partitioning the interior of a circle with chords

Partitioning the INTERIOR OF A CIRCLE with ChordsAuthor(s): Dennis ParkerSource: The Mathematics Teacher, Vol. 99, No. 2 (SEPTEMBER 2005), pp. 120-124Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27971890 .

Accessed: 24/04/2014 18:03

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

National Council of Teachers of Mathematics is collaborating with JSTOR to digitize, preserve and extendaccess to The Mathematics Teacher.

http://www.jstor.org

This content downloaded from 174.91.153.63 on Thu, 24 Apr 2014 18:03:53 PMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=nctm

http://www.jstor.org/stable/27971890?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


Partitioning the

INTERIOR OF

A CIRCLE

with Chords

Dennis Parker

attera recognition is one of the most

important devices that we use in under

standing mathematical concepts and

principles, as well as in solving mathe matics problems. In middle school and

high school mathematics classes, students com

monly study several specific cases of a problem situ ation and then attempt to generalize to the nth case. A problem about partitioning a circular region with

chords, sometimes called Moser's circle problem, il lustrates that one must be careful when using in ductive reasoning to generalize.

Canadian mathematician Leo Moser may have been the first to pose the problem in 1950. The statement of the problem is?

Find the number of nonoverlapping regions into which the interior of a circle is divided if

distinct points on the circle are joined by all

possible chords with no three intersecting at one point.

For = 1, 2, 3,4, and 5 the numbers of regions are 1, 2, 4, 8, and 16, respectively. Figure 1 shows the first five cases. On the basis of this pattern, it is

tempting to generalize to?

The number of regions for the nth case is 2*"1.

120 MATHEMATICS TEACHER | Vol. 99, No. 2 ? September 2005



However, for = 6, the number of regions is 31, not 32, as shown in figure 2. Students invariably think that they miscounted or drew the diagram in a way that led them to an incorrect answer. High school mathematics teachers may find this problem in their textbooks or other curricular materials, but often no explanation is given for the correct for mula for the number of regions.

This article develops a proof-based derivation of the formula (Murphy 1972) in a way that is ac

cessible to high school students. The following is a

general description of one day's lesson in an hon ors geometry class that consisted mostly of ninth

graders. In the previous class meeting, the stu

dents had drawn the diagrams for the first seven cases and had constructed the first seven terms of the sequence 1, 2, 4, 8, 16, 31, 57. They under stood that the "powers of 2" pattern was incorrect for > 6.

The method used to develop the correct formula is based on two major principles:

The counting formula for combinations

c= *!

n r r\(n-r)\

Euler's formula for polyhedra

Faces + Vertices - Edges = 2

PREREQUISITE PRINCIPLE 1: THE CALCULATION OF COMBINATIONS Whether the teacher is reviewing or introducing the calculation of combinations, students need suf ficient knowledge of the combinations 6C4 and MC4 in particular, since they will be used when deriving the formula. So the discussion of combinations

began with the following problem:

Suppose that from a set of six students {Ashanti, Bill, Carlos, Dharma, Elaine, Feng}, a committee of four students is to be chosen. How many dis tinct four-student committees are possible?

Many students were able to find a correct answer without knowing a combinations formula. The teacher later introduced the formula and used it to show the solution that they had found without it.

The number of possible committees is

6C4= ,6! =15. 6 4 4! (6-4)!

More generally, what is the formula for calculat

ing the number of distinct four-student committees chosen from a set of students (n > 4)? An expla

E

Fig. 1 First five cases of Moser's circle problem

Fig. 2 Moser's problem for = 6

nation was given to show that the number of possi ble committees is

n\ _ ftp-1)Q-2) Qz-3) n 4

" 4!0-4)!

~ 24

PREREQUISITE PRINCIPLE 2: EULER'S FORMULA (F + V - E = 2) The textbook used by the honors geometry class included Euler's formula, but the students had not

yet studied it. I gave students drawings of several

Vol. 99, No. 2 ? September 2005 | MATHEMATICS TEACHER



Fig. 3 A flattened cube with one face removed

Fig. 4 Six regions outside the hexagon but inside the circle

Ashanti Carlos

Feng Dharma

Elaine

Fig. 5 The committee combination problem for = 6

polyhedra and asked them to count the number of

faces, vertices, and edges. Then I asked them to make a conjecture about the relationship among the three. Following this activity, I discussed how

we could take a cube, remove one face (without changing the number of vertices or edges), and embed it in a plane by "flattening" it into a two-di mensional figure, as shown in figure 3. The dis cussion led the students to a two-dimensional (pla nar graph) version of Euler's formula. This formula is the same as the three-dimensional ver

sion, since the one face removed to embed the cube

may be replaced, when counting, by the un bounded exterior region. I also began referring to

regions instead of faces.

COUNTING THE REGIONS FOR SIX POINTS We are ready to take the students through the

process of counting all regions for six points placed on the circle, in a way that will easily generalize to a process for points.

With six points on the circle, six regions are outside hexagon ABCDEF but inside the circle, as shown in figure 4. So we are left to count the number of regions inside the hexagon. Using R for interior regions of the polygon instead of F for

faces, we modify the planar graph version of Euler's formula, F+V-E =

2,toR + V-E=l,

since we are not including the unbounded exterior

region in our count. We start by counting the total number of ver

tices. First, six vertices, the points labeled A

through F, determine the hexagon. How many in terior vertices exist? This question is easily con nected with the committee combination problem that was previously discussed, since they have the same underlying mathematical structure. Figure 5 depicts the committee combination problem. The number of vertices inside the hexagon is 6C4, since each interior vertex is determined by four

points chosen from the set of six that were placed on the circle. So the total number of vertices is 6 +

6C4. Substituting into Euler's formula gives us R +

(6 + 6C4)-E=l. To find the number of edges for the hexagon and

its interior, we will count the number of edges at each vertex, find the sum for all vertices, and divide

by 2, since two vertices are on each edge. Each of the six vertices A through F is an endpoint for five

edges. In the language of graph theory, each of the six vertices A through F is of degree 5. Each of the

6C4 vertices inside the hexagon is an endpoint for four edges, that is, each interior vertex is of degree 4. Thus the total number of edges is

M(s)+U)M 2 122 MATHEMATICS TEACHER | Vol. 99, No. 2 ? September 2005



Substituting this expression for E in Euler's for mula gives us

, , (6)(5) + (,C4)(4) 2

R = 25.

We next add the six regions that are outside the

hexagon but inside the circle to these twenty-five regions to get the thirty-one total regions.

COUNTING THE REGIONS FOR POINTS Clearly, for points on the circle, regions are out side the n-gon but inside the circle. The regions that remain to be counted are those inside the n-gon. As in the case for six points, the unbounded exterior

region is not one of the pieces that we are attempt ing to count, so we use R + V-E=\.

We first focus on the number of vertices on our ri

gori and in its interior. A total of vertices are on the

n-gon itself. The number of vertices inside the n-gon is

6C4, since each subset of four from the points on the circle uniquely determines an intersection point of two chords; and conversely, each intersection point is determined by exactly one subset of 4 from the

points (those at the ends of the two chords). So, V= + ?C4 To count the edges, we count the number of

edges at each vertex, sum for all vertices, and divide

by 2, since two vertices are on each edge. Each of the vertices on the ?z-gon is on - 1 edges, that is, the

vertices on the n-gon are of degree - 1. Each of the

nC4 vertices inside the n-gon is on exactly four edges, that is, each interior vertex is of degree 4. So

?(?-l) + 4(.cJ 2

, > n(?-l) + 4( Cj Thus, R +

(?+>C4)?i-;2 U

*'=1,

n(n-l) which is equivalent to R = ?^?-

+ wC4 +1 - n.

Combining this result with the regions outside the n-gon but inside the circle gives us

total number of regions

n(n-l) (1) =^?L+C4

+ l-n + n

(2) n{n-l) ̂^-1)^-2)^-3) ^ 2 24

(3) = ? (

4 - 6n3 + 23n2 - 18? + 24).

= Number of Pascal's Triangle Number of Regions Points on Circle

11 1 2 11 2

3 12 1 4 4 13 3 1 8

5 1464 1 16 6 1 5 10 10 5 / 1 31

7 1 6 15 20 15 / 6 1 57 8 1 7 21 35 35 /21 71 99

Fig. 6 The bold entries in a row add to the number of regions.

From step (1), an alternate form of the formula is

total number of regions = nC2 + MC4 + 1 =

?Q + nc2 + nc0.

We may specify that nCr= 0 for < r, so that either form (polynomial or combinatorial) of the formula holds for all positive integers n.

The derivation of the preceding formula, using combinations and Euler's formula, is a mathemati

cally valid proof (see Murphy 1972) that is much more accessible to high school students than a proof by mathematical induction.

THE PASCAL'S TRIANGLE CONNECTION Although the "powers of 2" pattern broke down, leading to a more complicated sequence-generating function, a fascinating yet simple connection to Pascal's triangle was pointed out by Moser himself. For each row of Pascal's triangle, the sum of all val ues equals a whole-number power of 2. The terms of the sequence 1, 2, 4, 8,16, 31, 57, 99,...., corre

spond to these sums for the first five rows and to the sums of the first five values in rows 6 and be

yond, as shown in figure 6.

THE METHOD OF FINITE DIFFERENCES Yet another exploration is associated with the se

quence generated by Moser's circle problem (Con way and Guy 1996; Kummel 2002). The method of finite differences reveals patterns that may be used to find a formula that fits a sequence of numbers.

We begin by writing the sequence of interest, fol lowed by a second row with the differences be

Sequence: 1 2 4 8 16 31 57 99

1st difference 1 2 4 8 15 26 42

2nd difference 1 2 4 7 11 16

3rd difference 1 2 3 4 5

4th difference 111 1

Fig. 7 Exploring Moser's problem using the method of

finite differences

Vol. 99, No. 2 ? September 2005 | MATHEMATICS TEACHER 123



P(l) P(2) P(3) P(4) P(5) a + 2? + c + d + cl6a + 82? + 4c + 2d + c Sia + 272? + 9c + 3d + c 256a + 642? + 16c + 4d + e 625a + 1252? + 25c + 5d + e

15a + 72? + 3c + d 65a + 192? + 5c + d 175a + 372? + 7c + d 369a + 612? + 9c + d

50a+122? +2c 110a+ 182? +2c 194a+ 242? +2c

60a + 62? 84a + 62?

24a

Fig. 8 Table of differences to find coefficients for P(n), = 1, 2, 3, 4, and 5

tween consecutive numbers above it, followed by a

third row with the differences of the differences, and so on. We continue until we get a row consist

ing of a constant value, as shown in the fourth row

of figure 7. The constant difference in the fourth difference

row suggests that the sequence-generating function is a fourth-degree polynomial:

To find the values of the coefficients, we evaluate thi

polynomial for = 1, 2, 3, 4, and 5, and construct a

table of differences, which appears in figure 8. Since the first entry of each row is equal to one

(see the original difference table), we can use the first algebraic expression in each row to set up the

following system to solve for a, b, c, d, and e:

a + b + c + d + e=l 15a + 72? + 3c + d= 1

50a + 1227 + 2c = 1 60a + 62? = 1

24a = 1

Starting with the last equation and using substi tution to work our way up we find that

Thus, this inductive method of finite differences

produces a formula that matches the one derived

using Euler's formula.

DISCUSSION AND CONCLUSION Moser's circle problem shows students that we

must be careful not to come to a final conclusion that is based on inductive reasoning alone. It also

provides an opportunity to use a variety of mathe matical tools to find a correct solution. Students can apply the topic of combinations in a context that is different from what they ordinarily see in a

typical textbook chapter on permutations and com

P[n) - an4 + bn3 + cn2 + dn + e

So

binations. Many students are exposed to Euler's formula as an enrichment activity. Here, they apply it in a problem-solving setting. Although students encounter Pascal's triangle when studying binomial

expansion or probability, Moser's circle problem provides one more opportunity to connect with its

intriguing patterns and applications. And if stu dents are studying sequences and series, this se

quence associated with Moser's circle problem of fers the often neglected (and pattern-rich) method of finite differences,

s For additional information on this problem, teachers may want to see Graening's (1971) solu

tion, which uses different methods and provides in

sights into other patterns connected to the problem. Also, Shultz, Shultz, and Brown (2003) discusses the use of interactive software to explore this problem.

The author wishes to thank Gay Matthews and her honors geometry students. He also wishes to thank the reviewers for their helpful comments.

BIBLIOGRAPHY Conway, John, and Richard Guy. The Book of

Numbers. New York: Copernicus, 1996.

Graening, Jay. "Induction: Fallible but Valuable." Mathematics Teacher 64 (February 1971): 127-31.

Kummel, Marc. "Fifteen Men on the Dead Man's

Chest." Unpublished paper (November 2002).

Murphy, Timothy. "The Dissection of a Circle by Chords." Mathematical Gazette 56 (May 1972): 113-15.

Shultz, Harris S., Janice W. Shultz, and Richard G. Brown. "Unexpected Answers." Mathematics

Teacher 96 (May 2003): 310-14.

Wyman, Max. "Leo Moser, 1921-1970." Canadian

Mathematics Bulletin 15 (March 1972): 1-3. oo

DENNIS PARKER, [email protected], ^^^^^^fl teaches at the University of the ^^^^^^^H Pacific in Stockton, CA 95219. He is ^^^^^^fl

^_ interested in public policy issues ^^^^^^^H concerning teacher credentialing. ^^^^^^^H

124 MATHEMATICS TEACHER | Vol. 99, No. 2 ? September 2005



partitioning the interior of a circle with chords

Documents