Jim Sauerberg and Alan Tarr

Transcription

1 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex Centering Jim Sauerberg and Alan Tarr Jim Sauerberg is an associate professor of mathematics and computer science at Saint Mary s College of California. His mathematical interests are in number theory, and his personal interests are currently centered around Elmo and vacuuming (thanks to his two young sons). He thanks his Spring 1999 Finite Mathematics students for inspiring this article. Alan Tarr (alantarr@hotmail.com) is currently a student at Campolindo High School, expecting to graduate in Spring His interests in mathematics include number theory. He also enjoys eating π and playing percussion in the jazz and symphonic bands at Campo. Introduction Four very egotistical friends are having a conversation. Each wants to be the center of attention so, simultaneously, each moves so as to be at the center of the circle determined by the locations of the other three friends. For example, suppose the friends started at the locations (3, 0), (0, 1), ( 1, 0) and (0, 1). The friend at (3, 0) knows that the center of the circle passing through (0, 1), (0, 1) and ( 1, 0) is (0, 0), and will move from (3, 0) to (0, 0). Likewise, the friend at (0, 1) knows that the center of the circle passing through (0, 1), (3, 0), and ( 1, 0) is (1, 1), and will move from (0, 1) to (1, 1). Similarly, the friends at ( 1, 0) and (0, 1), respectively, will move (4/3, 0) and (1, 1), respectively. (See Figure 1.) (0,1) (3,0) (-1,0) (0,0) (0,-1) (1,-1) Original Locations (1,1) (1,1) (4/3,0) (0,0) (4/3,0) (1,-1) Final Locations Figure 1. The friends finding their centers. Of course, since each friend moved, none becomes the center of attention and so the result satisfies none of them. So they do it again. And again. Will the process stop? Will it bring the friends closer together? Or push them further apart? Let us make a definition. 24 c THE MATHEMATICAL ASSOCIATION OF AMERICA page 24

2 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 25 Definition. Fix a set of points in the plane. A point centers itself by (1) Determining its three nearest neighbors (2) Finding the center of the circle those neighbors determine, and (3) Translating itself to that center. A set of points centers itself when all of its member points center themselves simultaneously. Most of this paper is devoted to the study of how centering affects sets of four points, or, equivalently, quadrilaterals. Our main results show how the interior angles, area, and edge-lengths of the original quadrilateral determine the angles, area, and edge-length of the centered quadrilateral. When the set contains more than four points the center in step (2) may or may not be well-defined. For example, although each vertex of a regular octagon has four nearest neighbors, any three of these define the same circle, and the octagon centers itself to its center. On the other hand, one cannot center {(0, 0), ( 1, 0), (1, 0), (0, 2), (0, 2)} for (0, 0) would be centered to either (0, 3/4) or (0, 3/4) depending on whether (0, 0) s nearest neighbors are chosen to be {( 1, 0), (1, 0), (0, 2)} or {( 1, 0), (1, 0), (0, 2)}. At the end of the paper we give several examples of sets of many points for which the centering map is well-defined and which have interesting behaviors under it. Simple examples We begin with a couple of simple examples to help the reader gain some familiarity with our map. The simplest way to construct graphically the new location A of a point A is to find the intersection of the perpendicular bisectors of the segments BC and CD,whereAs nearest neighbors are B, C, andd. This is because these bisectors contain the points equidistant from B and C, andc and D, respectively. Hence, their intersection is the point that is equidistant from B, C, andd, i.e., the center of the circle determined by B, C,andD. B B A C A C D D Figure 2. A quadrilateral ABCD centered to A B C D. The set consisting of the vertices of a square centers itself to the center of the square. In fact, the vertices of a regular polyhedron with at least four edges center themselves to the center of the polyhedron. More generally, any set of at least four points lying on a circle behave this way. As a slightly more complicated example, consider the rhombus with corners at (±x, 0), (0, ±y). It is easy to show that the centered object is also a rhombus with VOL. 33, NO. 1, JANUARY 2002 THE COLLEGE MATHEMATICS JOURNAL 25

3 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 26 corners at (±t, 0), (0, s),wheret = y2 x 2 and s = y2 x 2. Notice that x/y = s/t, 2x 2y so the centered rhombus has the same interior angles as the original, with the major and minor axes having switched. (0,y) (0, t) (-s,0) (-x,0) (x,0) (s,0) (0, -t) (0, -y) Figure 3. A rhombus and centered rhombus. The area of the original rhombus is 2xy while the area of the centered rhombus is (y 2 x 2 ) 2 /(2xy). If the area of the rhombus is preserved, then (x 2 x 2 ) 2 = (2xy) 2, so y = (1 + 2)x. In this case, both rhombi are equilateral, since one can check that edge-length is preserved if and only if y = (1 + 2)x. Hence, a rhombus centers itself to a congruent rhombus if and only if its interior angles are 45 and 135. Interior angles In this section we discuss the effect centering has on the interior angles of a quadrilateral. Proposition. Let ABC D be a quadrilateral that has been centered to A B C D. The angle at each corner of A B C D is equal to the supplement of the angle at the opposite corner in ABC D. The proof follows from the construction (see Figure 2). Point A is formed from the intersection of the perpendicular bisectors of segments CB and CD.ThatA lies on the quadrilateral formed by A, C and the feet of the two bisectors on BC and CD shows that A = 180 C,where A means the interior angle at the vertex A. We might say that a quadrilateral centers itself to a supplementary quadrilateral,by which we mean the angles are changed to their supplements. If we center a quadrilateral twice in succession then we get a quadrilateral with the same interior angles as the original. However, the new quadrilateral is not necessarily similar to the original since the edge-lengths do not change proportionally, as we will see in the next section. It is well-known that a quadrilateral may be inscribed in a circle if and only if its opposite angles are supplementary. We have seen that a quadrilateral inscribed in a circle is centered to the center of that circle. So if we have a quadrilateral whose angles consist of supplementary pairs but which does not become trivial upon being centered, then these supplementary pairs must be consecutive. That is, our original quadrilateral must be a parallelogram. 26 c THE MATHEMATICAL ASSOCIATION OF AMERICA

4 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 27 Corollary 1. A quadrilateral is centered to a non-trivial quadrilateral with the same interior angles if and only if it is a parallelogram. This result explains why the parallelogram with corners at (±x, 0), (0, ±y) was centered to one with the same interior angles. Area and edge-length We next consider the area and edge-lengths of a centered quadrilateral. Any nonrectangular convex quadrilateral ABCD must have an acute angle adjacent to an obtuse or right angle. So we may position the quadrilateral to have corners at (0, 0), (0, y 1 ), (x 2, y 2 ),and(x 3, y 3 ), with x 2, x 3, y 1, y 2 > 0, y 3 0, as in Figure 4. The area of the quadrilateral is (x 3 y 2 + x 2 y 1 x 2 y 3 )/2. (0,y1) B C (x2,y2) (0,0) A D (x3,y3) Figure 4. A general quadrilateral. After a bit of work, and with the help of Mathematica, we find that the area of the centered quadrilateral is (x 3 y 2 + x 2 (y 1 y 3 ))(x 2 x 3 (x 2 x 3 ) + x 3 y 2 (y 2 y 1 ) x 2 y 3 (y 3 y 1 )) 2. 8x 2 x 3 (x 2 (y 1 y 3 ) + x 3 (y 2 y 1 ))(x 2 y 3 x 3 y 2 ) The ratio of the area of the centered quadrilateral to the area of the original simplifies to new area old area = (x 2x 3 (x 2 x 3 ) + x 3 y 2 (y 2 y 1 ) + x 2 y 3 (y 3 y 1 )) 2 4x 2 x 3 (x 2 (y 1 y 3 ) + x 3 (y 2 y 1 ))(x 2 y 3 x 3 y 2 ). Next we compute the tangents of the interior angles of ABCD: tan( A) = x 3 y 3 tan( B) = x 2 y 1 y 2 tan( C) = tan( D) = x 2 (y 1 y 3 ) + x 3 (y 2 y 1 ) x 2 (x 2 x 3 ) + (y 2 y 3 )(y 2 y 1 ) x 2 y 3 x 3 y 2 x 3 (x 2 x 3 ) + y 3 (y 2 y 3 ). A simple computation then gives our main result. VOL. 33, NO. 1, JANUARY 2002 THE COLLEGE MATHEMATICS JOURNAL 27

5 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 28 Theorem 1. The ratio of the area of the centered quadrilateral to the area of the original quadrilateral is given by R = 1 ( )( 1 4 tan( A) + 1 tan( C) 1 tan( B) + 1 tan( D) ). (1) In words, when centered, the area of a quadrilateral changes proportionally with the inverse of the product of the harmonic means of the tangents of the opposite angles. We are unaware of a synthetic proof of Theorem 1, although one would be almost certainly preferable to the analytic proof given here. The ratio in (1) explains much of the centering process. First, notice that it is never positive. Clearly R = 0when A and C,or B and D are supplementary. If neither of the sums of the opposite angles is equal to 180, then obviously one is larger and one is smaller. However, if A + C > 180 and B + D < 180,then 1 tan( A) + 1 tan( C) > 0 and 1 tan( B) + 1 tan( D) < 0. Hence we always have R 0. This is why, as the observant reader will already have noticed, the centering process reverses the rotation of the quadrilateral. That is, if the order of the vertices of a quadrilateral, read clockwise, is ABCD, then the order of the vertices of the centered quadrilateral, read clockwise, will be A D C B. In effect, the centered quadrilateral is inverted, giving negative area. We may also deduce our earlier geometric results from this algebraic one. First, if our quadrilateral is a rectangle, then the tangents of the various angles are infinite. Hence (1) indicates that the area of the centered rectangle will be zero, as we have seen. Next, suppose one of the angles of the quadrilateral is 180, so the quadrilateral degenerates into a triangle. Then (1) implies that the centered object will have infinite area. This agrees with our geometric intuition, since centering will put the noncollinear corner of the quadrilateral at infinity. In a parallelogram angles A and C are the supplements of B and D. Hence ( tan( A) + 1 tan( C) )( ) 1 tan( B) + 1 tan( D) = 1 tan 2 ( A). (2) Since 1/ tan 2 ( A) >1 exactly when 45 < A, we see that a parallelogram whose smaller angle is larger than 45 will have its area grow when repeatedly centered, while those whose smaller angle is less than 45 will shrink toward nothingness. Further, a parallelogram with angles 45 and 135 will not have its area change. Finally, we also see in this result a proof of the fact that a quadrilateral may be inscribed in a circle if and only if its opposite angles are supplementary. For R will be 0 if and only if tan( A) = tan( C) or tan( B) = tan( D), and this is so if and only if A, C and B, D form supplementary pairs. Just as (1) relates the angles and the areas of our quadrilaterals, we may ask for a similar relationship between the angles and the edge-lengths. In Figure 5 we see two quadrilaterals, A 1 BCD and A 2 BCD, that are nearly identical. However, in the centered versions, the edges A B 1 and A B 2 have very different lengths. This suggests we cannot hope for a relationship relating only the lengths of AB and A B to the angles involved. 28 c THE MATHEMATICAL ASSOCIATION OF AMERICA

6 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 29 D C A 1 A 2 A B 1 B 2 Figure 5. Variations in edge-length. B Perhaps surprisingly, when we consider pairs of opposite edges simultaneously, there is such a relationship. We state without proof the following result, which is easily checked with Mathematica. Theorem 2. If ABC D is a quadrilateral and A B C D is the centered quadrilateral, then we have A B C D AB CD = A D B C AD BC = A C B D AC BD = R. In words, while it is not true in general that the individual edge-lengths grow in proportion, the products of the lengths of opposites edges do grow in such a manner. Further, this holds also for the diagonals. In the special case of parallelograms, since opposite edges have the same edgelength, we do have proportional growth, with A B = A D = 1. (Again the negative denotes the flipping of the figure.) Hence we AB AD tan( A) have Corollary 2. A parallelogram is centered to a similar parallelogram. It is not clear to us when individual edge-lengths are preserved. By Theorem 2, if the centering preserves edge-length, then it also preserves area, but the converse is not necessarily true. It also would be interesting to determine exactly which quadrilaterals have predecessors, i.e., which quadrilaterals are centered versions of other quadrilaterals. It is easy to see that an inscribed quadrilateral cannot have a predecessor, since by the Proposition its predecessor would also be inscribable. On the other hand, all parallelograms do have predecessors. Examples of other behavior When the set under consideration has more than four points a variety of interesting behaviors can result. In this section we give several of these. We continue to consider only sets of points for which centering is well-defined. Finite pointwise fixed sets. We have seen that no set of four points stays fixed when centered. The reader may demonstrate that this is also true for sets consisting of five and six points. The smallest pointwise fixed set is that of the seven points making up the vertices of a regular hexagon plus its center. In fact, any finite tiling such that every point is either the center or corner of a regular hexagon is pointwise fixed. VOL. 33, NO. 1, JANUARY 2002 THE COLLEGE MATHEMATICS JOURNAL 29

7 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 30 Plane-covering pointwise fixed sets. There are many tilings of the plane whose vertices remain pointwise fixed when centered. For example, any tiling composed of equilateral triangles, squares, and regular hexagons, all of whose edge-lengths are the same, will be pointwise fixed, as will any tiling made up of squares and their centers. Many of the well-known tilings of the plane will also be pointwise fixed. For example, all eleven of the Archimedean or uniform tilings of the plane (with the proper choice of edge-lengths for ( ), (4.6.12) and (4.8 2 ))[1, p. 63], and seven of the eleven Laves tilings of the plane ([3 6 ], [ ], [ ], [ ], [4 4 ], [4.8 2 ],and[6 3 ]) [1, p. 96] have this property. Infinite pointwise fixed sets with finite area. There are pointwise fixed sets with infinitely many points but whose convex hulls have finite area. One example is the self-similar figure we call a fan. Begin with a point A. Add to it three connecting lines of length one separated by 60 each and place points B, C, andd at the end of each of these lines. Shrink this three-segment prong by a factor of 1/r for any fixed r > 2 and append it to each of the end points. Continue repeating this process of appending a prong to the end of each line, shrinking the prong by a factor of 1/r at each step. (See Figure 6.) D C B E F A Figure 6. The Beginning of the Fan, and Several Generations of the Fan. Are the points in the fan actually pointwise fixed? Certainly it appears from our picture that the points B, C, andd are the closest to point A and so A should remain fixed when the set is centered. To prove this we will first show that all of the points in the family of points extending from B, C, andd remain outside of the unit circle centered at A, and then will show that the families of points emanating from B and C do not interact. First notice that the entire fan fits inside a circle of radius r/(r 1). So, traveling along the connecting lines of the prongs, the distance from point A to any end point is also r/(r 1). Because of the self-similarity, the distance from point B to any end point is (1/r) r/(r 1) = 1/(r 1), and from point E to any end point is 1/(r 2 r). If we place point A at the origin, point E has coordinates ( 3/2, 1/2 + 1/r) and is a distance of r 2 + r + 1/r from A. Hence the closest a point extending from B could r 2 +r+1 come to point A is. But it is easy to show that this is larger than 1 for r > 2. Hence the closest point to A in the family of points emanating from B is B. To see that the points from Bs andcs families stay separated, notice that the distance from E to the line midway between B and C is the same as the distance from 1 r r 2 r point F to the x-axis. Since the coordinates of F are ( 3 ( r+1 ), 1 ( r 1 )), and this is 2 r 2 r greater than 1/(r 2 r) when r > 2. Hence if we take, for instance, r = 3 then none 30 c THE MATHEMATICAL ASSOCIATION OF AMERICA

8 Integre Technical Publishing Co., Inc. College Mathematics Journal 33:1 September 28, :30 a.m. sauerberg.tex page 31 of the three families emanating from A can interfere with each other, and, by selfsimilarity, this holds for all the subsequent families. Therefore the fan is an infinite pointwise fixed set occupying finite area. Open questions We finish with a small list of unanswered questions. (1) The fundamental building blocks for the finite pointwise fixed structures we gave were regular hexagons, equilateral triangles and squares. Are there other ways of building finite pointwise fixed structures? (2) Our finite pointwise fixed examples were built from regular hexagons, equilateral triangles, and squares, so the distance from a point to its nearest neighbor(s) was independent of the point. Are there any finite pointwise fixed structures in which this is not so? Are there any tilings of the plane for which this is not so? (3) Is there a finite set of points that is set-wise fixed but not pointwise fixed? (4) It is possible to create four points so that two of them coalesce in the next centering (or, if you prefer, so that one disappears)? Is there a set of points that nicely vanishes, one by one, until three or fewer are left? (5) Is there a set of points that exhibits true movement? Our quadrilaterals all lurch around, and generally either grow or shrink, but do not really have directional motion. (6) What happens in the general case? Suppose we have n points on the plane and during each unit time interval each point finds its t nearest neighbors, determines the position so that the sum of the distances to those neighbors is minimal, and then at the end of the unit time interval translates itself to that position. If we iterate this process, how does the set of points behave? The boundary case t = 3 was considered in this paper. The other boundary case is easier to understand. If t = n then all points move to the same point (what we d probably call the center of mass of the set of points) at the end of the first time interval. In fact, if t is a sufficiently large fraction of n, we d similarly expect the points to lump together into one or more centers of mass. For example, if t = n/4 and the points were in four widely separated groups of n/4 points, then we d end up with four points at the end. What happens for t small in relation to n? Reference 1. Branko Grünbaum, Tilings and Patterns, W. H. Freeman, VOL. 33, NO. 1, JANUARY 2002 THE COLLEGE MATHEMATICS JOURNAL 31