Mathematical Journeys (Schumer/Mathematical Journeys) || Pizza Slicing, Map Coloring, Pointillism, and Jack-in-the-Box

10 Pizza Slicing, Map Coloring,Pointillism, and Jack-in-the-Box

In his book, The Teaching of Geometry (1911), D.E. Smith said, “Geometry isa mountain. Vigor is needed for its ascent. The views all along the paths aremagnificent. The effort of climbing is stimulating. A guide who points out thebeauties, the grandeur, and the special places of interest commands the admirationof his group of pilgrims.” Even though this quote might be a bit over the top,I invite you to be your own guide as we discuss four different problems, eachwith a different geometric view.

Problem #1 (Pizza Slicing Problem): How many regions can the plane beseparated into with n straight lines?

We can think of the plane as a gigantic pizza and the straight lines as the cutsmade across the pie. To get a feel for this problem, make some sketches withsmall values of n and see if a pattern emerges. This problem appears so natu-rally; if you’re like me you’ve probably doodled with this problem on place matsand scraps of paper since childhood. Its mathematical formulation was stated andsolved by the great synthetic geometer, Jakob Steiner (1796–1863), in an articlein Crelle’s Journal (1826). Steiner was Swiss, but was educated in Heidelberg andspent most of his career as a professor at the University of Berlin. It may be ofinterest to note that the man who created so many beautiful geometric theoremsdidn’t learn to read or write until he was 14 years old. To mention just one ofhis accomplishments: The Danish mathematician Georg Mohr (1640–1697) hadshown that all Euclidean constructions (those involving straightedge and compass)could actually be accomplished with compass alone, as long as we consider a lineto exist once two of its points are so constructed. In the other direction, Steinershowed that all Euclidean constructions can be accomplished with one fixed circleplus a straightedge, that is, with a straightedge and a compass that’s stuck!

Problem #2 (Map Coloring Problem): Slice up the plane with any numberof straight lines. How many colors are required so that adjoining regions havedifferent colors?

Mathematical Journeys, by Peter D. SchumerISBN 0-471-22066-3 Copyright c© 2004 John Wiley & Sons, Inc.

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92 MATHEMATICAL JOURNEYS

Problem #2 begins where Problem #1 left off. Once we have sliced up theplane, we now want to color the “map” that we’ve created.

For comparison’s sake, a much more difficult and appropriately celebratedproblem is the question of how many colors are needed to color any map inthe plane so that adjoining regions have different colors. Boundaries need nolonger be just straight lines. This significant problem confounded amateur andprofessional mathematicians alike from the mid-1800’s until 1976, when KennethAppel and Wolfgang Haken of the University of Illinois, assisted by an enormousamount of computer time (about 1,200 hours), established that four colors suffice.The fact that five colors suffice was established by P.J. Heawood in 1890 and hedidn’t need any computational assistance, but the step from five colors to fourcolors is a large one indeed.

Problem #3 (Pointillism Problem): If every point of the plane is colored eitherred, green, or blue, are there necessarily at least two points of the same color oneunit apart?

This problem reminds me of the paintings of Georges Seurat and hence theallusion to pointillism. Notice that we have not defined what our units are. Inter-estingly, it makes no difference. To get a feel for this type of problem, considerthe situation where every point of the plane is colored either red or green. Wouldthere be two points of the same color one unit apart? The answer is “yes” as canbe easily seen by considering any equilateral triangle in the plane with sides ofunit length. If one vertex is red and another green, what color could the thirdvertex be? If it is red, then there are two points one unit apart both of whichare red. If it is green, then there are two green points one unit apart. Either way,there are two points of the same color a unit apart.

Problem #4 (Jack-in-the-Box Problem): Consider Figure 10.1, where four unitcircles in the plane bound a smaller circle placed within them so that the innercircle is tangent to the four unit circles. The inner circle lies well within theblack square containing the four unit circles. In three-dimensions, eight unit ballssurround a ball that remains within the cube containing the eight unit balls. Whathappens in higher dimensions? Does the “inner ball” remain within the outerhypercube or does it somehow pop out like a wind-up jack-in-the-box?

Solution to Problem #1: Instead of working in the entire plane, we limit our viewto a large circle. Let Ln = the maximum number of regions created by drawingn straight lines. Certainly the lines must be in general position, that is, no threelines intersect at a point and no two lines are parallel to each other. Figure 10.2shows the situation with five straight lines. In this case, the circle is separatedinto 16 distinct regions. In Table 10.1 we chart the growth of Ln for n from 0 to 5.Analyzing Table 10.1, we see for 1 ≤ n ≤ 5 that Ln = Ln−1 + n. If we can showthat this pattern continues indefinitely, then it would follow that for all n ≥ 1 :

Ln = 1 + 1 + 2 + . . . + n = 1 + n(n+1 )2 = n2 +n+2

2 .

PIZZA SLICING, MAP COLORING, POINTILLISM, AND JACK-IN-THE-BOX 93

1

2

√2 −1

−1−2 1 2

−1

−2

y

x

Figure 10.1 Circle bounded by four unit circles.

2

4

15

3

1

2 34

5

6

7

89

10

11

12

13

1614

15

Figure 10.2 Five lines separate the plane into sixteen regions.

We now show that Ln = Ln−1 + n for all n ≥ 1 by induction on n.For n = 1 , L1 = 2 = L0 + 1 . Now assume the result holds up to n − 1

lines. Since a line is determined by two points, in order to place the n th line soas to maximize the number of new regions, it suffices to choose two points on


TABLE 10.1 Number of regions createdby n straight lines

n 0 1 2 3 4 5

Ln 1 2 4 7 11 16

2

1

1

2

Figure 10.3 Placement of n th line.

the circumference of the circle that lie directly between the same two lines (as inFig. 10.3), thus avoiding parallelism. Furthermore, we must make sure that thenew line does not go through any previous points of intersection. This can beeasily done since avoiding such points only reduces our infinitely many choiceson the circumference by a finite number.

By construction, the new line must intersect each and every one of the previouslines exactly once (since they are straight lines). But the new line divides previousregions into two parts for each consecutive pair of lines that it meets. Counting thecircumference of the circle among these lines, the new line creates n new regions.Hence Ln = Ln−1 + n as desired. �

Solution to Problem #2: The answer is two colors. To get a feel for this problem,look at Figure 10.4, where we have sliced up a large circle with four slices (linesegments) and then colored the resulting regions with just two shades (dark and


Figure 10.4 Four slices can be two-colored.

light). I invite you to check the cases with one, two, or three lines to check thattwo colors suffice in those cases.

Assume that for any map with n lines that two colors are all that is needed.Now consider Figure 10.5 with n + 1 lines. If the dotted line l is removed, thereremains n lines. By our inductive assumption, the remaining figure can be two-colored as shown. If we now add in the dotted line, it separates the diagram intotwo parts—call them top and bottom. Keep the top colored as before, but switchall the colors in the bottom region (Fig. 10.6). We have now two-colored the newmap. To see why this must work, we consider three possible cases in turn:

1. Regions that border line l used to be one color on each side, but now areseparated into two colors at the boundary of l.

2. Top regions not bordering line l have the same color as before, but theseregions only border other top regions and our previous map was properlytwo-colored.

3. Similarly, bottom regions not bordering line l have all colors flipped. Butthese regions only border other bottom regions and the previous two-color-ing is preserved by switching all such regions. �

Solution to Problem #3: Assume that the answer is “no,” namely that there issome way to color the plane with no two points of the same color exactly one unitapart. Hence, any unit equilateral triangle will have vertices of all three colors.Consider such a triangle, �RGB as in Figure 10.7. If we use line segment GB


Figure 10.5 Original coloring with n lines.

Figure 10.6 New coloring with bottom color switched.


1G

1 1

1 1

R∗

R

B

Figure 10.7 Two triangles with vertices red, green, and blue.

as one side of a new equilateral triangle, we can form another unit equilateraltriangle, �R∗GB with vertices colored red, green, and blue.

Now “bolt down” vertex R and spin the diamond-shaped region RGBR∗ coun-terclockwise as a rigid motion with R as the pivot point. Continue until R∗ movesto a point R1 that is one unit away. Call the points where vertices G and B end upG1 and B1 , respectively. The situation is diagrammed in Figure 10.8. The pointB1 must be either blue or green since B1 is one unit away from the red point R.Analogously, the point G1 must be either green or blue (whichever color B1 isnot). In either event, R1 must be red since it is one unit from both B1 and G1 . Butthen both R∗ and R1 are red and one unit apart. �

At this point, the following question is a natural one. Is there a coloring of theplane with any finite number of colors such that no two points a unit apart havethe same color? If so, by Problem #3, the answer must be greater than three. Theanswer is “yes” and it can be shown that seven colors suffice, as in Figure 10.9.The key idea is to tile the plane with regular hexagons of side length 2

5 and thencolor them with seven colors as shown. It can then be shown that no two pointsof distance d with 4

5 < d <√

285 have the same color. Since 4

5 < 1 <√

285 , this

seven-coloring of the plane does the trick.The next question, known as the chromatic number problem for the plane,

is still unresolved. What is the minimum number of colors needed to paint theplane so that no two points at unit distance have the same color? The answerconceivably could be four, five, six, or seven. It’s time to get out your paletteand start painting!


1

G

G1

1

1

1

1111

1

1

1

R∗

R

R1

B

B1

Figure 10.8 Rhombus RGR*B pivoted to RG, R, B.

5

2

4 7 6 2 1

1 5 3

3 4 7 6

2/5

Figure 10.9 Tessellation of the plane with black-bordered 18-sided polygon of sidelengths 2

5 .

Solution to Problem #4: Let Rn represent n-dimensional Euclidean space and

let rn denote the radius of the inner n-dimensional ball contained within the outerunit balls. Figure 10.1 shows the situation in the plane when n = 2 . Here the fourunit outer circles have centers at the points (1, 1), (1, −1 ), (−1 , 1). and (−1 , −1 ).They are contained within the square bounded by the lines x = 2 , x = −2 , y = 2 ,and y = −2 of side length four. Since the center of the outer balls lie at a distanceof

√2 from the origin and each has radius one, being tangent to the four outer


Figure 10.10 Ball bounded by eight unit balls.

circles, the inner circle has radius r2 = √2 − 1 and is certainly bounded within

the outer square.In R

3 the eight outer balls are centered at (±1 , ±1 , ±1 ). You can make amodel of the situation with eight tennis or lacrosse balls with a golf ball squeezedin the middle (Fig. 10.10). In this case, the outer cube is bounded by the planesx = ±2 , y = ±2 , and z = ±2 . The outer balls have radius one, and hence theinner ball has radius r3 = √

3 − 1 , keeping it well within the containing cube.By analogy, the general situation is easily described. In R

n there are 2 n unitouter balls centered at (±1 , ±1 , . . . , ±1 ). They are contained within a hypercubebounded by the 2 n hyperplanes x1 = ±2 , x2 = ±2 , . . . , and xn = ±2 . The innersphere has radius rn = √

n − 1 . Notice for n = 9 that r9 = √9 − 1 = 2 , and

so the “inner” ball just touches the hypercube on all its faces. When n ≥ 10 ,the inner ball actually pokes out of the hypercube. Pop goes the weasel! So indimension ten, although the round inner ball is surrounded by over a thousand unitballs, it somehow manages to squeeze out beyond them. If this seems pretty weird,it may be because all the experiences in our local world involve substantiallyfewer dimensions. Even so, the result is both perplexing and wonderful. Don’tyou agree? �

WORTH CONSIDERING

1. Consider n pizza slices in general position as in Problem #1. How manyof the regions border the circumference of the circle? In other words, howmany such pizza slices have a crust?


A B

C

D

E

F

Figure 10.11 Equilateral triangle with six internal slices.

2. If every point on a straight line is colored either red or blue, must there betwo points one unit apart having the same color?

3. (a) If all points in three-space are colored either blue, green, or red, mustthere be two points one unit apart that are the same color?

(b) What about four colors in three-space?

4. (Problem from Quantum, January, 1990) Consider a pizza in the shape of anequilateral triangle. Pick any point inside it and make six slices by cuttingfrom the chosen point to each of the three vertices and from the chosen pointperpendicularly to each side (Fig. 10.11). Show that if two people consumethe pizza by eating alternating slices, each gets exactly half the pie.

5. (Green Chicken Contest, 1986) Suppose six points are given, no threecollinear. If all 15 line segments are drawn joining pairs of points, howmany segments can be colored black without forming any black triangles?

6. You have a job promising a bar of gold for seven days’ work. However, youmay elect to stop working at the end of any one of the seven days. Can thebar be sliced at the beginning of the week with just two cuts to guaranteethat you can be paid the appropriate amount owed you?

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Mathematical Journeys (Schumer/Mathematical Journeys) || Pizza Slicing, Map Coloring, Pointillism, and Jack-in-the-Box