Let’s begin with a geometry problem that could be presented as a magic trick. On a separate sheet of paper, follow the steps below.
Step 1. Draw a four-sided figure where the sides don’t cross each other. (This is called a quadrilateral.) Label the four corners A, B, C, and D, in clockwise order. (See some examples below.)
C B A
D
A
B
C
D A
C
B
D
Three random quadrilaterals
Step 2. Label the midpoints of the four sides AB, BC, CD, and DA as E, F, G, and H, respectively.
Step 3. Connect the midpoints to form a quadrilateral EFGH, as shown in the examples below.
C
Connecting the midpoints of a quadrilateral always produces a parallelogram
Believe it or not, EFGH is guaranteed to be a parallelogram. In other words, EF will be parallel to GH; FG will be parallel to HE. (Also, EF and GH will have the same length, as will FG and HE.) This is illustrated in the figure above, but you should try some examples of your own.
Geometry is filled with surprises like this. Applying simple logi-cal arguments to the simplest of assumptions, one often winds up with beautiful results. Let’s take a short quiz to test your geometric intu-ition. Some of these questions have very intuitive answers, but some of these questions have answers that will surprise you, even after you have learned the appropriate geometry.
Question 1. A farmer wishes to build a rectangular fence with a perimeter of 52 feet. What should the dimensions of the rectangle be to maximize the rectangle’s area?
A) A square (13 feet per side).
B) Close to the proportions of the golden ratio 1.618 (say around 16 feet by 10 feet).
C) Make the width as long as possible (close to 26 feet by 0 feet).
D) All answers above give the same area.
Question 2. Consider the parallel lines below, with X and Y on the lower line. We wish to choose a third point on the upper line so that the triangle formed by X, Y, and the upper point has the smallest perimeter.
What upper point should be chosen?
A) Point A (above the point that is halfway between X and Y).
B) Point B (so the triangle formed by B, X, and Y is a right triangle).
C) As far away from X and Y as possible (like point C).
D) It doesn’t matter. All triangles will have the same perimeter.
X Y
A C B
Which point on the upper line results in a triangle (with points X and Y) with the smallest perimeter? Which point results in the greatest area?
Question 3. Using the same figure as above, what point P on the upper line should be chosen so that the triangle formed by X, Y, and P has the greatest area?
A) Point A.
B) Point B.
C) As far away from X and Y as possible.
D) It doesn’t matter. All triangles will have the same area.
Question 4. The distance between two goalposts on a football field is 360 feet (120 yards). A rope of length 360 feet is about to be tied tightly between the bottom of the two goalposts, when an extra foot of rope is added. How high can the rope be lifted in the middle of the field?
A) Less than one inch off the ground.
B) Just high enough to crawl under.
C) Just high enough to walk under.
D) High enough to drive a truck under.
A rope of length 361 feet is tied between goalposts 360 feet apart.
How high can we lift the rope in the middle of the field?
The answers to these questions are given below. I think the first two answers are pretty intuitive, but the other two answers will surprise most people. All of the answers will be explained later in the chapter.
Answer 1. A. For any given perimeter, to maximize the area of the rectangle, you should let all sides have equal length. Thus, the optimal shape will be a square.
Answer 2. A. Choosing the point A above the midpoint of X and Y will create the triangle XAY that has smallest perimeter.
Answer 3.D. All triangles will have the same area.
Answer 4. D. At the midpoint of the field, the rope can be lifted more than 13 feet in the air—high enough for most trucks to fit under.
We can explain the first answer using simple algebra. For a rectangle with top and bottom lengths b (b as in base) and left and right lengths h (h as in height), the perimeter of the rectangle is 2b+2h, which is the sum of the lengths of all four sides. The area measures what can fit into the rectangle and is the product bh. (We’ll say more about area later.) Since the perimeter is required to be 52 feet, we have 2b+2h = 52, or equivalently
b+h=26
And since h=26−b, the area bh that we wish to maximize is equal to b(26− b) =26b− b2
What value of b maximizes this quantity? We’ll see an easy way to do this with calculus in Chapter 11. But we can also find b by using the technique of completing the square, presented in Chapter 2. Notice that
26b− b2 =169−(b2−26b+169) =169−(b−13)2
is the area of our rectangle when base b is chosen. When b = 13 our rectangle has area 169−02=169. When b=13, our area is
169−(something not equal to 0)2
Since we are subtracting a positive quantity from 169, this will always be less than 169. Consequently, the area of the rectangle is maximized when b = 13 and h = 26−b = 13. One of the amazing things about geometry is that the fact that the farmer has 52 feet of fence is irrelevant.
To maximize the area of a rectangle with any given perimeter p, the same technique can be used to show that the optimal shape will be a square, where all sides are of length p/4.
In order to explain the other problems, we need to first look at some seemingly paradoxical results and explore some classics of geometry.
Why should a triangle have 180 degrees? What is the Pythagorean the-orem all about? How can you tell when two triangles will have the same shape, and why should we care?