Notes — Chapter 11: The Right Triangle
To the Student. As exercises in the Chapter 10 show, you can draw a reasonable map using a plane table without any calculations. But to map a township, state, or nation, you need measurements giving precise numbers. Similar right triangles and trigonometry are the tools for this job. They let you calculate lengths that can’t be measured directly, for example, the height of Mount Everest.
The Pythagorean Theorem’s a² + b² = c² lets us find the length of the third side of a right triangle given the length of two of the sides. Trigonometry extends these ideas to other triangles by using the Laws of Sines and Cosines (Lesson 7), allowing us to “solve” triangles, working out all sides and angles.
To apply these ideas, you need to be able to see right triangles in problems and to understand slopes. Many problems can be solved using special triangles: isosceles right triangles or 30°, 60°, 90° triangles. Learn to look for these special triangles, too. “
The altitude to the hypotenuse divides a right triangle into similar triangles, both similar to the original. This has many applications, explored in Lesson 1. In Lesson 2, we use this observation to prove the Pythagorean Theorem. We apply similarity and the Pythagorean Theorem in this lesson’s exercises. Lesson 3 looks at special triangles and their applications. We come to the trigonometric ratios in Lessons 4 and 5. In Lesson 6,
we study slope and its uses, returning to trigonometry in Lesson 7 on the Laws of Sines and Cosines.
This chapter shows how to solve triangles using trigonometry. Key skills are: Being able to identify similar right triangles in the figure of an application. Seeing how to apply the Pythagorean Theorem in these situations. Recognizing and using special triangles. Knowing the definitions and uses of sines, cosines, and tangents. Knowing that more general triangles can be solved using the Laws of Sines and Cosines. (Not all programs will hold students responsible for these laws.)
ll” Right Triangles, Maps, and Coordinates. Numbers rule in our age. Every point on the surface of the earth has a latitude and longitude (measured relative to the geodetic, geocentric, or astronomical systems). A GPS unit can tell you where you are within a few meters, and advanced units have an accuracy to within a few centimeters. US topographical maps show a Universal Transverse Mercator (UTM) grid, and a GPS will give you your coordinates relative to this.
Maps are flat and the earth is curved, even somewhat rough. To make a flat map of a curved thing, we must relate points in the landscape to points on an ideal surface (the Clark spheroid or variants are used). This is the surface of an ideal sea that can be adjusted to better match local terrain to reflect the bulge at the equator and flattening at the poles. Such a surface is called a datum. By looking at the grid ticks showing overlapping UTM grid patches on a map, you can see the distortions that are inevitable when we try to impose a square grid on a curved surface.
The refinements in mapping are many. From its beginning, geometry gave us tools to map farmer’s plots and to think about the shape of the earth. Recall Lesson 2 of Chapter 1, with Aristarchus’ estimate of the earth’s circumference and Lesson 5, with the Egyptian assessor’s formula. Geometry is about numbers as well as shapes, and its trigonometrical side goes back to the Greek astronomer Hipparchus (about 161-126 B.C.).
Lesson 1. Proportions in a Right Triangle. The nautilus shell has a self-similar spiral. The shell in the photograph is mapped onto itself by a counterclockwise rotation of 180° combined with a dilation by a factor of 1.75 about its center. Self-similar spirals are also called logarithmic spirals, and each cuts every ray from its center at a constant angle characteristic of that spiral. The closer this angle is to 90°, the closer the successive turns of the spiral are. (What if the angle is 90°?)
The figures at the top of page 429 show that the altitude to the hypotenuse of a right triangle splits the triangle into two triangles, both similar to each other and to the original triangle. This is easy to see. The three proportions that follow from this need to be studied and applied in different situations until using them becomes automatic. In the figures on this page, the segment that
is the geometric mean is always shown in red. These figures deserve careful study.
Set I. Chameleon Tail (1-7) builds on a second selfsimilar spiral. In Shadows (8-14) the idea of a projection is related to the projection of shadow by a light source; this helps give meaning to projections. These exercises use perpendicular projection. If the sun is not overhead, we get parallel projection, also a useful idea but not needed here. Find the Lengths (15-19), these provide practice with these proportions. Turn about is fair play, so teachers might ask their students to make up three problems, one using each of these proportionalities, and then collect these and use some in a quiz. Exercises 20-30 combine numerical calculation with geometric insight. The final two questions in this series can be answered using the numerical results in hand. Is there a way to see that AD/DB = (AC/CB)? in a flash of geometric intuition? Yes: AD/DB = [Area(△ACD)]/[Area(△CBD)] = (AC/CB)?, because these triangles are similar (△ACD ∼ △CBD). The ratios of their areas is equal to the square of the ratios of any pair of corresponding sides. The sides used here are the hypotenuses.
Rectangle and Square (31). When we reach Chapter 12, Circles, we will see that this observation gives us a way to construct the geometric mean of lengths a and 6 and thus to “Square a Rectangle” with sides a and 6, getting a square of side Vab having the same area (Euclid VI.13). This uses the result that a triangle inscribed on a semicircle with the diameter as its long side is always a right triangle. See also the rectangle inscribed in a circle in Exercises 32-37. This fact about angles inscribed in semicircles is Corollary 2 of Theorem 63 in Chapter 12.
Set II. Strongest Beam (32-34). In this sequence both exact and approximate calculations are used.
Zig Zag (38-45). What we have here is a series of triangles and segments that form various geometric series. Following the steps a student will be able to work all of this out easily, but there is another way to think about this.
▶ Zig Zag, Similarity, & Geometric Series. Consider the similarity T consisting of reflection in the bisector of ∠AOB followed by a dilation of magnitude r about O sends T(△AOB) = △BOC, T(△BOC) = △COD, and so on. When T is repeated to get T² we have a dilation of magnitude r² about O. Therefore, T?(AC) = CE and T² (CE): EG:
Draw the bisector of ∠AOB and let it intersect AB at Oi, BC at Og, and so on until it intersects FG at Og. The dilation of magnitude r about O sends Oto: Opes The similarities we are considering all leave O fixed and send this bisector onto itself. Looking only at points on this bisector, our mapping behaves as if it were a
dilation of magnitude r about the fixed point O. Using similarity, we can find S, the lengths of the segments:
(*) Su O,Oo ap 0203
- 0304 + O4O5 + O50¢6
= 1(0102) =F r(O,O²) ar r?(O10)
- 7°(0102) + r4(0102)
0h = ri ar r² ap r² ae r²)(O1Oo).
Looking at S another way, we get
S =0; 0; = OO.6 — OO;
= 7°(00;) =1(00;) = (7° — 1)(OO;).
Now use (O;02) = (OO2) — (O01) = r(OO;) 1(00,) = (r−1)(OO;) to replace OO; by Ga (0102) and get
(r= 1) (*x) S= (0102).
aia)
Equating the final expressions for S in (*) and (**), cancelling the (0,02) factors on both sides, and generalizing from 5 to n gives
ae = 1)
ee tr? toe ir) =
(r — 1)
This is a geometric derivation of the formula for the sum for a finite geometric series. This sum is calculated in Proposition 35 of Book IX of Euclid.
Using algebra we can get this result more easily than by Euclid’s geometric methods. For nice geometric proofs of this result, see Proof Without Words by Roger B. Nelson, Mathematical Association of America, washington, DC, 1993, pages 119 and 120. More generally, you will find many wonderful geometric derivations of algebraic results in the Nelson book and in The Book of Numbers by John H. Conway and Richard K. Guy, Springer-Verlag New York, 1996, especially in Chapter 2, “Figures from Figures: Doing Arithmetic and Algebra by Geometry.”
Set III. The Golden Rectangle exploration combines construction and calculation. The Golden Number or Golden Section and the Golden Rectangle are sources of endless study and speculation. Project: Look up material on the Golden Section in the library or on the Web. As a ratio, it goes by the names of Phi (¢) and Tau (7) and is the positive root of the equation eg Lei), It satisfies x = 1+ i, which suggests how it can be written as a continued fraction. See Chapter 14, Lesson 2, Set III notes. It is the limit of the quotients of successive Fibonacci numbers. See also Euclid, Proposition 30, Book VI.
Lesson 2. The Pythagorean Theorem Revisited. Geometry is a coherent whole, and argument by which we derive the Pythagorean Theorem by using proportions shows us one of the links binding the subject together.
In this lesson, we use proportions and c² = a² + 6? to work out lengths in problems. Note that many of the exercises in Sets I and II build familiarity with standard or easily computed right triangles, including Pythagorean Triples, Exercises 18 —-25. There are several ways to get all Pythagorean Triples. One is to consider all triples of the form 2k, k² + 2, k² — ℓ2, where k and # are distinct integers.
There is a wealth of material on Pythagorean Triples. Researching and reporting on them would be a good project. Another interesting project would be to research some of the ancient units referred to here and elsewhere in this book. In Geometry Civilized, Oxford University Press, Oxford, United Kingdom, 1998, John Heilbron wrote on page 153 that the “sacred cubit” is 25.025 inches long. By e-mail, Heilbron comments that this three-place precision was a joke. We really do not know exactly how long many of these units were, but it is a good guess that a cubit was about the length of a forearm. But whose forearm and how was it measured? One ch’ih is about a foot. An interesting resource on measurements is Measure for Measure by Richard A. Young and Thomas J. Glover, Sequoia Publishing, Littleton, Colorado, 1996, which gives conversion factors for all manner of measurements.
Distance Formula (28-30) uses coordinate methods as does Grid Exercise (37-41). The Set II Oil Well problem (47-50) has a surprisingly nice solution. The numbers involved will be simplified if students work in thousands of feet, but a calculator will do the job without this simplification.
Applications in the exercises includes figuring the sag in a suspension bridge, working out the turning radius of a car, and relativistic length contractions.
Set III is a gentle exploration of Fermat’s Last Theorem (FLT). Students who want to learn more can go to the Web. Try Eric Weisstein’s site mathworld.wolfram.com/FermatsLastTheorem.html. Weisstein notes that Marylin vos Savant’s book on FLT states that the solution of the problem by Andrew Wiles and Richard Taylor is faulty. Vos Savant faults Wiles and Taylor for using the theory of elliptic curves in their work. She identified elliptic curves with a type of non-Euclidean geometry (specifically, elliptic geometry, the Lobachevskian geometry we study in Chapter 16). Since vos Savant questions the validity of non-Euclidean geometry, she questioned Wiles and Taylor’s proof. But the elliptic curves of their work have nothing to do with elliptic geometry, and a proof based on non-Euclidean geometry is as correct as a proof using other.
Lesson 3. Isosceles and 30°-60° Right Triangles. Pythagorean triples and the special triangles of this lesson give us all of the examples that students can work out in exact algebraic form. These are the basis for all examples that can be handled easily without a calculator. More importantly for the development of geometry and trigonometry, special triangles kicked off an era of trigonometrical calculation that began over 3700 years ago, as the lead example of this lesson shows. I give a brief sketch of the development of such calculations in “To learn how to calculate better” found at the end of the notes for this chapter. These efforts gave us today’s scientific calculator — a boon for us and for students.
Even today it is important that students be familiar with these special triangles. This familiarity is the concrete basis for the trigonometric ideas that come in Lessons 4, 5, and 7. These special triangles often show up on standardized tests.
Set I gives students practice with special triangles and their applications. Set II. Not Quite Right (34-41) asks students to compare approximate and exact answers involving special triangles. More generally, there are many places in these exercises where students may get answers either in exact form or as decimal approximations. Class discussion of which form is most appropriate should help. The key is understanding what is being asked for or what is needed to proceed practically. If you want a diagonal brace for a square framework made of two-by-fours, you don’t need the exact answer of /2 meters, or even a decimal answer to four places. The best approach would be to lay the frame of two-by-fours on a pair of sawhorses, square it up, and cut the brace to fit. No calculation would be needed.
Trigonometric Angles (42-47) shows how ideas about angles are generalized to angles larger that 90°. Calculators are able to deal with a wide range of angles, expressed in degrees or radians. These exercises open the door to exploration of these features on the calculator and give access to material beyond the trigonometry of acute angle triangles that is the focus here.
Construction Exercise (48-52) involves construction, measuring, and exact and approximate calculations. Area Problem (53-56) involves problem solving and exact and approximate calculation.
Moscow Challenge, Exercise 57, requires thinking outside the box. If a student sees how to do it it, it is easy; otherwise, it isn’t.
Set III. A Big Mistake, invites students to think critically and nail a glaring error. This exercise is within reach of most students.
Lesson 4. The Tangent Ratio. Indirect measurement is a major use of trigonometry, and finding the height of Mount Everest is a striking example of this use. The Great Arc by John Keay, footnoted in this
lesson, outlines the work of the Great Trigonometrical Survey. Everest was just one height determined by this survey. See page 472 of Geometry, Triangulation, for more or look up the Great Trigonometrical Survey on the Web.
The calculator is essential here. In the exercises it works along with exact calculations and with what we learned earlier about special triangles. See Set I Exercises 17- 28. Note the first steps toward solving triangles in Exercises 29-34.
Set II. Fly Ball (35-38) shows that a baseball appears to rise steadily as it comes to a fielder. This seems surprising. If you check sites on the physics of baseball, you should find a Java applet that gives a more realistic and detailed picture. If you use the usual simple model for the flight of the ball, ignoring air resistance, you can find the ball’s trajectory. If te trajectory is projected into a plane perpendicular to the line from the batter to the fielder and passing through the batter, it appears as uniform linear upward motion. The calculation of this motion is the same as the calculation of the slope of the tangent as the limit of a secants passing through a point on a parabola. This latter calculation is usually done in the first weeks of a calculus class.
The lighthouse of Exercises 39-43 is the Pharos. See the map on page 8 of Geometry. Grid Triangle (48- 52) compares exact and approximate results. Students will see that exact calculations are needed to prove you have a right triangle, although sufficiently precise approximate ones suffice to show when an angle is not a right angle.
Set ITI. What Time Is It? This is an exercise in photographic interpretation. We are given that the camel’s shadows are three times their heights, and this gives the tangent of the sun’s elevation. Given information on sunrise and where the sun will be at noon, we can find the time the photograph was taken. But how could we find out that the shadows are three times as long as the camels are high? To do this we must know something about camels, namely that a full-grown camel is almost exactly as long, head-to-tail, as it is tall, hoofto-hump. Checking side views of camels in other books shows that this is about right.
Picking a camel in the lower left and one in the upper right that are fairly separate from the others and measuring their lengths and the lengths of their shadows in printer’s points, I got:
shadow length = 36 points camel length = 14 points tan = 0.3888 angle = 21.25° time after sunrise =( angle/90) x 6 hours + 6:00 am = 7:25 am
and
shadow length = 38 points
camel length = 16 points tan=“0/42 angle = 22.83° time after sunrise =( angle/90) x 6 hours + 6:00 am = 7:31 am
These measurements and calculations are in close agreement with the assumptions given in the exercise. Students might be asked how they could check whether the camel’s shadows were three times the camel’s height. If we know that this is morning, we can also figure out the compass bearing the camels are walking along. See page 661 for another photographic interpretation exercise.
Lesson 5. The Sine and Cosine Ratios. The glide path of a flying squirrel introduces this lesson on sines and cosines. These are introduced for acute triangles, but Sines from a Circle (23-35) puts sine and cosine in the context of points on a quarter circle. These exercises are based on work of Bashkara, in 7th century India. Extremely accurate trigonometric methods were developed by astronomers in the southern Indian region of Kerala in the 15th century. See the outline of the trigonometric calculations following the notes on this chapter. If a coordinate system is set up with point P in the figure on page 457 as the origin and using centimeters for the scales on both axes, the points on the circle shown have coordinates (10 cos θ, 10sin 6), where the angle θ is measured counterclockwise from the positive x-axis.
The Set I exercises for this lesson focus on using the sine and cosine and working from sines and cosines back to angles. Finding Lengths and Angles (23-28) takes us further toward solving triangles. The calculator is a key tool, and it is used to get approximate values for the angles in Pythagorean Triangles (13-22).
Set II turns to applications. Note the vector ideas in Force Components, (41 and 42). SAS Area (54 and 55) also has wide use: aAABC = SabsinC = LICK x CBI), where the final expression is half the length of the cross product of the vectors CA and CB. See, for example, Eric Weisstein’s World of Mathematics at mathworld.wolfram.com on vector operations. See also “Dot Products, Rotations, Matrices” in the supplementary material for Chapter 11 in these notes.
Set III. Star Distance. If a student use a calculators to find the answer to this question, they may give the distance as 25,240,298.6593 million miles. This is computational accuracy beyond the accuracy of the data. If the calculations are done in scientific notation, as Jacobs has them, you get an answer that is correct, more easily understood, and gives the right sense of accuracy. Assuming thata parallax of 0.76 seconds implies the figure is correct to +0.005 seconds, then the distance lies between 25.075 and 25.407 trillion miles or 4.27 and 4,32 light-years.
An interesting source on parallax measurements is Parallaz: The Race to Measure the Cosmos by Alan W. Hirshfeld, W. H. Freeman and Company, New York, 2001. The Hipparcos satellite (High Precision Parallax Collecting Satellite) launched in 1989 measured parallax to an average accuracy of a thousandth of an arc second. A satellite planned for launch in 2009, GAIA, is set to measure parallaxes down to 1/5,000,000ths of an arc second.
Lesson 6. Slope. The basic definition of slope and the relationships between slopes of parallel and perpendicular lines are the key mathematical ideas of this lesson. An intuitive feeling for meaning of slopes in physical situations and applications is the key nonmathematical subject.
The lesson begins with the slope of a line 2
Rise _ y—m
Sl ope o f f= — Ban = tainvA,
2 — LX
where the line ℓ passes through points P = (xj, y,) and Q = (x²,y²). Angle A is the angle of inclination of ℓ. On a calculator ∠A is the arctangent of the slope of ℓ, which may be found using the “atan” or “tan7!” buttons on the calculator. Often this requires using a second function button.
Geometrically, ∠A is the angle that 2 makes with the x-axis. Angle A is measured by starting at the intersection X of @ with the z-axis, and finding the counterclockwise rotation about X needed to bring the ray running from X in the increasing direction on the zaxis into line with @. For example, the line ℓ with equation y = —x has slope −1. This line intersects the x-axis at X = the origin = (0,0). My calculator gives atan(−1) = −45°, so ∠A = −45°. This means that the positive x-axis must be turned through a counterclockwise angle of −45° (that is, backwards 45°) about X the to bring it into line with 2. A clockwise rotation of 135° leaves the line at the same inclination as a counterclockwise rotation by −45°, because two rotations about X that differ by 180° result in the same inclination. Depending on the conventions we adopt, we may think of the inclination of this line as either −45° counterclockwise or 135° counterclockwise. (But keep in mind that for rotations −45° counterclockwise is equal to 45° clockwise.) Calculators give us angles of inclination that range from −90° to +90°, and for ease of use, we should stick to these. Students should see how slopes and angles of inclination are related by experimenting with their calculators as well as by looking at different lines in a coordinate system.
Theorem 52 tells us that nonvertical lines are parallel precisely if they have the same slope. Intuitively, the slope is the same if and only if the angle of inclination is the same, so either both lines are parallel to the z-axis or they both make the same angle with the z-axis. In
either case, they are parallel if and only if they have the same slope. Exercises 52-59 prove the forward implication based on similarity. To prove the converse, show that if my; = m², then △ABC ∼ △DEF, so ∠1 = 22, and so ℓ, ∥ ℓ9.
Theorem 53 tells us that two nonvertical lines are perpendicular if and only if the product of their slopes is −1. The coordinate example on page 463 can be generalized to prove this. A conventional proof of the forward implication is given in Theorem 53 (60-65). The converse is handled like that of Theorem 52.
The Point-Slope Equation of a Line (42-46), is another useful result found in this lesson. See the Algebra Review “Graphing Linear Equations” on pages 481 and 482 for slope-intercept form of a line and related material.
It will seem intuitive to students that lines have linear equations. This fact is so “blazingly obvious” that students are impatient if you take the time to prove it. This phenomenon was noted by Keith Henderson, and the phrase is his. The fact that lines have linear equations is easy to establish by arguments using similar triangles.
Set I exercises give practice with the basis ideas, Set II cover more applications and proofs, and Set III tells us something interesting about temperature scales. Set III is within reach of most students.
Lesson 7. The Laws of Sines and Cosines. Label AABC so that side a is opposite vertex A, side b is opposite vertex B, and side c is opposite vertex C. Then
sinA _ sinB sin C 9
= and c² = a² +b² — abcosC.
ti eG ‘6
The first set of equalities is the Law of Sines, and the second is the Law of Cosines. As you see in this lesson, the Law of Sines boils down to the observation that you get the same area for a triangle, no matter which pair of sides and included angle you use to find that area. See the SAS Area exercises on page 459.
The Law of Cosines is an extension of the Pythagorean Theorem that holds for arbitrary triangles. It is proved here for the acute case, but with such identities involving trigonometrical functions, if they hold over one range of variables, then they are universally valid. This is a general principle that students should find reasonable. We have looked at sines, cosines, and tangents for acute angles, but a calculator will give us the values of these functions for arbitrary angles. Experiment suggests that the formulas we use work for angles in general, and so they do.
This lesson gives us the tools needed to solve triangles in general. The Set I exercises cover the key ideas. Calculators are essential in working out answers.
Set II exercises include applications such as land surveying and triangulation (25-30) and triangulation extending to the moon (31-33). Note that although we know that side BM in the Distance to the Moon exercises is longer than side AM because BM is opposite the larger angle, this does not show up numerically, at least not on my calculator. Likewise, the Set III exercises are something of a challenge in precision calculation. Jacobs has a nice discussion of these in his answer section.
Chapter 11: Summary and Review
In addition to the results listed here, teachers may want to include some of the results on graphing lines from the Algebra Review on pages 473 and 474.
The cuneiform tablet in Number Mystery (65-66) is known as Plimpton 322. It has been extensively studied and analyzed. Down the years authors have concluded that it shows that Pythagorean Triples were known long before Pythagoras. But recent articles suggest other interpretations. Search the Web for the latest under Plimpton 322 and Eleanor Robson, of Oxford University, whose article “Words and Pictures: New light on Plimpton 322” in the February 2002 American Mathematical Monthly argues that this tablet was used as an aid for Babylonian mathematics teachers to create exercises that came out with simple answers.
▶ Similarity and Non-Euclidean Geometry. Similarities are unique to Euclidean geometry. In Euclidean geometry we can pick a center O and transform every shape in the plane by a dilation T of magnitude λ about O. This dilation about O has a simple expression if we use a Cartesian coordinate system with O as its origin. Expressed in such coordinates, T sends P = (z, y, z) to the point T(P) = P’ = (λx, λy, λz). Similarly, we write Q’ for T(Q). Using the coordinate from of this dilation or geometry, you can see that the distance between P’ and Q’ will be times the distance between P and ©;
Because the surface of a sphere is curved, geometry on the surface of a sphere is not Euclidean. The closest we can come to a straight line on a sphere is a great circle, the intersection of a plane passing through the center of the sphere with the surface of the sphere. The path of a hockey puck sliding without friction on a spherical planet will be a great circle. A spherical triangle is formed by arcs from three great circles. The Ruler Postulate must be modified in order to fit the geometry of the sphere because the lines of this geometry are great circles and have finite length and curve back on themselves. As you will see in Chapter 16, Non-Euclidean Geometries, spherical triangles have the ASA, SAS, and SSS congruence properties of Euclidean triangles. But there is a difference in behavior with respect to the sum of the angles in a triangle.
Spherical triangles with the same angle sum have the same area. T’wo spherical triangles whose corresponding angles are equal must be congruent; this is the AAA
congruence theorem for triangles on a given sphere. Similar triangles at different scales can’t exist on a given sphere. The spherical excess of a triangle is defined as the sum of its angles minus a straight angle. Spherical excess is proportional to area for triangles on a - given sphere. For general information on spherical triangles, see Eric Weisstein’s World of Mathematics at mathworld.wolfram.com. For areas of spherical triangles, read on in these notes.
Besides the geometry of the sphere, there is a second sort of non-Euclidean geometry, the geometry explored by Saccheri, Lobachevsky, Bolyai, and Gauss. See Lessons 2-4 of Chapter 16. This Lobachevskian geometry is like the geometry on a sphere — a sphere whose radius is an imaginary number. Johann Heinrich Lambert made this observation in his work on the parallel postulate published in 1786. See Morris Kline’s Mathematical Thought from Ancient to Modern Times, Oxford University Press, 1972, page 868 and pages 875-877, where you will see how using imaginary values for the radius gives a trigonometry that fits with Lobachevskian geometry.
The sum of the measures of the angles of a triangle in Lobachevskian geometry is always less than a straight angle. The area of a triangle is proportional to its Lobachevskian deficit, defined as 180° minus the angle sum. As in spherical geometry, we have the AAA Congruence Theorem and the congruence theorems of Euclidean geometry. In Lobachevskian geometry as in spherical geometry, two triangles cannot be similar without being congruent.
It may seem odd that the areas of triangles should be related to angular excesses or deficits in non-Euclidean geometry. But here is a way to see why this must be the case. If you draw a triangle and divide it into subtriangles in any way, you will see that these deficits (or excesses) add up so that the deficit (or excess) for a large triangle is always the sum of those for smaller triangles into which it can be decomposed, no matter how this decomposition is done. This is exactly the property we expect for areas, so it is not surprising that the angular deficit (or excess) of a triangle is a measure of its area in non-Euclidean geometry. The reason is that the standard non-Euclidean geometries are spaces of constant curvature. The spherical excess of a spherical triangle measures the total curvature over the area the triangle encloses. The deficit in a Lobachevskian triangle should be thought of as the total negative curvature of the interior of that triangle. Lobachevskian geometry is just the geometry of a surface with constant negative curvature.
lt Areas of Spherical Triangles. ‘The areas of spherical triangles can be worked out by methods independent of calculus. To do this we need to think of a spherical triangle T as the intersection of three hemispheres. We
also need the Principle of Inclusion and Exclusion explained below. This is a way to count objects in a union of sets or find the area of the union of these sets.
Let the notation #X stand for the number of things in the set X. The number of things in the union of a family of sets is given by the alternating sum: (sum of the number of things in the sets taken one at a time) — (sum of the number of things in the pairwise intersections of the sets) + (sum of the numbers of things in the intersections of the sets taken three at a time) —–., and so on. For example, for three sets S, U, and V, this gives us #(SUUUV) = [#8 + #U + #V] —- [(#(SN U)+#(UNV)+#(VNS)|+#(SNUNYV). This is the Principle of Inclusion and Exclusion, and it works for figuring areas as well as for counting elements, provided that all the numbers involved are finite.
Using the Principle of Inclusion and Exclusion, the area covered by the three hemispheres whose intersection is T is given by the sum of the areas of these three hemispheres, minus the areas of their pairwise intersections, plus the area of T, which is their intersection taken three at a time. If two hemispheres overlap forming angle A radians on a sphere of radius R, their area is 2AR?. The sphere has area 47R?. The union of the three hemispheres leaves the triangle T’ antipodal to T uncovered. Note that T and T’ are congruent. We denote their common area by A. The area of the sphere is the area covered by the three hemispheres plus the area of T’. Algebraically, 4rR² = (Area Covered by the Three Hemispheres) + Area of T’ = (27R? + 27R? + 27 R² — 2AR? — 2BR? −2CR² + A) +A, where the angles of T are A, B, and C in radians. Solving for the area A of T gives
Area of T= A= R?[((A+B+C) —q].
This result is known as Girard’s Theorem. You can find this proof in John C. Polking’s Geometry on the Sphere, available on the Web with pictures at mathforum.org in 2003. A nice proof is also found in both editions of Jeffrey Weeks’s book The Shape of Space, Marcel Dekker, New York, 1985, with a Second Edition in 2001. “
▶ Flatness. Euclidean geometry is the geometry of a flat surface, one whose curvature is zero. Hence the sum of the angles in a triangle is always 180° and the total curvature of the area inside the triangle is always 180° — 180° = 0 (or z in radians). This conclusion is equivalent to the Angle Sum Theorem for triangles.
▶ Equilateral Triangles. Let’s see how equilateral triangles behave in the spherical, Lobachevskian, and Euclidean geometries. If you look at larger and larger equilateral (and hence equiangular) triangles centered on the north pole, you will see that the small ones have an angle sum close to 180°, while those which are close to the equator have an angle sum close to 3 x 180°. There is a limit to the angle sum for spherical triangles and for the area of such triangles. None has an
area greater than that of a hemisphere. Similarly, in Lobachevskian geometry, all triangles have a positive angle sum, although this sum is always less than 180°. The limiting area for a triangle in Lobachevskian geometry is the area that would be associated with an asymptotic triangle having angle sum 0°. Only in Euclidean geometry can we have equilateral triangles of arbitrarily large area, and in this case the sum of the angles is always 180°.
▶ Equilateral Triangles and Uniformity of Curvature. The geometries we consider all have the usual congruence properties for triangles: SAS, ASA, SSS. Let us fix our unit length so that we can speak of segments of length s. Because of the congruence properties we assume, an equilateral triangle of side s has angles and an area that are determined by s and are independent of where the triangle is or how it is oriented. The angular deficit or excess depends only on s. The curvature of the surface within such a triangle is the same wherever the triangle is, and hence the curvature must be constant throughout the plane. This is geometrically clear for a sphere, whose curvature is the reciprocal of the square of the radius of the sphere. (For an n-sphere the curvature is r~”, where r is the sphere’s radius.) The Lobachevskian case of negative curvature cannot be visualized in such a simple way, but the same uniformity of curvature holds for it.
lt” Angle Sums: Subtriangles and Supertriangles. The congruence postulates imply that, whether it is positive or negative, the curvature must be same everywhere in the plane. Suppose the Angle Sum Theorem (AST) holds for some triangle T. Divide T into subtriangles. If you draw a diagram, you will see that the angular excess or deficit for T is the sum of the excesses or deficits for its subtriangles. If any of these excesses were positive (or negative), then they all are positive (or negative) and the excess for T would be positive (negative) contrary to AST for T. So if T satisfies AST, then so do all subtriangles of T. Putting four copies of T together, you can create new triangles with sides twice the lengths of the original, and this supertriangle will also have angular deficit or excess equal to 0. If a triangle T’ is congruent to some triangle contained in T, then T’ satisfies the Angle Sum Theorem. Otherwise, T’ will be congruent to some triangle inside some similar enlarged version of T created by successive doublings of T. By this we see that if any T satisfies the AST, then so does any other triangle T’.
▶ Similarity and the Angle Sum Theorem. Suppose we have two triangles △ABC ≅ AA‘B/C’, where the unprimed triangle has the larger sides. We may assume that the smaller triangle has been copied so that A and A’ coincide and B’ lies on AB and C’ lies on AC. Then the quadrilateral B’BCC’ must have angle sum 360°. Drawing the diagram and using the fact that the angle sum must be consistently greater or equal to 180° or less than or equal to 180°, you will see that ABB’/C and
ACC’B’ must both satisfy the Angle Sum Theorem. So if we have any pair of similar triangles that are not congruent, then every triangle satisfies the Angle Sum Theorem.
John Wallis (1616-1703) was a British mathematician who deciphered coded messages for the Cromwell government and for Charles II’s government after the Restoration. He proposed replacing the Euclidean parallel postulate with the postulate that given any △ABC and any segment EF, there is △EFG with △ABC ∼ △EFG. Wallis favored his postulate over Euclid’s or Playfair’s because his does not involve lines being prolonged arbitrarily. Wallis’s assumption is stronger than the existence of just one pair of noncongruent but similar triangles. We have shown above how the existence of just one such pair of triangles implies that the Angle Sum Theorem holds. As we will see below, the Angle Sum Theorem implies the usual Parallel Postulate. So Wallis’s postulate is stronger than is needed to derive the Parallel Postulate uding our reasoning. Wallis needed his stronger assumptions because he used simpler arguments, ones not based on ideas of curvature of space.
▶ Parallels, Pythagoras, and Similarity. The existence of similar figures is a distinguishing feature of Euclidean geometry. It is equivalent to the Parallel Postulate. With the Parallel Postulate, we were able to prove the Pythagorean Theorem. This reasoning can be reversed. If the Pythagorean Theorem is true, then we can construct an isosceles right triangle △ABC with hypotenuse BC. Say its legs have length a, so its hypotenuse has length a². The right angle is at A, and we let the hypotenuse be bisected by point A’. The segment AA’ divides △ABC into triangles △AA′C ≅ △AA′B, by SAS. These right triangles have hypotenuses of length a. Because one leg of each has length a//2, the Pythagorean Theorem ensures that the other leg also has length a/√2, hence they are isosceles.
The base angles of these triangles must then be 45°, because one pair of these four equal base angles adds up to give the right angle of the original right triangle. Hence the original triangle is a 45°-45°-90° triangle, and thus satisfies the Angle Sum Theorem. Now that we have one triangle T = △ABC that satisfies the Angle Sum Theorem, we know that all triangles must satisfy this theorem. tl
lt Angle Sum Theorem and Parallel Postulate. Now we show that if the Angle Sum Theorem is true, then the Parallel Postulate must hold. Given a line ℓ and a point P not on é, we drop the perpendicular k from P to @, meeting @ at some point we call O. Then the line m through P and perpendicular to the 2 must be parallel to ℓ. (See Construction 7, page 225.) We must show that any line n other than n and passing through P must cross 2.
Draw the diagram of ℓ, m, O, and P so that @ has the same directions as the bottom edge of the paper. Marking the points at distance OP from O on @ and from P on m, you get two squares that share OP as an edge. It is the Angle Sum Theorem that lets you prove that these * quadrilaterals are equilateral rectangles. If n is different from m, then it must slant downward toward @ as it enters one of these squares. Having entered that square, the line n must leave it. If n does this by crossing 2, then n is not parallel to and we are done. Otherwise n must cross the far side of this square at some distance 6 below line m. We continue our row of squares between lines ℓ2 and m and follow the line n in the same direction. Each time it crosses the far boundary of one of these squares, line n must be closer to line ℓ by distance 6. Because 6 is positive, eventually within one of these squares, n must cross from the upper side of @ to the lower side of ℓ, and hence n and ℓ cannot be parallel. “
▶ On Pythagoras, Angle Sums, and Parallels. The Pythagorean Theorem implies the Angle Sum Theorem, which implies the Parallel Postulate, and the existence of similar but noncongruent triangles implies the Angle Sum Theorem. We already know that the Parallel Postulate implies the Pythagorean Theorem, so each of these properties for a geometry implies the others. The possibility of finding similar but noncongruent figures is the distinguishing feature of Euclidean geometry, as contrasted to the non-Euclidean geometries discussed in Chapter 16.
▶ To Learn How to Calculate Better. In a conversation with Andrew M. Gleason about his suggestions on the draft of this edition of Geometry, we asked ourselves what we wanted students to understand. We were asking, “What is the purpose of mathematics?” One answer Andy gave was “to learn how to calculate better.” This idea is a shock when courses emphasize “understanding” and try to avoid getting bogged down with standard algorithms and fussy details. The importance of calculation seems particularly clear in trigonometry.
The definitions of trigonometric functions based on simple geometry, right triangles, and the Pythagorean relationship, sin? + cos? = 1, will let students work out the sides of various special triangles and the sines, cosines, and tangents of special angles. Knowing these things and having a calculator is enough for most practical purposes. More complicated triangle problems can be solved using the Laws of Sines and Cosines, if this material is covered.
The addition formulas are important for understanding how rotations are represented by matrices and for interpreting de Moivre’s formula, e² — cos@+isin§. These addition formulas are used to work out the derivatives of the trigonometric functions in calculus. But when trigonometry was young, the addition formulas and the half-angle formulas were used for calculation.
Hipparchus of Niczea (about 161-126 B.C.) was the first to work out an accurate table of trigonometric ratios and is considered the father of trigonometry. (See for instance Isaac Asimov’s Biographical Encyclopedia of Science and Technology, more details can be found in The Dictionary of Scientific Biography, Charles Coulston Gillispe, editor in chief, New York, Scribner, 1970-1980. The Supplement has a long article on Hipparchus.)
Hipparchus must have used geometric versions of the addition formulas and half angle formulas to work out his tables, much as Ptolemy did in his Almagest (around 150 AD). For a beautiful description of these methods, see “How Ptolemy Constructed Trigonometry Tables” by Brother T. Brendan in The Mathematics Teacher, February 1965, pages 141-149, reprinted in From Five Fingers to Infinity, edited by Frank J. Swetz, Open Court, Chicago and La Salle, Illinois, 1994. For material on trigonometry that is easy to find, see Eli Maor’s Trigonometrical Delights, Princeton University Press, 1989 (try http://pup.princeton.edu/books/maor/chapter-2).
Construction of trigonometric tables was driven in part by the desire to measure distances. For instance, Hipparchus determined that the moon was 30 earth diameters from the earth, which is about right (30 x 8000 miles = 240,000 miles). One method he might have used is outlined below. It depends only on observations that can be made from a single location plus knowing that the area covered by totality during a solar eclipse is quite small. For the geometry of eclipses see pages 516 and 518 of Geometry.
Experimenting with shadows cast by disks or spheres, the Greeks found that the length of the dark conical shadow region (the umbra) cast by the moon and by the earth is about 108 times the diameter of the body casting the shadow. One way of checking this is to hold a disk so that it just eclipses the full moon. Measure the distance from the disk to your eye as a multiple of the diameter of the disk. This is one way to get the multiple 108 used here.
The sun lies at a great and relatively constant distance from the earth-moon system, so this multiple of 108 is approximately correct for the earth’s and moon’s umbrae because these umbrae are approximately similar figures. Because the umbra of the moon during a solar eclipse is at most a very small spot, the Greeks knew that the moon was in an orbit about 108 moon diameters from the earth. In addition, observations during lunar eclipses showed the Greeks that the diameter of the earth’s umbra where the moon passes through it is about 2.5 lunar diameters. Lunar eclipses are fairly common and can be viewed over a large region, so this number is not too difficult to come by.
Now for the geometry. If x is the distance from the earth to the moon, then Lunar Diameter = 7Gg. Let D be the earth’s diameter. If we are at distance x from
the earth to the apex of its umbra, we are at distance 108D — x from that apex. By similarity, the width of the earth’s umbra at distance x from the earth is given by
108D
Earth Umbra Diameter = D
108D — x
Setting Earth Umbra Diameter = 2.5 x Moon Diameter gives us
108D x 108 (san ———— a;) = f z 5(+5) — 5 Pen ] i πre = g Ole
My numbers come from the online notes for Michael Fowler’s University of Virginia course, Galileo and Einstein, galileoandeinstein.physics.virginia.edu. Checking with a current astronomy text and computing the numbers, you will find the following bounds for the distance to the moon, 2, in terms of the earth’s diameter: 28.5 < x < 31.8. The length of the umbrae vary between 105 and 109 times the diameter of the sphere casting the shadow depending on the varying distance from the sun to the earth. See Earth Shadow, page 521 of Geometry, where the data give 108.6 earth diameters for the length of the earth’s shadow.
The Greeks knew that the distances in the earth-moonsun geometry varied as well as did the angles of the umbrae. One piece of evidence bearing on this is that some solar eclipses were total while at other times the moon was centered on the sun but left the rim of the sun uncovered (giving an annular eclipse). The angular size of the moon has been measured by various instruments over the millennia. Sosigines, in the 2nd century AD, noted that at some times the moon spans 11 fingerbreadths while at other times it spans 12 fingerbreadths, so the distance from earth to moon varies.
The Greeks knew that the distance they were trying to compute was not fixed, and they knew that their data were subject to errors. They would have taken any of their own numbers “with a grain of salt,” and we should do so too. Hipparchus used observations of the solar eclipse of 14 March 190 B.C. to estimate the distance to the moon by a second method. See NASA’s Goddard Space Flight Center website. Try www-istp.gsfc.nasa.gov/stargaze/Shipparc.htm or search on Hipparchos or Hipparchus and Distance to the Moon.
The Epilogue to the Introduction of Otto Neugebauer’s A History of Ancient Mathematical Astronomy, Springer-Verlag, New York, Heidelberg, Berlin, 1975 contrasts ancient and modern astronomy as follows:
“Both Babylonian and Greek astronomy are based on a set of relative few data, like period relations, orbital inclinations, nodes and apogees, etc. The selection of these data undoubtedly required a great number of observations and much experience to know what to look for. Nevertheless, a mathematical system constructed
at the earliest possible stage of the game was generally no longer systematically tested under modified conditions. Ancient observers were aware of the many sources of inaccuracies which made individual data very insecure…. On the other hand period relations, e.g. times between planetary oppositions and sidereal periods, can be established within a few decades with comparatively high accuracy……. If a theory was capable of guaranteeing correct periodic recurrence of the characteristic phenomena the intermediate deviations would matter little, in particular when one had no means to distinguish clearly between theoretical and observational errors. Thus we notice an outspoken tendency in the development of astronomy to take refuge behind mathematical schemes, rather than to embark on systematic observational programs, which are so characteristic for astronomy since the invention of the telescope…. Rightly, Tycho Brahe called his observational program the foundation of an Astronomia Nova. Ancient Astronomy is overwhelmingly mathematical astronomy.”
The ancient goal was to develop and explore models for the motion of the stars, sun, and planets. Much of this work was mathematical, and developing methods of calculation was an important part of it. The system developed by Claudius Ptolemy some 200 years after Hipparchus held sway for over 1300 years until the emergence of the ideas of Copernicus, Kepler, and Newton.
David Bressoud’s article “Was Calculus Invented in India?” in the January 2002 issue of The College Mathematics Journal shows why Hipparchus needed to go beyond similarity and develop a table of chords in order to work out the geometry of his model of the motions of the sun and earth. Hipparchus observed that spring and summer were longer than fall and winter, so he could not match these observations by placing the earth at the center of a circle and having the sun run around that circle at a uniform speed. In order to work out the details of Hipparchus’ improved model we need a table of chords (or of sines). See Digression on Ancient Astronomy in these notes, Chapter 12, Lesson 5. Bressoud finds that the Greek estimate for sin 1° was accurate to about 1 part in 10,000.
Pushing this work forward, Indian astronomers in the state of Kerala developed the power series expansions for sine, cosine, and the arctangent before 1500 AD. Bressoud estimated that using these expansions would have allowed the Keralans to reduce the error to 1 part in 10’*. The methods are themselves capable of accuracy to any finite degree of precision, but Bressoud considered that practical limits of computation would have prevented the Keralans from going beyond 1 part in 10!?. (Source, e-mail exchange with Bressoud) We think of such expansions as part of calculus, but they were used in Kerala 200 years before Newton and Leibnitz. The need for more precise calculation often gives us new mathematics.
Power Series. You can work the other way — going from calculus and the idea of power series back to trigonometry. Take a leap and consider e² = > az/k!, summed for 0 < k. Formal multiplication of power series and the Binomial Theorem shows that, ‘ term-by-term, eeY = etY, so this series behaves as an exponential function should. Differentiating termby-term, we find formally that A et =e”, as it should. Using de Moivre’s formula, e® = cos θ+isin6, and recognizing that cosine is an even function and sine is an odd function, we might hope that the sum of the even terms in the power series expansion of e*? would give a power series for cos 8, and it does. The sum of the terms with odd exponents gives us isin. Term-by-term differentiation of these series gives formulas agreeing with the expected
- sin’ = cos and cos’ = —sin.
Note that this tells us that the variable is in radians, not degrees.
Assuming the validity of these term-by-term operations with these trigonometric series, we note that sin?(0) + cos?(0) = 1. Why? Let S(@) = sin(@) + cos?(6). Note that ℓ5(0) = 4[sin?(6) + cos?(@)| = 0, by the usual rules of differentiation and () above, obtained by termby-term differentiation of the series. So S(@) is constant. Because S(0) = 1, S(@) = 1, and these series satisfy geometric condition implied by the Pythagorean Theorem. Our series for sine and cosine give points P(@) = (sin θ,cos θ) on the unit circle.
The arc P(@) covers as @ goes from 0 to z has length
/ ((dar/d0)? + (dy/d6)2)49 =
0
/ (sin? 6 + cos? 6)7 d6 = i 1A0 — a.
0 0
so these series give the sine and cosine in radians. This must be the case because the arc length traversed along the unit circle by P(@) as @ goes from 0 to = is just z, which is the radian measure of the angle from 0 to @ on the unit circle.
Next, using ee’ = e?+? and de Moivre’s formula we can work out the angle sum formulas for the trigonometric functions as follows:
cos(9 + ¢) + isin(6 + d) = ett?) = eel? = (cos θ + isin #)(cos ¢ + isin @) = (cos θcos ¢ — sin θsin ¢) + i(cos θsin ¢ + sin θ cos ¢).
Equating the real parts of the initial and final expressions above, we get cos(9 + ¢) = cos@ cos ¢ — sin θ sin ¢. Similarly, the angle sum formula for sine is obtained by equating the imaginary parts. These formulas agree with those you get from the geometric definitions of the sine and cosine functions.
To construct a set of trigonometric tables, we could follow Ptolemy. See the article by Brother Brendan, cited above, for details. What follows is a sketch of the method.
First, find accurate values for special angles found in regular polygons and differences of such angles, for example for angles 0°, 90°, 72°, 45°, 30°, 60°, 15°, 12°. Next use half angle formulas to get to smaller angles. We continue working in degrees. For 0 < ¢ < 8 < 90°, the area of the sector between angles 0 and @ on a unit circle is 76/360°. Call this area Sector(@). The area of the triangle spanned by the radii that bound this sector is 51 sin@. Call this area Triangle(@). Drawing the figure, you can see that the ratio Sector(@)/Triangle(@) increases with the angle in this range. Ptolemy gives a proof of this inequality by Euclidean methods. So,
m™p/360° _ Sector(¢)
tx1sing Triangle(¢)
Sector(@) 78 /360°
Triangle(6) is $x 1sing
Taking the left hand side and the right hand side of this inequality and looking at what these tell us about the sine we get
sin 0
<
sing O1D
We know Triangle(@) < Sector(@) as well, which gives
- c ° us sin@ < Tie Using values for the sine of + and a Ptolemy concludes sin(4”) = 0.008726851, where I used a calculator to convert from Ptolemy’s sexagesimal system to our decimal one. The same HP 48G+ gives sin(0.5°) = 0.0087265355. These values agree to better than six places.
Ptolemy used linear interpolation to fill in the chords for chords whose angles ran from 0° to 1°, going by intervals of one minute. These correspond to sines running from 0° to 1° tabulated for half minute increments. The rest of his table can be filled out using the values already computed, linear interpolation, and the addition formulas.
The astronomers of Kerala improved on the idea of using linear interpolation and developed a polynomial approach that anticipates the power series presented above. ‘The Kerala series were developed for a circle of radius R, the variable was the arc length, denoted by a@ in Bressoud’s article, so a/R = 0, where 0 is the corresponding angle expressed in radians. The numbers computed by the Kerala series were the “half bowstring” (ardha-jyd, later just jya), corresponding to our sines rather than to Ptolemy’s chords. Explicitly, jya a = Rsin(a/R) = Rsin@.
Using series is one means toward more accurate and easier calculation. Napier’s logarithms offered another means — and were also developed with trigonometric calculations in mind. Napier’s logarithms reduced the problems of multiplication and division to the simpler ones of addition and subtraction — once you had his tables. Briggs’s base 10 logarithms were much easier to use than Napier’s.
To see how bold Napier’s plan was and how different his logarithms were from ours look at “Napierian Logarithms and Natural Logarithms” by Howard Eves. This was published in The Mathematics Teacher, May 1960, and reprinted in From Five Fingers to Infinity, edited by Frank J. Swetz, Open Court, Lasalle, 1994. For Napier the logarithm of 10,000,000 was 0, and for him the logarithm of a product was not the sum of the logarithms of its factors.
Section 1.2.2 of Volume 1, Fundamental Algorithms of Donald E. Knuth’s series The Art of Computer Programming, Addison-Wesley, Reading, Massachusetts, First edition 1968 and Third Edition 1997 is devoted to “Numbers, Powers, and Logarithms.” It shows how logarithms and powers can be defined and computed by simple algorithms that do not depend on power series or integrals, not yet invented in the time of Napier and Briggs. In the answers given in the back of Knuth’s book, there are references to similar methods for calculating trigonometric functions.
My calculator whips out answers to 12 places, the 1 part in 10!* accuracy that David Bressoud felt was within reach using the methods of the Kerala astronomers around 1500 ap. These calculations are effortless for me, the mere pushing of a few buttons. The problem of carrying out trigonometric calculations is solved at this level by the microchip, but students still need to learn what to calculate and what the answer means.
How does your calculator or computer calculate sines, cosines, exponentials, and logarithms? Probably by completely different approach from those mentioned above. A common method is the CORDIC algorithm, and there are many web sites that explain it.
Learning how to calculate better is not simply a matter of being able to grind something out to a few more places accuracy. The world of high precision computing has reached previously inconceivable levels. In October of 1995, Takahashi and Kanada computed 7 to 6,422,450,938 decimal places. Let’s put this in context. In the 15th century Kerala astronomers did work Bressoud felt capable of 1 part in 10!2 accuracy. The 1995 work calculated 7 to an accuracy of 1 part in 10% 422,450,938 For late news on m, check the Web. Search on Jonathan and Peter Borwein (at Simon Frazer University), Simon Plouffe, David Bailey, Victor Adamchik, Stan Wagon, Richard Crandall, Yasumasa Kanada, and PiHex.
Surprisingly, it turns out that there are ways to find far off digits in the base sixteen (hexadecimal) expansion of pi without calculation of the earlier hexadecimal digits. This follows from work of David Bailey, Peter Borwein, and Simon Plouffe, who discovered the following formula:
“1 4 2 1 1 => ae oR 16° 8k+1 ≅ 8k+4 8k+ 8k ae Seraen.
Using this and related formulas, the quadrillionth binary digit of pi has been calculated. It’s 0. A quadrillion is 10!*, a large number indeed. See PiHex on the Web or the paper “Pi: A 2000-year Search Changes Direction” by Victor Adamchik and Stan Wagon, on the Web as of 2003. The suspicion is that no such “spigot algorithms” exist that will calculate any given decimal digit of 7 without calculating those before your target digit, but no one is sure about this.
A calculator gives /2 = 1.14142135624 = A. But paper and pencil shows:
A² = 1.141421356242 = 2.000 000 000 074 098 693 76.
So this A is a bit too large to be /2. Even without any arithmetic we knew that the twentieth and last decimal digit of A² would be a 6. Thinking along this line, we can see why the square of a finite decimal expression with a proper fractional part can never be an integer.
What else can we find out about roots? If (25 = 2, algebra gives m² = 2n?. Could both m and n be integers? No, because if they were, then the integer on both sides of the equation m² = 2n? would be an exception to the theorem that every integer has a unique factorization into primes. On the right side, the isolated factor of 2 together with whatever factors of 2 there might be in n² tells us that there are an odd number of factors of 2 in 2n?. On the left hand side, there must be an even number of factors of 2 in m² — perhaps 0. An integer cannot have both an even and an odd number of factors of 2. So V² is not the ratio of two integers; it is irrational.
Proving irrationality requires a different sort of calculation, a calculation that tells us what must be true if the prime factors on both sides of an equation are to balance. Working this out for each prime factor of ℓ, we can see that that an integer @ has a rational kth root only if @ is a perfect kth power of an integer, otherwise it is impossible to balance the prime factors in an equation of the form m² = én², where m and n are integers. So ¢!/” is either an integer (when ℓ is a perfect nth power of an integer) or it is irrational.
The argument that shows V² is irrational is based on the fact that a positive integer can be written as products of primes in only one way, disregarding order. After working with a few examples, we should be willing
to believe that positive integers have unique prime factorizations. The proof of this fact goes back to Euclid (Book VII.1 — 3 and Book IX.14). The key to the proof is the fact that if a prime p divides the product of two integers m and n, then p divides at least one of m and - n. The proof of this, in turn, is based on a computational method that writes the Greatest Common Divisor (G.C.D.), d, of integers k and @ as the least positive integer that can be written as mk + né, where m and nm are integers. This method is called the Euclidean algorithm. Generalized, this algorithm is the basis for breaking more elaborate algebraic structures into prime components in group theory and linear algebra.
Suppose you wanted to find the G.C.D. of 4293 and 2816. Rather than factoring them, you might observe that the common divisors 4293 and 2816 are exactly the same as the common divisors of 2816 and 1477 = 4293 −1 x 2816. The largest number in this new pair of numbers is the smaller of the two numbers in our initial pair. By subtracting multiples of the smaller of our two numbers from the larger we can get a new pair with the same G.C.D., and we can do this so that the largest number in this new pair will always be the smallest number in our previous pair. This decreasing chain of integer pairs must terminate with a pair d, 0. Thed in this pair is the G.C.D. Keeping track of the final result in terms of the original integers gives the G.C.D as an integral combination of those original numbers. This is the Euclidean algorithm. The only way to really understand it is to either do examples by hand or write a program to carry it out yourself. Applying this algorithm to 4293 and 2816 gives us 1151 x 2816 — 755 x 4293 = 1, so these two numbers have a G.C.D. of 1. Their prime factorizations also show that they are relatively prime: doe 3a ns and 2816-5 2.x 11.
Suppose prime p divides the product km. We want to show that p divides one of the factors, k and m. If p divides k evenly, then we are done for p divides the first of the two factors. If not then the G.C.D. of p and k is 1 because the only positive divisors of a prime are 1 and itself. So we can find integers s and t such that sp + tk = 1. Multiplying this equation on both sides by m we see that spm +tkm = m. But p divides spm and tkm, whose sum is m, so p divides m, the second of the two factors. In either case a prime that divides a product of integers must divide one of the factors. Knowing this, you can prove that positive integers have unique prime factorizations by using cancellation to show that no smallest counterexample to unique factorization could exist. From unique factorization it follows that V² is irrational.
The argument that /2 is irrational is based on prime factorization of integers, a result established by essentially computational methods. But we use computing in this analysis for something quite different than finding the decimal expansion of NG.
“The purpose of computing is insight, not numbers.” This is the maxim at the front of Richard Hamming’s classic Numerical Methods for Scientists and Engineers, Second Edition, McGraw Hill, New York, 1973. No matter how closely we measure the diagonal of a unit square, we will never be able to conclude from measurements that its length is irrational. In practical measurement, there is only finite precision; infinite decimals play no role. No matter how many decimal places we compute of /2, these numbers will never give evidence that /2 is irrational. More subtle reasoning, using algebra and number theory, is needed to show that quantities are irrational multiples of the unit.
Mathematics 2s a practical tool, and computation is essential. But computation also plays a role in mathematical theory. Reasons beyond the practical drive the pursuit of accuracy to 1 part in 10!%, and beyond —even to the insistence on exact algebraic answers. Without exact calculations, we would never understand that some numbers are irrational.
Without exact calculations we could never show that 12 congruent regular pentagons fit together exactly to form a regular dodecahedron. This result is given in Euclid, Book VIII, Proposition 17. The construction is found in Geometry in Dodecahedron Construction, page 681. All that is needed to match Euclid’s proof is to carry out the computations algebraically and to see that the faces fit as the figure suggests.
To understand how precise 1 part in 10! is, consider that the meter was intended to be 1/10,000,000th of the length of a quadrant on the earth’s surface — measured along meridian starting at the North Pole, passing through Paris, and ending on the equator. To measure this quadrant with an accuracy of 1 part in 101”, we would have to get the distance right to within +1/100th of a millimeter. Such accuracy is unattainable.
We don’t need the 1 part in 10!? accuracy unless we are following some intricate chain of calculations through many steps. Even in chains of triangulations, uncertainties in the initial data introduce errors larger than those likely to arise from errors in the trigonometric functions. But to decide whether the predictions of the Ptolemaic system match the movements of the sun, moon, and planets, we need to know that deviations between observations and predictions are real, and not due to errors in calculations. A theoretical model like this is supposed to work exactly, and so it is appropriate to ask for extreme, even absolute, precision in testing it. This is one reason that astronomers played an important part in trigonometrical advances. _
▶ A Colorful Contributor to Trigonometry. Some of those who worked on trigonometry led exciting lives. Consider Nasir ad-Din at-Ttist, who is described by Otto Neugebauer in Part One of his previously cited History of Ancient Mathematical Astronomy as follows:
“In 1255 the Mongol prince Hulagii Kahn, grandson of Genghis Kahn, succeeded in taking the mountain stronghold of the Assassins, al-Alamut, which had been a major obstacle for a Mongol domination of Persia and Iraq. Among the prisoners taken by Hulagii in al- Alamut was Nasir ad-Din at-Tiisi, who had been held as an honored prisoner and astrological advisor by the ‘Old man of the Mountain,’ the Grand Master of the Assassins. At-TiisI once more became the trusted advisor of his master, now Hulagii, and seems to have played in that capacity a somewhat sinister role in the end of the last Abbasid Caliph when Baghdad was taken by the Mongols in 1258.”
Neugebauer is more concerned with astronomy, so we turn to the Revised and Enlarged Edition from 1919 of Florian Cajori’s A History of Mathematics for an appraisal of at-Tisi’s mathematics:
“During the supremacy of Hulagu, lived Nasir-Eddin (1201-1274), a man of broad culture and an able astronomer. He persuaded Hulagu to build him and his associates a large observatory at Magara. Treatises on algebra, geometry, arithmetic, and a translation of Euclid’s Elements were prepared by him. He for the first time elaborated trigonometry independently from astronomy and to such great perfection that, had his work been known, Europeans of the fifteenth century might have been spared their labors.”
Nasir Eddin’s work on non-Euclidean geometry is mentioned on page 669 of Geometry.
▶ Lines, Vectors, Transformations, and Matrices. The notes below outline some other ways of looking at lines and their equations, in the plane, and in higher dimensions. For more on these matters, see standard texts or Web resources such as Eric Weisstein’s MathWorld.
▶ Equations of Lines, Distances. The general form for the equation of a line ℓ in the coordinate plane is
- Az + By =C, where A² + B² ℓ0.
The slope of ℓ is m = — 3 and is independent of C. Using this we can find the point on @ closest to the origin. The line k | ℓ through the origin has slope =1 = 5, so its equation is -Bx+ Ay = 0. Finding the intersection of k and é by algebra gives us the following coordinates for the point P closest to the origin on @:
AC BC
Lin Gian Baya a! Seat
The distance from ¢ to the origin O is the distance from P to O. By the distance formula this is
C² (**) Distance from ℓ to the origin = (A² +B)’
When C = 0 we get the line that is parallel to the general line ℓ and that passes through the origin. Call this line n. Other values of C give lines parallel to n and at varying distances from the origin. Using C in place of VC? in (**), we get a signed distance from ℓ to the origin O.
C
Signed Distance from ¢ to the O =
JER
This distance is positive when @ is on the same side of n as the point with coordinates (A, B), zero when? = n, and negative when @ is on the opposite side of n from the point (A,B). This analytic geometry ties in with the vectors as studied in linear algebra.
▶ Vectors and Lines. In Cartesian coordinates we can think of points in the coordinate plane as vectors, say v and w. These we conceptualize as arrows extending from the origin to the points defining the tips of the arrows. Suppose our v and w are specified by coordinate points as follows: v = (A,B) and w = (z,y). The length of a vector is written ||v|| and defined in Cartesian coordinates by
Length of vector w = ||w|| = /x² + y²,
which makes this length the distance from the origin to the point of the arrow, by the Pythagorean Theorem.
The dot or scalar product of these two vectors is written uv-w and defined by
Dot Product of v and w=v-w = |lv||||w]| cos,
where @ is the angle between v and w. In Cartesian coordinates the dot product is the sum of the products of the coordinates. For example, v- w = Ax + By.
If v # 0, the multiples of v, { Av | is a real number ie form a line k through the origin, O. The vectors { w | v-w = dllv|| } form a line ℓ perpendicular to the line k and at a directed distance of d from the origin. If d is positive, ℓ is displaced in the direction that v points. If d is negative, @ is displaced in the opposite direction from the origin. This agrees with the directed distance as developed above.
I” Dot Products and Cosines. As above, think of v and w as two vectors in the plane represented by arrows from the origin. Let 6 be the angle between these arrows. Then v - w = ||v|| ||w|| cos θ = Az + By, so
eee of the nae
=cos0
between v and w
ayes Vo MAG ER By
ell Nell etl feo}?
which gives a simple way to find cos 6, if we know the Cartesian coordinates of the points of these vectors.
The dot product is built in as a vector operation on calculators such as the TI-89 (under 2nd function, Math, Matrix, Vectr ops) or the HP-48G+ (under Math, Vectr). In Cartesian coordinates in n-dimensions the dot product of two vectors v = (v1,V²,…,Un) ° (01) Woes AW) SAW, Pi dowa + = Leo,wy = |||| ||w|| cos 6, where ||v|| is the length of the vector v and this length is defined by |lv|| = /v12 + vo2 + — + Un? and ||w|| is the similarly defined length of w. So this method for finding the angle between two vectors generalizes to higher dimensional Euclidean spaces. In order to have this work in our coordinates, we must use a rectangular system of coordinates with equal units along each axis. In this case, the generalization of the Pythagorean formula for lengths between points is valid. These are called orthonormal coordinates.
lt Dot Products, Rotations, Matrices. Suppose u is a unit vector whose tail is at the origin, O. Then its tip lies somewhere on the unit circle about O. Find the smallest positive counterclockwise angle θ so that a rotation about O through angle θ carries (1,0) (on the z-axis) onto the tip of u. Then the tip of u has coordinates (cos@,sin@). The same counterclockwise rotation carries the point (0,1) on the y-axis to (cos (@ + 90°), sin(@+ 90°) = (—sin@,cos6), because (1,0) is 90° counterclockwise from (1,0) on the unit circle about O. This gives us a second vector v whose tail is at O and whose tip is at (— sin 8, cos 0). I always draw a diagram to work these relations out. The words are only good if you back them with your own pictures.
We know u-u=v-v = 1 = |lu|| = |lv|| because these vectors run from O to points on the unit circle, and computation gives cos? 6 + sin? @ = 1 for u-u and v-v, confirming this. We expect u-v = 0 because these vectors are at right angles to each other and u-v = ||u|| ||v|| cos@, where @ = the angle between u and v, which is +90°. And computation confirms this, too.
Additivity. If T is a rotation through the counterclockwise angle of 6, what is T( (z, y) ), where these points are given in terms of our original coordinate system? The vector sum of vectors w and z is defined by the Parallelogram Law to be the point w + z such that O, w, w+ z, w form a parallelogram. Because rotations are isometries, they preserve distances and send parallel lines into parallel lines, hence T(w + z) = T(w)+ T(z), because these vector operations are defined in terms of parallelism. A mapping that distributes over addition the way that T does is said to be additive.
Scalar Multiplication. Multiplication of a vector, w, by a number, lambda, is defined by requiring that the point of the vector lie on the ray from O through the head of the arrow representing w at a distance A||w|| from O. This operation is scalar multiplication of vectors and is defined geometrically in terms of distances, given the origin, O. Hence, scalar multiplication is respected by
isometries that leave O fixed. That is, T(Aw) = (w). This paragraph and the previous one show us that
Linearity T(au+ 6v) = aT(u) + BT(v),
for all scalars a and @ and all vectors u and v. Mappings with this property are linear.
Using the linearity of T, we can express the rotation T using the coordinates x and y of our point to find the coordinates x’ and y’ of its image.
T((x,y)) = T(2(1,0) + y(0, 1) )
= ℓT(1,0) + yT(0, 1)
= xz (cos θ, sin #) + y(—sin 6, cos θ)
= (x cos@ — y sin, z sin + y cos θ)
=e (aug),
Matriz Notation. ‘This gives a compact way of writing T. We write our vectors as columns. The transformation T is represented by a 2 x 2 array of numbers. The matrix operates on the coordinate vectors following the Rows x Columns pattern seen in the right-hand equality below.
x’ _ ( cos@is— sind fer
Ulm sin Gaercos). y
cos 0 — sin θ
Bata +a cos 8:
The left-hand equality above shows how 2’ and y/ arise as the sum of column vectors representing the images of (1,0) and (0,1) in our coordinate system.
Using this pattern, we can define transformations using coordinates and matrix notation. For example, we may define S( (x,y) ) = (2’,y’) by
/ Gy) (on ae) (5) =#Caa) +o
Y Q21 422 y a21 a22
Here you can put in any numbers you like for the a,; and you will get some sort of transformation, S. Whatever you choose for these numbers, S( (0,0) ) = (0,0) so the origin is mapped onto itself. Furthermore, lines are mapped onto lines or collapsed to single points. Triangles may be distorted or flattened, but lines never become curved.
Area and the Determinant. A key geometric experiment is to see what S does to the square whose vertices are (0,0), (1,0), (1,1), (0,1), called the unit square. S will always send this square onto a parallelogram, although that parallelogram may be flattened onto a line. The signed area of that parallelogram is given by the determinant of the matrix A, written det |A| and defined by the following computation
Q11 412
det |A| = det = 411422 — 421421.
a
This area is counted as positive if a clockwise path around the unit square gives a clockwise tour around its image, and the signed area is negative if the orientation is reversed.
The determinant used to be studied in Algebra 2, in connection with Cramer’s Rule, where it was a formal device used in solving systems of linear equations. Now it has been recognized that Gauss-Jordan elimination is an easier, more efficient, and more comprehensible way to solve such systems. The determinant has moved to the linear algebra course, a new addition to the curriculum that you can read about at such sites as www.cde.ca.gov/ standards/math/html in 2003.
Why does det |A| give the area of S(UnitSquare)? In order to be able to see the geometry and work out the algebra, assume that w = S((1,0)) lies in the first quadrant and z = S((0,1)) lies in the second quadrant. The other possibilities can be handled similarly, with careful attention to the signs of all quantities. The UnitSquare has vertices (0,0), (1,0), (1,1), and (0,1) in the counterclockwise order. The corresponding vertices of S(UnitSquare) are: 0 (the vector zero), w, z, and w + z, and they are also in counterclockwise order, so the orientation is preserved, and the signed area of S(UnitSquare) will be positive.
If the matrix representing S is A = (a;;), we have the following, where we shift from row to column vector representation for w and z:
ee a 1 4 a ial a 11 a 12
21 a22 @21 422.
Because w is in the first quadrant, a,; and ag are both positive. Because z is in the second quadrant, ai2 is negative while a² is positive. S(UnitSquare) is a parallelogram with vertices 0, w, w+ z, and z. Drawing the figure, you will see that these lie on the sides of the rectangle R with vertices (a12,0), (ai1,0), (211, @21 + G22), (412, @21 + a²). The width and height of R are —aj2 + a1; and a²; + a², where the —ayj2 is needed because aj2 is negative while the other a’s are positive. From the diagram you see that
Area of R = (—aj2 + a11)(a21 + a22)
1 1 see a.(S(UnitSquare)) +2x zauidai +2x 3 022(—a21);
where a denotes the area and the last two summands account for the areas of the four right triangles that surround S(UnitSquare) in R. Solve this for aS(UnitSquare) and you will find a(S(UnitSquare)) = 411422 — @21@21, which is how det |A| was defined.
Note that the area found above can also be expressed by ||w]| ||z|| sin θ, where 6 is the counterclockwise angle between vector w and vector z. This tells us that
1 Wt 21 W122 — 21W2
sin 9 = ———— det
I|w | lz W² 22 ainda
The numerator on the left gives the magnitude of the cross product of the vectors w and z, denoted, w x v, an operation that can be found in standard sources such as Eric Weisstein’s MathWorld on the Web.
▶ Matrices and Computer Graphics. Software that displays material or sends it to a printer usually has the ability to resize, rotate, and shift the image. Matrix operations are built into this software. Postscript™ is a example of such software. It manipulates images by sending what was at the point (z, y) to a new point with coordinates (z’,y’) which are determined by a matrix as follows.
oe WO
(az Cie 1) = a, Y,; 1) c d 0
ean ee
We multiply using RowsxColumns, but Postscript™ uses row vectors, so these must be placed left of the matrix that operates on them. The matrix shown above is the Current Transformation Matrix (CTM) in Postscript™. Multiplying we get
a’ =ax+cyt+ty
y’ =br + dy + ty.
The third component is always 1, and the t, and t, are used to perform translations of the coordinates. The Postscript™ language is built into many graphics devices. It allows you to create many interesting geometric patterns by downloading Postscript™ programs directly to your printer. There is a wealth of material on Postscript™ and creation of designs both in books and on the Web. Search on Postscript™, designs, and, if you wish, fractals.
The matrices representing two sorts of transformations in this augmented vector pattern are given below.
Rotation about O aes Csi 63.0
through angle θ —sin@ cos@ 0
0 Osnd el
Dilation of magnitude A about ay 0 0 (Zo, yo) 0 ‘
(1—A)apo (1—A)yo 1