CHAPTER 2

Analytic Geometry

Euclid talks about geometry in Elements as if there is only one geometry. Today, some people think of there being several, and others think of there being infinitely many. Hopefully, after you get through this course, you will be in the second group. The people in the first group generally think of geometry as running from Euclid to Hilbert and then branching into Euclidean geometry, hyperbolic geometry and elliptic geometry. People in the second group understand that perfectly well, but include another, earlier brach starting with Rene Descartes (usually pronounced like “day KART”). This branch continues through people like Gauss and Riemann, and even people like Albert Einstein. In my mind, two of the major advances in the understanding of geometry were known to Descartes. One of these, was only known to Descartes, but Gauss, it seems, figured it out on his own, and we all followed him. The other you know very well, but you probably don’t know much about where it came from. This was the development of analytic geometry, geometry using coordinates. The other is not known very well at all, but Descartes noticed something that tells us that geometry should be built upon a general concept of curvature. What is generally called Modern geometry begins with Euclid and ends with Hilbert. The alternate path, the truly contemporary geometry, begins with Descartes and blossoms with Riemann. Note that, as is typical, modern usually means a long time ago. We’ll look at analytic geometry first, and Descartes’ other piece of insight will come later. You’ll often hear people say that Descartes invented analytic geometry, but they usually don’t go into much more detail than that. Descartes’ Discours de la Methode was first published in French, I believe, in 1637, and it is an important book in philosophy. An appendix to this book is known as La Geometrie [Descartes], or in English The Geometry. This, apparently, is where analytic geometry was invented. Many people associate axioms and proofs with the word geometry, and you may think that your only exposure to geometry was in your high school geometry class. On the

7 1. CONSTRUCTIONS WITH STRAIGHTEDGE AND COMPASS 8

other hand, most of what you know about geometry probably came from your high school algebra and college calculus classes. Think about that. This is a manifestation of the power of Descartes’ approach. I’ll be referring to the Dover publication of a translation of La Geometrie. A lot of the “classics” are available through Dover and other publishers, and they’re a lot cheaper than most of our textbooks. I paid $8.95 for my copy a few years ago. The book I have has copies of the pages from the originally published version along with a translation into English. There should be a copy in the library, and it wouldn’t be a bad idea to buy one for yourself. When you’re teaching calculus or algebra, and you want to say that analytic geometry and cartesian coordinates are due to Descartes, it would be nice to be able to wave the book around in front of class.

1. Constructions with straightedge and compass

Traditional geometry often deals with constructions with straightedge and com- pass. You can see this in Euclid’s First Postulate, “To draw a straight line from any point to any point,” and his third, “To describe a circle with any centre and distance.” With this, we allow ourselves only the ability to do the following. Given two points A and B, we can draw a straight line through the two points with the straightedge, and given a third point C, we can draw a circle with center C and radius equal to the distance between A and B with the compass. It should be emphasized that the straightedge is not a ruler, and so measuring lengths with it is against the rules. I don’t see this as a practical approach, but like driving with your left foot, it is exciting, and it will instill a greater appreciation for life. In some sense, Euclid’s Elements is as a very methodical description of the things you can do with straightedge and compass constructions. It actually makes more sense to think of it this way, as opposed to thinking of it as an axiom system. The first three postulates tell you how you can construct geometric figures, draw a line through any two points with the straightedge, extend a line you already have with the straight edge, and draw circles with the compass. The last two postulates give some basis for interpreting what the figures represent and that the things you see in a figure always behave the same way. Euclid’s Elements consist of thirteen Books, and these contain 432 Proposi- tions. The propositions are basically theorems that tell you that a certain kind of 1. CONSTRUCTIONS WITH STRAIGHTEDGE AND COMPASS 9

figure can be constructed (the proof tells you how) or some fact about a particular kind of figure. We’ll look at some of these propositions to get some sort of feeling for Euclid and to help motivate Descartes’ work in analytic geometry. Euclid’s Proposition I (from Book I) states [Euclid, p 241]

On a given finite straight line to construct an equilateral triangle.

Here Euclid is saying that if you have a line segment (finite straight line), then you can construct an equilateral triangle (a triangle with three equal-length sides) with this segment as one of the sides. Euclid’s proof goes something like this. Let’s say our segment has endpoints A and B, and we’ll call the segment AB. We then draw two circles each with radius AB, one with center at A and one with center at B. The circles will have two points of intersection, C and C0. Both 4ABC and 4ABC0 are equilateral triangles, since AC, AC0, BC, and BC0 are all radii of one of these two circles. See Figure 1.

C

AB

Figure 1. Given segment AB, we can construct an equilateral triangle 4ABC.

For some reason, Euclid used a collapsing compass. He could put one end at a point (the center) and the drawing end at another point, and then he could draw the circle. Once he picked it up, however, the length of the radius was lost. We’re obviously not talking about a real-world compass, and the motivations here are probably that Euclid was trying to start with the most basic assumptions possible. In his second proposition, Euclid shows that a collapsing compass is equivalent to a non-collapsing one. This is of no concern to me. I’m interested in how Euclid’s big geometric ideas got us to where we are today. 1. CONSTRUCTIONS WITH STRAIGHTEDGE AND COMPASS 10

Let’s skip up to Proposition 4 [Euclid, p 247]

If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

This proposition illustrates one of Euclid’s logical flaws as an axiom system. Eu- clid starts the Elements with some basic assumptions, and then seems to prove the propositions from these assumptions. The proof he gives for Proposition 4, however, says little more than “it’s true, because it’s obviously true.” If you read the statement carefully, you may recognize this as the side-angle-side criterion for the congruence of triangles or SAS. One of Hilbert’s fixes is to assume SAS as an axiom. Ge- ometrically, SAS tells us a couple of things. One is that a triangle only has three degrees of freedom. In other words, designating a angle and the two adjacent sides completely determines the triangle (the lengths of all its sides, the measures of its angles, and its area). It also expresses the uniformity of the Euclidean plane: the geometry of a triangle is the same no matter where it is. Before we move on, let’s look at two of Euclid’s early propositions and some of the constructions we’ll use in exploring Descartes’ work. Proposition 11 tells us how we can construct a perpendicular. For example, suppose that we have a line l and a point P on it. To construct a line through P that is perpendicular to l, we would do the following. Draw a circle (with any radius) centered at P . This would give us two points A and B. See Figure 2.

ABP

Figure 2. Given a point P on a line, we draw a circle centered at P to find points A and B.

Now we do what we did in Proposition 1, and draw circles centered at A and B, both with radii AB. Then we draw the perpendicular through the two points where 1. CONSTRUCTIONS WITH STRAIGHTEDGE AND COMPASS 11

the two circles intersect. Let’s call the two points C and D. See Figure 3. The triangle 4ABC is an equilateral triangle, which means that the three sides have the same length. It’s probably obvious to you that the three angles must be equal also (i.e., that the triangle is also equiangular), but it’s not terribly easy to prove this (actually it’s technically impossible) within Euclid’s postulate system. That’s much of what the previous ten propositions are trying to establish. In addition, we will find that similarly obvious “facts” are not necessarily true. We’ll skip over this, and once we accept that the three angles of an equilateral triangle are equal, it easily follows that ∠AP C = ∠BPC, and so they must be right angles (see Euclid’s Definition 10). Note that if we had started with the points A and B, then the line or segment CD gives us the point P and a perpendicular bisector. (Which proposition is that?).

C

ABP

D Figure 3. As in Proposition 1, we can then find the perpendicular through P .

Proposition 23 tells us how to construct the copy of an angle. Suppose we have an angle with vertex A and a line l with A0 on it. These are the lines in black in Figure 4. We want to construct another line through A0 creating the copied angle. We start by drawing any circle with A at its center, which produces points B and C on the two sides of the original angle. Next, draw a circle with the same length radius with A0 at its center (ours is not a collapsing compass). This will give us a point B0 on l. Now, draw a circle with center at B0 with radius equal to BC. The circles with centers at A0 and B0 will intersect at a point C0 (there are two choices). The line through A0 and C0 gives us a copy of the original angle. We know this because 4ABC and 4A0B0C0 are congruent triangles. Again, this isn’t as easy 2. GEOMETRIC ARITHMETIC 12

to prove as you might expect. It is easy to see that the corresponding sides of the two triangles are equal, but it takes some work to show that the corresponding angles are also equal. We will skip over this part too.

C C0

AB A0 B0

Figure 4. We can copy an angle onto a line at a given point.

1.1. Exercises.

–1– In constructing a perpendicular, we essentially also showed how we could find a perpendicular bisector for a segment. Which of Euclid’s propositions would this establish? –2– As we’re drawing our circles on the right side of Figure 4, how many different copies of the angle could we find? In other words, how many choices for B0 and C0 are available in the construction?

2. Geometric Arithmetic

In La Geometrie, Descartes begins by illustrating how we can talk about addition, subtraction, multiplication, division, and extraction of roots geometrically. That is, he shows how we can construct these things geometrically. This is not new with Descartes, but it will help us to see his motivations in developing analytic geometry. D b B C a A

Figure 5. We are given two line segments with lengths a and b. 3. ADDITION AND SUBTRACTION 13

3. Addition and Subtraction

Descartes does not discuss addition and subtraction of line segments in much detail. In the middle of the second sentence he says (I’m trying to copy the French, which was typeset in the 1600’s, as best I can) [Descartes,p3]

Ainsi n’at’on autre chose a faire en Geometrie touchant les lignes qu’on cherche, pour les prepareer a estre connu¨s, que leur en adiouster d’autres, ...

My Dover book, translated by David E. Smith and Marcia L. Latham, gives this in English as [Descartes,p2]

so in geometry, to find required lines it is merely necessary to add or subtract other lines, ...

Even though Descartes did not choose to explain it, let’s look at it here to get us started.1 Suppose we are given two line segments AB and CD as in Figure 5. Let’s say that AB has length a and CD has length b. We can extend the segment AB beyond B as in Figure 6 with our straightedge. We can set our compass to the distance b on the line CD, and then draw a circle with radius b with center B.We get two new points E and F . D F B C E A

Figure 6. Here we extend AB beyond B, and then we draw a circle with center B and radius b. This constructs lengths a + b and a − b.

3.1. Exercises.

–1– Which line segment is the sum a + b? –2– Which line segment is the difference a − b?

1It’s partly Descartes’ fault that we are not familiar with this sort of thing. 4. MULTIPLICATION AND DIVISION 14

4. Multiplication and Division

Descartes talks a bit more about multiplication and division. Continuing where the quote above left off [Descartes,p3]

ou en oster, Oubien en ayant vne, que ie nommeray l’vnit´epour la rapporter d’autant mieux aux nombres, & qui peut ordinairement estre prise a discretion, puis en ayant encore deux autres, en trouuer vne quatriesme, qui foit ´al’vne de ces deux, comme l’autre est, a l’vnit´e, ce qui est le mesme que la Multiplication; ... Soit par exemple A B l’vnit´e,& qu’il faille multiplier B D par B C, ie n’ay qu’a ioindre les poins A & C, puis tirer D E parallele a C A, & B E est le produit de cete Multiplication.

Smith and Latham translate this as [Descartes,p2]

or else, taking one line which I shall call unity in order to relate it as closely as possible to numbers, and which can in general be chosen arbitrarily, and having given two other lines, to find a fourth line which shall be to one of the given lines as the other is to unity (which is the same as multiplication); ...For example, let AB be taken as unity, and let it be required to multiply BD by BC. I have only to join the points A and C, and draw DE parallel to CA; Then BE is the product of BD and BC.

E

C

D AB Figure 7. A geometric construction of multiplication segment BD times BC relative to the unit AB.

Note that Descartes uses the word line (actually ligne) when we would probably say line segment. Let’s look at the last part of this quote first. In the original text, 4. MULTIPLICATION AND DIVISION 15

there is a picture that looks very much like Figure 7 (with the same letters). It starts out with, “AB is taken as unity,” so the length of the segment AB is 1. We want to multiply BD by BC. We can draw a line through A and C, and then draw a parallel to it through D. This gives us the point E. Now, triangle 4BAC is similar to triangle 4BDE,soBE is to BD as BC is to BA,or BE BC (1) = . BD BA We’ll talk more about similar triangles later, but I’m guessing you’ve seen them before. Now, notice that if BC = x (i.e., if BC has length x), BD = y, and BA =1 (the unity), then BE x (2) = , y 1 and (3) BE = xy. We’ve just constructed a segment of length xy, and we’ve done a geometric multi- plication. If we can do multiplication, then we should be able to do division also. We just use equation (1) differently. We could let BE = z, and then we have the equation z BC (4) = , y 1 and now z (5) = BC. y

E

D AB Figure 8. You are given the unit segment AB = 1 and the segments BD and BE. 4. MULTIPLICATION AND DIVISION 16

4.1. Exercises.

BE –1– Indicate how you would find the point C in Figure 8 so that BC = BD. You may be familiar with the idea of representing the product of two numbers geometrically with the area of a rectangle, since A = lw. What we’re doing here is very different, because the things we’re multiplying and their product are all lengths, and the product is directly comparable to the two factors. Both ways of thinking of multiplication are valuable, and they offer different opportunities. One big difference between these different views of multiplication is that here the choice of a unit segment is critical. We get different answers, if we use a different unit segment. Let’s explore this a bit more.

4.2. Exercises.

–1– Let’s suppose that we use a unit segment of length 1 foot. Then the product of a segment of length 2 feet and a segment of length 3 feet will have length 6 feet. What is the product of these two segments, if you use a unit segment of length 1 yard? Draw a picture illustrating these two multiplications. –2– If we do the product using the area of a rectangle, are the answers different geometrically? In other words, is the area of a rectangle with sides 2 feet 2 and 3 feet the same as the area of a rectangle with sides 3 yards and 1 yard? –3– Suppose we have a unit segment of length 1 cm, and we divide a segment of length 5 cm by a segment of length 2.5 cm. The “answer” segment will be how long in centimeters? –4– In the previous problem, suppose that the unit segment is 1.5 cm? What is the length of the answer segment?

Basic Principle 1. If we are going to convert segments to numbers using their lengths, we must do this relative to some unit length. The particular choice is not necessarily important, but we must be consistent, or we should at least be careful when we change units. If we multiply x units by y units, then the answer segment will have length xy units in our usual symbolic multiplications. If we divide x units by y units, x then the answer segment will have length y units. 5. SQUARE ROOTS 17

I

FGK H Figure 9. We wish to find the square root of GH relative to the unit segment FG.

5. Square Roots

Descartes also discusses the square root. He says, “Ou s’il faut tirer la racine quarr´eede G H ... [Descartes, p 4]” (that is, “If the square root of GH is required, ... [Descartes, p 5]”), and he describes how to find it in the same way he did multiplication and division. There is a figure much like Figure 9 (with the same letters) in the original text. We are given a unit segment FG, and we want to find the square root of GH. The segment FH is FG+ GH. We can find the midpoint of FH, which is K, and we can draw a circle of radius KH with K at the center. Finally, we draw a perpendicular at G. This gives us the point I. The segment IG is the square root of GH. Let’s explore this in modern notation. We can let FG= 1 and GH = x. Then we have that FH = x + 1. We can draw in a segment KI, which is the hypotenuse of a right triangle GKI. We know the length of KI, since it is the radius of the circle. In particular, it is x +1 (6) KI = KH = . 2 We also know the length of the base of 4GKI, because x +1 x − 1 (7) GK = FK − FG= − 1= . 2 2 The length of the third side of 4GKI is supposedly the square root of x, but let’s call it y for now. The Pythagorean theorem (which is Proposition 47 in Euclid’s Elements ) tells us that

(8) GK2 + IG2 = KI2, 6. DYNAMIC FIGURES 18

or using our letters

x − 1 2 x +1 2 (9)   + y2 =   . 2 2 √ Solving for y yields y2 = x, which means that y = x (the positive square root, since we’re talking about lengths).

5.1. Exercises.

–1– As accurately as you can, do the following. Draw a segment that will serve as your unit segment. Now draw a√ segment of length 3 cm. As Descartes did, construct a segment of length 2 cm. –2– We can construct all of these things with straightedge and compass. For example, given two segments, we can construct a segment whose length is the sum of the two given segments. We can also construct a circle from a given center point and radius (a segment that has length equal to the desired radius). There are two other constructions we need to find a square root. What are they?

6. Dynamic figures

Descartes talks about machines that represent dynamic geometric figures.As we will see, these dynamic figures are roughly equivalent to algebraic equations, and this is an indicator of the power of algebraic equations when compared to the static figures constructed by straightedge and compass. We can use the figure used for multiplication as an example. Look at Figure 10. The circles mark off unit lengths on the two black lines. The blue line hits the lower black line at one unit and the upper black line at two units. The red line hits at three units and six units, and represents the multiplication

(10) 2 · 3=6.

Moving the red line to the left or right without changing its slope gives other mul- tiplications of the form 2 · x =2x, or in other words, it represents multiplication by two. 6. DYNAMIC FIGURES 19

1 2 3 4 5 6

Figure 10. Moving the red line to the left or right performs multipli- cation by 2.

What we have here is a dynamic geometric figure that is equivalent to the equation z =2x. We can debate whether what the equation tells us about the figure is more important than what the figure tells us about the equation. I would claim, however, that what is most important is the linkage between the two. The figure- equation pair is more complex, interesting, and useful objectg than the sum of the figure and equation considered independently. An important insight of Descartes’ for our study of geometry lies in the fact that each of any two lengths of a dynamic figure in the plane with slopes held constant are completely determined by the other. As an example, let’s add two lines to Figure 10 to obtain the figure in Figure 12.

z y w x

Figure 11. A dynamic figure with slopes held constant is equivalent to an equation with two variables. 6. DYNAMIC FIGURES 20

We already have that (11) z =2x. The way I’ve drawn this figure, we have the relationship (12) y = x, and therefore, we also have (13) z =2y. A fourth segment is marked w. We’ll review the details of the Law of Cosines later, but if the angle between the two black lines is θ (all of the angles are constant, because we are keeping all of the slopes constant), then by the Law of Cosines (14) w2 = x2 + z2 − 2xz cos θ. Since z =2x, we can make a substitution to obtain (15) w2 = x2 +(2x)2 − 2x(2x) cos θ =5x2 − 4x2 cos θ = x2(5 − 4 cos θ). It follows that √ (16) w = x 5 − 4 cos θ, and w is a constant multiple of x. If we slide the blue lines to the positions of the red lines, all of these same relationships will hold, and once we know the length of any one of the new segments, we can use these equations to find the lengths of the others. Before moving on, note that what we’re doing here looks more familiar, if we make all the guidelines either vertical or horizontal.

y

x

Figure 12. Making the guide lines horizontal and vertical makes a dynamic figure look more familiar. 7. GEOMETER’S SKETCHPAD 21

7. Geometer’s Sketchpad

A nice place to play with dynamic figures is in Geometer’s Sketchpad. Let’s use our multiply-by-two machine as a simple example of how Geometer’s Sketchpad works. After you open a Geometer’s Sketchpad window, you will see six buttons. The first one has an arrow on it. This is called the selection arrow tool. The next button, the point tool, has a little dot on it. The third button, the compass tool, has a circle on it. The fourth button is for the straightedge tools. You’ll see a small triangle at the lower-right of this button, and if you click and hold it, you’ll see choices for a line segment, a ray, and a line. Geometer’s Sketchpad works as it looks like it should, for the most part, and playing with it is probably the best way to learn how to use it.

Figure 13. Our multiply-by-2 machine might look like this in Geome- ter’s Sketchpad

To build our multiply-by-two machine the way I did, choose the ray version of the straightedge tool. Click near the left side of the window, and then click again a little ways to the right. You should get a relatively horizontal ray pointing to the right. Next, click on your first point, and then again above and to the right. You should now have an angle with three points. Now, choose the segment straightedge tool. Click on the point on the lower side of the angle, and then on the point on the upper side. This should give you a segment between these two points. Next, choose the point tool, and put a point on the lower side of the angle further out than the point you already have. 8. SOLVING EQUATIONS 22

Choose the selection arrow tool. If you click on any of the objects in the picture, this will either highlight it, or turn it off. Highlight the segment and the last point. Along the top of the window, you will see headings for menus. Choose Construct and parallel. This should give you a line parallel to the segment through the point you chose. The parallel line intersects the upper side, but there is no point there. Put a point at the intersection. Alright, you should now have something that looks like Figure 13. Choose the selection arrow tool, and highlight the points in Figure 13 marked A and B. Then choose Measure and distance from the menu. The letters and a measurement should appear. Make sure that nothing (including the measurement boxes) are highlighted, and then do the same for A and C. Finally, measure AD and AE. OK. The numbers might be different, but otherwise, your picture should look like Figure 13. To see things easier, let’s make the unit segment AB 1cm long. Drag the point B around (if you want, you can highlight it and use the arrow buttons on the keyboard), and get AB as close to 1cm as you can. Now, drag C so that AC = 2cm. Now you can drag the line DE around and

(17) 2 · AD = AE, at least approximately, because of round-off error. It is very important to note that when you move objects, only those objects constructed later move. In our construction, the line DE was constructed to be parallel to the segment BC. Geometer’s Sketchpad will say that BC is a parent of DE, and also that DE is a child of BC. When you move an object, only its children will move.

8. Solving equations

If Euclid had written a book for middle school students, it might have included a problem like

Given two lines, to construct a third line so that the third and first equals the second.

In algebraic language, we might state the problem as

Given numbers a and b, find x so that x + a = b. 8. SOLVING EQUATIONS 23

Our minds would immediately jump to a solution x = b − a. Our teacher, of course, would demand that we show our work, and we might do something like (18) x + a = b (19) x + a − a = b − a (20) x +0=b − a (21) x = b − a We’ve proven a theorem, essentially. A very trivial theorem, but every theorem is trivial to someone.

Theorem 1. If x + a = b, then x = b − a.

Euclid would say something like this (the theorem would typically be part of the proof). Given lines AB and CD, draw a circle with center D and radius equal to AB. This circle intersects CD in a point E. The line CE is the desired third line. He would then explain how this is indeed the solution. Since we know how to construct additions, subtractions, multiplications, and di- visions, we have the tools necessary to construct the solution to any linear equation. Since the solution to the equation (22) ax + b = c is c − b (23) x = , a Euclid would start with a segment of length c, subtract b, and then divide that segment by a. Some might say that Euclid could do basic algebra and that he could solve linear equations. That’s not really true, however. He could find the solution to a particular problem that we would solve using algebra, and he could tell you how to find the solution geometrically, but he did not have a general theory that would solve all linear equations. For example, he would have a different solution description for the equation b − ax = c. What does Descartes have to say about all of this? He sees that in terms of geometric constructions, you might have twenty different problems that seem to have nothing to do with each other. If you convert these problems into algebraic form, you may see that all these problems are really all just one problem. 9. SOLVING QUADRATIC EQUATIONS 24

Basic Principle 2. Non-obvious structures in one representation might be ob- vious in another.

Do not take this observation to mean that algebra is superior to geometry. Some things, like relative extrema and gravity, are easier to understand geometrically.

9. Solving quadratic equations

Let’s take a look at what Descartes actually says. One thing he does is to show how a quadratic equation can be solved geometrically. He gives, essentially, a geometric version of the quadratic formula, although he does not derive it. In other words, he simply states the theorem. He states [Descartes, p 12]:

Car si i’ay par exemple z2 az + bb ie fais le triangle rectangle N L M, dont le cost´eL M est esgal `a b racine 1 quarr´eede la quantit´econnue bb, & l’autre L N est 2 a, la moiti´ede l’autre quantit´econnue, qui estoit multipli´eepar z que ie suppose estre la ligne inconnue, puis prolongeant M N la baze de ce triangle, insques a O, en sorte qu’ N O soit esgale a N L, la toute O M est z la ligne cherch´ee.Et elle s’exprime en cete forte

1 √1 z 2 a + 4 aa + bb. I’ve drawn in the symbol, because I don’t have the correct character easily avail- able. It’s used for the equal sign, and I believe that it was pretty standard at the time and eventually became the “=” we use today. In the original, the radical symbol looks drawn in by hand. What you see here has been electronically typeset, and I don’t have all the old French characters either, but you can see how that the mathematical notation is not as foreign looking as you might expect. A lot of the mathematics as we know it today is starting to form here. Here’s Smith and Latham’s translation [Descartes, p 13]:

For example, if I have z2 = az + b2, I construct a right triangle NLM with one side LM, equal to b, the square root of the known quantity b2, and the other side, LN, equal to 9. SOLVING QUADRATIC EQUATIONS 25

1 2 a, that is, to half the other known quantity which was multiplied by z, which I supposed to be the unknown line. Then prolonging MN, the hypotenuse of this triangle, to O, so that NO is equal to NL, the whole line OM is the required line z. This is expressed in the following way:

1 q 1 2 2 z = 2 a + 4 a + b .

O N

L M Figure 14. 1 Given lengths LM = b and LN = 2 a, MO = z is the solution to the quadratic equation z2 = az + b2.

Descartes doesn’t tell us how he gets his solution, he is only telling us how to construct it geometrically. We know how to find it algebraically from the equation, however. The quadratic equation is z2 = az + b2, and we can write it in the form we normally do today as

(24) z2 − az − b2 =0.

The quadratic formula gives us √ −(−a) ± p(−a)2 − 4(1)(−b2) a ± a2 +4b2 (25) z = = , 2(1) 2 which simplifies to what Descartes has, if you ignore the − in the ±. So how does Descartes’ construction work? The two constants in the equation are a and b2. The number b2 is given, and he is just calling this number b2, because he wants to use b, the square root of b2. From these two numbers, he would like to construct a segment of length z such that

(26) z2 = az + b2.

We have already seen how to construct square roots and midpoints, so we can con- 1 struct segments of length 2a and b. We can also construct perpendiculars, so we can 4 1 make a right triangle NLM with sides LN = 2 a and LM = b. Look at Figure 14. 9. SOLVING QUADRATIC EQUATIONS 26

1 We next draw a circle of radius LN = 2a, and then we extend the hypotenuse MN out to the circle. This gives us the point O. The segment MO gives us the length z that is the solution of the quadratic equation. Let’s check that. The Pythagorean theorem says that

(27) LM 2 + LN 2 = MN2,

so 1 2 (28) b2 +  a  = MN2. 2

Therefore,

r 1 (29) MN = b2 + a2. 4 Now we know that

1 r 1 (30) MO = NO + MN = a + b2 + a2 = z. 2 4

9.1. Geometer’s Sketchpad. What we have here is a machine that solves qua- dratic equations of the form z2 = az + b2. You can see some of Descartes’ thinking in his book, where he talks about actual machines with bars and hinges representing segments and angles. This would work even better in Geometer’s Sketchpad. Let’s do that. Geometer’s Sketchpad can construct perpendiculars and midpoints, but we’ll also need to construct a square root, so I’ve done that in the figure. The stuff that we put into the picture first should be the basic stuff, and the solution segments should be last, so I started with taking the square root of b2. I’m going to do a picture like Figure 9, but on its side. Start with a ray BM. This will be our unit segment. Put C on this ray. MC will be the length b2, and we want the square root of this. The other point not marked is the midpoint between B and C. One way to do this is to highlight the points B and C, choose Construct and Segment from the menu, and then choose Construct and Midpoint from the menu. Now make a circle with center at the midpoint and radius out to B or C. 9. SOLVING QUADRATIC EQUATIONS 27

Figure 15. A Geometer’s Sketchpad quadratic equation machine.

Next, we need to make a perpendicular to BC at M. Highlight the segment BC and the point M, and then choose Construct and Perpendicular Line from the menu. Put the point L at the intersection. ML has length b. We’ve done our square root construction. Let’s check that first. Measure the segments BM, MC, and ML. Move B around so that MB√ = 1 as close as you can get it. Then play with C. You should see that ML = MC (approximately, since we don’t have precise control over the lengths of our segments).

1 Now, we’re ready to implement Figure 14. We need the point N next, so NL = 2 a, one of our imputs. Highlight the point L and the segment ML, and then choose Construct and Perpendicular Line from the menu. Put the point N on this line. Put a ray through M and N. Also put a circle with center at N and radius out to L. Put the points O and X at the appropriate intersections. MO = z, and for convenience, LX = a. As pictured, I have a = 5 and b2 = 4. This would go into the equation z2 = az +b2 as

(31) z2 =5z +4, or z2 − 5z − 4 = 0. In the quadratic formula, we would have √ −(−5) ± p(−5)2 − 4(1)(−4) 5 ± 41 (32) z = = =5.70156 ... 2(1) 2 10. MORE EQUATION SOLVING MACHINES 28

10. More equation solving machines

You can perhaps imagine applications for these kinds of machines. I hear people talking sometimes about the geometry of a car’s suspension. I don’t know, but perhaps the forces exerted by the suspension spring do not optimally match the forces you see where the tires meet the road. The suspension pieces act like one of our figures, and we can maybe control the forces as the suspension moves. I’m just throwing that out to give you some ideas. The main mathematical idea that I want get across is that Descartes embraced the idea that if you look at these problems algebraically, what looks like a complex array of geometric problems becomes one or two relatively simple algebraic ones, and the geometric constructions are translations of the algebraic solutions. Anyway, after the constructions we’ve just done, Descartes goes on to discuss the solutions of y2 = −ay+b2 and x4 = −ax2+b2. The stuff covered so far was not exactly new, and Descartes discusses this before moving on. I’ll just give the translation for part of it [Descartes, p 17].

These same roots can be found by many other methods, I have given these very simple ones to show that it is possible to construct all the problems of ordinary geometry by doing no more than the little cov- ered in the four figures that I have explained. This is one thing which I believe the ancient mathematicians did not observe, for otherwise they would not have put so much labor into writing so many books in which the very sequence of the propositions shows that they did not have a sure method of finding all, but rather gathered together those propositions on which they had happened by accident.

Again, if you think only geometrically, only in terms of pictures and constructions, then it can be difficult to see a common thread in different problems. Translating the problem into algebraic terms can often make the problem easier, and it can also make it more obvious that a certain class of problems can all be solved the same way. Today, we think of all quadratic equations as being essentially the same, but Descartes points out that the ancient Greek mathematicians could only see a large number of different problems. 11. QUADRATIC EQUATIONS THE GREEKS COULD SOLVE 29

10.1. Exercises.

–1– Consider the class of all problems that come down to solving a quadratic equation. Once we have the quadratic equation, can we always solve it? If so, is there one technique that will always work? If so, what is that technique? –2– According to Descartes, did the ancient Greeks have one technique that would solve all quadratic equations?

11. Quadratic Equations the Greeks could solve

Descartes’ work probably sounds somewhat trivial and complicated at the same time. To try to get a perspective on the advances he proposes, let’s look at one of Euclid’s propositions and the way the Greeks did things. In the T.L. Heath translation of Euclid’s The Elements, we have Proposition 5 of Book II [Euclid, p 251].

If a straight line be cut into equal and unequal segments, the rectan- gle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half. D

ABC Figure 16. The line segment AB is cut at C, and the rectangle is contained by the two unequal segments of AB.

Let’s figure out what this means. We’ll start with the straight line (segment) cut into unequal segments and the rectangle contained by the unequal segments. In Figure 16, we have a segment AB cut into unequal segments by C. If we flip the segment CB up so that it’s perpendicular to the segment AC, this gives us a segment CD. The segments AC and CD contain a rectangle (completed with dotted lines). We cut AB into equal segments with the point E (i.e., E is the midpoint of AB). The points of section are C and E, and I have added a square on the segment EC with dotted lines in Figure 17. Euclid claims that the rectangle (AD) and the small 11. QUADRATIC EQUATIONS THE GREEKS COULD SOLVE 30 D

ABCE

Figure 17. The point E cuts AB into equal segments, and the seg- ment EC is the segment between the points of section . square (with side EC) together are equal to the square on the half (the square with side AE). K J D I

123

B A E 45C FG H

Figure 18. I’ve added a few more lines for the proof.

Euclid’s proof starts with the big square FHIJ, and it treats the individual squares and rectangles like puzzle pieces. I’ll use P 1 for “piece 1,” which is the rectangle AEJK. Since P 3 and P 4 are squares, P 2 and P 5 are congruent rectangles. Since E cuts AB into equal segments, P 1 and P 2+P 3 are congruent rectangles. It follows that P 1 ∼= P 2+P 3 =∼ P 3+P 5. In terms of areas (33) FHIJ = P 2+P 3+P 5+P 4=P 2+P 1+P 4=ACDK + FGCE.

If you look through Euclid’s Elements, you see a long list of propositions. We can interpret many, if not all, of the propositions as a solution technique for a certain class of problems. Here, the problem may go something like this. Given a rectangle and square with the same perimeter, what is the difference in the areas? This is the same problem, since if two adjacent sides add up to the same length, then the perimeters are the same. Euclid’s answer is that the square is bigger by the area of the little square FGCE,orEC2. 12. REVIEW OF TRIGONOMETRY 31

That’s kind of cool, but Euclid’s approach to how would we do that problem. If a+b we have a rectangle with sides a and b, then the square would have sides 2 . The difference in the areas would be a + b 2 a2 +2ab + b2 4ab a2 − 2ab + b2 (34)   − ab = − = . 2 4 4 4

a−b We would probably also notice that this difference is the square of 2 . Given this problem, using Euclid’s Proposition 5 is actually a bit easier, but note that we would have to know Proposition 5. Using our algebraic techniques, we would just figure it out from scratch. You could easily solve problems like this when you were a kid, and I hope this gives you some feeling for what Descartes is saying about the ancient Greek mathe- maticians. If you were a student of Euclid, and you couldn’t figure this out on your own, you would look through all the theorems of Euclid (his propositions), and find the one that applied. Thanks to people like Descartes, this problem is now just a very simple application of basic algebra. In Heath’s translation of The Elements, he puts forward the interpretation that Proposition 5 gives the solution to a quadratic equation. This is true, but as Descartes indicates, while Euclid provides the solution to a quadratic equation, he fails to realize that he has done so. It’s certainly understandable that Euclid would not see this, of course. We take the quadratic formula for granted, but it unified and simplified a great number of disjoint geometric solutions. With the advancement of mathematical techniques, we don’t have to be as smart to solve the same problems.

12. Review of trigonometry

I want to continue looking at Descartes’ La Geometrie. Descartes’ development of analytic geometry appears to be motivated by a problem of interest to the ancient Greeks, where they investigate curves described by their relationship to a number of reference lines. Certainly, basic ideas related to the use of coordinate systems goes back at least as far as the ancient Greeks, but it is Descartes who seems to have focused on the right things. It appears to me that the two key insights due to Descartes are as follows: 1) With the help of algebra, we can reduce the description of a curve to its relationship with any two reference lines. 2) By adding two particular lines, we can always use the same two reference lines for every curve in the plane. It is 13. THE PYTHAGOREAN THEOREM 32

standard practice today to describe a curve in the plane in reference to two particular lines, the x- and y-axes of our cartesian coordinate system. I found Descartes’ explanation to be difficult to digest, so let’s look at Descartes’ first example, and then figure out what it is. We’ll just be working through a couple of simple cases to get some of the flavor. A lot of the difficulty comes from the old terminology, and we’re also not accustomed to thinking about problems like this. We could say, actually, that Descartes made problems like this obsolete. Before doing that, I want to review a little trigonometry. A

c b

CBa Figure 19. The Pythagorean theorem states that if C =90◦, then c2 = a2 + b2.

13. The Pythagorean Theorem

The Pythagorean theorem is older than Euclid’s Elements, but we’re using Euclid as our starting point, so let’s look at how he states it. We have Proposition 47 in Book I [Euclid, p 349].

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Here “the side subtending the right angle” is the side opposite the right angle. The Greek word that Heath is translating as subtending is υπoτινoυσης´ (upsilon pi omi- cron tau epsilon iota nu omicron upsilon sigma eta sigma). In this case, the first upsilon, the υ, is pronounced with a breathing sound like an “h,” so the thing that was translated to subtending reads kind of like hupoteinouses. To us, the hypotenuse is the side opposite the right angle in a right triangle. In Figure 20, we’ll use the letters A, B, and C as both the names of the vertices and the measures of the angles at these vertices, and the theorem looks like this in our modern language. 13. THE PYTHAGOREAN THEOREM 33

Pythagorean Theorem. For a triangle with angles A, B, and C and opposite sides a, b, and c,ifC =90◦, then

(35) c2 = a2 + b2.

E F

D A

CB Figure 20. The Pythagorean theorem states that if C =90◦, then c2 = a2 + b2.

There are lots of proofs of the Pythagorean theorem, but Euclid’s proof is pretty cute, so I want to at least give the basic idea. If we put squares on each of the sides, the areas of the two smaller squares (a2 and b2) add up to the big square (c2). We “drop” a perpendicular from C to the hypotenuse and continue it to the point F . We also add the lines CE and BD, as shown in Figure 20. Euclid shows that the b2-square CD has the same area as the rectangle AF . (The translation, and I’m assuming Euclid, indicate rectangles and parallelograms with a pair of opposite corners.) The rest of the c2-square, the rectangle BF, is equal to a2 by a similar argument. Euclid first shows that triangles 4DAB and 4CAE are congruent. We will assume all the basic facts of plane geometry including the SAS criterion: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then all parts of the two triangles (including the area) are congruent. Note that ∠DAC =90◦ +∠CAB. Note also that ∠DAB = ∠CAE, since they are both ∠CAB plus a right angle. We also know that DA = CA, since they’re sides of a square. For the same reason, AB = AE. By SAS, 4DAB and 4CAE are congruent, and so they have the same area. Recall that the area of a parallelogram (or rectangle, or square) is base-times- height, and the area of a triangle is half of the base-times-height. We can think of 14. THE LAW OF COSINES 34

the segment DA as the base of triangle 4DAB and also as the base of the square DC. The heights are also the same. Therefore, the areas are related as 1 1 (36) 4DAB = squareDC = b2, 2 2 and this is also equal to the area of 4CAE Now, we also have that the segment AE is the base of the 4CAE and the rectangle AF . And furthermore, their heights are the same. Therefore, the area of the rectangle AF must be the same as the area of square DC, in particular, it is b2. Repeating this argument will show that the square on the side CB must be equal to the area of rectangle BF. Is that not cool?

A

c b0 b

D a0 CBa

Figure 21. If we slide vertex C to the right, the angle C is now obtuse.

14. The Law of Cosines

The law of cosines is also in Euclid’s Elements. In Book II, Proposition 12 states [Euclid, p 403]

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

If we take Figure 19, and slide vertex C to the right, then angle C becomes obtuse (i.e., larger than 90◦), as in Figure 21. Euclid’s Proposition 12 states that a2 + b2 is now smaller than c2, instead of being exactly equal, as it is in the Pythagorean 14. THE LAW OF COSINES 35

Theorem. Euclid goes on to say how much bigger c2 is. The difference is “twice the rectangle” of (i.e., the product of) a and a0. If you haven’t noticed this yet, this is the Law of Cosines. Let’s reconcile Proposition 12 and the Law of Cosines first. In the main triangle 4ACB, we will say that the measure of angle ∠ACB = C, using C both as the name of the vertex and the measure of the angle. We’re also using ∠ACB as the name of the angle and as its measure. The angle ∠DCA is exterior to the triangle, and it measures ∠DCA = 180◦ − C. The cosine of this angle in the small right triangle is adjacent over the hypotenuse, so we have then that a0 (37) cos(∠DCA) = cos(180◦ − C)= . b Recall that if you shift the cosine function by 180◦, the graph looks upside-down, so for any angle θ,

(38) cos(180◦ + θ)=− cos(θ).

If you change the sign of the thing inside the cosine function, everything gets reflected left-to-right, but cosine is symmetric in this direction, so

(39) cos(θ) = cos(−θ).

Using these trig identities, (37) can be rewritten a0 (40) = cos(180◦ − C)=− cos(−C)=− cos(C), b and

(41) a0 = −b cos(C).

The rectangle, therefore, is (a)(−b cos(C)), and this quantity is positive, since C> 90◦. Saying that c2 is bigger than a2 + b2 by twice the rectangle comes out to

(42) c2 = a2 + b2 − 2ab cos(C)

which is the formula from the Law of Cosines. We can prove this using the Pythagorean theorem using the right triangles shown in Figure 21. In one right triangle we have

(43) b2 =(b0)2 +(a0)2, 14. THE LAW OF COSINES 36

so (44) (b0)2 = b2 − (a0)2 (45) = b2 − ( −b cos(C))2 (46) = b2 1 − cos2(C)
(47) = b2 sin2(C). In the other right triangle, the big one, we have (48) c2 =(a + a0)2 +(b0)2. Therefore, (49) c2 =(a + a0)2 +(b0)2 (50) =(a +(−b cos(C)))2 + b2 sin2(C) (51) = a2 − 2ab cos(C)+b2 cos2(C)+b2 sin2(C) (52) = a2 − 2ab cos(C)+b2 cos2(C)+sin2(C)
(53) = a2 + b2 − 2ab cos(C).

This is an algebraic proof of Euclid’s Proposition 12. Note again that the “squares of the sides containing the obtuse angle” are the quantities a2 and b2. More difficult to see, “the straight line cut off outside by the perpendicular towards the obtuse angle” is the segment DC, which we labeled a0, and which is equal to −b cos(C), and the side “on which the perpendicular falls” is the segment CB, which we labeled a in Figure 21. The “twice the rectangle,” therefore, is double the product of these two quantities, which is −2ab cos(C). If we had used AC as the base, and dropped a perpendicular from B, then the a and b would have exchanged roles, but the formula would have come out the same. A

b c b0

0 CBa D a Figure 22. If we slide vertex C to the left, we can make the angle C acute. Note that CB = a 14. THE LAW OF COSINES 37

Euclid’s Proposition 13 says basically the same thing for acute-angled triangles (acute angles are less than 90◦). Here is Proposition 13 [Euclid, p 406].

In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.

The proof for Proposition 13 is similar to the one for Propostion 12. We have a right triangle to the left, so (54) b2 =(a0)2 +(b0)2. There is also a right triangle to the right. Note that a represents the entire segment CB,so (55) c2 =(a − a0)2 +(b0)2. This time, we have cos C directly, and a0 (56) cos C = . b In either case, the formula comes out the same, so it doesn’t really matter if the angles are acute or obtuse. This formula is known as the Law of Cosines.

Law of Cosines. For any triangle with sides a, b, and c with angles A, B, and C opposite each side, (57) c2 = a2 + b2 − 2ab cos C.

As long as you label a opposite of A, b opposite B, and c opposite C, the Law of Cosines holds true for any labeling. There isn’t anything special about the angle C, therefore. You could have a2 = b2 + c2 − 2bc cos(A), for example.

14.1. Exercises.

–1– Go through the same steps as we did in the proof of Proposition 12. –2– Algebraically, how are Propositions 12 and 13 related? –3– In the Law of Cosines, if C =90◦, then what is cos C? In this case, what famous theorem do we have? 15. LAW OF SINES 38

15. Law of Sines

Take a look back at Figure 22. Note that b0 (58) sin(C)= . b Using the right triangle on the other side of the figure, we see that b0 (59) sin(B)= . c Solving equations (58) and (59) for b0,weget

(60) b sin(C)=c sin(B).

We could also write this as b c (61) = . sin(B) sin(C) We get the same relationships for the triangles in Figure 21 and Figure 19. For the right triangle in Figure 19, we get

(62) sin(C)=1, since sin(90◦) = 1, and b (63) sin(B)= . c Equation (63) becomes b (64) c = , sin(B) and since sin(C) = 1, we can write this as c c b (65) = = . 1 sin(C) sin(B) We can take any triangle and use any of the three sides as the base, and the triangle will look something like Figure 19, Figure 21, or Figure 22, so there must not be a anything special about B and C, and sin(A) must be equal to the things in equation (61) as well. This relationship is known as the Law of Sines. 16. SIMILARITY 39

Law of Sines. For any triangle with sides a, b, and c with angles A, B, and C opposite each side, a b c (66) = = . sin(A) sin(B) sin(C)

15.1. Exercises.

–1– Find equations similar to (58), (59) and (61) but for Figure 21. You may need the trig identities sin(180◦ + θ)=− sin(θ), and sin(θ)=− sin(−θ).

C0

C b0 c0 b c

A a B A0 a0 B0

Figure 23. These are similar triangles.

16. Similarity

We will want to say that two objects that have the same shape, but not necessarily the same size, are similar. For two triangles, this will mean that corresponding angles are congruent (have the same measure). In Figure 23, for example, A = A0, B = B0, and C = C0 (using these letters as the measures of the angles), so these triangles are similar. In this case the sides of triangle 4A0B0C0 are twice as long as the sides of 4ABC. In general, for two similar triangles, there is no special relationship between the lengths of a and a0. One can be a little bigger than the other, or a lot bigger. There is a useful relationship, if we consider all parts of the triangles together, however. The Law of Sines gives us the following, for example. a sin A sin A0 a0 (67) = = = . b sin B sin B0 b0 This says that the ratio between sides a and b is the same as the ratio between the sides a0 and b0. In other words, if a is k times longer than b, then a0 will also be k 17. CONGRUENCE 40

times longer than b0. The same holds for any pair of sides, so a a0 a a0 b b0 (68) = , = , and = . b b0 c c0 c c0 We can also flip these equations around to get a b a c b c (69) = , = , and = , a0 b0 a0 c0 b0 c0 or more compactly a b c (70) = = . a0 b0 c0

17. Congruence

If two similar triangles are the same size, we’ll say that they are congruent. Con- gruent triangles are the same size and shape. This means that all corresponding parts are the same, corresponding sides have the same length, corresponding angles have the same measure, and the areas are the same, as we’ve said before. If all three angles match the three angles of another triangle, then they are similar. Certainly, this is not enough to guarantee congruence between the two triangles. Note however, if two pairs of corresponding angles have the same measure (i.e., are congruent), then the third pair must also, since the angle sum of a triangle is always 180◦. If we have two pairs of congruent angles, this guarantees similarity. If in addition, one pair of corresponding sides have the same length (e.g., a = a0), by equation (70), all pairs of sides must have the same length. We have the following theorem.

Theorem 2. (AAS, ASA,orAAS) If we have two triangles, and two pairs of corresponding angles are congruent, and one pair of corresponding sides are congruent, then the triangles are congruent.

We have already mentioned the SAS criterion for the congruence of triangles several times. This states that if two sides and the included angle in one triangle are congruent to two sides and the included angle of another triangle, then the two triangles must be congruent. It is called SAS, because the angle must be between the two sides, side-angle-side (e.g. a, C, and b). If these three pieces of information are sufficient to guarantee congruence, they must also determine all the measurements of 17. CONGRUENCE 41

the triangle. We can use the Law of Cosines to do this. For example, given a, C, and b, we can find c using (71) c = pa2 + b2 − 2ab cos(C). We can then find A, for example, using the equation (72) a2 = b2 + c2 − 2bc cos(A). In fact, once you have all three sides of a triangle, you can use the Law of Cosines to find values for any of the angles. As long as the angles are unique, having three sides congruent is sufficient to guarantee that the triangles are congruent. Theorem 3. (SAS and SSS) If two sides and the included angle (SAS) match on two triangles, then they are congruent. If all three sides match (SSS), then the two triangles are congruent.

OK. We have SSS, SAS, AAS, ASA, and SAA criteria for congruence. AAA only guarantees similarity. The two remaining possibilities, are ASS and SSA. These guarantee neither congruence nor similarity. A

b b0 c

CBC0

Figure 24. Triangles 4ABC and 4ABC0, satisfy SSA, but are not congruent.

17.1. Exercises.

–1– Sketch a graph of the cosine function in degrees over −180 ≤ x ≤ 360. It should pass through (−180, −1), (−90, 0), (0, 1), (90, 0) and (180, −1). Now √ draw a horizontal line y = 3. You should see three points of intersection. 2 √ 3 These correspond to solutions to the equation cos(x)= 2 . What are the three values of x? (In general, for 0

–2– For −1

18. Descartes’ Example

Now, let’s get back to Descartes’ La Geometrie. As I said before, I found Descartes’ explanation to be difficult to digest, so let’s look at Descartes’ example, and then fig- ure out what he’s saying. OK. Here’s Descartes’ example. I’ll just refer to the translated version, but I will use the same letters that Descartes used for the points and segments (which he called lines,orlignes in French) [Descartes, p 26].

Let AB, AD, EF, GH, ... be any number of straight lines given in position, and let it be required to find a point C, from which straight 18. DESCARTES’ EXAMPLE 43

T

S R E AB G

H

F C

D

Figure 25. Redrawn copy of a figure from page 27 of The Geometry of Descartes (page 309 of original).

lines CB, CD, CF, CH, ... can be drawn, making given angles CBA, CDA, CFE, CHG, ... respectively, with the given lines, and such that the product of certain of them is equal to the product of the rest, or at least such that these two products shall have a given ratio, for this condition does not make the problem any more difficult. First, I suppose the thing done, and since so many lines are confus- ing, I may simplify matters by considering one of the given lines and one of those to be drawn (as, for example AB and BC) as the principal lines, to which I shall try to refer all the others.

As we try to understand this, let’s start with where this problem comes from. Descartes quotes the Greek mathematician Pappus (who lived around 300 A.D.). Pappus, in turn, talks about earlier Greek mathematicians Euclid (who lived around 300 B.C.) and Apollonius (who lived from about 260 B.C. to about 190 B.C.). He quotes Pappus (for some reason in Latin), and Smith and Latham give this translation in a footnote [Descartes, p 18].

Moreover, he (Apollonius) says that the problem of the locus related to three or four lines was not entirely solved by Euclid, and that neither he himself, nor any one else has been able to solve it completely, nor were 19. CURVES DESCRIBED IN TERMS OF TWO REFERENCE LINES 44

they able to add anything at all to those things which Euclid had writ- ten by means of the conic sections only which had been demonstrated before Euclid.

Descartes’ example and some snippy words from Pappus (which probably have lost, or perhaps gained, something in the translations) are seen a bit later [Descartes, p 21].

The problem of the locus related to three or four lines, about which he (Apollonius) boasts so proudly, giving no credit to the writer who has preceded him, is of this nature: If three straight lines are given in position, and if straight lines be drawn from one and the same point, making given angles with the three given lines; and if there be given the ratio of the rectangle contained by two of the lines so drawn to the square of the other, the point lies on a solid locus given in position, namely, one of the three conic sections. 2

19. Curves described in terms of two reference lines

The basic problem we’re trying to understand has a lot of things in it, and it’s hard to keep track of it all. Let’s step back and look at a simpler problem. That’s always a good idea. We wouldn’t use this language today, but we probably need some practice reading Descartes’ words.

D

AB Figure 26. We are given the two lines AB and AD, the angles at B and D, but not the positions of B and D on their lines.

2Note that Descartes mentions the “three conic sections.” Apollonius used these words υπρβoλη ((h)upsilon pi epsilon rho beta omicron lambda eta), λλιψις (epsilon lambda lambda epsilon iota psi iota sigma), and παραβoλη (pi alpha rho alpha beta omicron lambda eta). Apparently, these became hyperbola, ellipse, and parabola. 19. CURVES DESCRIBED IN TERMS OF TWO REFERENCE LINES 45

Question 1. Let AB and AD be two straight lines given in position, and let it be required to find a point C, from which straight lines CB and CD can be drawn, making given angles CBA and CDA with the given lines.

When Descartes says, let AB and AD be two straight lines given in position, he is meaning only that we are given two lines, and the points A, B, and D are somewhere on these lines. In some sense, the points can move as in our geometric machines. We want to find points A, B, and D that satisfy the requirements of the problem. One of these requirements is that A and B lie on one of the lines, and A and D lie on the other. The requirement that A lie on both lines forces it to be at the intersection of the two lines, and sliding the lines around doesn’t really change the location of A in a substantive way, so we can think of A as being fixed. Now, B and D can lie anywhere on their lines, but we are also given angles CBA and CDA. Descartes means for us to find the points B, C, and D so that these angles have a given measure. If you look at Figure 26, the two solid lines are fixed, the angle measures indicated are fixed, but the lengths of segments AB and AD are free to vary. If in Figure 26, we extend the two solid lines and the two dotted lines to infinity in both directions, we can put C anywhere in the plane by varying the positions of B and D. The Greeks had a coordinate system staring them in the face. Did they actually see it? I don’t know. They didn’t have an algebra as we know it, so even if they did see the coordinate system, I don’t think they had the ability to see the coordinate system in the way that we do today.

19.1. Exercises.

–1– In Figure 26, the dotted rays will intersect if extended long enough. This point of intersection will be the point C. Copy Figure 26 without the points B and C. Shade in the possible locations of the point C, if the lengths of AB and AD are allowed to range through all postive numbers. –2– If we allow the length of BC to be zero or negative (this would put C below the line AB), but require that DC be positive, then where can C be? 19. CURVES DESCRIBED IN TERMS OF TWO REFERENCE LINES 46

D C

AB Figure 27. We are looking for positions on the two reference lines for B and D so that CB = CD.

OK. We’ve got a picture like Figure 27, and we can slide the angles ∠ABC and ∠ADC along the two solid lines. This moves C around, and C can be anywhere. Now let’s add another condition to the problem, so now it looks like this.

Question 2. Let AB and AD be two straight lines given in position, and let it be required to find a point C, from which straight lines CB and CD can be drawn, making given angles CBA and CDA with the given lines, and such that CB and CD have a given ratio.

We now have a new given in the problem. Essentially, we are given a number k such that CB (73) = k. CD Let’s take the simplest case, and say that the given ratio is k = 1. This means that CB = CD. Figure 27 shows a position for B, D, and C corresponding to k =1. Imagine sliding B a bit to the right. This move alone, would make CD a bit longer, so to keep the ratio at k = 1, we would have to slide D up. With these constraints, the points B, C, and D are dependent on any one of the others, and we slide them around, C traces out a curve. If you really look at the picture, this curve might even be a straight line. Let’s investigate that. We know more about triangles than other shapes, so I’ve extended the figure a bit in Figure 28 to make triangles. We can interpret the problem as finding values for x and z that will make CB and CD equal to each other. The length of CB and CD will be a third variable y. Now, this isn’t the easiest problem to solve, but it should be reasonable to assume that you can solve it, but it isn’t so obvious how. If we know what y is supposed to be, then finding C is easy. Mark off a distance x from A to get B, and then mark 19. CURVES DESCRIBED IN TERMS OF TWO REFERENCE LINES 47 E u v D y C z y ABx Figure 28. We have added additional line segments CE and DE so that we can work with triangles. off the distance y to get C. This is essentially what Descartes uses for the general problem. Recall that in plane geometry, the angle sum of a triangle is always 180◦. Now in our problem, the lines AB and AD are given, so the measure of angle ∠BAD is also given. These are fixed. The measure of angles ∠ABC and ∠ADC are also given, so these are fixed, too. Since ∠BAD, ∠ABC, and ∠BEA are the three angles of a triangle, we must have

(74) ∠BAD + ∠ABC + ∠BEA = 180◦.

Therefore,

(75) ∠BEA = 180◦ − ∠BAD − ∠ABC.

This means that for any value of x and z, the measure of ∠BEA will always be the same, even though the position of point E will move around. In this same way, we can show that all the angles in the figure are fixed. Now once we see that all the angles stay the same as we slide the points around, the triangles in the figure remain similar. That is, as we find different solutions for C, Figure 28 gets bigger and smaller, but the triangle 4ABE always stays the same shape. In other words, all possible triangles 4ABE are similar to each other. The same goes for triangle 4DCE. Since the shape, or in other words the proportions, of triangle 4ABE stays the same, the ratios between any two sides in the figure stay constant. For example, the ratio

y + v (76) = q x 19. CURVES DESCRIBED IN TERMS OF TWO REFERENCE LINES 48

stays the same, even though x, y, and v will vary. The value q is fixed. Triangle DCE also maintains its shape, so v (77) = r y

stays the same, and the value r is fixed. These equations together give us y + ry (78) = q. x We can solve this for y to get q (79) y = x, 1+r and y is linear function of x, since q and r are constants. These constants are deter- mined by the given angles. In fact, this determination is captured by the sine function in the law of sines In Figure 28, let’s say that ∠BAD = α and ∠BEA = , then by the law of sines, y + v x (80) = . sin(α) sin()

With equation (76) therefore, the constant q would be

y + v sin() (81) q = = . x sin(α)

We can do something similar for r.

19.2. Exercises.

–1– Find a formula like equation (75) for ∠BCD that will show that this angle is also fixed. You should be able to get it down to something involving only numbers and the angles ∠BAD, ∠ABC, and ∠ADC, which are all given. –2– If ∠CDE = δ, find r in terms of δ and . –3– Suppose ∠BAD =30◦, ∠ABC = 120◦, and ∠ADC = 135◦. Find y in terms of x. 20. MORE COMPLEX CURVES 49

K N L C y B

E x

GA

Figure 29. A hyperbola drawing machine.

20. More complex curves

At the beginning of Book II in La Geometrie, Descartes says [Descartes, p 40-43]

The ancients were familiar with the fact that the problems of geometry may be divided into three classes, ... This is equivalent to saying that some problems require only circles and straight lines for their construc- tion, while others require a conic section and still others require more complex curves. ...It is not because the other instruments, being more complicated than the ruler and compasses, are therefore less accurate ...

Descartes goes on to categorize the conic sections and “more complex curves” in algebraic terms. For example, if the resulting equation “contains no term of higher degree than the rectangle of two unknown quantities, or the square of one, the curve belongs to the first and simplest class” [Descartes, p 48]. We would call these second degree equations. For geometric reasons, the second class of curves had third and fourth degree terms, and the third class had fifth and sixth degree terms. Most importantly, I think, these classes continued indefinitely. To get a flavor of how the non-linear curves can come about, let’s look at a simple machine that Descartes discusses in Book II. I have drawn a geometric figure similar to Descartes’ machine in Figure 29. This is different from the machines discussed previously in that the line GL is allowed to pivot at G and L. This will allow the curve traced by C to be something other than a straight line. The segment GA is 20. MORE COMPLEX CURVES 50

fixed, and the triangle 4NKL can slide up and down, but stays congruent as it moves. The letters in Figure 29 are the same ones used by Descartes. Although we’re probably more used to x being a horizontal measurement and y a vertical one, this is starting to look like what we do today. Descartes is after the relationship between y and x. Essentially, he does the following. The triangles 4NKL and 4CKB are similar, so LK BK (82) = . NL y Therefore, LK · y (83) BK = . NL Then, LK (84) BL = BK − LK = · y − LK, NL and LK (85) AL = x + BL = x + · y − LK. NL The triangles 4GLA and 4CLB are similar so AG y (86) = , AL BL or (87) BL · AG = AL · y. Substituting for BL and AL, we get LK LK (88)  · y − LK  · AG =  x + · y − LK  · y. NL NL Multiplying this out, we get LK · AG LK (89) · y − LK · AG = xy + · y2 − LK · y. NL NL Solving for y2, we get LK · AG LK (90) LK · y − xy + · y − LK · AG = · y2 NL NL NL (91) NL· y − · xy + AG · y − NL · AG = y2. LK 21. FINAL THOUGHTS ON ANALYTIC GEOMETRY 51

Descartes used AG = a, LK = b, and NL = c, and his final equation is cx (92) yy cy -- y + ay -- ac, b which is the same as the one we got [Descartes, p 54]. This equation is clearly of second degree, and the curve turns out to be a hyperbola.

21. Final thoughts on analytic geometry

We have only skimmed through about the first quarter of La Geometrie, and af- terwards, Descartes goes on to analyze more complicated curves and higher degree equations. Included in these discussions is Descartes’ Rule of Signs [Descartes,p 160] and constructing solutions to equations as the intersections of particular curves. Most of this is not related directly to the geometry we’re going to explore, but it illus- trates the power contained in the interplay between geometric figures and algebraic equations. This is the simple idea we’re going to keep coming back to. Overall, Descartes tells us that if we consider measurements along reference lines in a geometric figure, they behave according to simple algebraic relationships, and these algebraic relationships reflect fundamental geometric structures. Equations of a certain type (second degree equations, for example) group together certain geometric objects (the conic sections), and help us to see geometric relationships that are not obvious on their own. The geometry, on the other hand, can help us to see algebraic structures as well. Once we see that the solutions of an equation like 3x2 − 2x +7= 3x + 12 correspond to the intersections between a parabola y =3x2 − 2x + 7 and a line y =3x + 12, we know immediately what to expect in terms of solutions (0, 1, or 2 possible solutions). The everyday result of Descartes’ insights is our use of coordinates in analytic geometry. We typically describe the positions of points in terms of distances along reference lines (the x- and y-axes, for example), and geometric objects (lines, curves, surfaces, etc.) are described by equations. The terms cartesian plane and cartesian coordinates honor Descartes’ contributions. CHAPTER 3

Euclidean Geometry

As I’ve mentioned earlier, my opinion on Euclid is that The Elements is an im- portant early step in the development of modern mathematics, and the object he is studying, what we now call Euclidean geometry, is an important mathematical object. He lived around 300 B.C., however, and there were a lot of things he did not know. From this alone, it seems unlikely that his approach to Euclidean geometry is going to be the best one. I think it’s pretty clear that it isn’t, although many would disagree with me. What is perhaps most important is the idea that something like Euclidean geometry can be described starting with a few undefined terms (points, lines, etc.) and relationships between these. This is the basis of what is sometimes called synthetic geometry, the axiomatic approach to geometry, and axiom systems in general. Virtually every modern geometer, including Hilbert, identifies Euclidean (plane) geometry with something equivalent to the xy-plane you studied in calculus or high school geometry. We still work within an axiom system with a few undefined terms, but the undefined terms are in an area of mathematics called set theory. Essentially, element, set, and belongs to are the undefined terms, and there are axioms re- lating these concepts. Again, I want to stress that the modern axiom systems for set theory are ultimately the result of Euclid’s work. From these we can define the counting numbers, whole numbers, integers, rational numbers, etc. We can also de- fine operations like addition and multiplications in terms of set theory. This is all very interesting, but the important thing for us to know is that we can axiomatize the real numbers very nicely. From here, we can define Euclidean geometry, and now geometry is on the same solid footing as real analysis, modern algebra, topology, and all the other areas of modern mathematics. What we’re going to do is to look at an analytic geometric approach to Euclidean geometry and compare it to Euclid’s geometry. This is not standard terminology, but for us, the logical system of axioms and propositions in The Elements will be

52