- Email: [email protected]
An axiomatic look at a windmill
Available online at www.sciencedirect.com
ScienceDirect Indagationes Mathematicae 25 (2014) 113–121 www.elsevier.com/locate/indag
An axiomatic look at a windmill Victor Pambuccian ∗ School of Mathematical and Natural Sciences (MC 2352), Arizona State University - West Campus, P. O. Box 37100, Phoenix, AZ 85069-7100, USA Received 15 September 2012; accepted 7 August 2013 Communicated by J.W. Klop
Abstract We present the problem stated in intuitive language as problem 2 at the 52nd International Mathematical Olympiad as a formal statement, and prove that it is valid in ordered regular incidence planes, the weakest ordered geometry whose models can be embedded in projective ordered planes. c 2013 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. ⃝ Keywords: Ordered regular incidence planes; Windmill problem; Purity of the method
Meinem verehrten Lehrer Andreas Blass zum 65. Geburtstag in Dankbarkeit gewidmet 1. Introduction Proposed by Geoffrey Smith of the University of Bath, the second problem on the first day of the 52nd International Mathematical Olympiad, held in Amsterdam, reads as follows (see [2] for the statements and proofs of all problems). Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the pivot P until the first time that the line meets some other point belonging to S. This point, Q, takes over as the new pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely, with the pivot always being a point from S. ∗ Tel.: +1 6025433021.
E-mail address: [email protected]. c 2013 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights 0019-3577/$ - see front matter ⃝ reserved. http://dx.doi.org/10.1016/j.indag.2013.08.002
114
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
Show that we can choose a point P in S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times. As stated, the problem appears on a first reading to be describing a process in Euclidean geometry or at any rate a geometry with a metric, as it appears to require the existence of rotations, without really belonging to Euclidean geometry proper, given its strong combinatorial flavor. The aim of this note is to state it as a theorem of ordered regular incidence planes, i.e. that it can be proved inside a very weak axiom system. We will proceed as follows. First, we will provide an axiom system for planar ordered domains (without the lower-dimension axiom). We will then state the windmill problem inside that formalism, which is perhaps the main achievement of this note, and provide a proof for it. The proof we provide inside that axiom system is not new, as it follows the same idea as the proof in [2]. What is new is the fact that it can be carried out inside the restricted language we have available. Next we will introduce axiomatically ordered regular incidence planes, state the windmill problem in that formalism, in which it turns out to be a universal statement (i.e. it does not contain any existential quantifier), and finally provide the rationale for the validity of the windmill problem inside that axiom system. We do not actually provide a syntactic proof inside the theory of ordered regular incidence planes, as our proof does not carry through inside that theory. What we prove is that such a proof actually exists. 2. The axiomatic framework for planar ordered domains The axiomatic framework is that of a very general two-dimensional theory of betweenness, the models of which will be referred to as planar ordered domains (see also [6]), axiomatized in terms of points as individual variables and the strict betweenness ternary predicate Z , with Z (abc) to be read as ‘b lies between a and c’ (and the order is strict, i.e. b is different from a and c). The axiom system consists of the axioms A1–A5 axiomatizing L, the universal theory of linear order, the lower-dimension axiom, stating that there are three non-collinear points, and the Pasch axiom. The linear order axioms A1–A5 are the following statements (throughout the paper we omit universal quantifiers in universal sentences): A1. A2. A3. A4. A5.
Z (abc) → Z (cba), Z (abc) → ¬Z (acb), Z (acb) ∧ Z (abd) → Z (cbd), Z (cab) ∧ Z (abd) → Z (cbd), c ̸= d ∧ Z (abc) ∧ Z (abd) → (Z (bcd) ∨ Z (bdc)).
With L standing for the collinearity predicate, defined by L(x yz) :⇔ Z (x yz) ∨ Z (yzx) ∨ Z (zx y) ∨ x = y ∨ y = z ∨ z = x, we can state the lower-dimension axiom as (∃abc) ¬L(abc), and the Pasch axiom as A6. (∀abcde)(∃ f ) [¬L(abc)∧ Z (adc)∧¬L(ace)∧¬L(edb) → (Z (a f b)∨ Z (b f c))∧L(ed f )]. Although we do not have the concept of a ‘line’ in our language, we will refer to lines, saying that the point p lies on the line ⟨a, b⟩, for two distinct points a and b, if p = a or p = b or − → Z (apb) or Z ( pba) or Z (bap). We also say that point p lies on ray ab if p = b, Z (apb), or Z (abp). Notice that we do not ask the order to be dense or unending, and we also will not need the lower-dimension axiom to state the windmill theorem or to prove it. It is shown [1] that, based on an axiom system weaker than ours (as it contains two forms of the Pasch axiom which also hold in higher-dimensional spaces), all the main results of convexity
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
115
hold; in particular, there exists a convex hull for any finite set of points. Given our version of the Pasch axiom, which forces the geometry to be two-dimensional, we can also prove the Crossbar Theorem (see [3, p. 116]), and thus we can define the interior of a convex polygon, and we can prove that a line passing through an interior point of a convex polygon must intersect that polygon in two points. 3. The windmill statement Let S = {a1 , . . . , an }, rm (k) denote the remainder of the division of k by m, and let ϵ ϕ stand for ϕ if ϵ = 1, and for ¬ϕ if ϵ = 0. To help with the readability of the formal statement of the windmill problem, we will introduce the defined predicates λ, with λ(abc) to be read ‘a, b, c are three different collinear points’, δ, with δ(abuv) to be read — in the case where neither a, b, u nor a, b, v are collinear — ‘u and v lie on different sides of ⟨a, b⟩’, and πS ,0 , πS .1 , with πS ,k (a, b, p, c) to be read, for k = 1, as ‘(a, b, c) and (a, b, p) are two non-collinear triples of points, p and c lie on different sides of ⟨a, b⟩, and there are no points from S inside the angle − → → formed by the rays ab and − ac and no points from S inside the angle formed by the rays opposite − → − → to ab and to ac’, and, for k = 0, as ‘(a, b, c) and (a, b, p) are two non-collinear triples of points, p and c lie on the same side of ⟨a, b⟩, and there are no points from S inside the angle formed by − → → the rays ab and the ray opposite to − ac, and no points from S inside the angle formed by the ray − → − → opposite to ab and ac’. The formal definitions of these predicates are λ(x yz) ⇔ Z (x yz) ∨ Z (yzx) ∨ Z (zx y), δ(abuv) ⇔ (∃t) λ(abt) ∧ Z (utv), πS ,k (a, b, p, c) ⇔ ¬δ(acbp) ∧ k δ(abcp) ∧
(δ(acbai ) ↔ 1−k δ(abcai )).
ai ∈S \{a,b,c, p}
Let α(n) = n(n − 1)/2. For odd values of n ≥ 3, let K n = { f | (∃k f (n)) k f (n) ≤ α(n), f : {1, 2, . . . , k f (n)+2} → {1, 2. . . . , n}, and f {1, 2, . . . , k f (n)} is onto, f (i) ̸∈ { f (i −1), f (i − 2)} for all i with 3 ≤ i ≤ k f (n) + 2, f (k f (n) + 1) = f (2), f (k f (n) + 2) = f (1)}. For even values of n ≥ 4, let K n = { f | (∃k f (n)) k f (n) ≤ 2α(n), k f (n) even, f : {1, 2, . . . , k f (n)+2} → {1, 2. . . . , n}, and f {1, 2, . . . , k f (n)/2} and f {k f (n)/2 + 1, k f (n)/2 + 2, . . . , k f (n)} are onto, f (i) ̸∈ { f (i − 1), f (i − 2)} for all i with 3 ≤ i ≤ k f (n) + 2, f (k f (n) + 1) = f (2), f (k f (n) + 2) = f (1)}. Let A 2 stand for the set of all functions from A to {0, 1}, and in the case A = {3, 4, . . . , m}, we will write m 2 for A 2. With these definitions, we are ready to express the windmill theorem for S as the following statement (it is a different statement for each individual value of n ≥ 3), reading, for all f ∈ K n the value of f (0) — which occurs in the windmill statement for i = 3 in f (i − 3) — as rn+1 ( f (1) + f (2)): k f (n)+2 WMn . 1≤i< j
116
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
Fig. 1. πS ,1 (a, b, p, c1 ) (c2 does not exist), πS ,0 (a, b, p, c2 ) (c1 does not exist).
a person swimming from a to b’, i.e. that R is not definable in terms of Z ). The point p is essential in this context, as it fixes the sense of the “rotation”, and the subscript k is there to tell − → us whether c will be met by ab or by the opposite ray during its “rotation” (see Fig. 1 for details). If we look at the “windmill” process of the original statement by Geoffrey Smith, we notice that the “windmill” consists of lines through single points of the given set S = {a1 , . . . , an }, and that pivots are obtained during what one could call “windmill stops”. Instead of focusing on the “windmill” as consisting of lines through single points of S, we decided to look at the “windmill” as a collection of “windmill stops”, i.e. a collection of lines passing through exactly two points of S. We shall also write ca for the windmill stop ⟨a, c⟩, in which c is the ‘new’ pivot and a the ‘old’ pivot (or a is the pivot and ⟨a, c⟩ is the starting position of the windmill), i.e. if a and c are in the πS ,k (a, b, p, c) relation for some k ∈ {0, 1} and some points b and p. The statement WMn makes is that, given a set S of points a1 , . . . , a2 , such that no three are collinear, one can find a map f in K n , such that the first windmill stop has pivot a f (2) and goes through a f (1) (a f (0) being chosen as a point different from a f (1) and a f (2) , playing the role of p in the formula for π , i.e. determining the “clockwise” direction for the entire windmill process; that f (0) ̸∈ { f (1), f (2)} is ensured by our choice of f (0) as rn+1 ( f (1) + f (2)), as else we would have, for some i ∈ {1, 2}, f (i) ≡ f (1) + f (2)(mod n + 1), i.e. f (1) or f (2) would need to be multiples of n + 1, impossible as f (1), f (2) ∈ {1, . . . , n}), with windmill pivots a f (i) , with i in {2, 3, . . . , k f (n) + 2}, which, given the definition of K n , exhaust S, and such that the last windmill stop has pivot a f (2) , and goes through a f (1) , just like the stop we started with. That this process must be completed in at most α(n) + 1 steps in case n is odd, and in at most 2α(n) + 1 steps in case n is even, follows from our proof. The proof also shows why the parity of n determines two different outcomes, and forced us to define K n according to the parity of n. The number k f (n) in the definition of K n stands, for odd n, for the total number of different lines the windmill passes through, and, for even n, for twice the total number of different lines the windmill passes through. During the whole process, the point p determining the orientation changes (whenever c becomes the pivot, and the windmill stop is ca, the new point p is chosen to be the point b from the windmill stop ab that “rotated about a” to stop at c, i.e. the b for which πS ,k (a, b, p, c) holds), but the orientation itself, and thus the “clockwise” sense of the “rotation”, stays the same. The reason we cannot stay with the same p is that the windmill can stop at p, i.e. we can have c = p in πS ,k (a, b, p, c). The g(i) are there to tell us on which side we encountered the pivot a f (i) . If a f (i) was on the same side of ⟨a f (i−1) , a f (i−2) ⟩ as a f (i−3) , then g(i) = 0, if on the opposite side, then g(i) = 1.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
117
4. The proof Since the proof in [2] is carried out inside the Euclidean plane, we need to provide a variant thereof that would hold inside planar ordered domains. We treat the case in which n, the number of points in the set S, is even and the case in which it is odd separately. If n is even, then let as be any vertex of the convex hull of S (these elementary notions of convex geometry can all be defined and have the usual properties, as shown in [1]). Among the lines ⟨as , ai ⟩ with i ∈ {1, . . . , n} \ {s} there must be one having an equal number of points on each of its sides. We denote that particular value of i by t, and start the windmill process with the first pivot in as and the windmill stop going through at , i.e. f (1) = t, f (2) = s. The first value of p will be arn+1 ( f (1)+ f (2)) (choosing the index to be rn+1 ( f (1) + f (2)) had no other function −−−→ than making sure it is different from f (1) and f (2)). We define the direction a−−f − (2) a f (1) as the − − − − − − → East (and thus a f (1) a f (2) as the West), the half-plane determined by ⟨a f (1) , a f (2) ⟩ in which a f (0) lies as the Northern half-plane. These directions change in the course of the windmill process as − → → follows: if πS ,k (a, b, p, c), then ab and − ac point in the same “direction” (i.e. both East or both West) if k = 1 and in different directions if k = 0, whereas the half-plane determined by ⟨a, b⟩ in which p lies has the same name as the half-plane determined by ⟨a, c⟩ in which b lies. The Southern half-plane determined by a windmill stop ab will be denoted by σab , and the Northern half-plane by νab . We now look at the possible changes in the difference δ(ab) = N (ab) − S(ab) between the number N (ab) of points in νab and the number S(ab) of points in σab during the windmill process. We want to show that δ can take only the values 0 and 2. The next stop will be a point a f (3) , with the property that there is no point in S between the −−−→ −−−−−−→ rays a−−f − (2) a f (1) and a f (2) a f (3) . Given that a f (2) is a vertex of the convex hull of S, and that the sense of the “rotation about a f (2) ” is “clock-wise”, i.e. towards the Southern half-plane, a f (3) must lie in σa f (2) a f (1) , thus δ(a f (3) a f (2) ) must be 2, as the Northern half-plane gains a point, namely a f (1) (and the Southern half-plane loses one, namely a f (3) ). Let us look at the following situations: (i) ca is a windmill stop, ab is the previous windmill stop, c is in σab , and δ(ca) = 2; (ii) ca is a windmill stop, ab is the previous windmill stop, c is in νab , and δ(ca) = 0. Let us see what happens with δ in each case at the next windmill stop, which will be denoted by dc for some point d in S. In case (i), if d is in νca , then νdc has one point less than νca , namely d, so δ(dc) = 0, which means that, at windmill stop dc we are in a situation of type (ii); if d is in σca , then νdc has the same points as νca , thus δ(dc) = δ(ca) = 2, which means that at windmill stop dc we are back in a situation of type (i). In case (ii), if d is in νca , then νdc has the same number of points as νca , as it loses one point, namely d, and gains another point, namely a, so δ(dc) = 0, which means that at windmill stop dc we are back in a situation of type (ii); if d is in σca , then νdc has one point more than νca , namely a, thus δ(dc) = 2, which means that at windmill stop dc we are in a situation of type (i). Since at windmill stop a f (3) a f (2) we are in a type (i) situation, by our above analysis, only type (i) and type (ii) situations can occur all through the windmill process, in particular δ can take only the values 0 and 2. At every stage of the windmill process, the imprint of the “East” on the convex hull of S, i.e. the point where the Eastward pointing ray of that windmill stop intersects the convex hull of S, moves in a clockwise direction towards a f (2) , except when the pivot is a vertex of the convex hull of S, in which case the imprint stays put for that one step in the process, but will have to move on in the next, as the pivot changes at that step. After a finite number of steps, in fact in
118
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
no more than n(n − 1)/2 steps (given that this is the total number of lines that can be formed by joining two points in S, and thus the upper bound on the number of windmill stops), the imprint will be for the last time in σa f (2) a f (1) , in the sense that at the next windmill stop the imprint of the “East” will have to be either a f (2) or be in νa f (2) a f (1) . However, it cannot be in νa f (2) a f (1) , as a f (2) would have lain between the two windmill stops, contradicting the definition of π . The pivot a of the last windmill stop ab for which the Eastward imprint is in σa f (2) a f (1) cannot lie in σa f (2) a f (1) , as if it did, then a f (2) a would be the next windmill stop, and δ(a f (2) a) would have to be ≤ −2, contradicting the fact that δ takes only non-negative values. If a were to lie on ⟨a f (1) , a f (2) ⟩, then a would have to be a f (1) , but that is not possible, for in that case δ(ab) would have to be ≤ −2. If a lies in νa f (2) a f (1) , then, since a f (2) is the next pivot, a f (2) a will be the next windmill stop after ba. Since a lies in νa f (2) a f (1) , δ(a f (2) a) would have to be positive, and since 2 is the only −−−→ positive value it is allowed to take, there can be no point in S lying between the rays a−−f − (2) a f (1) − − − → and a f (2) a, so the next windmill stop, after a f (2) a, has to be a f (1) a f (2) . We are not quite back where we started from, for although the line of windmill stop is ⟨a f (1) , a f (2) ⟩, the pivot is now a f (1) , and not a f (2) as it was at the start. However, by the same reasoning that showed us that, in the case where we start with a line with δ = 0 as the first windmill stop, we arrive in at most n(n − 1)/2 + 1 steps back to the line we started with, we conclude that, starting from the windmill stop a f (1) a f (2) as our first step, we will be at the original windmill stop a f (2) a f (1) in at most n(n − 1)/2 + 1 steps, as the current pivot a f (1) will become a f (2) in the same way as the original pivot a f (2) became a f (1) at windmill stop a f (1) a f (2) . That each point in S must have become a pivot already by the time the windmill stop becomes a f (1) a f (2) is easily seen by noticing that, at this stage, ν(a f (1) a f (2) ) has become what σ (a f (2) a f (1) ) was at the start of the process, and that a point in S can move from the Southern to the Northern half-plane only by having been touched by a windmill stop. In the case where n is odd, say n = 2k + 1, we let as be any vertex of the convex hull of S and choose among the lines ⟨as , ai ⟩ with i ∈ {1, . . . , n} \ {s} the one having k + 1 points on one side and k points on the other side. Just like in the case in which n is even, we denote that particular value of i by t, and start the windmill process with the first pivot in as and windmill stop as at , i.e. f (1) = t, f (2) = s. We now distinguish two possibilities: the first value of p chosen to start the windmill process, i.e. arn+1 ( f (1)+ f (2)) , can be (i) in the half-plane with k points or (ii) in the half-plane with k + 1 points. In case (i), we notice, as in the n even case treated earlier, that during the windmill process δ can take on only two values, namely −1 and 1. If we follow the path of the imprint of the East on the convex hull of S, we notice that it moves in the “clockwise” direction (i.e. moving inside νa f (2) a f (1) ) at every step of the process, unless the pivot is a vertex of the convex hull of S, in which case it rests for one step of the windmill process, but will have to move afterwards. After a finite number of steps, no more than the total number of lines that can be formed by two points in S, the imprint of the East has to come back to its original location, a f (2) (it cannot jump over it, the reason being the same as in the already discussed n even case). Now, the other point a of the windmill stop a f (2) a we arrive at, when the imprint of the East is back at a f (2) , must be a f (1) . To see this, notice that, if a were in σa f (2) a f (1) , then δ(a f (2)a ) would have to be ≤ −1, so at the next windmill stop ba f (2) , we would have δ(ba f (2) ) ≤ −3, since b would have to lie in σa f (2) a . If a were in νa f (1) a f (2) , then δ(aa f (1) ) would have to be >1, which is impossible, and thus a = a f (2) . Case (ii) is treated analogously, by noticing that throughout the windmill process δ takes on only the values −1 and 1. For reasons similar to those mentioned in the n even case, each point in S must have become a pivot during the windmill process.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
119
5. Validity in ordered regular incidence planes There is a weaker axiom system, for ordered regular incidence planes, introduced by Joussen in [4], from which WMn can be derived. It cannot be expressed in terms of points and Z , as it is based on the notion of sides of a line in a plane, put forward by Sperner in [9]. The notion Z of betweenness can be defined in terms of the sides-notion, but that Z does not satisfy enough conditions to allow for a definition of the sides-notion. The axiom system can be expressed in a twosorted language, with variables for points (to be represented by lower-case Latin characters) and for lines (to be represented by lower-case Gothic characters), with two relation symbols, I , with I (ag) to be read as ‘point a is incident with line g’, and D, with D(agb) to be read as ‘the points a and b lie on different sides of line g’. With µ(abgh) :⇔ [(D(agb) ∧ D(ahb)) ∨ (¬D(agb) ∧ ¬D(ahb))] and ϵ µ standing for µ if ϵ = 1 and for ¬µ if ϵ = 0, the axioms are (see [4]) J1. J2. J3. J4. J5. J6. J7. J8.
(∀ab)(∃=1 g) a ̸= b → I (ag) ∧ I (bg), 4 (∀g)(∃a1 a2 a3 a4 ) 1≤i< j≤4 ai ̸= a j ∧ i=1 I (ai g), (∃abc)(∀g) ¬(I (ag) ∧ I (bg) ∧ I (cg)), D(agb) → ¬I (ag), D(agb) → D(bga), ¬I (cg) ∧ D(agb) → (D(agc) ∨ D(bgc)), ¬(D(agb) ∧ D(bgc) ∧ D(cga)), 4 4 4 [ 1≤i< j≤4 ai ̸= a j ∧hi ̸= h j ∧ i=1 I (ai hi )∧hi ̸= g∧(( i=1 I (ai g))∨( i=1 I (ohi )))] → ϵ 1 ϵ ϵ 2 3 [ µ(a3 a4 h1 h2 ) ∧ µ(a2 a4 h1 h3 ) ∧ µ(a2 a3 h1 h4 )]. ϵi ∈{0,1} ϵ1 +ϵ2 +ϵ3 =2
J6 is a weak variant of Pasch’s axiom, stating that if a line g does not pass through any of the points a, b, and c, and a and b are on different sides of g then so is at least one of the pairs {a, c} and {b, c}. J7 is a variant of Pasch’s theorem, stating that a line cannot separate all three pairs {a, b}, {b, c}, and {c, a}. One of its special cases, when a = b = c, implies that a and b can be on different sides of g only if a ̸= b. That these versions are called “weak” stems from the fact that, if a line g separates the points a and b, it no longer means that there is a point on g which is between a and b. Indeed, the line g and the line determined by a and b may have no point in common (a simple example is provided by the submodel of the ordered affine plane over Q whose points have coordinates whose denominators are powers of 2, with the plane separation relation inherited from the ordered affine plane over Q; see [5] for other examples). The meaning of J8 is best understood in terms of the notion of separation // (with ab//cd to be read as the point-pair (a, b) separates the point-pair (c, d)), defined by a1 a2 //a3 a4 :⇔ (∃ghk)
4 i=1
I (ai g) ∧
ai ̸= a j ∧ I (a1 h) ∧ I (a2 k)
1≤i< j≤4
∧h ̸= g ∧ k ̸= g ∧ ¬µ(a3 a4 hk). (1) 4 One part of it (corresponding to the i=1 I (ai g) disjunct) states that, if a1 , a2 , a3 , a4 are four different collinear points, then exactly one of the separation relations a1 a2 //a3 a4 , a1 a3 //a2 a4 , 4 a1 a4 //a2 a3 holds. Its other part (corresponding to the i=1 I (ohi ) disjunct) is the dual statement (in the sense of projective geometry). Joussen [4] showed that any model M of J1–J8 can be embedded in a projective ordered plane P, whose separation relation / /P is an extension of the separation relation / /M , defined in M terms of IM and DM by (1).
120
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
The windmill statement WMn remains the same, if we change the definition of the defined notions L and δ occurring in it to δ(abuv) ⇔ (∃g) a ̸= b ∧ I (ag) ∧ I (bg) ∧ D(ugv), L(abc) ⇔ (∃g) (I (ag) ∧ I (bg) ∧ I (cg)) ∨ a = b ∨ b = c ∨ c = a. To see that WMn is true in ordered regular incidence planes, suppose WMn were not derivable from J1–J8. Then there would have to exist a model M of J1–J8 in which WMn is false, i.e. in which ¬WMn holds. Now notice that ¬WMn can be expressed as an existential statement in the following way: (∃ai )1≤i≤n (∃gi j )1≤i< j≤n ai ̸= a j ∧ I (ai gi j ) ∧ I (a j gi j ) 1≤i< j≤n
∧
1≤i< j
gi j ̸= gkl
(i, j)̸=(k,l),1≤i< j≤n,1≤k
∧
kf (n)+2
k f (n)+2
f ∈K n , g∈
2
¬πS ,g(i) (a f (i−1) , a f (i−2) , a f (i−3) , a f (i) )
i=3
in which the δ(ai a j uv) occurring in the π ’s are just D(ugi j v). Then ¬WMn , as an existential statement, would have to hold in the ordered projective plane P, in which M can be embedded, as well. If we remove from the projective plane P any line which does not contain any of the points ai which ¬WMn claims to exist in M, such that the windmill process does not close regardless of the choice of its starting position, we obtain a model N of A1–A6 in which ¬WMn holds, a contradiction. Theorem 1. WMn , expressed in terms of points, lines, I , and D, holds in any model of {J1–J2, J4–J8}. By defining Z in terms of I and D by Z (abc) :⇔ (∃gh) h ̸= g ∧ I (ag) ∧ I (bg) ∧ I (cg) ∧ I (bh) ∧ D(ahc),
(2)
one can compare the set of Z -consequences of the axiom system {J1–J2, J4–J8} to {A1–A6}. It turns out that the Z defined by (2) satisfies A1–A5, but does not need to satisfy A6. On the other hand, J2 does not follow from {A1–A6}. Although formally incomparable, intuitively {J1–J2, J4–J8} is the weaker axiom system. Acknowledgments Thanks are due to two anonymous referees, who have suggested several improvements of an earlier version, leading to significant simplifications and an improved readability. References [1] W.A. Coppel, Foundations of Convex Geometry, in: Australian Mathematical Society Lecture Series, vol. 12, Cambridge University Press, Cambridge, 1998. [2] Z. Feng, Y. Sun, 52nd international mathematical olympiad, Math. Mag. 84 (2011) 316–319. [3] M.J. Greenberg, Euclidean and Non-Euclidean Geometries, fourth ed., W. H. Freeeman, San Francisco, 2006. [4] J. Joussen, Die Anordnungsf¨ahigkeit der freien Ebenen, Abh. Math. Semin. Univ. Hambg. 29 (1966) 137–184.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121 [5] [6] [7] [8] [9]
121
V. Pambuccian, Weakly ordered plane geometry, Ann. Univ. Ferrara 56 (2010) 91–96. V. Pambuccian, The axiomatics of ordered geometry, I. Ordered incidence spaces, Expo. Math. 29 (2011) 24–66. A.R. Schweitzer, On a fundamental relation in abstract geometry, Bull. Amer. Math. Soc. (2) 13 (1906) 79–81. A.R. Schweitzer, Concerning linear projective order, Amer. J. Math. 34 (1912) 169–176. E. Sperner, Die Ordnungsfunktionen einer Geometrie, Math. Ann. 121 (1949) 107–130.
ScienceDirect Indagationes Mathematicae 25 (2014) 113–121 www.elsevier.com/locate/indag
An axiomatic look at a windmill Victor Pambuccian ∗ School of Mathematical and Natural Sciences (MC 2352), Arizona State University - West Campus, P. O. Box 37100, Phoenix, AZ 85069-7100, USA Received 15 September 2012; accepted 7 August 2013 Communicated by J.W. Klop
Abstract We present the problem stated in intuitive language as problem 2 at the 52nd International Mathematical Olympiad as a formal statement, and prove that it is valid in ordered regular incidence planes, the weakest ordered geometry whose models can be embedded in projective ordered planes. c 2013 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. ⃝ Keywords: Ordered regular incidence planes; Windmill problem; Purity of the method
Meinem verehrten Lehrer Andreas Blass zum 65. Geburtstag in Dankbarkeit gewidmet 1. Introduction Proposed by Geoffrey Smith of the University of Bath, the second problem on the first day of the 52nd International Mathematical Olympiad, held in Amsterdam, reads as follows (see [2] for the statements and proofs of all problems). Let S be a finite set of at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P ∈ S. The line rotates clockwise about the pivot P until the first time that the line meets some other point belonging to S. This point, Q, takes over as the new pivot, and the line now rotates clockwise about Q, until it next meets a point of S. This process continues indefinitely, with the pivot always being a point from S. ∗ Tel.: +1 6025433021.
E-mail address: [email protected]. c 2013 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights 0019-3577/$ - see front matter ⃝ reserved. http://dx.doi.org/10.1016/j.indag.2013.08.002
114
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
Show that we can choose a point P in S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times. As stated, the problem appears on a first reading to be describing a process in Euclidean geometry or at any rate a geometry with a metric, as it appears to require the existence of rotations, without really belonging to Euclidean geometry proper, given its strong combinatorial flavor. The aim of this note is to state it as a theorem of ordered regular incidence planes, i.e. that it can be proved inside a very weak axiom system. We will proceed as follows. First, we will provide an axiom system for planar ordered domains (without the lower-dimension axiom). We will then state the windmill problem inside that formalism, which is perhaps the main achievement of this note, and provide a proof for it. The proof we provide inside that axiom system is not new, as it follows the same idea as the proof in [2]. What is new is the fact that it can be carried out inside the restricted language we have available. Next we will introduce axiomatically ordered regular incidence planes, state the windmill problem in that formalism, in which it turns out to be a universal statement (i.e. it does not contain any existential quantifier), and finally provide the rationale for the validity of the windmill problem inside that axiom system. We do not actually provide a syntactic proof inside the theory of ordered regular incidence planes, as our proof does not carry through inside that theory. What we prove is that such a proof actually exists. 2. The axiomatic framework for planar ordered domains The axiomatic framework is that of a very general two-dimensional theory of betweenness, the models of which will be referred to as planar ordered domains (see also [6]), axiomatized in terms of points as individual variables and the strict betweenness ternary predicate Z , with Z (abc) to be read as ‘b lies between a and c’ (and the order is strict, i.e. b is different from a and c). The axiom system consists of the axioms A1–A5 axiomatizing L, the universal theory of linear order, the lower-dimension axiom, stating that there are three non-collinear points, and the Pasch axiom. The linear order axioms A1–A5 are the following statements (throughout the paper we omit universal quantifiers in universal sentences): A1. A2. A3. A4. A5.
Z (abc) → Z (cba), Z (abc) → ¬Z (acb), Z (acb) ∧ Z (abd) → Z (cbd), Z (cab) ∧ Z (abd) → Z (cbd), c ̸= d ∧ Z (abc) ∧ Z (abd) → (Z (bcd) ∨ Z (bdc)).
With L standing for the collinearity predicate, defined by L(x yz) :⇔ Z (x yz) ∨ Z (yzx) ∨ Z (zx y) ∨ x = y ∨ y = z ∨ z = x, we can state the lower-dimension axiom as (∃abc) ¬L(abc), and the Pasch axiom as A6. (∀abcde)(∃ f ) [¬L(abc)∧ Z (adc)∧¬L(ace)∧¬L(edb) → (Z (a f b)∨ Z (b f c))∧L(ed f )]. Although we do not have the concept of a ‘line’ in our language, we will refer to lines, saying that the point p lies on the line ⟨a, b⟩, for two distinct points a and b, if p = a or p = b or − → Z (apb) or Z ( pba) or Z (bap). We also say that point p lies on ray ab if p = b, Z (apb), or Z (abp). Notice that we do not ask the order to be dense or unending, and we also will not need the lower-dimension axiom to state the windmill theorem or to prove it. It is shown [1] that, based on an axiom system weaker than ours (as it contains two forms of the Pasch axiom which also hold in higher-dimensional spaces), all the main results of convexity
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
115
hold; in particular, there exists a convex hull for any finite set of points. Given our version of the Pasch axiom, which forces the geometry to be two-dimensional, we can also prove the Crossbar Theorem (see [3, p. 116]), and thus we can define the interior of a convex polygon, and we can prove that a line passing through an interior point of a convex polygon must intersect that polygon in two points. 3. The windmill statement Let S = {a1 , . . . , an }, rm (k) denote the remainder of the division of k by m, and let ϵ ϕ stand for ϕ if ϵ = 1, and for ¬ϕ if ϵ = 0. To help with the readability of the formal statement of the windmill problem, we will introduce the defined predicates λ, with λ(abc) to be read ‘a, b, c are three different collinear points’, δ, with δ(abuv) to be read — in the case where neither a, b, u nor a, b, v are collinear — ‘u and v lie on different sides of ⟨a, b⟩’, and πS ,0 , πS .1 , with πS ,k (a, b, p, c) to be read, for k = 1, as ‘(a, b, c) and (a, b, p) are two non-collinear triples of points, p and c lie on different sides of ⟨a, b⟩, and there are no points from S inside the angle − → → formed by the rays ab and − ac and no points from S inside the angle formed by the rays opposite − → − → to ab and to ac’, and, for k = 0, as ‘(a, b, c) and (a, b, p) are two non-collinear triples of points, p and c lie on the same side of ⟨a, b⟩, and there are no points from S inside the angle formed by − → → the rays ab and the ray opposite to − ac, and no points from S inside the angle formed by the ray − → − → opposite to ab and ac’. The formal definitions of these predicates are λ(x yz) ⇔ Z (x yz) ∨ Z (yzx) ∨ Z (zx y), δ(abuv) ⇔ (∃t) λ(abt) ∧ Z (utv), πS ,k (a, b, p, c) ⇔ ¬δ(acbp) ∧ k δ(abcp) ∧
(δ(acbai ) ↔ 1−k δ(abcai )).
ai ∈S \{a,b,c, p}
Let α(n) = n(n − 1)/2. For odd values of n ≥ 3, let K n = { f | (∃k f (n)) k f (n) ≤ α(n), f : {1, 2, . . . , k f (n)+2} → {1, 2. . . . , n}, and f {1, 2, . . . , k f (n)} is onto, f (i) ̸∈ { f (i −1), f (i − 2)} for all i with 3 ≤ i ≤ k f (n) + 2, f (k f (n) + 1) = f (2), f (k f (n) + 2) = f (1)}. For even values of n ≥ 4, let K n = { f | (∃k f (n)) k f (n) ≤ 2α(n), k f (n) even, f : {1, 2, . . . , k f (n)+2} → {1, 2. . . . , n}, and f {1, 2, . . . , k f (n)/2} and f {k f (n)/2 + 1, k f (n)/2 + 2, . . . , k f (n)} are onto, f (i) ̸∈ { f (i − 1), f (i − 2)} for all i with 3 ≤ i ≤ k f (n) + 2, f (k f (n) + 1) = f (2), f (k f (n) + 2) = f (1)}. Let A 2 stand for the set of all functions from A to {0, 1}, and in the case A = {3, 4, . . . , m}, we will write m 2 for A 2. With these definitions, we are ready to express the windmill theorem for S as the following statement (it is a different statement for each individual value of n ≥ 3), reading, for all f ∈ K n the value of f (0) — which occurs in the windmill statement for i = 3 in f (i − 3) — as rn+1 ( f (1) + f (2)): k f (n)+2 WMn . 1≤i< j
116
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
Fig. 1. πS ,1 (a, b, p, c1 ) (c2 does not exist), πS ,0 (a, b, p, c2 ) (c1 does not exist).
a person swimming from a to b’, i.e. that R is not definable in terms of Z ). The point p is essential in this context, as it fixes the sense of the “rotation”, and the subscript k is there to tell − → us whether c will be met by ab or by the opposite ray during its “rotation” (see Fig. 1 for details). If we look at the “windmill” process of the original statement by Geoffrey Smith, we notice that the “windmill” consists of lines through single points of the given set S = {a1 , . . . , an }, and that pivots are obtained during what one could call “windmill stops”. Instead of focusing on the “windmill” as consisting of lines through single points of S, we decided to look at the “windmill” as a collection of “windmill stops”, i.e. a collection of lines passing through exactly two points of S. We shall also write ca for the windmill stop ⟨a, c⟩, in which c is the ‘new’ pivot and a the ‘old’ pivot (or a is the pivot and ⟨a, c⟩ is the starting position of the windmill), i.e. if a and c are in the πS ,k (a, b, p, c) relation for some k ∈ {0, 1} and some points b and p. The statement WMn makes is that, given a set S of points a1 , . . . , a2 , such that no three are collinear, one can find a map f in K n , such that the first windmill stop has pivot a f (2) and goes through a f (1) (a f (0) being chosen as a point different from a f (1) and a f (2) , playing the role of p in the formula for π , i.e. determining the “clockwise” direction for the entire windmill process; that f (0) ̸∈ { f (1), f (2)} is ensured by our choice of f (0) as rn+1 ( f (1) + f (2)), as else we would have, for some i ∈ {1, 2}, f (i) ≡ f (1) + f (2)(mod n + 1), i.e. f (1) or f (2) would need to be multiples of n + 1, impossible as f (1), f (2) ∈ {1, . . . , n}), with windmill pivots a f (i) , with i in {2, 3, . . . , k f (n) + 2}, which, given the definition of K n , exhaust S, and such that the last windmill stop has pivot a f (2) , and goes through a f (1) , just like the stop we started with. That this process must be completed in at most α(n) + 1 steps in case n is odd, and in at most 2α(n) + 1 steps in case n is even, follows from our proof. The proof also shows why the parity of n determines two different outcomes, and forced us to define K n according to the parity of n. The number k f (n) in the definition of K n stands, for odd n, for the total number of different lines the windmill passes through, and, for even n, for twice the total number of different lines the windmill passes through. During the whole process, the point p determining the orientation changes (whenever c becomes the pivot, and the windmill stop is ca, the new point p is chosen to be the point b from the windmill stop ab that “rotated about a” to stop at c, i.e. the b for which πS ,k (a, b, p, c) holds), but the orientation itself, and thus the “clockwise” sense of the “rotation”, stays the same. The reason we cannot stay with the same p is that the windmill can stop at p, i.e. we can have c = p in πS ,k (a, b, p, c). The g(i) are there to tell us on which side we encountered the pivot a f (i) . If a f (i) was on the same side of ⟨a f (i−1) , a f (i−2) ⟩ as a f (i−3) , then g(i) = 0, if on the opposite side, then g(i) = 1.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
117
4. The proof Since the proof in [2] is carried out inside the Euclidean plane, we need to provide a variant thereof that would hold inside planar ordered domains. We treat the case in which n, the number of points in the set S, is even and the case in which it is odd separately. If n is even, then let as be any vertex of the convex hull of S (these elementary notions of convex geometry can all be defined and have the usual properties, as shown in [1]). Among the lines ⟨as , ai ⟩ with i ∈ {1, . . . , n} \ {s} there must be one having an equal number of points on each of its sides. We denote that particular value of i by t, and start the windmill process with the first pivot in as and the windmill stop going through at , i.e. f (1) = t, f (2) = s. The first value of p will be arn+1 ( f (1)+ f (2)) (choosing the index to be rn+1 ( f (1) + f (2)) had no other function −−−→ than making sure it is different from f (1) and f (2)). We define the direction a−−f − (2) a f (1) as the − − − − − − → East (and thus a f (1) a f (2) as the West), the half-plane determined by ⟨a f (1) , a f (2) ⟩ in which a f (0) lies as the Northern half-plane. These directions change in the course of the windmill process as − → → follows: if πS ,k (a, b, p, c), then ab and − ac point in the same “direction” (i.e. both East or both West) if k = 1 and in different directions if k = 0, whereas the half-plane determined by ⟨a, b⟩ in which p lies has the same name as the half-plane determined by ⟨a, c⟩ in which b lies. The Southern half-plane determined by a windmill stop ab will be denoted by σab , and the Northern half-plane by νab . We now look at the possible changes in the difference δ(ab) = N (ab) − S(ab) between the number N (ab) of points in νab and the number S(ab) of points in σab during the windmill process. We want to show that δ can take only the values 0 and 2. The next stop will be a point a f (3) , with the property that there is no point in S between the −−−→ −−−−−−→ rays a−−f − (2) a f (1) and a f (2) a f (3) . Given that a f (2) is a vertex of the convex hull of S, and that the sense of the “rotation about a f (2) ” is “clock-wise”, i.e. towards the Southern half-plane, a f (3) must lie in σa f (2) a f (1) , thus δ(a f (3) a f (2) ) must be 2, as the Northern half-plane gains a point, namely a f (1) (and the Southern half-plane loses one, namely a f (3) ). Let us look at the following situations: (i) ca is a windmill stop, ab is the previous windmill stop, c is in σab , and δ(ca) = 2; (ii) ca is a windmill stop, ab is the previous windmill stop, c is in νab , and δ(ca) = 0. Let us see what happens with δ in each case at the next windmill stop, which will be denoted by dc for some point d in S. In case (i), if d is in νca , then νdc has one point less than νca , namely d, so δ(dc) = 0, which means that, at windmill stop dc we are in a situation of type (ii); if d is in σca , then νdc has the same points as νca , thus δ(dc) = δ(ca) = 2, which means that at windmill stop dc we are back in a situation of type (i). In case (ii), if d is in νca , then νdc has the same number of points as νca , as it loses one point, namely d, and gains another point, namely a, so δ(dc) = 0, which means that at windmill stop dc we are back in a situation of type (ii); if d is in σca , then νdc has one point more than νca , namely a, thus δ(dc) = 2, which means that at windmill stop dc we are in a situation of type (i). Since at windmill stop a f (3) a f (2) we are in a type (i) situation, by our above analysis, only type (i) and type (ii) situations can occur all through the windmill process, in particular δ can take only the values 0 and 2. At every stage of the windmill process, the imprint of the “East” on the convex hull of S, i.e. the point where the Eastward pointing ray of that windmill stop intersects the convex hull of S, moves in a clockwise direction towards a f (2) , except when the pivot is a vertex of the convex hull of S, in which case the imprint stays put for that one step in the process, but will have to move on in the next, as the pivot changes at that step. After a finite number of steps, in fact in
118
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
no more than n(n − 1)/2 steps (given that this is the total number of lines that can be formed by joining two points in S, and thus the upper bound on the number of windmill stops), the imprint will be for the last time in σa f (2) a f (1) , in the sense that at the next windmill stop the imprint of the “East” will have to be either a f (2) or be in νa f (2) a f (1) . However, it cannot be in νa f (2) a f (1) , as a f (2) would have lain between the two windmill stops, contradicting the definition of π . The pivot a of the last windmill stop ab for which the Eastward imprint is in σa f (2) a f (1) cannot lie in σa f (2) a f (1) , as if it did, then a f (2) a would be the next windmill stop, and δ(a f (2) a) would have to be ≤ −2, contradicting the fact that δ takes only non-negative values. If a were to lie on ⟨a f (1) , a f (2) ⟩, then a would have to be a f (1) , but that is not possible, for in that case δ(ab) would have to be ≤ −2. If a lies in νa f (2) a f (1) , then, since a f (2) is the next pivot, a f (2) a will be the next windmill stop after ba. Since a lies in νa f (2) a f (1) , δ(a f (2) a) would have to be positive, and since 2 is the only −−−→ positive value it is allowed to take, there can be no point in S lying between the rays a−−f − (2) a f (1) − − − → and a f (2) a, so the next windmill stop, after a f (2) a, has to be a f (1) a f (2) . We are not quite back where we started from, for although the line of windmill stop is ⟨a f (1) , a f (2) ⟩, the pivot is now a f (1) , and not a f (2) as it was at the start. However, by the same reasoning that showed us that, in the case where we start with a line with δ = 0 as the first windmill stop, we arrive in at most n(n − 1)/2 + 1 steps back to the line we started with, we conclude that, starting from the windmill stop a f (1) a f (2) as our first step, we will be at the original windmill stop a f (2) a f (1) in at most n(n − 1)/2 + 1 steps, as the current pivot a f (1) will become a f (2) in the same way as the original pivot a f (2) became a f (1) at windmill stop a f (1) a f (2) . That each point in S must have become a pivot already by the time the windmill stop becomes a f (1) a f (2) is easily seen by noticing that, at this stage, ν(a f (1) a f (2) ) has become what σ (a f (2) a f (1) ) was at the start of the process, and that a point in S can move from the Southern to the Northern half-plane only by having been touched by a windmill stop. In the case where n is odd, say n = 2k + 1, we let as be any vertex of the convex hull of S and choose among the lines ⟨as , ai ⟩ with i ∈ {1, . . . , n} \ {s} the one having k + 1 points on one side and k points on the other side. Just like in the case in which n is even, we denote that particular value of i by t, and start the windmill process with the first pivot in as and windmill stop as at , i.e. f (1) = t, f (2) = s. We now distinguish two possibilities: the first value of p chosen to start the windmill process, i.e. arn+1 ( f (1)+ f (2)) , can be (i) in the half-plane with k points or (ii) in the half-plane with k + 1 points. In case (i), we notice, as in the n even case treated earlier, that during the windmill process δ can take on only two values, namely −1 and 1. If we follow the path of the imprint of the East on the convex hull of S, we notice that it moves in the “clockwise” direction (i.e. moving inside νa f (2) a f (1) ) at every step of the process, unless the pivot is a vertex of the convex hull of S, in which case it rests for one step of the windmill process, but will have to move afterwards. After a finite number of steps, no more than the total number of lines that can be formed by two points in S, the imprint of the East has to come back to its original location, a f (2) (it cannot jump over it, the reason being the same as in the already discussed n even case). Now, the other point a of the windmill stop a f (2) a we arrive at, when the imprint of the East is back at a f (2) , must be a f (1) . To see this, notice that, if a were in σa f (2) a f (1) , then δ(a f (2)a ) would have to be ≤ −1, so at the next windmill stop ba f (2) , we would have δ(ba f (2) ) ≤ −3, since b would have to lie in σa f (2) a . If a were in νa f (1) a f (2) , then δ(aa f (1) ) would have to be >1, which is impossible, and thus a = a f (2) . Case (ii) is treated analogously, by noticing that throughout the windmill process δ takes on only the values −1 and 1. For reasons similar to those mentioned in the n even case, each point in S must have become a pivot during the windmill process.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
119
5. Validity in ordered regular incidence planes There is a weaker axiom system, for ordered regular incidence planes, introduced by Joussen in [4], from which WMn can be derived. It cannot be expressed in terms of points and Z , as it is based on the notion of sides of a line in a plane, put forward by Sperner in [9]. The notion Z of betweenness can be defined in terms of the sides-notion, but that Z does not satisfy enough conditions to allow for a definition of the sides-notion. The axiom system can be expressed in a twosorted language, with variables for points (to be represented by lower-case Latin characters) and for lines (to be represented by lower-case Gothic characters), with two relation symbols, I , with I (ag) to be read as ‘point a is incident with line g’, and D, with D(agb) to be read as ‘the points a and b lie on different sides of line g’. With µ(abgh) :⇔ [(D(agb) ∧ D(ahb)) ∨ (¬D(agb) ∧ ¬D(ahb))] and ϵ µ standing for µ if ϵ = 1 and for ¬µ if ϵ = 0, the axioms are (see [4]) J1. J2. J3. J4. J5. J6. J7. J8.
(∀ab)(∃=1 g) a ̸= b → I (ag) ∧ I (bg), 4 (∀g)(∃a1 a2 a3 a4 ) 1≤i< j≤4 ai ̸= a j ∧ i=1 I (ai g), (∃abc)(∀g) ¬(I (ag) ∧ I (bg) ∧ I (cg)), D(agb) → ¬I (ag), D(agb) → D(bga), ¬I (cg) ∧ D(agb) → (D(agc) ∨ D(bgc)), ¬(D(agb) ∧ D(bgc) ∧ D(cga)), 4 4 4 [ 1≤i< j≤4 ai ̸= a j ∧hi ̸= h j ∧ i=1 I (ai hi )∧hi ̸= g∧(( i=1 I (ai g))∨( i=1 I (ohi )))] → ϵ 1 ϵ ϵ 2 3 [ µ(a3 a4 h1 h2 ) ∧ µ(a2 a4 h1 h3 ) ∧ µ(a2 a3 h1 h4 )]. ϵi ∈{0,1} ϵ1 +ϵ2 +ϵ3 =2
J6 is a weak variant of Pasch’s axiom, stating that if a line g does not pass through any of the points a, b, and c, and a and b are on different sides of g then so is at least one of the pairs {a, c} and {b, c}. J7 is a variant of Pasch’s theorem, stating that a line cannot separate all three pairs {a, b}, {b, c}, and {c, a}. One of its special cases, when a = b = c, implies that a and b can be on different sides of g only if a ̸= b. That these versions are called “weak” stems from the fact that, if a line g separates the points a and b, it no longer means that there is a point on g which is between a and b. Indeed, the line g and the line determined by a and b may have no point in common (a simple example is provided by the submodel of the ordered affine plane over Q whose points have coordinates whose denominators are powers of 2, with the plane separation relation inherited from the ordered affine plane over Q; see [5] for other examples). The meaning of J8 is best understood in terms of the notion of separation // (with ab//cd to be read as the point-pair (a, b) separates the point-pair (c, d)), defined by a1 a2 //a3 a4 :⇔ (∃ghk)
4 i=1
I (ai g) ∧
ai ̸= a j ∧ I (a1 h) ∧ I (a2 k)
1≤i< j≤4
∧h ̸= g ∧ k ̸= g ∧ ¬µ(a3 a4 hk). (1) 4 One part of it (corresponding to the i=1 I (ai g) disjunct) states that, if a1 , a2 , a3 , a4 are four different collinear points, then exactly one of the separation relations a1 a2 //a3 a4 , a1 a3 //a2 a4 , 4 a1 a4 //a2 a3 holds. Its other part (corresponding to the i=1 I (ohi ) disjunct) is the dual statement (in the sense of projective geometry). Joussen [4] showed that any model M of J1–J8 can be embedded in a projective ordered plane P, whose separation relation / /P is an extension of the separation relation / /M , defined in M terms of IM and DM by (1).
120
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121
The windmill statement WMn remains the same, if we change the definition of the defined notions L and δ occurring in it to δ(abuv) ⇔ (∃g) a ̸= b ∧ I (ag) ∧ I (bg) ∧ D(ugv), L(abc) ⇔ (∃g) (I (ag) ∧ I (bg) ∧ I (cg)) ∨ a = b ∨ b = c ∨ c = a. To see that WMn is true in ordered regular incidence planes, suppose WMn were not derivable from J1–J8. Then there would have to exist a model M of J1–J8 in which WMn is false, i.e. in which ¬WMn holds. Now notice that ¬WMn can be expressed as an existential statement in the following way: (∃ai )1≤i≤n (∃gi j )1≤i< j≤n ai ̸= a j ∧ I (ai gi j ) ∧ I (a j gi j ) 1≤i< j≤n
∧
1≤i< j
gi j ̸= gkl
(i, j)̸=(k,l),1≤i< j≤n,1≤k
∧
kf (n)+2
k f (n)+2
f ∈K n , g∈
2
¬πS ,g(i) (a f (i−1) , a f (i−2) , a f (i−3) , a f (i) )
i=3
in which the δ(ai a j uv) occurring in the π ’s are just D(ugi j v). Then ¬WMn , as an existential statement, would have to hold in the ordered projective plane P, in which M can be embedded, as well. If we remove from the projective plane P any line which does not contain any of the points ai which ¬WMn claims to exist in M, such that the windmill process does not close regardless of the choice of its starting position, we obtain a model N of A1–A6 in which ¬WMn holds, a contradiction. Theorem 1. WMn , expressed in terms of points, lines, I , and D, holds in any model of {J1–J2, J4–J8}. By defining Z in terms of I and D by Z (abc) :⇔ (∃gh) h ̸= g ∧ I (ag) ∧ I (bg) ∧ I (cg) ∧ I (bh) ∧ D(ahc),
(2)
one can compare the set of Z -consequences of the axiom system {J1–J2, J4–J8} to {A1–A6}. It turns out that the Z defined by (2) satisfies A1–A5, but does not need to satisfy A6. On the other hand, J2 does not follow from {A1–A6}. Although formally incomparable, intuitively {J1–J2, J4–J8} is the weaker axiom system. Acknowledgments Thanks are due to two anonymous referees, who have suggested several improvements of an earlier version, leading to significant simplifications and an improved readability. References [1] W.A. Coppel, Foundations of Convex Geometry, in: Australian Mathematical Society Lecture Series, vol. 12, Cambridge University Press, Cambridge, 1998. [2] Z. Feng, Y. Sun, 52nd international mathematical olympiad, Math. Mag. 84 (2011) 316–319. [3] M.J. Greenberg, Euclidean and Non-Euclidean Geometries, fourth ed., W. H. Freeeman, San Francisco, 2006. [4] J. Joussen, Die Anordnungsf¨ahigkeit der freien Ebenen, Abh. Math. Semin. Univ. Hambg. 29 (1966) 137–184.
V. Pambuccian / Indagationes Mathematicae 25 (2014) 113–121 [5] [6] [7] [8] [9]
121
V. Pambuccian, Weakly ordered plane geometry, Ann. Univ. Ferrara 56 (2010) 91–96. V. Pambuccian, The axiomatics of ordered geometry, I. Ordered incidence spaces, Expo. Math. 29 (2011) 24–66. A.R. Schweitzer, On a fundamental relation in abstract geometry, Bull. Amer. Math. Soc. (2) 13 (1906) 79–81. A.R. Schweitzer, Concerning linear projective order, Amer. J. Math. 34 (1912) 169–176. E. Sperner, Die Ordnungsfunktionen einer Geometrie, Math. Ann. 121 (1949) 107–130.