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Lesson Plans


Fw: 3 projects in color and math

[ Thread ][ Subject ][ Author ][ Date ]
Larry Cox (L_J_Cox)
Tue, 16 Feb 1999 10:17:34 -0700


>Subject Terms: Secondary school students
> Middle school students
> Mathematics education
> Geometry
> Color
>Three activities in discrete mathematics that involve the coloration of
>geometric objects--counting colored regions of overlapping simple closed
>curves, counting colored triangulations of polygons and determining the
>number of colors either necessary or sufficient to paint the points of
>the plane while satisfying a distance rule--are presented for students
>in grades 6-12.
>
>TEACHER'S GUIDE
>
>In this article, we suggest three activities in discrete mathematies, all
>of which involve the coloration of geometric objects: counting colored
>regions of overlapping simple closed curves, counting colored
triangulations
>of polygons, and determining the number of colors either necessary or
sufficient
>to paint the points of the plane while satisfying a distance rule. They
>are designed to be used in small groups and can also serve as topies for
>whole-class discussions, offering opportunities for generating, collecting,
>and organizing data and examples; identifying, describing, and studying
>patterns; and making, testing, and verifying conjectures.
>
>The activities are independent of one another and need not be done
consecutively.
>Sheet 1 is very accessible and makes a nice stand-alone investigation.
>Sheet 2 makes connections to the angle-sum formula for polygons and to
>Euler's formula. Sheet 3 enhances work on tiling and distance.
>
>Coloring problems are not a staple of the secondary curriculum, but they
>have an important place in mathematics. The most famous coloring problem
>is the four-color problem: If a map is to be colored so that countries
>sharing a border must receive contrasting colow, will more than four colors
>ever be needed? The problem originated in 1850 and eluded solution until
>1976 when Kenneth Appel and Wolfgang Haken proved that four colors suffice.
>Other coloring problems originate in the classification of border and
wallpaper
>patterns in which motifs are distinguished by color as well as by geometric
>shape. Although not a focus of these activities, a variety of real-world
>problems in scheduling, communications, traffic flow, and inventory control
>and storage can be interpreted and solved using concepts of color.
>
>Grade levels: 6-12. These activities, or portions thereof, have been used
>in grades 6 through 12. The discovery of the patterns does not depend on
>any specific formulas or theorems from other sources. The slightly of meat
>topics are real attention grabbers. Indeed, it has been our experience
>that even some of the more average students rise to the occasion and become
>major contributors to the class discussion. Especially gratifying has been
>the request from students to pursue the question of why these unexpected
>patterns hold, giving the teacher a wonderful opportunity to work through
>a proof or derivation that is truly appreciated by the class. These proofs
>are outlined subsequently. Each teacher should assess the readiness of
>students to tackle these proofs.
>
>Materials: Scratch paper; activity sheets; rulers; colored pencils in at
>least four distinct colors, preferably red, yellow, green, and blue;
compasses
>(optional)
>
>Prerequisites: Students should be familiar with the following terms and
>concepts from elementary geometry: polygon, vertex of a polygon, interior
>angle of a polygon, regular hexagon, and polygon sum property (the sum
>of the interior angles of an n-gon is (n - 2)180).
>
>Sheet 1: Before handing out the sheet, review or introduce the definition
>of a simple closed curve: a closed curve that does not intersect itself.
>Have students draw on the chalkboard with colored markers or chalk several
>examples and nonexamples, as shown in figure 1. (Figure 1 omitted) Mention
>that any simple closed curve separates the plane into two disjoint regions,
>the interior and the exterior. This important result is called the Jordan
>curve theorem, which was developed by Camille Jordan (1838-1922). If
necessary,
>display and discuss one or two examples before letting students investigate
>the questions in small groups.
>
>If sheets of yellow and blue plastic are available--report covers work
>nicely--use scissors to cut along fairly complex, but still simple, closed
>curves from each sheet to create two regions. When the yellow and blue
>shapes are overlapped on the overhead projector, four types of regions
>are created: white (outside both curves), yellow (inside the yellow curve
>only), blue (inside the blue curve only)l and green (inside both the yellow
>and the blue curves). The colored plastic sheets can be used to emphasize
>the crossing rules of intersection and are easy to reposition to recount
>the number of regions.
>
>Students should discover that all properly overlapping simple closed curves
>have the following properties:
>
>i. The number of yellow regions is always equal to the number of blue
regions.
>
>ii. The number of green regions is always equal to the number of white
>regions.
>
>A nice extension can be used to challenge cooperative groups that find
>these properties quickly and want to look further. Suggest that they also
>consider the number of points of intersection of the two curves and how
>that number compares with the number of colored regions. They will find
>this property:
>
>iii. The number of intersection points is always even, say 2n, and is 2
>less than the total number of regions. If y, b, g, and w denote,
respectively,
>the numbers of yellow, blue, green, and white regions, then we have y =
>b, g = w, and b + w = n + 1.
>
>An informal but convincing argument, the rubberband proof, can be given
>to justify these properties. Imagine that the blue curve is allowed to
>shrink to a small circle that intersects the yellow curve in just two
points.
>This case is much like the two-circle case shown in the example, which
>clearly shows that each of the four colors has one region and that n equals
>two points of intersection. Next imagine that the blue circle grows and
>distorts as it returns to its original shape. Each time the growing blue
>curve intersects the yellow curve at two new points of intersection, two
>new regions are formed. Either one new region is blue and one is yellow
>or one new region is green and the other is white. Either way, the counts
>of blue and yellow regions, and of green and white regions, are kept equal
>at each step, and the relationship b + w = n + 1 is maintained. When the
>blue curve is stretched to coincide with its original shape, properties
>i, ii, and iii follow.
>
>Sheets 2A and 2B: The first three explorations on sheet 2 lay the
groundwork
>for exploration 4. However, they are interesting in their own right and
>can be pursued even if exploration 4 is not used. Exploration 1 is
strightforward
>and should be accessible to almost all students. For exploration 2,
polygons
>with n equal to 5, 6, 7, and 8 work well. The result of exploration 2 is
>surprising; an explanation follows.
>
>CONJECTURE. The parity of the number of vertices of each type and the
parity
>of the total number of vertices must be the same. Inegers have the same
>parity ifall are even or all are odd.
>
>PROOF. Each side of the polygon contains two vertices. Hence, the number
>of red-green (rg) vertices plus the number of blue-green (bg) vertices
>is equal to twice the number of green sides and is, therefore, an even
>number. Thus the parity of rg must be the same as the parity of bg. By
>a similar argument, the parity of rg must be the same as the parity of
>rb. Since the parity of all three types is the same, and since the sum
>of three odd or even numbers is odd or even, respectively, the total number
>of vertices is odd if all three types are odd and is even if all three
>types are even.
>
>Before assigning exploration 3, discuss the definition of a triangulated
>polygon and review what is meant by a vertex of the polygon and how it
>differs from an interior vertex, which was introduced in the triangulation
>process. Also make sure that the notation is clear.
>
>If students have trouble discovering the pattern in the table, suggest
>that they draw figures that maintain one variable constant to see the
relationship
>between the other two. For example, they can use i = 1, vary n, and see
>what happens to t, then keep n constant while they vary i. If students
>get stuck, suggest that an extra row be added to the table, showing 2i.
>The formula is t = n + 2i - 2. An outline of the proof follows.
>
>Suppose that the degree measures of all interior angles of all t triangles
>in a triangulation are added. Why is this sum equal to 180t? Why is this
>sum also given by 360i + 180(n - 2) for a triangulated n-gon having i
interior
>vertices? What formula connecting n, i, and t results when these
expressions
>are equated? Is the formula consistent with the data?
>
>Another proof of this formula, based on Euler's formula in graph theory,
>is presented subsequently. This formula leads to a nice proof of Pick's
>formula for the area of a lattice polygon; for details, request a copy
>of the first reference from the author.
>
>Before assigning exploration 4, discuss the coloring rules carefully. All
>interior vertices should be checked to see if all three colors mistakenly
>appear; if so, the error must be corrected. Illustrate how to color the
>example on the sheet. It is actually instructive to make an error, point
>it out, and then show how some edges can be recolored to correct the error.
>A proof of the final result requires familiarity with the basics of graph
>theory. A copy of the proof is available from the author. However, students
>can do the exploration and make a conjecture even if they cannot understand
>the proof. The new results entered in the table from exploration 2 often
>amaze them] The last three questions consolidate an understanding of the
>conjecture.
>
>Note that a valid coloring always exists: one permissible color can always
>be assigned to any edge. For example, the edge between a red-blue and a
>blue-green vertex can--in fact, must--be colored blue.
>
>If geometry-drawing software is available, the polygons and triangulations
>can first be drawn in black; edges are next selected and colors are
assigned.
>Using this software is a convenient way to draw the figures and correct
>easily the inevitable coloring errors.
>
>If students are familiar with the basic definitions and concepts of graph
>theory, this section will help the teacher connect those ideas with sheet
>2. Recall that a graph is a set of vertices, denoted V sub 0 , V sub 1
>, ..., v sub n , and a set of eges that join certain pairs of distinct
>vertices. A graph is depicted by drawing dots for vertices and joining
>pairs of dots to indicate edges.
>
>An alternative proof of the exploration-3 formula follows. Euleis formula,
>V-E + R = 1, relates the number of vertices V, edges E, and bounded regions
>R in a connected plane network. For a triangulated n-gon with i interior
>vertices and t triangles, V = n + i and R = t. Since each triangle is
bounded
>by three edges, 3t counts each inner edge of the triangulation twice and
>each n edge of the polygon once. Thus 2E = 3t + n. Rewriting Euler's
formula
>as 2V - 2E + 2R = 2 gives 2 (n + i) - (3t + n) + 2t = 2. This equation
>is easily solved for t, resulting in t = n + 2i - 2.
>
>Sheets 3A and 3B: Be sure that the one-inch color rule is understood
correctly.
>Explain that the smallest number of colors needed to paint the plane
according
>to the rule is not known and that the work on this sheet aims to show how
>a lower and an upper bound for this number have been found.
>
>A compass set at a one-inch opening can be useful, but an inch ruler also
>works. Instead of coloring the hexagons, it is easier to put a letter or
>number inside a hexagon to indicate its color.
>
>Sheets 3A and 3B prove the following theorem:
>
>THEOREM. The minimum number of colors needed to paint the points of the
>plane so that no two points one inch apart are of the same color is at
>least four but no more than seven.
>
>The required number of colors is four, five, six, or seven, but determining
>which of these four numbers is the precise minimum value is still an
unsolved
>problem of geometry some thirty-five years after it was proposed by Gardner
>(1960) and Hadwiger (1960). A discussion of the problem, and other painting
>problems in the plane, can be found in Klee and Wagon (1991).
>
>Answers to selected problems: Sheet 1
>
>Example: One region of each color exists. The other figure given as an
>example has four yellow and four blue regions and three white and three
>green regions.
>
>Exploration. See the teacher's guide.
>
>Sheets 2A and 2B
>
>Exploration 1. When n is even, two colors are enough, alternating about
>the edges of the polygon. When n is odd, alternating two colors will no
>longer satisfy rule 1, and a third color, say, for the last edge, is
required.
>
>Exploration 2. The first four numbers in each column will be all even or
>all odd. That is, the number of vertices of each type, and the total number
>of vertices, all have the same parity.
>
>Exploration 3. The relationship is t = n + 2i - 2. See the teacher's guide
>for a proof.
>
>Exploration 4. The six numbers in each column have the same parity. See
>the teacher's guide for a proof.
>
>Answers to follow-up questions
>
>1. If the number of red-blue vertices is 7, then the parity of all the
>entrie in that column is odd. Since an odd positive integer is 1 or larger,
>at least one blue-green veTtex of the polygon must exist.
>
>2. Since all the entries in a column have the same parity, the number of
>complete triangles is also odd. In particular, at least one complete
triangle
>always exists.
>
>3. The number 13 is odd, so all numbers in the column are odd and, as in
>question 2, this fact alone indicates that at least one complete triangle
>exists.
>
>Sheet 3A
>
>Exploration 1. Let point A be painted any color, say, red. Since B is one
>inch from A, it cannot be red, so paint it blue. Point C, being one inch
>from both A and B, cannot be red or blue, and so a third color is required.
>Thus three, or perhaps more, colors are required to paint the points of
>the plane to satisfy the one-inch color rule.
>
>Exploration 2. From the foregoing discussion, we know that three colors
>are necessary to paint A, B, and C. Suppose that A is red, B is blue, and
>C is green. Point D, since it is one inch from B and C, cannot be blue
>or green. But since D is more that one inch from A, it can be red. Only
>three colors are needed to paint A, B, C, and D, but then A and D must
>have the same color.
>
>Exploration 3. Suppose that A is red. From exploration 1, we know that
>B and C must be painted two other colors, say, blue and green. Similarly
>B' and C' must be blue and green or green and blue. By exploration 2, if
>we are attempting to use just three colors, then poinC D would have to
>be red, the color of A. But the same reasoning shows that D' would also
>need to be red. Since D and D' are one inch apart and both are red, we
>see that it is impossible to use just three colors to color the points
>A, B, C, D, B: C: and D' according to the one-inch rule. A fourth color,
>say, yellow, is needed for D'. The following conclusion can be reached:
>If each point of the plane is painted a color such that no two points one
>inch apart are painted the same color, then at least four colors are
needed.
>
>Note that we do not claim that four colors are enough; perhaps some more
>ingenious set of points would show that even four colors are not enough.
>
>Sheet 3B
>
>Exploration 1. No two points of any one hexagon are one inch apart, so
>painting a hexagon and its interior all the sme color is permissible. If
>hexagon 1 is, say, red, then hexagon 2 must be of a different color, say,
>blue, since it has points that are one inch from points in hexagon 1.
Similarly,
>hexagon 3 must get a different color, say, green, than either hexagon 1
>or hexagon 2. Hexagon 4 cannot be blue or green, but since all its points
>are more than one inch from all the points in hexagon 1, it is permissible
>to paint hexagon 4 red, the same color as hexagon 1. Notice that the points
>along the edges of the hexagons can be painted with either color of the
>adjoining hexagons.
>
>Exploration 2. Every one of the seven hexagons contains points that are
>exactly one inch from some point in any other hexagon. Thus, if all the
>points of the hexagon are to be of the same color, we need seven colors.
>
>Exploration 3. The seven adjoining hexagons, each with a different color,
>form a tile that can cover the plane. The following conclusion can be
reached:
>If each point of the plane is assigned a color such that no two points
>one inch apart receive the same color, then seven colors are sufficient.
>