.1.
LetP1,P2,P3,P4 in an n-point isosceles set 𝒫={P1,…,Pn} 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 forn=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.
The number of points satisfying (i)The number of points satisfying (ii)〈1〉07〈2〉16〈3〉25〈4〉34〈5〉43〈6〉52〈7〉61We observe〈1〉–〈3〉 in Table 4. We see that another point Pi of an 11-point isosceles set 𝒫={P1,…,P11} which satisfies (ii) in Proposition 13.1 is on one of four circles.LetS1 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.When𝒫 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 chooseS1 as the first circle because of symmetry. Now we separate the choice of the second circle into two cases whether Si 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 pointPi 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 26/3.Proof.
We may assume thatPi=(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 Pj=(0,1/2,z,w) on S3, where z2+w2=3/4. We consider ΔP1PiPj. Since P1Pi=1 and P1Pj=2, we have PiPj=1 or 2. When PiPj=1, Pj is (0,1/2,3/2,0). When PiPj=2, Pj is (0,1/2,3/6,6/3) or (0,1/2,3/6,-6/3).
Therefore the possible situations forPj 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 26/3.We consider the other case. We may suppose that the choice of the second circle isS2 because of symmetry.Proposition 13.3.
One considers that the choice of the second circle isS2. One fixes a point Pi=(0,-1/2,zi,wi) on S1, where zi2+wi2=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.
LetPj=(1/2,0,z,w) on S2, where z2+w2=3/4. We consider ΔP1PiPj. Since P1Pi=1 and P1Pj=2, we have PiPj=1 or 2. Then (z-zi)2+(w-wi)2=1/2 or (z-zi)2+(w-wi)2=3/2 holds. From them, we have ziz+wiw-1/2=0 or ziz+wiw=0. And z and w satisfy z2+w2=3/4. Hence the possible situations for Pj are at most four.
Since (1/2,0,0,0) is on ziz+wiw=0, the distance between (1/2,0,0,0) and ziz+wiw-1/2=0 is |-1/2|/zi2+wi2=1/2/3/2=3/3 in spite of the way to fix Pi. So let Q1,Q2,Q3,Q4 be the four possible points for Pj, the distances between Q-points are in Figure 5 in spite of the way to fix Pi. Looking at Figure 5, all triangles that we choose from Q-points are scalene.
Therefore if we fix a pointPi 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 5For the supposition of Proposition13.3, moreover we suppose that there is a point on S4, too. Then the distance between two points on S2 must be 1 or 26/3 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 inℝ4 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 set𝒫={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 lineL which satisfies az+w3/4-a2=1/4, or(II)
at some ofR1,…,R4, where
(13.1)R1=(0,0,-2a-5(3-4a2)6,25a-3-4a26),R2=(0,0,-2a+5(3-4a2)6,-25a-3-4a26),R3=(0,0,2a-3-4a22,2a+3-4a22),R4=(0,0,2a+3-4a22,-2a+3-4a22).Proof.
Fori=8,…,11, let Pi=(0,0,z,w). We consider ΔP1P5Pi. 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 P1Pi=1,(a-2)
(z-a)2+(w-3/4-a2)2=3/4 when P5Pi=1,(a-3)
az+w3/4-a2=1/4 when P1Pi=P5Pi.
On the other hand, we considerΔP3P5Pi. 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 P3Pi=2,(b-2)
(z-a)2+(w-3/4-a2)2=7/4 when P5Pi=2,(b-3)
az+w3/4-a2=1/4 when P3Pi=P5Pi.
Hence conbining one of (a-1)–(a-3) and one of (b-1)–(b-3), we see that the possible situations forPi must be in the list of the proposition.Proposition 13.6.
Any 11-point isoseles set inℝ4 cannot satisfy one of 〈5〉–〈7〉 in Table 4.Proof.
By Proposition13.5, 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 ofP8 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 inℝ4 satisfying 〈4〉 in Table 4, then it is isomorphic to X or Y in Theorem 1.1.Proof.
IfP5,…,P8 are distributed on S1, then they are all points of a square or four points of a regular pentagon.
We consider whenP5,…,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.
IfP5,…,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 Proposition13.2, 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 Proposition13.5 to P5, then P9, P10, and P11 in the plane x=y=0 must satisfy one of the following situations:(13.2)(I)onthelinewhichsatisfiesz=36,(II)atsomeof(0,0,-36,156),(0,0,-36,-156),(0,0,32,32),(0,0,32,-32).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 Propositions13.2 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 inℝ4 satisfying 〈4〉 in Table 4, then it is isomorphic to X or Y in Theorem 1.1.We remark thatY 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 inℝ4 containing a square. This is X in Theorem 1.1.


## 14. Completion of the Proofs of Theorem1.1 and Corollary 1.2

First, Lemma3.1 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 setsX 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.



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