GEOMETRY FOR PIE LOVERSAuthor(s): WILLIAM FISHERSource: The Mathematics Teacher, Vol. 75, No. 5 (May 1982), pp. 416-419Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27962971 .

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GEOMETRY FOR PIE LOVERS

Get the most from your pies by choosing your cuts.

By WILLIAM FISHER California State University, Chico

Chico, CA 95929

It is unfortunate, but students in high school algebra and geometry classes of tentimes do not get to see some very useful applications of their studies. I'm

thinking specifically of problems that in volve finding maxima and minima. These

types of problems are generally thought of as part of calculus and are usually avoided until then. Let me illustrate by way of an

example:

Consider a rectangular pen that is to be built along a stream so that only three sides need fencing (see fig. 1). If there are 600 m of available fencing, find the dimensions that will make the pen of maximum area.

Fig. l

This problem, or one just like it, is found in every calculus book. Granted, the derivative is easy to find, set equal to zero and solve, but the problem itself

belongs in the high school algebra class.

Unfortunately, in some calculus books almost half the word problems that in volve finding maxima or minima end up to be simple quadratic expressions. These are beautiful examples of a place where all the work done with completing the square can be put to good use.

If A represents the area, we have

A = Iw

A(w) = (600 -

2w)w = -2w2 + 600w =

-2(h>2 -

300w) = -2(w2

- 300w> + (150)2 -

(150)2) = -2(w

- 150)2 + 2(150)2.

From this it is clear that the maximum of the area function is obtained when the

negative term is zero, that is, when w =

150. It is also immediate that the maxi mum area is 2(150)2.

This solution, along with the solution to

any other quadratic expression, could also be explained graphically since finding the maximum or minimum corresponds to

finding the vertex of the associated parab ola. The preceding problem is not the only type of that problem that can be solved

without the aid of calculus. The Mathe matics Teacher has had several articles

dealing with elementary, noncalculus so lutions to seemingly difficult calculus problems. A partial list is contained in the

bibliography at the end of this article. However, one of my favorites (because

of a love for geometry and, probably more

so, because of a love for pies) is a classic cake- or pie-cutting problem (Engel 1971) that can be explained as follows.

Suppose you and a friend go into a pie shop and buy a pie. You may cut off a

piece as big as you can?the remainder is left for your friend. There are only two conditions: first, your cut must be

straight; and second, your cut must pass through a point O which has been chosen

by your friend. From this simple pie problem arise

some very interesting geometric ques tions. First, suppose the pie is circular. If

your friend puts the point O in the center of the circle, there is not much you can

do?every possible cut will be a diameter

416 Mathematics Teacher

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and will divide the pie exactly in half. But now suppose the point O is not put in the center. Clearly, you can obtain more than one-half the pie. But where should you cut the pie to obtain as much as possible (see fig. 2)?

Fig. 2

After a few attempts, I'm sure that most

people discover that the cut ?F, perpen dicular to the diameter through 0, seems to be the best possible one (see fig. 3). Is this easily proved?

that you actually have found the optimal cut?

Let me explain one method that finds the optimal cut and, very quickly, prove that it is the optimal cut. In fact, this

method will work for any convex pie, not

just the rectangle or the circle. However, let's go back and demonstrate the method

with the rectangle. Rotate the figure 180? about the point O so that the resulting configuration has O as the point of central

symmetry (fig. 5). I claim that the cut AB connecting the points of intersection of the rectangle and its rotated image is

optimal. Suppose that CD is any cut

through O. Let's compare the cut CD to the cut AB by making a list of gains and losses for the cut CD. Looking at the cut

CD and assuming we take the bigger piece, the only thing that is not also in the

Fig. 3

Suppose now that the pie is rectangular. Again, if the point O is put in the "center" of the rectangle, that is, at the center of

symmetry, no matter what cut you make, you will divide the pie exactly in half. But what happens if O is not at the center? where is the optimal cut (see fig. 4)? And then, once you find it, how can you prove

Fig. 4

Fig. 5

bigger part determined by the cut ?B is the region labeled 1. But by symmetry, region 1 is balanced out by the region labeled 2, which is in the cut AB butnot in the cut CD. Then finally, the cut ?B also includes the shaded region; but the cut

CD has nothing to counter with, so this is the net loss of cut CD. When the very same method is applied

to the circular pie, we see that our belief?

May 1982 417

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that the perpendicular chord was the opti mal cut?is indeed true as the net loss of another cut CD is shaded in figure 6. Note that chord AB is bisected by the point O and hence perpendicular to the diameter

through O.

Fig. 6

This beautiful and surprisingly simple method of finding optimal cuts still holds some surprises for other convex figures. Consider a semicircle with the point O chosen so that O is the midpoint of the radius that is perpendicular to the diame ter (fig. 7). By our method of rotation

Fig. 7

about the point O, we now obtain two

cuts, AB and CD. Cut 2* is indeed the maximal cut we want, but cut CD is the

"equalizer" cut that divides the figure exactly in half (fig. 8).

Fig. 8

Now that we have seen that more than one cut can be created by this method, it

may seem possible that for other convex

figures the maximal cut cannot be found in this way. However, the following theorem

proves that this method works all the time.

Theorem 1. Every convex figure has a maximum cut through any given point O.

Proof. Consider the continuous func tion A defined on the convex figure by

A(X) = (area of shaded side of cut ) -

(area of clear side of cut XOY). Now the extreme value theorem guarantees that a maximum cut exists.

The fact that there is a maximum comes as no surprise; however, we have still not shown that our method of rotation will find it. This will be established once we show the following.

Theorem 2. If AOB is a maximum cut

for a convex region R, then AO = OB.

Proof. Suppose that AOB is a maximum cut and AO OB. Assume the shaded

side contains the maximum area and AO < OB. Now we can find an angle a

through which we can rotate the cut line such that the area of the clear wedge BOY is larger than the area of the shaded wedge

XOA. (If not, this would imply that OY < XO for all angles a, which is impossible.) But now the cut XOY defines an area

larger than the cut AOB, and hence, AOB is not maximal.

Indeed, we have also established our

first claim, since the only points that are

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equidistant from the point O are points that lie both on the region and on the rotated image. It should be mentioned that the aforementioned proofs are actually only the sketches of the main ideas and have avoided many of the sticky details. These omissions are warranted since it is

only the method of reflecting that is in tended for a high school audience any

way. The problem is interesting and in

structive; students can easily understand the question involved and make some

attempt at a solution on their own. It is an

excellent application of transformation ge ometry?a topic that is many times includ ed in a high school geometry class.

BIBLIOGRAPHY

Engel, Arthur. "Geometrical Activities." The

Teaching of Geometry at the Pre-College Level, edited by Hans-Georg Steiner. Dordrecht, Hol land: D. Reidel Publishing Co., 1971.

Fletcher, T. J. "Doing without Calculus." Mathe matical Gazette 50 (February 1971):4?17.

Moulton, J. Paul. "Experiments Leading to Figures

of Maximum Area." Mathematics Teacher 68

(May 1975):356-63.

Nannini, Amos. "Maxima and Minima by Elemen tary Methods." Mathematics Teacher 60 (January 1967):31-32.

Turney, John A. "Elementary Techniques in Maxi ma and Minima." Mathematics Teacher 46 (No vember 1953):484-86.

Zimmerman, Wayne J. "Maxima and Minima by Algebraic Methods." Mathematics Teacher 65

(December 1972):748-49.

(Continued from page 393)

BIBLIOGRAPHY Bidwell, James . "A Physical Model for Factoring

Quadratic Polynomials." Mathematics Teacher 65

(March 1972):201-5.

Bruner, Jerome S. Towards a Theory of Instruction. New York: W. W. Norton & Co., 1968.

Flax, Rosabel. "A Squeeze Play on Quadratic Equa tions." Mathematics Teacher 75 (February 1982): 132-34.

Hendrickson, A. Dean. "Discovery in Advanced

Algebra with Concrete Models." Mathematics Teacher 74 (May 1981):353-58.

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May 1982 419

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