"FIGURING" OUT A JIGSAW PUZZLE

"FIGURING" OUT A JIGSAW PUZZLEAuthor(s): Ken IrbySource: The Mathematics Teacher, Vol. 82, No. 4 (APRIL 1989), pp. 260-263Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27966236 .

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SHARING TEACHING IDEAS

|J INDIRECT PROOF: THE TOMATO STORY Sometimes we can elicit attention from our students by supplementing textbook exam

ples with personal illustrations. As a case in point, when teaching indirect proof in

geometry, I tell the "tomato story." This story is true. I was three years old

and seated in my high chair in the break fast room. My parents were in the kitchen breakfast-room area. My mother was fixing dinner, and my father was just observing.

My mother placed a sliced tomato in front of me. I stared at it and then announced with conviction, "I am not going to eat this!"

Now my mother would have been happy to whisk away the plate of sliced tomato. But my father valued the ?ating of vegeta bles. He simply asked, patiently and re

spectfully, "Why not, Philinda?"

Apparently I immediately realized that ? needed a reason for not eating it. So I stared at the tomato long and hard and fi

nally answered, "Because it's so red." My father remained patient and respectful but

now, with even a bit more interest, repeated what I had just said: "Because it's so red" He then walked over to the refrigerator, opened the door, and examined its contents. In a minute he brought out a rather large glass bowl of red Jello (which he called

"wiggle-waggle" to amuse me). He held the bowl up, shook it gently, and said, "This

wiggle-waggle is red, and you like it."

Well, of course, he had won this little round of logic, and nothing remained for me to do but try the tomato. And do you

know what? It was good, and I've liked to matoes just fine ever since.

Now I believe we can fit my father's clever persuasive technique into our formal

indirect-proof-format :

1. My father wanted to prove to me that I might like red-colored food (in this situ ation a tomato).

2. The assumption to the exact contrary was that I would not like red-colored food.

3. This assumption led to the conclusion that I would not like red Jello.

4. But this assumption was a contradic tion to the known fact that I did like red Jello.

5. Therefore, I might like the tomato. I didn't have a valid excuse not to try it.

My students certainly paid attention, and they have forgotten neither the tomato

story nor the indirect proof.

BIBLIOGRAPHY

Moise, Edwin, and Floyd Downs. Geometry. Menlo Park, Calif. : Addison-Wesley Publishing Co., 1982.

Philinda Stern Denson Rosemead High School Rosemead, CA 91770

"FIGURING" OUT A JIGSAW PUZZLE This past year was textbook-adoption time at our school, and one of the textbook rep resentatives gave each of the mathematics teachers a small jigsaw puzzle having five

brightly colored pieces (fig. l[a]). The

challenge was to arrange the five pieces to

"Sharing Teaching Ideas "

offers practical tips on the teaching of topics related to the secondary school cur

riculum. We hope to include classroom-tested approaches that offer new slants on familiar subjects for the

beginning and the experienced teacher. See the masthead page for details on submitting manuscripts for review.

260 Mathematics Teacher



IV

(a)

(b)

Fig. 1

form a square. At first I wasn't interested in

the puzzle. It didn't seem that it could be

much of a challenge?with only five pieces, how difficult could it be? Months later

when I rediscovered the puzzle buried in my desk drawer and decided to work on it, I

was struck by two observations: (1) the

puzzle was more difficult than I had antici

pated, and (2) although the five pieces didn't readily fall into a square as I had

expected, four of the pieces did form a

square and the remaining piece was itself a

square. After about ten or fifteen minutes

of shifting pieces around, I was finally able to arrange all five of the pieces into one

large square, thereby solving the puzzle. As I thought about this puzzle and how

the five pieces not only formed a large square but could also be arranged into two

smaller squares, it seemed to me that a fair amount of mathematics must be involved in

the relationship of the pieces to the whole.

The question that ultimately came to mind

was whether a mathematical verification of

the solution to the puzzle might be possible. To answer this question I began by com

paring the sides of the five pieces with each

other and found that only four different

lengths were represented, which I specified as a, b, c, and d; a designated the smallest side and d the largest (fig. l[b]). Futher

more, by comparing the angles of the cor

ners of the pieces I determined that all the

corners were 45-degree or 90-degree angles with the exception of piece IV, which con

tained two 135-degree angles. From the Pythagorean theorem, the fol

lowing relationships are evident:

? = (from piece IV)

c = byj2 (from piece III) d = (from piece II)

It follows from this system, by substitu

tion, that c = 2a

and

d = 2b.

Thus the ratio of the sides

a:b:c:d

forms the geometric sequence

1:^/2 : 2: 2^/2 .

As a first attempt at deducing a math

ematical verification of the solution to the

puzzle, I decided to start with the fact that

the five pieces could be arranged into two

April 1989



separate squares as shown in figure 2.

Somehow, it seemed, this interesting fact

might have a bearing on the solution.

V

Fig. 2

The area of the solution square must be

equivalent to the area of these two smaller

squares combined. Therefore, the area is

d2 + a2. Because the area of any square is

equal to the length of its side squared, it

follows that the side of the solution square must be ^Jd2 + a2. By substitution,

Id2 +a2 = J{2b)2 + a2

? Ji^?)2

+ a2 =

y/9a~2 = Sa.

Thus, each of the four sides of the de sired square must be composed of either three a lengths or the equivalent, an a

length and a c length, inasmuch as c = 2a.

From figure 1(b) it is evident that since

only four a sides are available to contribute to the solution square (two sides on piece IV and two on piece V) and since four c

sides are available (two sides on piece II and one each on pieces I and III), the de

sired square must have sides formed by combining one a length and one c length (fig. 3). Furthermore, because piece II con

c + a

c + a c + a

c + a

Fig. 3

tains two of the four c sides, it must be a corner of the solution square with the re

mainder of the two sides reserved for a

lengths (fig. 4). As all four a sides are con

tained on pieces IV and V, both these pieces must also be corner pieces and fit as shown in figure 5. (Note: interchanging pieces IV and V has no effect on the eventual out

come.)

c + a

Fig. 5

We also see from figure 5 that a puzzle

piece is needed that has adjacent c and b sides. Of the remaining pieces, I and III,

piece III has this characteristic and hence must fit as shown in figure 6. It is hoped that piece I, the single piece left, will fill the remaining space. Using figure 1(b) and

figure 6, it is easy to verify a match be

Fig. 6

262 Mathematics Teacher



tween the respective sides of piece I and the

confines of the space available, given the

fact that d = 2b.

Thus, the fit of the last piece is verified, and the analysis of the solution of the

puzzle is complete. This puzzle is easy to reproduce on

paper or cardboard and is an enjoyable ma

nipulative for students. In addition, the

mathematical aspects of the puzzle supply . the basis for an interesting lesson for al

gebra or geometry classes, vividly demon

strating an interrelationship between the

two branches of mathematics.

Ken Irby Powhatan High School

Powhatan, VA 23139

SOLVING EQUATIONS OF THE TYPE a cos w + b sin u = c.

Three often-used methods for solving

(1) a cos u + fesin u = c

are as follows :

Express (1) in the form Rcos(u ?

v) = c

with

R = Ja2 + b2

and l> given by tan = fe/a. Solve R cos ?

v) = c.

Use t = tan w/2 ; then

1-t2 cos i? =

1 + t2

and

2t

l + t2' sin w =

which reduces (1) to the quadratic equa tion

a(l-*2) + fe(2i) = c(l + *2).

Solve for t to obtain u/2 and finally u.

Draw the graph of a cos u + fe sin w =

and find the intersection with = c.

My approach is different from these

methods and is adaptable to quite a differ

ent graphical method. To solve (1), let

cos u = , sin u = y. Then

(i) 2 + y2 = 1,

and the given equation becomes

(ii) ax + by = c.

Solve (i) and (ii) as a pair of simultaneous

equations to obtain solution pairs (jc, y). De

termine u from these pairs.

Example

^3 cos u + sin u = 1

yields 2 + y2

= 1

and

Fig. 1. ()i and 02 are solutions.

April 1989 263



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"FIGURING" OUT A JIGSAW PUZZLE