Let $S$ be a set of $n\geq 7$ points in the plane, no three of which are
collinear. Suppose that $S$ determines $n+1$ directions. That is to say, the
segments whose endpoints are in $S$ form $n+1$ distinct slopes. We prove that
$S$ is, up to an affine transformation, equal to $n$ of the vertices of a
regular $(n+1)$-gon. This result was conjectured in 1986 by R. E. Jamison.
  In an addendum to the paper, we show that a much stronger result can be
obtained as a corollary of a structure theorem of Green and Tao on point sets
spanning few ordinary lines.Comment: Paper: 7 pages, 5 figures. Addendum: 3 page

Pilatte, Cédric

English

arXiv

peer reviewedLet S be a set of n > 7 points in the plane, no  three of which are collinear. Suppose that S determines n + 1 directions. That is to say, the segments whose endpoints are in S form n + 1 distinct slopes. We prove that S is, up to an affine transformation, equal to n of the vertices of a regular (n + 1)-gon. This result was conjectured in 1986 by R. E. Jamison

ORBi UMONS

On the sets of n points forming n + 1 directionsCédric PilatteDepartment of MathematicsUniversity of MONS (UMONS)Place du Parc, 207000 Mons, Belgiumcedric.pilatte@umons.ac.beSubmitted: Nov 11, 2018; Accepted: Dec 22, 2019; Published: Jan 24, 2020c©The author. Released under the CC BY-ND license (International 4.0).AbstractLet S be a set of n > 7 points in the plane, no three of which are collinear.Suppose that S determines n + 1 directions. That is to say, the segments whoseendpoints are in S form n + 1 distinct slopes. We prove that S is, up to an affinetransformation, equal to n of the vertices of a regular (n + 1)-gon. This result wasconjectured in 1986 by R. E. Jamison.Mathematics Subject Classifications: 52C10, 52C30, 52C351 IntroductionIn 1970, inspired by a problem of Erdős, Scott [15] asked the following question, nowknown as the slope problem: what is the minimum number of directions determined by aset of n points in R2, not all on the same line? By the number of directions (or slopes) ofa set S, we mean the size of the quotient set {PQ | P,Q ∈ S, P 6= Q}/∼, where ∼ is theequivalence relation given by parallelism: P1Q1 ∼ P2Q2 ⇐⇒ P1Q1 ‖ P2Q2.Scott conjectured that n points, not all collinear, determine at least 2bn2c slopes. Thisbound can be achieved, for even n, by a regular n-gon; and for odd n, by a regular (n−1)-gon with its center. After some initial results of Burton and Purdy [2], this conjecture wasproven by Ungar [16] in 1982, using techniques of Goodman and Pollack [5]. His beautifulproof is also exposed in the famous Proofs from the Book [1, Chapter 11]. Recently, Pach,Pinchasi and Sharir solved the tree-dimensional analogue of this problem, see[12, 13].A lot of work has been done to determine the configurations where equality in Ungar’stheorem is achieved. A critical set (respectively near-critical set) is a set of n non-collinear points forming n− 1 slopes (respectively n slopes). Jamison and Hill describedfour infinite families and 102 sporadic critical configurations [6, 7, 10]. It is conjecturedthe electronic journal of combinatorics 27(1) (2020), #P1.24 1that this classification is accurate for n > 49. No classification is known in the near-criticalcase. See [8] for a survey of these questions, and other related ones.In this paper, we suppose that no three points of S are collinear (we say that S isin general position). This situation was first investigated by Jamison [9], who provedthat S must determine at least n slopes. As above, equality is possible with a regularn-gon. It is a well-known fact that affine transformations preserve parallelism. Therefore,the image of a regular n-gon under an affine transformation also determines exactly nslopes.1 Jamison proved the converse, i.e. that the affinely regular polygons are the onlyconfigurations forming exactly n slopes.A much more general statement is believed to be true: for some constant c1, if a set ofn points in general position forms m = 2n − c1 slopes, then it is affinely equivalent to nof the vertices of a regular m-gon (see [9]). This would imply, in particular, that for everyc > 0 and n sufficiently large, every simple configuration of n points determining n + cslopes arises from an affinely regular (n + c)-gon, after deletion of c points. Jamison’sresult thus shows it for c = 0. Here, we will prove the case c = 1. The general conjectureis still open. In fact, for c > 2, it is not even known whether the points of S form a convexpolygon.Every affinely regular polygon is inscribed in an ellipse. Conics will play an importantrole in our proof. Another problem of Elekes [3] is the following: for all m > 6 and C > 0,there exists some n0(m,C) such that every set S ⊂ R2 with |S| > n0(m,C) forming atmost C|S| slopes contains m points on a (possibly degenerate) conic. It is still unsolved,even for m = 6.2 Preliminary RemarksLet S be a set of n points in the plane, in general position, that determines exactly n+ 1slopes. If S had a point lying strictly inside its convex hull, there would be at least n+ 2slopes, as was proved by Jamison [9, Theorem 7]. Therefore, we know that we can labelthe points of S as A1, . . . , An, such that A1A2 . . . An is a convex polygon.For every point Ai ∈ S, there are n − 1 segments, with distinct slopes, joining Ai tothe other points of S. We will say that a slope is forbidden at Ai if it is not the slopeof any segment AiAj, for j 6= i. Since S determines n + 1 slopes, there are exactly twoforbidden slopes at each point of S.We will denote by ∇AiAj the slope of the line AiAj. Thus, an equality like ∇Ai1Ai2 =∇Ai3Ai4 is equivalent to Ai1Ai2 ‖ Ai3Ai4 . Throughout our main proof, we will repeatedlymake use of the next lemma. It will be particularly useful to prove that a slope is forbiddenat a point or that two slopes are equal. As an obvious corollary, we have that ∇Ai−1Ai+1is forbidden at Ai for all i ∈ Z. Throughout the paper, when we say “for all i ∈ Z”, weconsider the indices modulo n, so that An+1 := A1, and so on.1A polygon obtained as the image of a regular polygon by an affine transformation is sometimes calledan affine-regular or affinely regular polygon.the electronic journal of combinatorics 27(1) (2020), #P1.24 2Lemma 1. Let 1 6 i < j < k 6 n. Exactly one of the following is true:• the slope of AiAk is forbidden at Aj;• ∃p, i < p < k such that AiAk ‖ AjAp.Moreover, in the second case, ∇AjAp /∈ {∇AiAl | l 6= j, k} ∪ {∇AlAj | l 6= i, k}.Proof. This is almost immediate from the definition of a forbidden slope. In the secondcase, if p ∈ {1, . . . , n} were not between i and k, the segments AiAk and AjAp wouldintersect. Finally, if ∇AjAp were equal to some ∇AiAl, then AiAk ‖ AjAp ‖ AiAl, soAi, Ak and Al would be aligned, a contradiction. The same is true for the segmentsAlAj.We will also need the following result, which can be found in [14, Chapter 1].Proposition 2. Let C be a non-degenerate conic and O a point on C. If P,Q are twopoints on C, define P + Q to be the unique point R on C such that RO ‖ PQ (with theconvention that XX is the tangent to C at X, for X ∈ C). This addition turns C into anabelian group, of which O is the identity element.In particular, for P,Q,R, S four points on C, we have P + Q = R + S if and onlyif PQ ‖ RS. Lemma 3 will enable us to introduce conics in the proof, in order to useproposition 2.Lemma 3. Suppose P1, . . . , P6 are points in the plane such that P1P6 ‖ P2P5, P2P3 ‖ P1P4and P4P5 ‖ P3P6. Then P1, . . . , P6 lie on a common conic.Proof. This follows immediately from Pascal’s theorem applied to the hexagon H =P1P4P5P2P3P6. Indeed, the intersections of the opposite sides of H are collinear on theline at infinity.P1P6P2P5P4P3Figure 1: Illustration of lemma 3.For the reader’s convenience, we reproduce here a result of Korchmáros [11] (which isalso discussed in [4]), that we will use twice in the proof.Lemma 4. Let P1, . . . , Pn be distinct points on a non-degenerate conic. Suppose that, forall j ∈ Z, Pj+1Pj+2 ‖ PjPj+3. Then, P is affinely equivalent to a regular n-gon.the electronic journal of combinatorics 27(1) (2020), #P1.24 33 Main TheoremIn this section, we prove the following theorem, using the results from section 2.Theorem 5. Any set S of n > 7 points in the plane, in general position, that determinesexactly n + 1 slopes, is affinely equivalent to n of the vertices of a regular (n + 1)-gon.Proof. We use the notations of section 2: S = {A1, A2, . . . , An} where A1A2 . . . An is aconvex polygon. We will split the proof into two cases. In the first case, we suppose that,for every i ∈ Z, Ai+1Ai+2 ‖ AiAi+3. If this fails for some i, we can assume that this i is 1.Case 1 For every i ∈ Z, Ai+1Ai+2 ‖ AiAi+3.We will distinguish subcases according to which segments are parallel to AiAi+5. Aswe will see, none of the subcases are actually possible.Case 1.1 For all i ∈ Z, AiAi+5 ‖ Ai+1Ai+4.Let Ak+1, . . . , Ak+6 be any six consecutive points of S. We have Ak+1Ak+6 ‖ Ak+2Ak+5,Ak+2Ak+3 ‖ Ak+1Ak+4 and Ak+4Ak+5 ‖ Ak+3Ak+6 from our two assumptions. Thus, lemma3 implies that the six points lie on a common conic. As this is true for any six consecutivepoints, and since five points in general position determine a unique conic, all the Ai’s lieon the same conic. Together with the fact that ∀i, Ai+1Ai+2 ‖ AiAi+3, this implies thatA1A2 . . . An is affinely equivalent to a regular n-gon, by lemma 4. Therefore, S determinesexactly n directions, which is a contradiction.Case 1.2 For some i ∈ Z, we have AiAi+5 ‖ Ai+2Ai+4.A1A2A3A4A5A6A7Figure 2: Case 1.2.Say i = 1, meaning A1A6 ‖ A3A5. By lemma 1 applied three times, we see that∇A2A6 is forbidden at A3, A4 and A5 (here, we have used that A2A6 ∦ A3A5 and, forA4, that A3A4 ‖ A2A5 and A4A5 ‖ A3A6). For l = 3, 4, 5, we know that ∇A2A6 and∇Al−1Al+1 are exactly the two forbidden slopes at Al. Therefore, ∇A1A5 is not forbiddenthe electronic journal of combinatorics 27(1) (2020), #P1.24 4at A4, hence, by lemma 1 again, we conclude that A1A5 ‖ A2A4. Similarly, ∇A3A7 isnot forbidden at A4, so A3A7 ‖ A4A6. As the slope of A2A7 is not forbidden at A5, weconclude A2A7 ‖ A4A5(‖ A3A6). We have A3A4 ‖ A2A5, A3A6 ‖ A2A7 and we just showedthat A2A7 ‖ A3A6. By lemma 3, A2, A3, . . . , A7 lie on a common conic.We will equip this conic with the group structure descibed in lemma 2, with A7 thezero element. We will write A7 = 0 and A6 = x. Then, A5A6 ‖ A4A7, A4A5 ‖ A3A6 andA4A6 ‖ A3A7 together imply A5 = 2x, A4 = 3x and A3 = 4x. Also, A3A4 ‖ A2A5 givesA2 = 5x. Let B be the point on the conic with B = 6x. We thus have A2A3 ‖ BA4and A2A4 ‖ BA5. However, there can only be one point P with A2A3 ‖ PA4 andA2A4 ‖ PA5. As A1 is such a point, A1 = B = 6x. This contradicts A1A6 ‖ A3A5, asA1 + A6 = 6x + x 6= 4x + 2x = A3 + A5.Case 1.3 For some i ∈ Z, we have AiAi+5 ‖ Ai+1Ai+3.This is exactly the previous case after having relabelled every Ai as An+1−i.Case 1.4 The previous cases do not apply.If none of the previous cases is possible, there must be some i, say i = 1, for whichA1A6 is not parallel to any of A2A5, A3A5 and A2A4. Then, ∇A1A6 is forbidden atA2, A3, A4 and A5. Once again, we deduce that the forbidden slopes at Al, 2 6 l 6 5, are∇A1A6 and ∇Al−1Al+1. We use lemma 1 to find A2A6 ‖ A3A5 (applied with Ak = A4)and A1A5 ‖ A2A4 (Ak = A2).Let C be the conic passing through A1, A2, . . . , A5. We use lemma 2 to define agroup structure on C, with A1 = 0. Let A2 = x and A3 = y. From A2A3 ‖ A1A4 andA3A4 ‖ A2A5, we have A4 = x + y and A5 = 2y. But A2A4 ‖ A1A5 implies y = 2x,so Ai = (i − 1)x for 1 6 i 6 5. We use the same argument as before. Let B = 5x,then A4A5 ‖ A3B and A3A5 ‖ A2B, so B = A6 = 5x. We deduce A1A6 ‖ A2A5, acontradiction.A6A5A4A3A2A1Figure 3: Case 1.4.the electronic journal of combinatorics 27(1) (2020), #P1.24 5Case 2 We have A2A3 ∦ A1A4 (without loss of generality).Without loss of generality, we can also suppose that the point A4 is closer to the lineA2A3 than is A1. In this situation, the line parallel to A2A3 passing through A4 intersectsthe segment [A1A2] in its relative interior, and the line parallel to A2A3 passing throughA1 does not intersect the segment [A3A4].From A2A3 ∦ A1A4, we deduce that the forbidden slopes at A2 and A3 are ∇A1A3,∇A1A4 and∇A2A4,∇A1A4, respectively. Thus, A1A2 ‖ AnA3 and A2A5 ‖ A3A4. We nowshow that A2A3 is forbidden at A4. Suppose, for some k, that A2A3 ‖ A4Ak. Then, k hasto be between 5 and n, so A1A2A3A4Ak must be a convex polygon, with A2A3 ‖ A4Ak.We can see that this contradicts the fact that A4 is closer than A1 to the line A2A3.Case 2.1 An−1A2 ‖ AnA1.We want to show that this case is impossible. From lemma 1, we find An−1A3 ‖ AnA2.When we apply this lemma again with the slope of AnA4, we find that AnA4 is parallelto A1A3, because A2A3 is forbidden at A4. In the same way, we get An−1A4 ‖ AnA3.A5A4A3A2A1AnAn−1Figure 4: Case 2.1.Let C be the conic passing through A3, A2, A1, An and An−1. Again, we use propo-sition 2, setting An−1 = 0. Let An = x and A2 = y. From An−1A3 ‖ AnA2 we deduceA3 = x+y, and from An−1A4 ‖ AnA3 we get A4 = y+2x. Let B = 2x. Then AnA4 ‖ BA3and AnA3 ‖ BA2. This means that B belongs to the line parallel to AnA4 through A3and to the line parallel to AnA3 through A2. So B = A1, i.e. A1 = 2x. On the one hand,the relation An−1A2 ‖ AnA1 gives 0 + y = x+ 2x. On the other hand, A2A3 ∦ A1A4 yieldsy + (y + x) 6= 2x + (y + 2x). This is a contradiction.Case 2.2 An−1A2 ∦ AnA1.This is the last case of the proof, and the only case that produces valid configurationsof points. As An−1A2 ∦ AnA1, ∇An−1A2 is forbidden at A1. With ∇A0A2, those are thetwo forbidden slopes at A1. Therefore, none of ∇A2Ai, 3 6 i 6 n− 2 is forbidden at A1.So, every ∇A2Ai, 3 6 i 6 n− 2, corresponds to a unique ∇A1Aj for some j.the electronic journal of combinatorics 27(1) (2020), #P1.24 6A1A2A3A4A5AnAn−1An−2Ai−2AiFigure 5: Case 2.2.A simple but important observation is that, for all 3 6 i1, i2 6 n−2 and 4 6 j1, j2 6 n,{A2Ai1 ‖ A1Aj1A2Ai2 ‖ A1Aj2=⇒ (i1 < i2 ⇐⇒ j1 < j2) .That is, the assignment f that maps every 3 6 i 6 n − 2 to the unique 4 6 j 6 n suchthat A2Ai ‖ A1Aj must be strictly increasing. Moreover, it has to satisfy f(3) 6= 4 as weassumed A2A3 ∦ A1A4. The unique possibility is then f(i) = i + 2 for every i. We haveproven that, for 5 6 i 6 n, A2Ai−2 ‖ A1Ai.Claim 2.2.1. For every i ∈ {5, . . . , n},1. Ai−2Ai and A2Ai−2 are the two forbidden slopes at Ai−1, and;2. ∀k ∈ {3, . . . , i− 2}, Ai−1Ak ‖ AiAk−1.Proof of claim. For i = 5, we have already proven those two statements. We will provethem for i = j, assuming it has already been proven for all 5 6 i 6 j − 1.1. We have to show that A2Aj−2 is forbidden at Aj−1. This is clear as A2Aj−2 ‖ A1Ajand there is no point of S between A1 and A2.2. Since we know the forbidden slopes at Aj−1, we can use lemma 1 at the point Aj−1several times, with different slopes. First, ∇AjAj−3 is not forbidden, so AjAj−3and Aj−1Aj−2 are parallel. Then ∇AjAj−4 is not forbidden, and is distinct from∇Aj−1Aj−2 = ∇AjAj−3, so AjAj−4 ‖ Aj−1Aj−3. We can continue this way, until weget AjA2 ‖ Aj−1A3. This concludes the proof of the claim.In particular, for every i ∈ {6, . . . , n − 1}, we have A3Ai ‖ A2Ai+1, A5Ai ‖ A4Ai+1.As A3A4 ‖ A2A5, we can use lemma 1, which shows that A2, A3, A4, A5, Ai and Ai+1 lieon a conic. As this is true for every 6 6 i 6 n− 1, we know that the Ai’s, for 2 6 i 6 n,the electronic journal of combinatorics 27(1) (2020), #P1.24 7all lie on a common conic (because there is a unique conic passing through five points ingeneral position).As we have done several times in this proof, we use the group structure on the conicgiven by parallelism. Choose A2 to be the identity element, let A3 = x. SolvingA3A4 ‖ A2A5A3A6 ‖ A4A5A3A5 ‖ A2A6gives A4 = 2x, A5 = 3x and A6 = 4x. Then, a simple induction (using Ai−1A3 ‖ AiA2)gives Ai = (i− 2)x for all i ∈ {2, . . . , n}. Let B be the point on the conic with B = −2x.Then A2A3 ‖ BA5 and A2A4 ‖ BA6. However, we proved before that A2A3 ‖ A1A5 andA2A4 ‖ A1A6, so A1 = B = −2x.To summarize, we know that all the n points of S are on a conic, Ai = (i − 2)x fori ∈ {2, . . . , n} and A1 = −2x. We use the group structure one last time: A3An ‖ A1A2implies x + (n− 2)x = −2x + 0, so (n + 1)x = 0. Therefore, the subgroup generated byA3 = x is a finite cyclic group of order n + 1:〈A3〉 ={A2︸︷︷︸0, A3︸︷︷︸x, A4︸︷︷︸2x, . . . , An−1︸ ︷︷ ︸(n−3)x, An︸︷︷︸(n−2)x, A1︸︷︷︸(n−1)x,−x}.To finish the proof, we use the more convenient notations Pj := jx for 0 6 j 6 n (sothat every Ai is a Pj). If the indices are considered modulo n + 1, we have, for all j ∈ Z,Pj+1Pj+2 ‖ PjPj+3, because (j+1)x+(j+2)x = jx+(j+3)x. By lemma 4, P0P1P2 . . . Pnis, up to an affine transformation, a regular (n + 1)-gon.AcknowledgementsI would like to thank Christian Michaux, without whom this work would not have beenpossible.References[1] Martin Aigner and Günter Ziegler. Proofs from the Book, volume 274. Springer,2010.[2] G. R. Burton and G. B. Purdy. The directions determined by n points in the plane.Journal of the London Mathematical Society, 2(1):109–114, 1979.[3] György Elekes. On linear combinatorics III. Few directions and distorted lattices.Combinatorica, 19(1):43–53, 1999.[4] J. Chris Fisher and Robert E. Jamison. Properties of affinely regular polygons.Geometriae Dedicata, 69(3):241–259, 1998.[5] Jacob E. Goodman and Richard Pollack. A combinatorial perspective on some prob-lems in geometry. Congressus Numerantium, 32(0981):383–394, 1981.the electronic journal of combinatorics 27(1) (2020), #P1.24 8[6] Robert E. Jamison. Planar configurations which determine few slopes. GeometriaeDedicata, 16(1):17–34, 1984.[7] Robert E. Jamison. Structure of slope-critical configurations. Geometriae Dedicata,16(3):249–277, 1984.[8] Robert E. Jamison. A survey of the slope problem. Annals of the New York Academyof Sciences, 440(1):34–51, 1985.[9] Robert E. Jamison. Few slopes without collinearity. Discrete Mathematics, 60:199–206, 1986.[10] Robert E. Jamison and Dale Hill. A catalogue of sporadic slope-critical configura-tions. Congressus Numerantium, 40:101–125, 1983.[11] G. Korchmáros. Poligoni affin-regolari dei piani di galois d’ordine dispari. Atti Accad.Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur.(8), 56(5):690–697, 1974.[12] János Pach, Rom Pinchasi, and Micha Sharir. On the number of directions deter-mined by a three-dimensional points set. Journal of Combinatorial Theory, SeriesA, 108(1):1–16, 2004.[13] János Pach, Rom Pinchasi, and Micha Sharir. Solution of scott’s problem on thenumber of directions determined by a point set in 3-space. Discrete & ComputationalGeometry, 38(2):399–441, 2007.[14] Viktor V. Prasolov and Yuri P. Solovyev. Elliptic functions and elliptic integrals,volume 170. American Mathematical Soc., 1997.[15] Paul R. Scott. On the sets of directions determined by n points. The AmericanMathematical Monthly, 77(5):502–505, 1970.[16] Peter Ungar. 2n noncollinear points determine at least 2n directions. Journal ofCombinatorial Theory, Series A, 33(3):343–347, 1982.the electronic journal of combinatorics 27(1) (2020), #P1.24 9

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