three distinct distances in the plane

Geometriae Dedicata 61:315-327, 1996. 315 (~) 1996 Kluwer Academic Publishers. Printed in the Netherlands.

Three Distinct Distances in the Plane

Dedicated to Professor Dr. Laszlo Fejes T6th on the occasion of his eightieth birthday

HEIKO HARBORTH and LOTHAR PIEPMEYER Diskrete Mathematik Technische Universittit Braunschweig, Pockelsstrafle 14, D-38106 Braunschweig, Germany

(Received: 18 January 1995)

Abstract. For n > 6 all sets of n points in the plane with three distinct distances are determined.

Mathematics Subject Classifications (1991): 52C 10, 51 K99.

Key words: Erd6s problems, distance geometry.

I. Introduction

For n points in the plane it is easy to see that all n ( n - 1)/2 distances can be pairwise different. However, what is the minimum number f (n) of distinct distances which are determined by n points? In 1946, R ErdOs [6] gave the upper bound

f (n) < cln/ lx/~n ,

and he still offers $500 for a proof or disproof that this bound is the asymptotical exact value for f(n). The best known lower bound

n4/5/(logn) c2 < f (n)

is due to Chung, Szemer6di, and Trotter [3]. Related problems are surveyed in [9]. In the literature the known exact values for f (n) are

f ( 3 ) = 1, f ( 4 ) = f ( 5 ) = 2 , and f ( 6 ) = f ( 7 ) = 3

(see F1 in [4]) and the known numbers r(n) of different configurations of n points with f (n) different distances are

r ( 3 ) = 1, r ( 4 ) = 6 , and r ( 5 ) = 1.

In [4] the set of all configurations for n = 4 is incomplete, see [5]. In this paper we will construct all configurations of six points with three distinct

distances. From this we deduce the values f (8 ) = f (9 ) = 4, and r(6) = 9, r(7) = 2. These results are part of the dissertation [10]. After preparation of this

316 HEIKO HARBORTH AND LOTHAR PIEPMEYER

6"

i t • I •

Fig. 1.

paper we received [7] where also f (8) = f (9) = 4 and r(7) --- 2 are proved and in addition f (10) = f ( l l ) = f (12) = 5, f (13) = 6, and r(9) - 4. However, our main result, r(6) = 9, is missing in [7].

It may be mentioned that n points as vertices of a convex polygon determine at least [n/2J distinct distances (see [1], [2]), and the extremal configurations are characterized in [8].

2. All Three-Distance Sets with Six Points

Those r(4) + r(5) = 7 configurations with f (4) -- f (5) = 2 distinct distances are given in Figure 1. For n = 6 points f (6 ) = 3 distinct distances are required.

THEOREM 1. A configuration of six points in the plane determines exactly three distinct distances if and only if it is similar to one of the r(6) = 9 examples shown in Figure 2.

Proof We partition the set S(k, n) of all point sets with n points and exactly k distinct distances into three classes. By Sl(k, n) we denote all sets of S(k, n) which contain the vertices of at least one equilateral triangle. By S2(k, n) we denote all sets of S(k, n) which neither contain three points as vertices of an equilateral triangle nor four points of S(2, 4). The remaining sets of S(k, n) are denoted by s3(k, n).

$1 (3, 6) It may be assumed that a, b, c are points in a set A4 of S1(3, n), 4 < n _< 6, which are vertices of a unit triangle (abc). Let c~ and/3 denote the two distinct distances in .A4 besides the unit distance. Any point s in A4 different from a, b, c has at least one of the three following properties:

A: At least one unit distance exists from s to a, b, or c.

B: At least two equal distances exist from s to a, b, and c.

THREE DISTINCT DISTANCES IN THE PLANE 317

• . / ¢ " " : . : " " : X

. : . : ' . . ; ..... I : : I

(1) (2) (3)

(4) (5)

(7) (8)

Fig. 2.

/ f Z x x x

(~

(9)

C: The distance from s to the center of the triangle (abc) equals ex /3 /3 with e = O, 1 , a , or f l .

A necessary condition for the existence of a, b, c, and s in the plane with the prescribed distances is the vanishing of the volume of the tetrahedron with vertices a, b, c, and s. This implies for A with distances 1, x, and y from s to (abc)

fA(x , y) = fA(Y, x) = x 4 -- 2x 2 -- x2y 2 + y4 _ 2y2 + 1 = 0,

for B with distances x, y, and y from s to (abc)

fB(x, y) = 374 - x 2 - 2x2y 2 + y4 _ 2y2 + 1 = 0,

and for C with distances x and y from s to two vertices of the unit triangle, and with distance zx/~/3 from s to the center of (abc)

f c (x , y, z) = 3 x 4 + 3 x 2 y 2 - 3 x 2 z 2 - 3 x 2

+ 3 y 4 -- 3y 2 -- 3y2z 2 + Z 4 7 t- Z 2 -~- 1 = 0.

L E M M A 1. A point set of Sl (3, 5) which contains exactly one point p of property B is congruent to one of the six configurations shown in Figure 3.

Proof. The fifth point q besides a, b, c, and p has distances 1, a , and/3 to (abc). Assume that p is the center of (abc). Then the distance from q to p is equal to one of the three distinct distances from q to (abc), say to a. However, no point on


Fig. 3.

qg e

p ~ b a

q q ~

the perpendicular bisector of a and p has distance v'~/3 to exactly one vertex of (abc), a contradiction, and p can be assumed to have exactly two equal distances to ( abc ).

One of the two distinct distances from p to (abc) has to be 1 since fB(x, y) = 0 for p and fn(x , y) = 0 for q have common positive solutions only for x = 1 or y = 1. Thus (abcp) is a two-distance set with a unit triangle. There exist only three such configurations (see Figure 1), and the second distances are o~ = x/3, a = (v/2 + x/~)/2, and a = ( - v ~ + x/~)/2. From fA(x, a) = 0 for q we obtain the third distances/3 = 2,/3 = v~ , and/3 = v/2, respectively. There are six possibilities for q to have distances 1, a, and/3 to the unit triangle (abc). In each of the three cases only those two configurations of Figure 3 remain for which q has distance 1, a, or/3 to p. []

LEMMA 2. A point of property C can have at most two distinct distances to the vertices of the unit triangle (abc), that is, it is also of property B. All possibilities for these two distances 7 , 6, 6 are

(X/'3 V"3 2_~33 ~ v/'3 1) 3 ' 3 ) ' ( ' 3 ) ' ( - 2 - " 2 '

( l + v / 3 V~ l + x / ~ _X/if) 2 ' 2 ) ' ° r ( - 2 ' 2

with

e = 2, l + x / ~ , - - l + x / ~ , O, 1, 1 I + V ~ - - l+v / -3 2 , 2 ' o r 2 '

respectively. ..


Proof We assume a point p has three distinct distances 1, a, and/3 to (abe). We can choose y = 1 for fc (x , y, z) = 0. If z = 1 we have a condition f c (x , l, 1) = 0. If z = a or/3 we can choose those two vertices of (abc) which imply the condition f c (x , 1, x) -- 0. The only positive solution for both equations is x = 1, a contradiction. Thus any point of property C has property B, too.

To determine all pairs of distances from p to (abc) we first assume that p is of property A, too. Then (abcp) again is a two-distance set with a unit triangle, and there exist three possible configurations in Figure 1 with a = v/3, a = (v/2 + v ~ ) / 2 , and a = ( - v / 2 + v/-6)/2. These configurations determine the first three pairs (7, 5) of Lemma 2 and the distances e v ~ / 3 from p to the center of (abc) give the first three values of E.

It remains that p has no unit distance to (abc). We have three necessary conditions for the distances in (abcp), namely, f~(x, y) = 0, fc(Y, Y, z) = 0, and fc(x , y, z) = 0 if appropriate pairs of vertices of (abe) are considered.

For z = 0, z = 1, and z = y, the second condition determines y = x/3/3, y = v ~ / 3 , and y = 1/2, respectively. The first condition yields x = x/3/3 and x = 2v/3/3 for y = v ~ / 3 and x = v ~ / 2 for y = 1/2. Since x = y can occur only if z -- 0, there remain the three possibilities (x, y, z) = (v '~/3, v ~ / 3 , 0), (2v/-3/3, v/3/3, 1 ), (v/3/2, 1/2, 1/2) which can be checked to be realizable. Thus in Lemma 2 the next three pairs of (7, 5) and the next three values of e are determined. It remains z = x. Now also the third of the three necessary conditions will be used, and we obtain for fB( x, y) = O, fC( Y, Y, x) = O, and fc( x, y, x) = 0 only the solutions (x ,y) = ((1 + v/3)/2, v/2/2) and (x ,y) = ( ( - 1 + v ~ ) / 2 , v/2/2). Both possibilities can be checked to be realizable and with the last two pairs (7, 5) and the last two values of ~ the proof of Lemma 2 is complete. []

For the proofs of the following two lemmas we note that exactly six points exist which are not of property B and which have the three distinct distances 1, a, and/3 to the vertices of (abc). These six points belong to two congruent regular triangles both having its center in the center of (abc).

LEMMA 3. Every point set of S1 (3, 6) contains a point of property B. Proof If a point of property B does not exist then two points belong to one of

the two regular triangles mentioned above. Since the side length of this triangle is 1, c~, or fl both points have property C. This contradicts Lemma 2. []

LEMMA 4. All point sets of $1 (3, 6) which contain exactly one point of property B are similar to one of the configurations (3), (5), (7), and (9) in Figure 2.

Proof Five of the six points which include the point of property B determine by Lemma 1 only the three possible pairs (a,/3) given in Figure 3. The two points without property B are vertex points of the two regular triangles mentioned above. If both points would belong to one triangle, both would have property C and then by Lemma 2 also property B. Thus each point belongs to one of the triangles


c

a b a b

Fig. 4.

and they form together with two points of (abc) an isosceles trapezium with one parallel of unit length. For the opposite parallel of length e we get e -- 7 2 -/~2 by Ptolomy's theorem where 7 and 6 denote the lengths of the legs and the diagonal, respectively.

All nine possibilities to choose 7 and ~ from 1 and the possible values a and fl give values for e from 1, a, or/~ only in the four cases shown in Figure 4.

The sixth point which is of property B has to correspond to point p in Figure 3. As combinations of Figures 3 and 4 only the configurations (3), (7), and (9) of Figure 2 are possible. []

LEMMA 5. Any point set of S1 (3, 6) which contains exactly two points of property B is similar to configuration (7) in Figure 2.

Proof Let Pl, P2 denote the points of property B and q the remaining one. Since (abcplq) and (abcp2q) belong to S 1 (3, 5) we conclude from Lemma 1 that (abepl) and (abcp2) are congruent. Only for (a , fl) = (x/3, 2) the points Pl and P2 do not determine a fourth distance (see Figure 3). Checking all possibilities for the point q, only configuration (7) remains. []

LEMMA 6. If (abcpl) and (abcp2) both are point sets of Sl(2, 4) and (abcplp2) belongs to $1 (3, 5) then they are congruent.

Proof If equilateral triangles of different sets of S1 (2, 4) (see four sets of Fig- ure 1) are identified in all possible ways then always four different distances occur, and thus none of these sets of five points belongs to S1 (3, 5). []

LEMMA 7. Let A4 be a set of $1 (3, 6) with all three points of property B. Then - (7.1) one of the configurations (4), (6), (7), (8), or (9) in Figure 2 is similar

to ;k4 if no point of.A4 is of property A, - (7.2) it is impossible that exactly one point in A4 is ofpropertyA, - (7.3) one of the configurations (3) or (5) in Figure 2 is similarto .M if exactly

two points in ./k4 are of property A, - (7.4) one of the configurations (4), (6), or (9) in Figure 2 is similar to A4 if

all three points in ./k4 are of property A.

Proof. (7.1) Two distinct points of property B in .A4, say Pl and P2, one with distances a , a, and fl and the other one with distances a , fl, and fl to (abc), cannot


exist since f a (x , y) = 0 and fB(Y, x) = 0 only have the common solution x = y. Thus (abcpl) and (abcp2) are congruent if Pl and P2 only have distances a and 3 to (abc). From the isosceles triangle of Pl, P2, and the center of (abe) it can be deduced that Pl and P2 also have property C.

Since no point in .A4 is of property A, that is, Pl, P2, and P3 have distances c~ or /3 to (abc), all three points Pl,P2,P3 have properties B and C.

If one point, say pl, is the center of (abc) then from Lemma 2 only the distances 2x/3/3 and v '~ /3 are possible for the congruent sets (abcpi) for i - 2, 3 and this determines configuration (7).

Otherwise only the last four pairs of distances in Lemma 2 are possible for the three congruent sets (abcpi), i = 1,2, 3, and configurations (4), (6), (8), and (9) are determined.

(7.2) A point, say Pl, of properties A and B has distances 1 and one of v'~, (x/~+ v/-6)/2, ( - v ' ~ + v/-6)/2 to (abc). If P2 or P3, say P2, is the center of (abc) then (abcpl) and (abcpz) both are point sets of St(2, 4) and are congruent by Lemma 6 in contradiction to Pl ~ P2. Thus P2 and P3 have distances a and/3 to (abc) and, as in the first paragraph of (7.1), it follows that P2 and P3 are of property C. Since no pair (7, iS) with distance v/-3, (x/2 + v/6)/2, or ( - x / 2 + x/6)/2 and a second distance ~ 1 occurs in Lemma 2, a configuration cannot exist.

(7.3) For the two points Pl, P2 of property A and B, Lemma 6 implies that (abcpl) and (abcp2) are congruent. Thus the distances t~ and/3 are fixed already. They can be calculated or found in the first three triples (7, 6, e) of Lemma 2 since, corresponding to the first paragraph of the proof of (7.1), it follows that Pl and P2 are also of property C. The third point p3 is not of property A and thus can only have the two distances to (abc) which are different from 1. Since fB(2, v/3) 0, fB(V'~, 2) # 0, fB((V/2+X/~)/2, 1 +V/J) ¢ 0, fB((--V"-2+V"6)/2, --1 +V'~) ¢ 0, and fB(1 + X/-3, (x/r2 + X/~)/2) = 0, fB(--1 + X/~, (--X/~ + X/-6)/2) = 0on ly the two configurations similar to (3) and (5) are possible.

(7.4) Lemma 6 implies that (abcpi) are pairwise congruent for i = 1,2, 3 and the three different triples of points of properties A and B are similar to configurations (4), (6), and (9). []

$2(3,6) and Sz(3, 5) In Figure 5, four graphs F1, F2, F3, and F4 with four vertices and three different types of edges, are given. If x, y, and z are used as distances for the three edge types , , and . . . . . , respectively, then a necessary condition for the realization of an Fi in the plane is the vanishing of the volume of the tetrahedron with vertices corresponding to the vertices of Fi. This implies

f f l (x ,y ,z) = x 6 -- 2x4y 2 + x2y 4 -- 3x2y2z 2 + y2z4 = O,

F2(x,y ,z ) = (x 4 -- 2z2y 2 -- x2z 2 + y4)(x2 + 2y 2 -- z 2) = O,


F 3 ( x , y , z ) = 2x 4 - 2 x 2 y 2 - 2 x 2 z 2 + y4 _ 2y2z2 + z 4 _- 0,

F4(x, y, z) = x2y 2 - 4x2z 2 + Z 4 ---- 0.

LEMMA 8. Every point in a set o f $2(3,5) has equal distances to at most two points ofS2(3, 5).

Proof. Assume that in the graph of a configuration in $2(3,5) the subgraph F4 of Figure 5 occurs. If three vertices with only one edge type and four vertices with only two edge types are avoided then only the seven possibilities in Figure 6 remain for five vertices and three edge types.

.....--...- w - v -

Fig. 5.

X . ~ "" i X 1¢ I t l S "%

S 'm .

li E; 1~,

Fig. 6.

""...,

For Fs, F6, and F7 a rhombus of side length x, x, and z exists, respectively, and thus 4x 2 = y2 + z2 and 4z 2 = x 2 + y2 holds. Since both equations together with F4(x, y, z) = 0 do not have nontrivial common solutions, that is, with x, y, z > 0, pairwise different and y, z ~ 1, and thus realizations of Fs, F6, and F7 cannot exist.

For F8 the vertices (1235) and (2345) induce F1 and F3. The nontrivial common solutions of F l ( x , y , z ) = 0 and F 3 ( z , x , y ) = 0 fulfill y = xv/'2 and 2z 4 - 6z2x 2 + x 4 = 0. The last equation, however, does not have common solutions with F4(x, y, z) = O.

For F9 the vertices (2345) and (1235) induce two Fl 's and nontrivial common solutions of F 1 ( x , y , z ) = O, F l ( y , x , z ) = O, and F 4 ( x , y , z ) = 0 do not exist.

For F~0 two F1 's are induced by (1235) and (1345). Nontrivial common solutions of Fl(x, z, y) = 0, Fl(z , x, y) = O, and Fa(x, y, z) = 0 do not exist.


O) (2) (3)

i " . : ". - t

• ". : • • ".. :" • , : ' . . . " . ,

(4) (5) (6)

Fig. 7.

5 3 . ..

.: • • '.,

1 2

. t , ,

Fig. 8.

For Fll the vertices (1235) and (1234) both induce F2 so that F2(x, y, z) = 0 and Fa(z, y, x) = 0. If both second factors of these equations or the first of one and the second of the other equation vanish, then y = 0. If both first factors vanish and F4(x, y, z) - 0 then nontrivial common solutions do not exist. []

LEMMA 9. Every point set ofS2(3, 5) is similar to one configuration of Figure 7. Proof In Figure 8 all graphs with five vertices and three edge types are con-

strutted such that no equilateral triangle occurs, no four vertices determine at most two edge types, and no vertex is incident to more than two edges of the same type (see Lemma 8). For the construction it may be noticed that one edge type has to occur five or four times. This is possible as a cycle with five vertices, as a path with five vertices, or as a cycle with four vertices. The cycles determine F12, F19, and F20. If the vertex in the center of the path is incident to two edges of type 2 or to one edge of type 2 and one of type 3, then only/'13, El4, and F15, or El6, FI7, and F18 are possible, respectively.


For F12 the vertices (1235), (1345), and (2345) induce F2, F2, and Fz so that F2(x ,y ,z ) = O, F2(x ,z ,y) = O, and F l (x , y , z ) = 0. All nontrivial solutions, y = x /2 and z = (x/2)x/~, correspond to configurat!ons similar to Figure 7(I).

For F13 the vertices (2345) and (1245) yield F1 (x, z, y) = 0 and F1 (z, x, y) = 0, respectively. Assuming x = 1, both nontrivial solutions,

(y ,z ) = (~/(1 + v f i 3 ) / 2 , ~ / ( 3 + v / ~ ) / 2 ) ,

(V/(5 - v'q-3)/2, V/( -3 + v / ~ ) / 2 ) ,

yield configurations similar to Figure 7(2). For /'14 the vertices (1345) and (1234) determine Fl (x , z , y ) = 0 and

Fl(z, x, y) = 0 which implies y4 = ( x 2 _ Z2)2. This contradicts F3(y, x, z) = 0 from vertices (1245).

For F15 the vertices (2345) and (1235) yield F2(x,y ,z) = 0 and F2(z, x, y) = 0. If the first two factors of Fz(x, y, z) and F2(z, x, y) vanish then for x = 1 thepolynomialy6-5y4+6y 2 - 1 = 0hasthree solut ions(1.80. . . , 1 .24 . . . , and 0 .44 . . . ) by which configurations similar to (4), (5), and (6) in Figure 7 are determined. Only one other solution occurs if the first factor of F2(x, y, z) and the second of F2(z, x, y) vanish where X 2 ---- 2y 2 corresponds to configuration (3) in Figure 7.

For F16 the vertices (2345), (1245), and (1235) imply Fl (x , y , z ) = O, F l ( y , x , z) -- 0, and F2(z, x, y) = 0, respectively. Subtraction of the first two equations with x = 1 gives z 4 = (y2 _ 1)2. Together with the third equation it follows that no configuration can occur.

For Fl7 the vertices (1245) and (1235) imply F3(y, x, z) = 0 and F3(x, y, z) = 0. Thus no configuration is possible.

For F18 the vertices (2345), (1345), and (1245) are used to get F2(x, z, y) = 0, Fl(x, y ,z) --- 0, and F3(y,x, z) = 0, respectively, so that configurations are impossible.

For El9 the vertices (2345) and (1345) determine F3(z ,x ,y) = 0 and F3(y, x, z) = 0 yielding y4 = Z 4 SO that no configuration can occur.

For F20 the vertices (2345) and (1345) imply Fl ( z , x , y ) = 0 and FI(y, x, z) --- 0. The vertices (1235) determine the vertex points of a rhombus so that y2 + z 2 = 4x 2. Substitution of y2 = 4 - z 2 into F l ( z , l , y ) = 0 and F1 (y, 1, z) = 0 does not give a solution, and thus no configuration exists. []

LEMMA 10. Any point set ofS2(3, 6) is similar to configuration (2) of Figure 2. Proof If any point of a set of $2(3,6) is deleted then a point set of $2(3,5) in

Figure 7 remains which for all six points determines the same three distances. For each of the configurations (1), (2), or (3) of Figure 7, two of those distances which occur only two or three times cannot be identified to give six points with only three distances.


The remaining three configurations of Figure 7 are concyclic with the same radius. By rotation only configuration (2) of Figure 2 is possible. []

5'3(3, 6) LEMMA 11. If every set of five vertices in a graph G with six vertices contains four independent vertices then all edges of G are incident to exactly one vertex of G.

Proof Every graph with five vertices and two independent edges does not contain four independent vertices. []

LEMMA 12. In any point set ofS3(3, 6) all point sets of S(2, 4) are vertex points of a regular pentagon.

Proof At least one point set of 5'(2,4) (see Figure 1) is contained in any point set of 5'3(3, 6). Since no equilateral triangle is allowed only the first two configurations of Figure 1 remain. If the square is proved to be impossible then Lemma 12 is proved.

If a unit square exists then the fifth and sixth point only can lie on one of the four axes of symmetry of the square.

One point which is not the center of the square and which lies on an axis

parallel to two square sides has distances v ~ and V/3 + v/ff or V/3 - v/ff since unit triangles do not occur in $3(3,6). By definition all six five-point subsets only can belong to $2(3,5) or $3(3,5). Since no triple of distances in Figure 7 is similar to

the distances 1, v~ , and V/3 + v/ff all five-point subsets belong to 5'3(3, 5), that is, they contain a four-point two-distance set. Checking the ratios of the two distances in Figure 1 it follows that only v ~ / 2 is possible for these sets of four points. Then

by Lemma 11 all distances V/3 + v/ff are incident to one point, and five points only

with distances 1 and V~ should exist, which is impossible. Any point on an axis of symmetry through opposite vertices of the unit square

determines two distances to the vertices of the square besides 1 and v/2 so that more than three distances occur. []

LEMMA 13. In any point set of $3(3,6) all point sets of S(2,4) are pairwise congruent.

Proof It follows from Lemma 12 that all point sets of S(2, 4) are similar to four vertex points of a regular pentagon. At most the two incongruent regular pentagons with sides 1 and r = ( 1 + v ~ ) / 2 are possible since then already the three distances 1, r, and ~_2 are determined. Assume the existence of both possible point sets of S(2, 4), say T1 and T~, with distances 1, r, and r , r 2, respectively. Both points being disjoint from T1 have at least one distance 7 -2 to Tl since T2 exists. At most one of the two points can have distance r to T1 since 1 + r = 7 -2. Then the second point has distances r 2 to all four points of T1 since a triangle with sides of length


Fig. 9.

1, 1, and 7-2 does not exist. However, the circumradius of the unit pentagon is less than 7-2, a contradiction. []

LEMMA 14. Any point set of $3 (3, 6) is similar to configuration (1) of Figure 2. Proof. Because of Lemma 12, two distances can be assumed to be 1 and r. If

any point is deleted then the remaining set of five points belongs to $3(3,5) since no set of $2(3,5) has two distances in the ratio 1:7- (see Figure 7). Thus every subset of five points contains four points with distances 1 and 7- only. Then by Lemma 11 all third distances, say a, are incident to one point p. The remaining five points have distances 1 and 7- only so that they have to be the vertex points of a regular pentagon (see Figure 1). If p is incident to at least three distances a then configuration (1) of Figure 2 is determined. Otherwise three vertex points of the pentagon together with p have distances 1 and 7- only and thus, by Lemma 12, they also have to be vertex points of a regular pentagon, which is impossible. []

Now S(3, 6) is discussed completely and Theorem 1 follows from Lemmas 4, 5, 7, 10, and 14. []

3. All Three-Distance Sets with Seven Points

For seven points with three different distances every subset of six points has to be one of Figure 2. Because of the different distances, only configuration (1), configuration (2), configurations (3) or (4), configurations (5) or (6), and configurations (7), (8), or (9) can be combined, respectively. The following theorem is obtained if all five-point subsets within these five classes of configurations are compared.

THEOREM 2. Any configuration of seven points in the plane determines exactly three distinct distances if and only if it is similar to one of the r(7) = 2 examples shown in Figure 9.

Since the configurations of Figure 9 cannot be combined to give a three-distance set and since the regular nonagon determines four distinct distances the following corollary results.

COROLLARY. f(8) = f (9) = 4.

For further investigations it might be of interest to determine all sets of eight points with four distances [9]. We conjecture that r(8) = 13.


References

1. Altman, E.: On a problem of E ErdOs, Amer. Math. Monthly 70 (1963), 148-157. 2. Altman, E.: Some theorems on convex polygons, Canad. Math. Bull. 15 (1972), 329-340. 3. Chung, E R. K., Szemer6di, E. and Trotter, W. T.: The number of different distances determined

by a set of points in the Euclidean plane, Discrete Comput. Geom. 7 (1992), 1-11. 4. Croft, H. T., Falconer, K. J. and Guy, R. K.: Unsolved Problems in Geometry, Springer-Vedag,

Berlin, Heidelberg, New York, 1991. 5. Einhorn, S. J. and Sch.oenberg, I. J.: On Euclidean sets having only two distances between points

II, lndag. Math. 28 (1966), 479-504. 6. ErdOs, P.: On sets of distances of n points, Amer. Math. Monthly 53 (1946), 248-250. 7. ErdOs, P. and Fishburn, P.: Maximum planar sets that determine k distances, Preprint, 1994. 8. Fishburn, P.: Convexpolygonswithfewintervertexdistances,DIMACS, TechnicalReport92-18. 9. Moser, W. and Pach, J.: Recent developments in combinatorial geometry, in: J. Pach (ed.),

New Trends in Discrete and Computational Geometry, Springer-Vedag, Berlin, Heidelberg, New York, 1993, pp. 281-302.

10. Piepmeyer, L.: Punktmengen mit minimaler Anzahl verschiedener Abst~nde, Dissertation, Techn. Univ. Braunschweig, 1992.

three distinct distances in the plane

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