Báo cáo toán học: "Movable (n4) Configurations" pps

Movable (n 4 ) Configurations Leah Wrenn Berman Submitted: Jul 17, 2006; Accepted: Nov 8, 2006; Published: Nov 17, 2006 Mathematics Subject Classification: 51A20, 52C35 Abstract An (n k ) configuration is a collection of points and straight lines, usually in the Euclidean plane, so that each point lies on k lines and each line passes through k points; such a configuration will be called symmetric if it possesses non-trivial geometric symmetry. Although examples of symmetric (n 3 ) con- figurations with continuous parameters are known, to this point, all known connected infinite families of (n 4 ) configurations with non-trivial geometric symmetry had the property that each set of discrete parameters describing the configuration corresponded to a single (n 4 ) configuration. This paper presents several new classes of highly symmetric (n 4 ) configurations which have at least one continuous parameter; that is, the configurations are movable. 1 Introduction A geometric (p q , n k ) configuration is a collection of points and straight lines, usually in the Euclidean plane, so that every point lies on q lines and every line passes through k points. By counting incidences (the number of point-line incidences must be equal to the number of line-point incidences), if p = n then q = k. Usually, an (n k , n k ) configuration is abbreviated as (n k ). Although (n 3 ) configurations have been studied since the late 1800s, (n 4 ) configurations have been studied for a much shorter time, initially in a series of papers by Branko Gr¨unbaum [10, 11, 13]. Recently, there has been a flurry of activity concerning various kinds of geometric (n 4 ) configurations (e.g., [2, 3, 5, 6, 7, 8] ). There exist connected symmetric (n 3 ) configurations (for example, astral (n 3 ) configurations with dihedral symmetry) describable by a set of discrete and con- tinuous parameters [9, 12]. However, all of the connected (n 4 ) configurations with the electronic journal of combinatorics 13 (2006), #R104 1 non-trivial rotational symmetry described previously in the literature (see, for exam- ple, [5, 6, 10]) have the property that for a single set of discrete parameters, there is only a single configuration corresponding to those parameters. This paper presents a large family of (n 4 ) configurations with non-trivial rotational symmetry—they all have m-fold rotational symmetry for some m ≥ 8—where a single set of discrete parameters corresponds to uncountably many (n 4 ) configurations. That is, the con- figurations are movable, meaning that they admit a continuous family of realizations fixing four points in general position but moving at least one other point. It should be noted that it is straightforward to construct highly non-symmetric (n k ) configurations—that is, configurations with no non-trivial geometric symmetry– with a continuous parameter by taking k copies of an (n k ) configuration, each with the same single line removed, translating them in a direction different from those determined by any line of the configuration, and then connecting the points which lie on the removed line with k new lines. Figure 1 shows a (96 4 ) configuration formed by translating four copies of a (24 4 ) configuration with a single line deleted from each copy and adding four new lines connecting the necessary points. Parts of the resulting (96 4 ) configuration corresponding to the original modified (24 4 ) configurations may be moved (translated) independently. Figure 1: A non-symmetric movable (96 4 ) configuration, formed by translating four modified copies (each missing a single line) of a (24 4 ) configuration (with red and blue lines) and appropriately connecting the four copies with four parallel (green) lines. The copies of the configuration may be moved back and forth along the green lines. the electronic journal of combinatorics 13 (2006), #R104 2 1.1 Preliminary definitions Label the vertices of a regular convex m-gon consecutively as w 0 , . . . , w m−1 . A di- agonal of the m-gon is of span c if it connects vertices w i and w i+c , where indices are taken modulo m. Given a regular polygon and a diagonal of span c, label the intersection points of the diagonal with other span c diagonals as c 1 , c 2 , . . . , c  m 2  , counted from the midpoint of the diagonal and travelling in one direction (usually, to the left). Following Gr¨unbaum [8], any point on a span s i line which is the t i -th intersection (that is, the intersection point with label s i t i ) is given the label [[s i , t i ]]. For an example of this labelling, see Figure 2. [[4,5]] [[4,3]] [[4,2]] [[4,1]] [[4,4]] Figure 2: An example of the notation [[s i , t i ]]. Here, m = 12 and s i = 4. 2 Celestial configurations All of the movable configurations which will be constructed later in the paper are based on a class of (n 4 ) configurations, originally developed by Branko Gr¨unbaum [12] and further studied by Marko Boben and Tomaˇz Pisanski [6] and Gr¨unbaum [8], which have the property that every point has precisely two lines from each of two symmetry classes passing through them, and if there are m points in a symme- try class, then the configuration has the dihedral symmetries of an m-gon. These are such a useful class of configurations that having a name to call them would the electronic journal of combinatorics 13 (2006), #R104 3 be helpful; I propose calling them celestial configurations. To date, celestial con- figurations were just discussed by referring to their symbol; in [8] Gr¨unbaum uses the symbol m#(s 1 , t 1 , s 2 , t 2 , . . . , s h , t h ), while in [6], the symbol C 4 (m, (s 1 , s 2 , . . . , s h ), (t 1 , t 2 , . . . t h ), t) corresponds to the same configuration. In this paper I am using a modification of the symbol from [8], m#(s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ), to refer to a celes- tial configuration. Note that a celestial configuration m#(s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ) is an (mh 4 ) configuration. The earliest drawing of a celestial configuration appeared in a paper by Gr¨unbaum and Rigby [13], where they presented a (21 4 ) configuration; other early drawings of celestial configurations appeared as examples in a paper by Maruˇsiˇc and Pisanksi [14]. The following description of celestial configurations is closely based on that given in [8], although it differs slightly in some choices of labelling and point of view. For the sequence (s 1 , t 1 ; . . . ; s h , t h ) to be valid, no two consecutive elements can be equal (and s 1 = t h also). To construct a celestial configuration m#(s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ) do the following: 1. Begin with m points forming the vertices of a regular m-gon; these vertices will be labelled v 0,0 , v 0,1 . . . , v 0,m−1 . Collectively, these vertices will be referred to as v 0 . 2. Draw in lines L 0,1 , L 0,2 , . . . , L 0,m−1 of span s 1 connecting these points, so that L 0,j connects points v 0,j and v 0,j+s 1 . These lines will be known collectively as L 0 . 3. Choose the t 1 -st intersection of the span s 1 lines, counting from the center and moving to the left. Label these vertices as v 1,0 , v 1,1 , . . . , v 1,m−1 , collectively known as v 1 , where v 1,0 is the t 1 -st intersection point on line L 0,0 . Note that each of the points v 1 has symbol [[s 1 , t 1 ]]. 4. The points v 1 form the vertices of a regular m-gon; using these vertices, draw in diagonals of span s 2 and label them as L 1,0 , L 1,1 , . . . , L 1,m−1 as above (and collectively as L 1 ). 5. Choose points on these lines which are the t 2 -nd intersection of the span s 2 lines, counting from the center (with label [[s 2 , t 2 ]]) and label them as v 2,0 , v 2,1 , . . . , v 2,m−1 , or collectively as v 2 , with v 2,0 the t 2 -nd intersection point on line L 1,0 . 6. Continue in this fashion until lines of span s h are constructed using points with label v h ; if the symbol correctly identifies an (n 4 ) configuration, then the points v h with label [[s h , t h ]] will coincide with the original m points labelled v 0 . the electronic journal of combinatorics 13 (2006), #R104 4 It is important to note that several symbols may correspond to the same geometric configuration, although the labelling of the points and lines depends on the precise choice of symbol. For an example of this, see Figure 3. In particular, the points labelled v 0 need not be the outermost ring of points, as in Figure 3(b). v 2,5 v 2,4 v 2,3 v 2,2 v 2,1 v 2,0 v 2,7 v 2,6 v 1,0 v 1,7 v 1,6 v 1,5 v 1,4 v 1,3 v 1,2 v 1,1 v 0,7 v 0,6 v 0,5 v 0,4 v 0,3 v 0,2 v 0,1 v 0,0 v 1,6 v 1,5 v 1,4 v 1,3 v 1,2 v 1,1 v 1,0 v 1,7 v 0,1 v 0,0 v 0,7 v 0,6 v 0,5 v 0,4 v 0,3 v 0,2 v 2,2 v 2,1 v 2,0 v 2,7 v 2,5 v 2,4 v 2,3 v 2,6 (a) (b) Figure 3: Two symbols corresponding to the same celestial configuration, with labels. (a) 8#(2, 1; 3, 2; 1, 3); (b) 8#(3, 2; 1, 3; 2, 1). In each configuration, lines L 0 are blue, L 1 are red, and L 2 are green. In [8] and [6] it is shown that given a symbol m#(s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ), reversing the sequence or cyclically permuting the sequence (s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ) using per- mutations that advance the sequence an even number of places leads to an equivalent configuration, while advancing an odd number of places yields a polar configuration. Gr¨unbaum listed two conditions, labelled (*) and (**) below, that must hold in order for a celestial configuration to exist (taken from [8]): s 1 + t 1 + s 2 + t 2 · · · + s h + t h is even (*) cos  πs 1 m  cos  πt 1 m  · cos  πs 2 m  cos  πt 2 m  · · · · · cos  πs h m  cos  πt h m  = 1 (**) In addition, for the configuration to be connected, if m, s 1 , s 2 , . . . s h , t 1 , t 2 , . . . t h have a common factor f, then the symbol m f #  s 1 f , t 1 f ; . . . ; s h f , t h f  must not satisfy conditions (*) and (**) (***) the electronic journal of combinatorics 13 (2006), #R104 5 (or else the original configuration consists of f concentric copies of the smaller con- figuration m f #  s 1 f , t 1 f ; . . . ; s h f , t h f  , rotated so the copies are evenly spaced). Of particular utility are the trivial configurations, where the unordered set of s j ’s is the same as the unordered set of t j ’s, so that conditions (*) and (**) are automatically satisfied; the configuration in Figure 3 is a trivial configuration. 3 Removing half of a symmetry class The movable (n 4 ) configurations will be constructed by “nesting” modified celestial configurations so that two sets of vertices of one configuration lie on the two sets of lines of a second configuration. However, simply nesting two configurations in this fashion would lead to sets of points with five lines passing through them and sets of lines with five points on them. In order to end up with an (n 4 ) configuration, we must delete half the points in one symmetry class of one configuration and half the lines in one symmetry class in the second configuration. Note that to be able to remove half the objects in a symmetry class, the number of objects, namely m, must be even! It is helpful to analyze carefully the notation for celestial configurations presented in the previous section. Consider a configuration m#(s 1 , t 1 ; s 2 , t 2 ; . . . ; s h , t h ). It has points v 0 = [[s h , t h ]], v 1 = [[s 1 , t 1 ]], . . . , v h−1 = [[s h−1 , t h−1 ]]. It also has lines L 0 , L 1 , . . . , L h−1 , where each class of lines L i contains points with label v i and v i+1 , so each point v i has lines L i and L i−1 passing through it, of spans as presented in Table 1. point line span line span v 0 L 0 s 1 L h−1 t h v 1 L 1 s 2 L 0 t 1 . . . . . . . . . . . . . . . v i−1 L i−1 s i L i−2 t i−1 v i L i s i+1 L i−1 t i . . . . . . . . . . . . . . . v h−1 L h−1 s 0 L h−2 t h−1 Table 1: Spans of lines, with their labels, passing through points v i , from the point of view of that class of points. When considering how to delete half the points or lines in a particular symmetry the electronic journal of combinatorics 13 (2006), #R104 6 class, we will look at the interaction of points and lines carefully, from the point of view of different classes of points and lines. The statement “half the lines in a symmetry class may be removed” means that if the lines in the symmetry class are labelled L i,0 , L i,1 , . . . , L i,m−1 , where each line has four points lying on it, then removing the every other line in the symmetry class—for example, removing lines L i,0 , L i,2 , . . . with even index— leaves one line of the symmetry class passing through each point in the symmetry class v i+1 , rather than some points having two lines of the symmetry class incident and others having none. Lemma 1. Half of the lines of span s i passing through the points labelled [[s i , t i ]] may be removed precisely when s i and t i are both odd. Proof. Following the notation in the section on celestial configurations, label the vertices [[s i−1 , t i−1 ]] as v i−1,0 , v i−1,1 , . . . , v i−1,m−1 , and label the lines of span i as L i,0 , L i,1 , . . . , L i,m−1 ; note that L i,j contains points v i−1,j and v i−1,j−s i . Label the vertices [[s i , t i ]] as v i,0 , v i,1 , . . . , v i,m−1 , where v i,0 is the t i -th intersection of line L i,0 with other span s i lines. Then line L i,j also contains points v i,j and v i,j−t i ; that is, they are span t i lines with respect to the v i,j . Suppose that lines L i,0 , L i,2 , . . . are removed. Point v i−1,j contains lines L i,j and L i,j−s i . If s i is even, then whenever j is even, both lines L i,j and L i,j−s i will be removed from point v i,j . On the other hand, if s i is odd, then each v i−1,j will have a single line L i removed. Similarly, considering the lines L i to be lines of span t i passing through the points v i , each v i,j will have a single line L i passing through it precisely when t i is odd. Therefore, if lines L i,0 , L i,2 , . . . are removed, the points v i−1,j and v i,j will each have a single line L i passing through them when both s i and t i are odd. A similar argument holds if lines with odd index L i,1 , L i,3 , . . . are removed. The statement “half the points in a symmetry class may be removed” means the following: Choose one symmetry class of points [[s i , t i ]]; they have labels v i,0 , v i,1 , . . . , v i,m−1 . Each point has four lines passing through it, two of span t i and two of span s i+1 . We wish to remove every other point in a symmetry class—for example, points v i,0 , v i,2 , . . . with even index—in such a way that each of the lines of span t i−1 and span s i have a single point labelled v i,j , rather than some lines having two points labelled v i,j and some lines having none. Lemma 2. Half the points in the symmetry class [[s i , t i ]] may be removed precisely when t i and s i+1 are both odd. the electronic journal of combinatorics 13 (2006), #R104 7 Proof. Consider the points [[s i+1 , t i+1 ]]. Label them v i+1,0 , v i+1,1 , v i+1,2 , . . . , v i+1,m−1 . The v i+1 lie on lines of span s i+1 , labelled as L i+1 , where line L i+1,j contains points v i+1,j and v i+1,j−s i+1 . Now remove the points v i+1,q where q is even. If s i+1 is even, then in line L i+1,j if j is even, both points v i+1,j and v i+1,j−s i+1 will be removed, while if j is odd, neither point will be removed. On the the other hand, if s i+1 is odd, in line L i+1,j , exactly one of the points v i+1,j or v i+1,j−s i+1 will be removed, since their indices are of different parity. Similarly, in the situation where points v i+1,q where q is odd, are removed, if s i+1 is odd, then in line L i+1,j , exactly one of the points v i+1,j or v i+1,j−s i+1 will be removed, while if s i+1 is even, both points or neither point will be removed. The points v i+1 also lie on lines L i , as the t i -th intersection of span s i lines. From the point of view of the points v i+1 , the lines L i are span t i lines; that is, the points v i+1 are the s i -th intersection of the span t i lines. Using this point of view, the previous argument shows that half of these points may be removed from the span t i lines, leaving only one point labelled v i+1 on each line, precisely when t i is odd. The method of constructing movable configurations which will be presented here will begin with one modifed celestial configuration C which has a symmetry class of points, called S, with half its points removed, and a second modified celestial configuration D, with the same value for m, which has had half of an appropriate symmetry class of lines removed. These configurations must be chosen so that when an arbitrary point is placed on a line in C containing points in S and then rotated to form an m-gon, and D is constructed using these points as its starting m-gon, one of the symmetry classes of points in D lies on a class of lines in C. In Figure 4, diagonals of span 2 and span 3 are shown (in blue and red, respec- tively), along with an arbitrary point placed on a blue line and rotated around to form another 10-gon, and a set of green lines which are diagonals of span 3 for the second 10-gon. Note that the second intersection point of the green lines with each other lies on the red lines. Theorem 3. Given a regular m-gon with vertices u 0 , u 1 , . . . , u m−1 and diagonals of span a and span b, suppose that w 0 is an arbitrary point on the 0-th diagonal of span a, which joins u 0 and u a and is denoted by u 0 , u a , and the other w i are formed by rotating w 0 by 2πi m . Let q i be the intersection of the line w i , w i+b  with the span b diagonal u i , u i+b , and let q  i be the intersection of the lines w i−a , w i−a+b  and u i , u i+b . Then q i = q  i . That is, if you begin with a set of diagonals of span a and span b of an m-gon M , construct another m-gon N whose vertices are the rotated images of a point placed the electronic journal of combinatorics 13 (2006), #R104 8 [4,5] [4,3] [4,2] [4,1] [4,4] Figure 4: Beginning with a regular m-gon with diagonals of span a (blue lines) and b (red lines), constructing a second m-gon with diagonals of span b (green lines) whose vertices are the rotated images of a point placed arbitrarily on a diagonals of span a leads to these diagonals of span b intersecting the original diagonals of span b. In this figure, m = 10, a = 2 and b = 3. arbitrarily on a diagonal of span a, and construct diagonals of span b using N, then these diagonals intersect the span b diagonals of M , and the intersection points are precisely the points labelled [[b, a]] in N. To prove this lemma, I will need two geometric results. Lemma 4. If quadrilateral QRST has the property that ∠RQS = ∠RT S, then the points Q, R, S, and T are concyclic. S R Q T This lemma is a well-known result from Euclidean geometry; see, for example, [1, p. 127] Lemma 5. Let M and N be two concentric m-gons with center O and vertices u 0 , u 1 , . . . , u m−1 and w 0 , w 1 , . . . , w m−1 , respectively. Let  1 be the line u 0 , u s , a diagonal of M of span s, and let  2 be the line w i , w i+s , a diagonal of N of span s. Let q be the intersection of  1 and  2 . Then the four points u 0 , w i , O and q are concyclic. Proof. See Figure 5(a) for an illustration. Let M u be the midpoint of u 0 u s and let the electronic journal of combinatorics 13 (2006), #R104 9 M w be the midpoint of w i w i+s . Then ∠u 0 OM u = πs m , ∠w i OM w = πs m , ∠u 0 M u O = π 2 , and ∠w i M w O = π 2 , since the angle between endpoints of a segment of span s is 2πs m and the segments M u O and M w O are perpendicular to the diagonals  1 and  2 respectively. It follows that ∠M u u 0 O = ∠M w w i O. But q lies on  1 and  2 , so ∠qu 0 O = ∠qw i O. By Lemma 4, q, w i , u 0 and O are concyclic. We will apply this result several times to prove Theorem 3; see Figure 5(b). Mw Mu q wi+s us O u0 wi q0 w-a wb wb-a ub ua O u0 w0 q' 0 (a) (b) Figure 5: Illustrations for the proofs of (a) Lemma 5 and (b) Theorem 3. Proof of Theorem 3. Let M be an m-gon with center O, vertices u 0 , u 1 , . . . u m−1 and diagonals of span a and span b, let w 0 be an arbitrary point on line u 0 , u a , and let the other w i be formed by rotating w 0 by 2πi m . Let q i be the intersection of line w i , w i+b  with the span b diagonal u i , u i+b , and let q  i be the intersection of the lines w i−a , w i−a+b  and u i , u i+b . By symmetry, it suffices to show that q 0 = q  0 . Since q 0 is the intersection of w 0 , w b  and u 0 , u b , by Lemma 5, q 0 , w 0 , u 0 and O are concyclic. Similarly, q  0 lies on the same circle as O, u 0 , and w −a , since it lies on the intersection of span b diagonals w −a , w −a+b  and u 0 , u b . Finally, since the lines w −a , w 0  and u 0 , u a  are both of span a and intersect at the point w 0 , it follows that the points w 0 , w −a , u 0 , and O are concyclic as well, again by Lemma 5. Since a circle is uniquely determined the electronic journal of combinatorics 13 (2006), #R104 10 [...]... odd; note that m#(a, b; c, a; b, c) and m#(c, b; a, c; b, a) represent the same geometric configuration, but the labelling is reversed The smallest such example is given in Figure 7 4.4 Using two astral (n4) configurations In [5], astral (n4 ) configurations — that is, (n4 ) configurations with precisely two symmetry classes each of points and lines — were completely characterized Some astral (n4 ) configurations,