# HIGHER GEOMETRY. 1. Notation. Below is some notation I will use. KEN RICHARDSON

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1 HIGHER GEOMETRY KEN RICHARDSON Contents. Notation. What is rigorous math? 3. Introduction to Euclidean plane geometry 3 4. Facts about lines, angles, triangles 6 5. Interlude: logic and proofs 9 6. Quadrilaterals 7. Circles and constructions 8. Computing areas, the Pythagorean Theorem 5 9. Constructible lengths and the Rigid Compass Theorem 8 0. Homotheties and similarity 9. Cyclic polygons 0. Centers of Triangles 3. Introduction to Analytic Geometry 4 4. Equations of Lines 5 5. Using Analytic Geometry 7 6. Trigonometry 9 7. Trigonometry and Geometry Vectors and vector operations Matrices, Linear and Affine transformations Spherical Geometry 56. Hyperbolic Geometry 59. Complex Numbers 6 Below is some notation I will use.. Notation

2 KEN RICHARDSON Notation meaning g : A B g is a function with domain A and codomain B R the set of all real numbers Z the set of all integers C the set of all complex numbers x A x is an element of the set A. C D The set C is a subset of the set D. exp x e x log x log e x ln x R the set of ordered pairs x, y such that x, y R A B If A and B are two sets, A B is the set of points p such that p A and p B. A B If A and B are two sets, A B is the set of points p such that p A or p B or both.. What is rigorous math? In my view, there are a couple of reasons why mathematicians learn to use rigorous mathematics to state problems and theorems, to give solutions to problems and proofs of theorems. Reasons for doing mathematics rigorously: Mathematicians want to be sure that their results are correct, so that mathematics is exact and precise. One feature of mathematics that distinguishes the subject among all other disciplines is that if a result of mathematics is proved, then it is correct for all time. Results of mathematics proved 000 years ago are still and will always be correct. This is not like other fields, where new information or better models can sometimes show that what was previously believed to be true is really not true. This is not to say that mathematicians don t make mistakes; there are in fact some interesting historical examples of this, but in those cases, it was discovered that assumptions made in the logic were not correct i.e. there was a mistake in the proof. And another thing: opinion or prejudice do not change the truth of mathematical statements. Mathematicians want to communicate to others to show why results are true. Part of the function of a good proof is to allow mathematicians and students of mathematics to understand a complicated result in terms of much more rudimentary and believable results. The idea is to check a theorem in a precise way so that conceivably a computer could check it. So rigorous mathematics has an educational value in that it cuts through the mystery and allows us to connect more difficult concepts with simpler ones. Keeping these things in mind, we need to make sure that when we state things and prove things, we do it exactly, and we don t allow any loopholes. And also, we want to make sure that the ways that we state things and prove things are done in the simplest and cleanest ways possible. Things to think about when doing rigorous math: When making statements: Is there any way this could be misinterpreted? Are all my variables and symbols defined? Is the mathematical meaning of every word precise? Make sure that everything is written in complete sentences. When proving and calculating: In the back of your head, there is an evil person saying I don t believe you! after every sentence. Make sure that there is absolutely no doubt that everything you have written is true, and that the logical flow leaves no room for error. Make sure that everything is written in complete sentences.

3 HIGHER GEOMETRY 3 3 After proof or calculation is complete: Read over it again. Can anything be written more clearly? Is it easy to understand? Are there any loopholes? Special note for geometric proofs: In no cases can we make a picture a rigorous proof. So it is good to provide pictures to make clear what objects are considered in a proof, but it should be the case that the proof could stand on its own without the picture. In other words, the person reading the proof without seeing the picture should be able to create the picture entirely only by using the words in the proof. Consequently, writing a clear geometric proof is difficult! 3. Introduction to Euclidean plane geometry The basic objects we work with in the plane are points, line segments, rays, and lines. In the earliest formalisms of geometry such as in Euclid s Elements, the definitions of these things we said to be axioms statements that are assumed and that are not proven. In later formulations of geometry, the plane is the Cartesian plane seen as defined as a set of ordered pairs of real numbers, and all of the properties of sets in the plane follow from properties of real numbers, which have their own axioms that are used to define them. We first start with the approach of Euclidean geometry. The basic notions / objects of Euclidean geometry: Points usually denoted by letters like A, B, C,...m Lines usually denoted by script letters like l or l, l or by using two different points on the line and an arrow on top, like AB, P Q Line Segments and their endpoints; we denote the segment connecting points A and B by AB Rays and its vertex; if P is the vertex of a ray and Q is another point on the ray, then P Q is the symbol denoting that ray Axioms are statements that we do not prove but that we accept as true. All mathematical facts should follow ultimately from axioms. Euclid s Axioms paraphrased: There is a unique line through any two different points in the plane. [Also, there is a unique line segment connecting these points.] The word unique means only one. Given any line segment AB, we can extend it to a ray AB or a line AB uniquely. Two rays joined at a vertex makes an angle. [We will also say that the angle ABC between the line segments AB and BC is the same as the angle between the rays BA and BC. The middle letter is the vertex in the notation.] When we draw pictures of angles on the board, we will use circular arcs to denote angles. When there is no ambiguity, we will often use the notation B for the angle ABC. A rigid motion or isometry or Euclidean motion is a transformation of the plane a function F : R R that preserves distances between points so that distance between A and B is the same as the distance from F A to F B. It turns out that every isometry is a rotation, reflection, translation, or a combination thereof. Two objects such as a line segment or an angle are called congruent or isometric if there exists a rigid motion of the plane that takes one of the objectssets exactly to the other. Thus we can say that we have congruent line segments or congruent angles or... when this happens. If S and T are two sets in the plane, we write S T to say S is congruent to T. We say that a

6 6 KEN RICHARDSON the set of points that are a distance from the origin in R.] Since the circumference of the unit circle is πr π π, the radian measure π 360, so that for example π 80 and π 90, π Note that radian measures of angles do not have any units like degrees. When direction is important, we measure positive angles as counterclockwise and negative angles as clockwise. Radian measures of angles are important in trigonometry and calculus, because for example the simple derivative formulas such as sin θ cos θ cos θ sin θ only work when θ is measured in radians. If we used degrees to measure θ, we would have to have formulas like sin θ π cos θ. 80 To convert angle measurements from degrees to radians or vice versa, use the scale factors π π, so for examples, π 80 π 5, π Now we see that A and B are supplementary if and C and D are complementary if m A + m B 80 π m C + m D 90 π. We have the important fact that if and only if A B m A m B. 4. Facts about lines, angles, triangles Note that the two different angles that occur when two lines intersect are supplementary, because they add to the straight line angle. Points on the plane are called collinear if they lie on a single line. A triangle is a set of three points in the plane that are not collinear and the three line segments connecting them. We denote the triangle formed using the three points vertices A, B, C as ABC. When we say that ABC is congruent to DEF, we mean that there is an isometry that takes A to D, B to E, and C to F. Note that the order is important, so that when we write ABC DEF, we also mean AB DE, BAC EDF, etc. A polygon is defined in a similar way to a triangle. A polygon with n sides is defined by n points vertices in order, connected in order. We require that the points are not collinear, and sometimes

7 HIGHER GEOMETRY 7 we require that no three of the points are collinear. Here are the common name for the different kinds of polygons. Number of sides Name 3 triangle 4 quadrilateral 5 pentagon 6 hexagon 7 heptagon 8 octagon 9 nonagon 0 decagon -gon dodecagon 7 7-gon n n-gon We now prove an important theorem. Theorem 4.. The sum of the measures of the interior angles of any triangle is 80. Proof. Let ABC be any triangle in the plane. Using Euclid s second axiom, extend the sides AB, BC, AC to lines AB, BC, and AC, respectively. Using the Playfair Postulate, construct a line l parallel to AB through the point C. Denote by D the alternate interior angle to A between l and the transversal AC. Denote by E the alternate interior angle to B between l and the transversal BC. Since D + C + E is the straight angle corresponding to the line l at the point C, m D + m C + m E 80. Since alternate interior angles are congruent, m A m D and m B m E. Substituting, m A + m C + m B 80. There are several triangle congruence theorems. The point of these theorems is that if we know that a few parts of two triangles are congruent, then the entire triangles are congruent. At this point, the only way we have of determining that two triangles are congruent is by proving that there exists an isometry of the plane taking one triangle exactly to the other. After we have the triangle congruence theorems, there are much simpler ways of determining when two triangles are congruent. Theorem 4.. SAS Triangle Congruence Theorem Let ABC and DEF be two triangles in the plane. If AB DE, ABC DEF, and BC EF, then ABC DEF. Proof. Given the hypothesis, we will show that there exists an isometry of the plane that maps ABC exactly onto DEF. First, by translating, we may move the plane so that the point B moves to E. That is, we may translate ABC to A B C, where B E. Next, we rotate around the point B E to move A B C to A B C so that B A ED and B B E. Then, since B B E and A B AB DE, the point A must coincide with the point D otherwise the lengths A B and DE would be different. At this point, there are two possibilities for where C is. Either it

8 8 KEN RICHARDSON is on the same side of DE as F or on the opposite side of DE as F. If it is on the opposite side, we then reflect across DE to move A B C to A B C so that C is on the same side of DE as F. If C is already on the same side of DE as F, we do no rigid motion and simply have A B C A B C. Now, since A B C ABC DEF and A B DE, it must be the case that EF B C ; otherwise one of the two angles A B C, DEF would strictly contain the other and would then have a larger measure, a contradiction. Also, since B C EF, the points C and F must coincide, because otherwise the lengths of B C and EF would be different. Therefore, since A D, B E, C F, and since the line segments connecting these points are unique, A B C DEF. Then the sequence of rigid motions mapping ABC to A B C is an isometry of the plane that maps ABC exactly to DEF. Therefore, ABC DEF. Theorem 4.3. ASA Triangle Congruence Theorem Let ABC and DEF be two triangles in the plane. If ABC DEF, BC EF, and BCA EF D, then ABC DEF. Proof. Left as an exercise. A similar plan to the proof of the SAS theorem will be needed, but of course the details will be different since the give information is different. Theorem 4.4. SAA Triangle Congruence Theorem... Proof. Left as an exercise. It turns out that we can prove this using the ASA Triangle Congruence Theorem. Theorem 4.5. SSS Triangle Congruence Theorem... Proof. Left as an exercise. Notice that there is no SSA triangle congruence theorem. In an exercise, it can be shown that there is a specific example of two triangles that would satisfy the SSA hypothesis but are not congruent. Also, there is no AAA triangle congruence theorem, no SS triangle congruence theorem, etc. Again, showing this amounts to providing a counterexample in each case. We postpone that at the moment. We can now prove several results using these congruence theorems. An isosceles triangle is a triangle with two congruent sides. An equilateral triangle is a triangle in which all sides have the same length and are thus congruent to each other. Theorem 4.6. Isosceles Triangle Theorem Let ABC be an isosceles triangle, such that BA BC. Then BAC BCA. Proof. Given the hypothesis, observe that since AB BC, ABC CBA, and BC BA, by the SAS Triangle Congruence Theorem we have that ABC CBA. Therefore, since corresponding parts of congruent shapes are congruent, BAC BCA. Corollary 4.7. Every angle of an equilateral triangle measures 60. Proof. Let P QR be an equilateral triangle. By the Isosceles Triangle Theorem, P QR P RQ. Also, by the Isosceles Triangle Theorem, QRP QP R. Thus, all the angles are congruent and thus have the same measure. Since the sum of the measures of the interior angles is 80, each angle measures Note that the converse of the Isosceles Triangle Theorem is also true. A statement of the form A implies B or A B has a converse, which is B implies A or B A. The converse and

9 HIGHER GEOMETRY 9 the original statement are not logically equivalent, so that it is not always true that the converse of a true statement is true. For example, If S is a square, then S is a rectangle. is a true statement, but the converse If S is a rectangle, then S is a square. is false. Theorem 4.8. Converse of Isosceles Triangle Theorem Let ABC be a triangle such that BAC BCA. Then BA BC, so that ABC is isosceles. Proof. Given the hypothesis, we see that since BAC BCA, AC CA, and BCA BAC, by the ASA Triangle Congruence Theorem, we have BAC BCA. Then, since corresponding parts of congruent shapes are congruent, BA BC. Here are some other important converses to our results involving parallel lines. Proposition 4.9. Suppose that l and l are two lines, and l T is a transversal across both lines, intersecting l and l at different points.. If A and B are the two interior angles on the same side of the transversal, and if A + B a straight angle, then l l. Proof. Let C be the point of intersection of l and l T. By the stronger version of Playfair s postulate, there exists a line L through C that is parallel to l. Assume L is not the same line as l, then L makes an angle with l, so that on one side of l T, the interior angles of L and l on one side of l T add up to less than a straight angle. By Euclid s Fifth Postulate, l and L must then intersect, which is a contradiction. Therefore, the assumption that L is not the same line as l is wrong, so that l L is parallel to l. Proposition 4.0. Suppose that l and l are two lines, and l T is a transversal across both lines. If A is an angle made by l and l T that is an interior angle, and if B is the alternate interior angle to A made by l and l T on the other side of l T, and if A B, then l l. Proof. Let l, l, l T, A, B be as given, and let B be the other interior angle at l l T that is supplementary to B, then A and B are supplementary and are interior angles on the same side of l T. By the previous proposition, l l. The following facts concern the lengths of the sides of a triangle. Theorem 4.. The triangle inequality Given any triangle, if the longest side has length z and the other two sides have length x and y, then z < x + y. Proof. Postponed Note that if z > x + y, the sides of length x and y would not be able to have a common vertex! Theorem 4.. Converse of the triangle inequality Given three positive real numbers x, y, z such that x < y < z and z < x + y, there is a triangle whose side lengths are x, y, and z. Proof. Postponed 5. Interlude: logic and proofs We have already considered some aspects of logic in the proofs we have done previously. We review: A statement of the form A implies B or A B has a converse, which is B implies A or B A. The converse and the original statement are not logically equivalent, so that it is not always true that the converse of a true statement is true. For example, If S is a square, then S

10 0 KEN RICHARDSON is a rectangle. is a true statement, but the converse If S is a rectangle, then S is a square. is false.a statement of the form A implies B or A B has a converse, which is B implies A or B A. The converse and the original statement are not logically equivalent, so that it is not always true that the converse of a true statement is true. For example, If S is a square, then S is a rectangle. is a true statement, but the converse If S is a rectangle, then S is a square. is false. Another type of logical statement is A if and only if B A iff B A B. This means A B and B A. Therefore, for example S is a square S is a rectangle is false, and ABC has two congruent sides ABC has two congruent angles is true. The negation of a statement A is the statement A is false and is written A and is read not A. Proposition 5.. Let A and B be two statements. the statement A B is logically equivalent to B A. Proof. If A B, then if B is not true, then A can t be true, so B A. On the other hand, if we assume that B A, then if A is actually true, then B must be false, so B is true. Thus, A B. An important kind of proof is proof by contradiction. Here is how this works. Suppose we want to prove that A is true. Here are the steps to use proof by contradiction. Assume A is true. Show that through logical steps we reach an impossibility called a contradiction. Therefore the assumption is wrong, so A is true. Another version of proof by contradiction: Suppose we want to show A B. Assume A is true. Assume B is actually false. Through logical steps, show we reach an impossibility like contradicting our assumptions the contradiction. Therefore the assumption that B is false must be wrong. So B is true. Thus A B. Here is a typical example of a proof by contradiction. Note that a rational number is a number x that can be expressed as x p, where p and q are q integers and q 0. Lemma 5.. Every rational number x can be expressed as x p, where p, q Z and q 0, and q where p and q have no common factors other than or. Proof. Suppose that p and q have a greatest common factor k, so that p km and q ks for some positive integer k > and integers m and s. Then p q km ks m s, and m and s are both strictly smaller in absolute value than p and q, respectively. If m and s have a common factor a >, then m au and s av for some integers u and v. But then p kau and q kav, so that ka is a positive common factor of p and q, and that is a contradiction to the fact that k is the greatest common factor of p and q. Proposition 5.3. The number is irrational.

13 HIGHER GEOMETRY 3 3. Third Axiom of Euclid Given a point P and a point Q, we may construct the circle centered at P with radius P Q. Along with the first and second axioms, this gives us the basis of doing constructions with unmarked straight edge and collapsing compass. That is, we may connect two given points in the plane with a line segment and could extend that to a line, and similarly, we could use a compass to draw circles as long as we are given the two points from which to start. The word unmarked means that we are not allowed to make measurements with the straight edge, and the word collapsing means that the compass used to draw a circle of radius R cannot be picked up from one spot and then used at a different center with the remembered R radius. Here are some examples of constructions we can do. Proposition 7.3. Given a line segment AB, we can construct an equilateral triangle such that one side of the triangle is AB. Proof. Given AB, construct the circle centered at B with radius AB and the circle centered at A with radius AB. Find one of the intersection points of the two circles, and label it C. Construct AC and BC. Then ABC is an equilateral triangle. Proof that the construction works: Let C be constructed as above. Then AB AC since B and C are both on the circle of radius AB centered at A. Also AB BC since both A and C are on the circle of radius AB centered at B. Thus ABC is equilateral. Before getting to more constructions, we need to make a few more definitions. Definition 7.4. Given a line segment AB, a midpoint C of AB is a point C on AB such that AC BC. Lemma 7.5. The midpoint of a line segment is uniquely determined. Proof. Let C and C be two midpoints of the line segment AB, and suppose C is not C and is on the segment AC without loss of generality; if it were on BC we could switch the labels A and B. Then AC < AC and C B > CB, and since AC CB we have AC < AC CB < C B, so in fact C is not a midpoint, a contradiction. Thus really C C and midpoints are unique. As a result of the lemma above, we may refer to the midpoint of a line segment. Definition 7.6. Given a line segment AB, a perpendicular bisector of AB is a line that is perpendicular to AB and that contains the midpoint of AB. Proposition 7.7. Given a line segment AB, a point X in the plane is on the perpendicular bisector of AB if and only if AX BX. Proof. If X is on the perpendicular bisector of AB, we construct the line segments AX, BX, and MX, where M is the midpoint of AB. Then MX is the perpendicular bisector of AB, so that AMX and BMX are both right angles and are thus congruent. Because M is the midpoint, AM BM. Also, MX MX, so by the SAS Triangle Congruence Theorem, AMX BMX. Then AX BX since they are corresponding distances on the two triangles. If X is in the plane and AX BX, there are two cases. First, if X is on AB, then by definition X is the midpoint, so it is definitely on the perpendicular bisector. Next, if X is not on AB, construct AX, BX, MX, where M is the midpoint of AB. Then since M is the midpoint, AM BM. Clearly MX MX, and also we are given AX BX, so AMX BMX by the SSS Triangle Congruence Theorem. Then m AM X m BM X, since they are corresponding

14 4 KEN RICHARDSON angles on the congruent triangles, so since they add to a straight angle, they add to 80. Then m AMX m BMX 90, so MX is the perpendicular bisector of AB. Proposition 7.8. Given a line segment AB in the plane, its perpendicular bisector can be constructed using an unmarked straight edge and collapsing compass. [In particular, this means that the midpoint of AB can be constructed. Proof. Construction: Construct the circle of radius AB centered at A and the circle of radius AB centered at B. Let P and Q be the two intersection points of the two circles. Then P Q is the perpendicular bisector of AB. Proof that the construction works: If we do the above, we observe that AP AQ BP BQ AB since they are all radii of circles of radius AB. By the previous proposition, P and Q are both on the perpendicular bisector of AB, so that P Q is the perpendicular bisector of AB. Remark 7.9. There is a subtle point missing from the proof of the constructions of the equilateral triangle and of the perpendicular bisector above: How do we know that the two circles actually intersect in two different points? We actually need a little more than the axioms of Euclid to completely justify this. Here is a brief description: For any X on the perpendicular bisector of AB, AX BX. In the particular case when X is the midpoint M of AB, AX AB. When X is chosen to be on the perpendicular bisector but 4 times as far from M as M is from A, then AM < AX and AM AB. Then, using a fact about the real numbers, since AB < AB < AB there must exist a position P on the perpendicular bisector where AP is exactly AB. And then Q is the point that is the reflection of P across AB. Thus, since AB AP AQ and BA BP BQ, P and Q are two different intersection points of the two circles. Proposition 7.0. Given a line l and a point P on l, we can construct a line L containing the point P that is perpendicular to l. Proof. Construction: Given l and P on l, let Q be any other point on l. Construct the circle with center P and radius P Q. This circle intersects the line l at Q and at another point R. Construct the circle with center Q and radius QR and the circle with center R and radius QR, and label their two intersection points as A and B. Then construct the line L AB. Then L is the desired line contains P and is perpendicular to l. Proof that the construction works: Since in the above construction, P is the midpoint of QR, since P Q P R, the desired perpendicular line is the perpendicular bisector of QR. The rest of the construction follows the construction of the perpendicular bisector previous proposition. Proposition 7.. Given a line l and a point Q not on l, we may construct a line through Q that is perpendicular to l in one unique point on l, using only an unmarked straight edge and a collapsing compass. Proof. Left as an exercise Proposition 7.. Given a line l and a point P not on l, we may construct the line through P that is parallel to l, using an unmarked straight edge and a collapsing compass.

15 Proof. Left as an exercise HIGHER GEOMETRY 5 Proposition 7.3. Given a line segment AB and a point P in the plane, we may construct a rectangle ABCD such that AB one side and such that AD AP, using an unmarked straight edge and a collapsing compass. Proof. Left as an exercise 8. Computing areas, the Pythagorean Theorem Definition 8.. The area of a rectangle is defined to be the positive real number that is the length times the width of the rectangle. A Lw The area of other regions of the plane is defined by assuming that area is cumulative, meaning that if a region is subdivided into pieces, the sum of the areas of the pieces is the area of the whole region. A general set S in the plane is said to have area A if for every positive number ε, there are sets S ε and S ε with known areas such that S ε S S ε and A ε < area S ε area S ε < A + ε. That is, the area of S is defined if it can be approximated as closely as desired on the inside and outside by sets that have known area. Remark 8.. It is a fact that even though a set can be subdivided in multiple ways, there is at most one possible area of the set. We will next examine the area of a parallelogram. Given a parallelogram ABCD, let one side say AB be designated the base of the parallelogram. Then, if we construct the line perpendicular to AB that contains C and call the intersection point X [see Proposition 7.]. We define the corresponding height of the parallelogram to be CX. Proposition 8.3. The area of a parallelogram is the product of the length of a base and the corresponding height. A bh. Proof. Sketch Using the X constructed above, we have several different situations that can occur. In one scenario, C is inside AB. Then the triangle AXC can be removed and translated so that AC BD. The resulting figure is a rectangle with length and width being b AB XB + AX and h CX. A similar calculation also works for the other scenarios. Remark 8.4. Observe that this means that the area can be computed in two different ways using different choices of base. We will next find the area of a triangle. Given a triangle ABC, let one side say AB be designated the base of the triangle. Then, if we construct the line perpendicular to AB that contains C and call the intersection point X [again we use Proposition 7.]. The point X is called the foot of the altitude from C to AB. We define the corresponding height of the triangle to be CX. Proposition 8.5. The area of a triangle with base length b and the corresponding height h is A bh.

16 6 KEN RICHARDSON Proof. Sketch Let ABC be the triangle, with X being the foot of the altitude from C to AB. Let M be the midpoint of AC. We rotate a copy of ABC 80 around M to produce A B C, where by construction A C and C A. The union of the two triangles is the quadrilateral ABCB. Using congruent alternate interior angles, we can prove that AB B C and BC AB, so that ABCB is a parallelogram whose area is twice the area of ABC. Then area ABC area ABCB AB CX bh. Proposition 8.6. The area of a trapezoid with two parallel sides of lengths b and b and corresponding height h is Area b + b h Proof. Sketch Given ABCD with parallel sides AB and CD such that AB b and CD b, we construct the diagonal AC, which subdivides the quadrilateral into two triangles, ABC with base and CDA. Both triangles have a height h, which is the distance between the parallel lines AB and CD. By the formula for the area of a triangle, Area ABCD Area ABC + Area CDA b h + b h b + b h. Before continuing, we need to learn the definition of π. Given any circle, the ratio of its circumference C to its diameter D is C D π. It may be surprising that the ratio is the same for any size circle. The reason is that both circumference and diameter are length measurements, i.e. -dimensional measurements, so rescaling causes these measurements to change at the same rate. Since D r, this definition implies that C πd πr. Proposition 8.7. The area of a disk of radius r is A πr. Proof. We first imagine an approximation of the disk from outside or in by a set of N skinny isosceles triangles with one vertex at the center of the circle and the other two vertices at or near the circumference of the circle. Then the area of each triangle is bh br. Adding up all of these areas, the area of the whole disk is A N br Nb r.

17 HIGHER GEOMETRY 7 If N is really large, Nb approaches the circumference πr of the circle, so as N, A πr r πr. Remark 8.8. Technically, a circle does not contain its interior, so really its area is zero. If one wishes to include the interior, the word disk should be used. We can now use our computations of area to prove something about right triangles. First, the long side of a right triangle is called the hypothenuse; it is the side opposite the right angle. The two shorter sides are called the legs of the right triangle. Theorem 8.9. If a right triangle has a hypothenuse of length c and legs of length a and b, then a + b c. Proof. Construct a line segment of length a + b that is subdivided into two line segments of length a and b. We then construct a square with sides of length a + b, all of which are subdivided into two line segments of length a and b in clockwise order so that each vertex of the square is a vertex of two sides of lengths a and b. We then connect the middle vertices of the sides in clockwise order to make 4 right triangles. Then each of these right triangles has sides a and b with a right angle between them. Thus, by the SAS triangle theorem, each of these four triangles is congruent to the right triangle with sides a, b, c. The sides of length c form a rhombus on the interior of the big a + b a + b square. Then at one of the middle vertices of the sides of the big square, let D be the interior angle of the rhombus of side length c. Let A and B be the other two angles, coming from angles opposite the sides of length a and b from the corner right triangles. Then, because A + D + B is a straight angle, m A + m B + m D 80. Also, from the sum of interior angles of the right triangle, m A + m B , so that m A + m B 90. Substituting in the previous equation, we get that 90 + m D 80, so that m D 90. Thus, each interior angle of the rhombus is a right angle, so the rhombus is a square. Now, using this information, we can compute the area of the large square in two different ways. On one hand, Area a + b a + b a + ab + b. By computing the area by adding up the areas of the corner triangles and the interior square, we get Area 4 ab + c ab + c. Thus, a + ab + b ab + c, so after subtracting ab from both sides, a + b c. Theorem 8.0. Converse of the Pythagorean Theorem Given any triangle whose sides have lengths a, b, c, if a + b c, then the angle opposite the side of length c is a right triangle.

18 8 KEN RICHARDSON 9. Constructible lengths and the Rigid Compass Theorem We say that a number x is constructible if we can construct a line segment with that length starting from a given line segment of length, using only an unmarked straight edge and collapsing compass. For example, any positive integer is constructible by the following construction. Given a line segment AB of length, extend it to a ray AB. Construct a circle centered at B with radius AB, and it will intersect AB at A and at a new point B. Similarly, construct a circle centered at B with radius BB AB, and let the new point of intersection with AB be the point B 3. Then, construct a circle centered at B 3 with radius B 3 B, and let the new point of intersection with AB be the point B 4. This process could be continued indefinitely, creating B n+ after B n in the same way. By construction, the line segment AB n has length n. Using the Pythagorean Theorem, we can determine how to construct many other lengths. For example, given AB of length, we extend it to a line AB. We may then create B as above and then construct the perpendicular bisector of AB. By constructing the circle with center B and radius AB, and letting one of the intersection points with the perpendicular bisector be C, BC as well, and so it we construct AC, by the Pythagorean theorem, AC will be AC AB + BC +. Thus, is constructible. And we can continue. As above, we can construct a right triangle with legs of length and, and then the hypothenuse would have length c + + 3, so 3 is constructible. We could continue this process indefinitely to produce 4, 5, etc., so that for any positive integer n, n is constructible. Also, we can in fact construct any rational number, using the following technique. Start with the line segment AB of length, and extend it to a ray AB. Construct another ray AC so that A is an acute angle measure doesn t matter. Draw a circle centered at A with radius AB, and let P on be the intersection point on AC. Then, using the circle technique as above, construct points P, P 3,..., P n so that AP P P P P 3... P n P n. Then we construct the line segment P n B. Finally, using the technique found in our lab, we construct a line through P so that it is parallel to P n B, and let its intersection with AB be the point Q. Then, as we will soon find out after discussing similar triangles in the next section, AQ. n We have constructed a segment of length, and by adding it to itself using the circles as above, n we can construct any number of the form m, where m is a positive integer. Thus, every positive n rational number is constructible. At this point, you may ask, is any positive real number constructible? It turns out that the answer is no. For example the number cos

20 0 KEN RICHARDSON has side lengths, 3, 4 and another triangle has side lengths 8,, 6, then those two triangles are similar scale factor c 4. Thus, if two sets are similar, all corresponding distance measurements are proportional by the same rescaling constant, and all corresponding angle measurements are the same for the two sets. We now will state some triangle similarity theorems. These theorems allow us to prove that given only a small amount of information, pairs of triangles are similar without constructing the homothety that maps one to the other. Theorem 0.. SAS Triangle Similarity Theorem If ABC and DEF satisfy AB and B E, then ABC DEF. BC DE EF Theorem 0.. SSS Triangle Similarity Theorem If ABC and DEF satisfy AB BC, then ABC DEF. AC EF DF DE Remark 0.3. The example we gave with one triangle having side lengths, 3, 4 and another triangle having side lengths 8,, 6: the SSS triangle similarity theorem proves that these triangles are similar, so there exists a homothety taking one to the other. The scale factor is 4. Theorem 0.4. AA Triangle Similarity Theorem If ABC and DEF satisfy A D and B E, then ABC DEF. We may now derive some consequences of these similarity theorems. Proposition 0.5. Triangle Midsection TheoremGiven triangle ABC, let M be the midpoint of AB, and let N be the midpoint of BC. Then, if we construct MN, then MN AC. Proof. Given the constructions above, since BM BN and B BA BC B, by the SAS Triangle Similarity Theorem, MBN ABC. Then MBN ABC since they are corresponding angles. Since AB is a transversal for MN and AC, and since the corresponding angles are congruent, MN AC. In the previous chapter, we constructed a line segment of length from a given line segment of n length. If you look back at the construction, the reason the construction works is through the AA Similarity Theorem, and the two relevant triangles differ by a homothety with scale factor. n Below is a very interesting fact about quadrilaterals. Proposition 0.6. Given any quadrilateral ABCD, the midpoints of the sides of the quadrilateral when connected either are collinear or form the vertices of a parallelogram. Proof. Sketch For the non-collinear case, construct the midpoints and the diagonal AC. Then apply Proposition 0.5 to the two triangles ACB and ACD formed. Then we can prove that those two midsections are parallel; a similar argument works to show the other midsections are parallel.. Cyclic polygons Recall that a cyclic polygon is a polygon such that all of its vertices lie on a fixed circle. Theorem.. Cyclic Quadrilateral Theorem The opposite angles of a simple cyclic quadrilateral are supplementary.

22 KEN RICHARDSON Theorem.5. Thales Theorem If AB is a diameter of a circle, if C is any other point on the circle, and if we construct AC and BC, then m C 90. Proof. Let O be the midpoint of AB, the center of the circle. By the Inscribed Angle Theorem, m C m AOB Centers of Triangles In this section we examine the intersections of various geometrically defined lines associated to triangles. In many cases, these lines are concurrent, meaning that they intersect in a single point. Proposition.. The circumcenter theorem The perpendicular bisectors of the three sides of any triangle intersect in a single point, and this point is the center of the circle that contains all three vertices of the triangle. Proof. Given a triangle ABC, consider the point O that is the intersection of the perpendicular bisectors of AB and BC. Since O is on the perpendicular bisector of AB, OA OB as in Proposition 7.7, and since it is on the perpendicular bisector of BC, OB OC. Thus, OA OC, and so O is on the perpendicular bisector of AC, so that the perpendicular bisectors of the three sides intersect in O. Also, since OA OB OC, O is the center of the circle of radius OA that contains the vertices of ABC. Remark.. The point O in the proposition above is called the circumcenter and is often denoted O. The circle of radius OA and center O is said to circumscribe the triangle. Definition.3. A median of a triangle is a line segment connecting a vertex with the midpoint of the opposite side of the triangle. Proposition.4. The centroid theorem The medians of any triangle are concurrent. Proof. Left as an exercise. Remark.5. The point of intersection above is called the centroid of the triangle and is often denoted G. This is also the center of mass of the triangle if it has a uniform density per unit area. Definition.6. Given an angle XY Z, its angle bisector is the line Y P such that XY P P Y Z. Proposition.7. The incenter theorem The angle bisectors of any triangle are concurrent, and the point of intersection is the center of the circle that is tangent to all three sides of the triangle. Proof. Left as an exercise. Remark.8. The point of intersection above is called the incenter and is often denoted I, and we say that the circle that is centered at that point and that is tangent to the sides of the triangle inscribes the triangle. Definition.9. The altitude from a vertex of a triangle is the line containing the vertex that is also perpendicular to the line containing the opposite side. That is, given ABC, for instance the altitude from A is a line containing A that is perpendicular to BC. Proposition.0. The orthocenter theorem The altitudes of any triangle are concurrent.

23 Proof. Postponed until we discuss analytic geometry HIGHER GEOMETRY 3 Remark.. The point of intersection of the altitudes is called the orthocenter and is often denoted H. Definition.. A set of points in the plane is called collinear if the set of points is contained in a single line. Theorem.3. For any given triangle, the centers O, G, H are collinear, and G is always between O and H, and GH OH. Remark.4. The line containing O, G, H is called the Euler line of the triangle. Definition.5. A cevian is a line segment from the vertex of a triangle to a point on the opposite side of the triangle. Remark.6. A median is a particular example of a cevian. Theorem.7. Ceva s Theorem Let ABC be given, and let P, Q, R be points on BC, AC, and AB. Construct the cevians AP, BQ, and CR. If the three cevians are concurrent, then AR BP CQ AQ CP BR. Remark.8. One particular example of this occurs when P, Q, R are the midpoints of the opposite sides, and AP, BQ, and CR are the medians that meet at the centroid. In that case AR BR, BP CP, and CQ AQ, so the ratio is. Theorem.9. Converse to Ceva s Theorem Let ABC be given, and let P, Q, R be points on BC, AC, and AB. Construct the cevians AP, BQ, and CR. If AR BP CQ AQ CP BR, then the three cevians are concurrent. Theorem.0. Menelaus Theorem Let ABC be given, and suppose that all the sides are extended as lines. Let X, Y, Z be points on AB, BC, and AC. Suppose that X, Y, and Z are collinear. Then AX BY CZ AZ CY BX. Theorem.. Partial Converse to Menelaus Theorem Let ABC be given, and suppose that all the sides are extended as lines. Let X, Y, Z be points on AB, BC, and AC, respectively, but where Z is not on AC. Suppose that AX BY CZ AZ CY BX. Then X, Y, and Z are collinear. A similar fact is true when X, Y, Z are points on AB, BC, and AC, respectively, but not on AB, BC, and AC, respectively.

24 4 KEN RICHARDSON 3. Introduction to Analytic Geometry The subject of analytic geometry comes from the modern view of geometry, where instead of Euclid s Axioms we accept as given: the Euclidean plane R {x, y : x, y R} also denoted E the distance D between any two points x, y and x, y is as in the Pythagorean theorem: D x x + y y. Many times it is more convenient to square this equation: D x x + y y. For any two points x, y and x, y that are horizontal to each other, note that D x x + y y x x x x. Similarly, if x, y and x, y are on the same vertical line, then D x x + y y y y y y In any case, Euclid s axioms follow automatically from the given hypotheses for Euclidean R, so we actually still have all the theorems we have proved previously. In some cases, we gain some understanding by considering the geometric figures placed in the x, y coordinate setting. In the analytic geometry setting, we do not worry if we can or cannot construct something with a straight edge and compass. Then the fundamental objects are given by algebraic equations. For example, the circle of radius r centered at 0, 0 is defined to be the set of points x, y of distance r from 0, 0, so it has the equation x + y r. More generally, the circle of radius r centered at h, k is the set of points x, y of distance r from h, k, so the equation is x h + y k r. Lines have important features in analytic geometry. Let l be any line in the plane. Given two different points x, y and x, y on the line, we define the slope m between those two points as m y x y y x x. We first observe that we can reverse the roles of the first and the second point, because for example y y y + y y y, and then y y y y x x x x y y. x x Note that if the line is a vertical line, the values of x are always the same, so the denominator is 0, and we say m undefined no matter what points are chosen. Notice that if a line is horizontal, the values of y at every point on the line are the same, and so the slope is 0 for any two points chosen on the line.

25 HIGHER GEOMETRY 5 Note that we automatically assume that horizontal lines are parallel to each other and that vertical lines are parallel to each other, because vertical and horizontal lines are automatically perpendicular by the converse to the Pythagorean theorem. We have the following important theorem for slopes. Proposition 3.. Slope theorem On any given line in R, the slope between any two different points on the line is the same. Proof. We have already shown that this is the case for horizontal and vertical lines. Otherwise, for a line that is neither horizontal nor vertical, suppose that we have computed the slope m between A x, y and B x, y on a given line. We will show that if we change B x, y to a third point C x 3, y 3 on that line, then the slope between A and C is also m. If we can show that is true, then we could change the roles of first and third point and then change the first point as well, so the theorem would be proven. Now, given the three points, we also construct the points D x, y and E x 3, y, and we construct AD, AE, BD, CE. Then BD CE since they are both vertical, so ABD ACE since they are either corresponding angles or alternate interior angles for the transversal AC. Since also ADB AEC since they are both right angles, by AA Triangle Similarity Theorem, ABD ACE. Thus, y 3 y x 3 x EC EA DB DA, or y y x x. So now we have that the slopes between A and B and between A and C have the same absolute value. Next, we note that if C is on AB, then y 3 y and y y have the same sign, and also x 3 x and x x have the same sign, so in that case the slopes between A and B and between A and C are the same. If C is on AB but not on AB, then y 3 y and y y have opposite signs, and also x 3 x and x x have opposite signs, so also the slopes between A and B and between A and C are the same in this other case. Therefore, in all cases, the slopes are the same. Proposition 3.. Converse to Slope theorem Let A, B, C be any three different points in the plane. If the slope between A and B is the same as the slope between A and C, then A, B, and C are collinear. Proof. Left as an exercise. theorem is used. Again, similar triangles are used, but this time the SAS similarity 4. Equations of Lines Because of the slope theorem of its converse, we can write precise equations for sets of points that are on a line. There are various versions of these equations. Vertical line equation: The set of points x, y that goes through a vertical line through x 0, y 0 is exactly the solution set of the equation x x 0, since x, y is on the vertical line if and only if x x 0.

26 6 KEN RICHARDSON Point-Slope equation of a line: Given a point x 0, y 0 and a nonvertical line of slope m, by the slope theorem any other point x, y on the line satisfies y y 0 x x 0 m. Then, multiplying by the denominator on both sides, we get y y 0 m x x 0. Note that the set of points x, y satisfying the equation above is exactly the set of points on the line; even x 0, y 0 works in the equation. A slight problem with this equation is the nonuniqueness. That is, there is more than one possible version of this equation. If x, y is any other point on the line, then the expression y y m x x also gives a point-slope equation of this line. The remedy for this is as follows. Slope-Intercept equation of a line: Given a point 0, b on the y-axis and a nonvertical line through that point of slope m, by the equation above, or after add b to both sides and simplifying, y b m x 0, y mx + b. Note that there is only one possible equation like this. The number b is called the y-intercept. Intercept-Intercept equation of a line: Given a nonvertical, nonhorizontal line that hits the x-axis at a, 0 and the y-axis at 0, b [so that a is the x-intercept and b is the y-intercept. We also must assume that the line does not go through the origin, so that both a and b are nonzero. Then by the point-slope equation we have y b b 0 x 0, 0 a so we get a new equation after a little algebra: y b b a x y + b a x b b y + b b a x b b x a + y. b This is the intercept-intercept form of the equation of the line. When such an equation exists, its form is unique. Standard Form of the equation of a line: Every line even vertical lines can be written in the standard form Ax + By + C 0, for some constants A, B, C. Note that the vertical lines x x 0 have this form with A, B 0, C x 0, and nonvertical lines also have the same form with A m, B, C b from the slope-intercept form.

27 HIGHER GEOMETRY 7 This standard form has the same problem as the point-slope form, in that the equation of a given line is not unique. That is, we could multiply the equation by a constant like and still get the same line. 5. Using Analytic Geometry The slopes of lines are very useful to us. For example, we can tell when two lines are parallel or perpendicular. Proposition 5.. Two lines in R are parallel if and only if their slopes defined or undefined are the same. Proof. Left as an exercise. Forward direction: For vertical and horizontal lines, the proof is simple. Otherwise, given two parallel lines, cut them both with horizontal and vertical transversals. Using facts about angles and similar triangles, we can show that the slopes are the same. The backward direction is similar. Recall that if x is a real number, the number is called its reciprocal. In the special case where x x a is a fraction, note that the reciprocal is b x a/b b a b a. Proposition 5.. Two lines in R are perpendicular if and only if their slopes are negative reciprocals. [For the purpose of including horizontal and vertical lines, we let undefined, and 0 0.] undefined Proof. By the verbage in brackets, the result is true for vertical and horizontal lines. So we only need to prove it when the slope is a nonzero real number. Suppose two lines l and l intersect at A x 0, y 0, and let B x, y be any other point of l such that x > x 0 without loss of generality, we choose l to be the line with positive slope. We intersect the vertical line through B with l to get the point C x, y on l, and we construct AC. Also, let D x, y 0 be the point on AC that is at the same vertical position as A, and construct AD. By construction, ABD and CAD are right triangles, with both angles at D being right angles. Since BAC is a right angle, m DAB + m DAC 90, but also m DAC + m DCA , so m DCA 90 m DAC m DAB. Thus, by the AA Triangle Similarity Theorem, ABD CAD. Then BD AD AD CD. Then observe that the slopes m of l and m of l satisfy BD CD m m AD AD AD CD. CD AD

28 8 KEN RICHARDSON Thus the slopes are negative reciprocals. Sketch With the same points constructed as above, we assume the equation first, and then use the SAS Triangle Similarity Theorem to prove that the triangles are similar, and then we can show that BAC is a right angle. We now can derive another important formula that we will use. Proposition 5.3. Midpoint formula Given line segment AB in the plane with A x, y, B x, y, the midpoint M of the line segment is x + x M, y + y. Proof. We will show a that M is on AB and b that AM MB. a The slope from A to B is The slope from A to M is y +y y x +x x m y y x x. y +y y x +x x y +y y x +x x y y x x m. y y x x Thus, M is on the same line as AB, and since x +x is between x and x, M is between A and B. b Observe that x + x y + y AM x + y x + x x y + y + y x x y y + and MB x x + x + y y + y x x + x y + y + y x x y y + AM.

29 HIGHER GEOMETRY 9 We may now prove results using analytic geometry. The idea of proof is this. We place the objects in R, and then we use the R computations to prove whatever is needed. Here is a basic tip: by using an isometry, we may place an object conveniently in a particular location in R. For example, given a triangle, we may after an isometry place it so that it has one vertex at 0, 0 and one side on the x-axis. Below are some examples of proofs by analytic geometry. Proposition 5.4. In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices. Proof. Given any right triangle ABC with right angle B, by applying an isometry, we may assume B 0, 0, A a, 0, and C 0, c for some nonzero real numbers a, c. Then the midpoint of the hypotenus AC is M a, c. Then clearly a c a c AM CM + c + a c + Also, BM a 0 + c 0 CM. Proposition 5.5. In any given triangle, the altitudes from the three vertices are concurrent. Proof. Given any triangle, by an isometry we orient it so that its longest side is on the x-axis and the opposite vertex is on the positive y-axis at 0, b with b > 0. Let a, 0 and c, 0 be the other two vertices on the x-axis. The altitude from 0, b is x 0. The altitude from a, 0 should have slope the negative reciprocal of the slope from c, 0 to 0, b, so the slope should be Thus, the equation of the altitude from a, 0 is b c b. c y c b x a c b x ac b. Similarly, the equation from the altitude from c, 0 is y a b x c a b x ac b. The intersection of all three altitudes at x 0 is H 0, ac b 6. Trigonometry. The unit circle S is the set of points in R that are a distance from the origin 0, 0. Thus, x, y S if and only if x + y. The counterclockwise angle θ measured between the x-axis and the ray through the origin and x, y S is called the angle corresponding to x, y. If it is positive, it is the same as arclength along S from, 0 to x, y counterclockwise. Note that given a point x, y corresponding to angle θ, that same point corresponds to θ + πk for any k Z.

30 30 KEN RICHARDSON The definitions of the trig functions of the angle θ corresponding to the point x, y S : cos θ x sin θ y tan θ y sin θ slope of the line through 0, 0 and x, y x cos θ sec θ x cos θ csc θ y sin θ cot θ x y tan θ. To find actual values of trig functions, we make a few observations using geometry. The radian measure of an angle is the arclength along the unit circle, so for instance π radians is 360 at the 3 point, 0, π radians is 90 at the point 0,, π radians is 6 30 at the point,, 3π 4 radians is 35 at the point,. The last two come from the two standard right triangles from geometry: the triangle whose sides are, 3, ; and the triangle whose sides are,. Note that in all cases the hypothenuse is, coinciding with the radius of the unit, circle. Examples of values of trig functions, from the information above: cos π, cos π 0, cos π 6 3, sin π 0, sin π, sin π 6, tan π 0 0 ; csc π 6 ; sec π undefined ; etc. 0 More examples: θ 3π corresponds to, 4 Thus, 3 and θ π corresponds to,. 6 cos 3π 4 cos π 6 3 sin 3π 4 sin π 6 tan 3π 4 tan π 6 3 sec 3π 4 sec π 6 3 csc 3π 4 csc π 6 cot 3π 4 cot π 6 3 For other values of trig functions that are not easily obtained by geometric figures, one must use a calculator, which ultimately uses the Taylor series to calculate the values of the trig functions. From calculus we know cos x x! + x4 4!... k x k k!, k0 sin x x x3 3! + x5 5!... k x k+ k +!, k0

31 HIGHER GEOMETRY 3 and these Taylor series converge to the exact values of the functions at all x. For example, we have cos! + 4 4! 6 6! + 8 8! 0 0! + k k k!. k0 We may calculate! + 4 4! 6 6! + 8 8! 0 0! cos i.e. after more terms are added. Note that because of these Taylor series facts, we may derive the famous Euler identity. In plotting trig functions as graphs in R, for example y sin x, realize that the angle is plotted on the horizontal axis. Here are some examples of plots of trig functions.

32 3 KEN RICHARDSON From the definitions, we see immediately a few things that must be true called trig identities: and same for all the other trig functions. sin θ + πk sin θ for any k Z cos θ + sin θ Pythagorean identity This means cos θ +sin θ lazy mathematician notation. From the Pythagorean identity, we get two other identities by dividing by cos θ and by sin θ, respectively: + tan θ sec θ, cot θ + csc θ. By looking at the angles on the unit circle, you can see what happens when you change an angle to its negative. Here are the results, the negative angle identities: Thus you can get all the others. For example, cos A cos A sin A sin A tan A sin A cos A sin A cos A tan A, and so on. The angle sum identities are very useful but hard to derive. Here they are: sin A + B sin A cos B + sin B cos A cos A + B cos A cos B sin A sin B.

33 HIGHER GEOMETRY 33 By replacing B with B and using the negative angle identities, you get the angle difference identities: sin A B sin A cos B sin B cos A cos A B cos A cos B + sin A sin B. Using the angle sum identities, you can replace B with A to get the double angle identities: sin A sin A + A... sin A cos A cos A cos A + A... cos A sin A. By either substituting sin A cos A or cos A sin A in the cos A identity, we actually get cos A cos A sin A cos A sin A. So you get to choose one of three identities for cos A. Finally, using the equation above, we can solve for cos A or sin A in terms of cos A. Then the results are: cos A + cos A, sin A cos A. The trigonometric identities are often used in calculus for simplifying algebraic expressions in different ways. Here are some examples: cot C csc C cot C csc C cos x sin x sec x cos C sin C sin C cos C sin C sin C cos C. cos x sin x sec C cos x sin x cos x cos x sin x cos x sin x cos x + sin x 3 cot B sin π B + cos B cot B sin π B + cos B cos B sin π cos B sin B cos π + cos B sin B cos B sin B + cos B sin B cos B + cos B cos B

34 34 KEN RICHARDSON 4 tan x cscx cosx sinx secx 5 tan x csc x cos x sin x sec x cosa sin a cosa cos a sin a cos a sin x cos x sinx cosx cos x sin x sin x cos x sin x cos x cos x sin x sin3 x cos x sin x cos x sin x. sin a sin a sin a sin a sin a + sin a sin a 4 sin a sin a sin a. The inverse trig functions are used to compute angles such as θ in the equation below: cos θ. We see that this occurs at two possible points on the unit circle, the two points where x cos θ. That is, x, y, or x, y,. We compute the possible values for θ for each of these cases: θ 3π + πk for some k Z, or 4 θ 3π + πm for some m Z. 4 We see that since there are more than one solution to cos θ in fact, an infinite number, there is no exact inverse function to cos or to any of the other 5 trig functions. In order to get a function θ out of the relation cos θ x, we restrict the domain of cos to the branch of principal angles, a certain interval of allowed angles, such that the restriction of cos to that domain actually has an inverse. Every trig function does have such a principle angle domain. The rule of thumb is that the interval [ ] 0, π the first quadrant is always included for the positive values of all the trig functions, and then either [ π, π] or [ π, 0] is added to allow negative values. Either choice would work in the cases of tan and cot, so in those cases the choice where the asymptote is not in the middle is used.

35 HIGHER GEOMETRY 35 Here are the results: Principle angle domains: sin θ, csc θ, tan θ : π θ π cos θ, sec θ, cot θ : 0 θ π In all cases, the inverse function for the given trig function is arcfcn, where fcn is the function, and it means the principal angle whose fcn is. For examples, arcsin t principal angle whose sin value is t arcsin principal angle whose sin value is π 6 arccot p principal angle whose cot value is p arccot 3 principal angle whose cot value is 3 5π 6 arccsc a principal angle whose csc value is a arccsc principal angle whose csc value is π Trigonometry and Geometry Suppose that ABC is a right triangle with C being a right angle. Let a, b, c be the lengths of the sides opposite the angles A, B, C, respectively. By the Pythagorean theorem, a + b c. Note that we can place this triangle in the unit circle if we first rescale the triangle by a factor of. Then we have a picture where the point B could be placed at 0, 0, C could be placed at a, 0, c c and A could be placed at a, b c c. Then from trigonometry we have cos B a c sin B b c tan B b a

36 36 KEN RICHARDSON and so on, so there is a relation with the sides and all the trig functions. Many students and teachers remember these relationships by the word SOHCAHTOA: sin θ O H cos θ A H tan θ O A, where θ stands for one of the acute angles in a right triangle, and O length of side Opposite to θ, A length of side Adjacent to θ, H length of the Hypotenuse. Note that the calculations above only work for right triangles. However, there are two other laws that allow us to relate the sides, angles, and trig functions on any triangle. Theorem 7.. Law of Sines Let ABC be any triangle, with a, b, c the lengths of the sides opposite the angles A, B, C, respectively. Then sin A sin B sin C. a b c Proof. First, suppose that A or C is a right angle. Let s assume without loss of generality that A is the right angle. Then by the calculations for right triangles, sin C c a, so sin C sin A. c a a Next, suppose that both A and C are acute angles. Let the point X on AC be the foot of the altitude from vertex B, and let h BX. Then AXB and CXB are right triangles, with both angles at X being the right angles. By the right triangle trigonometry, we have so or sin A h c, sin C h a, h c sin A a sin C, sin A a sin C. c Finally, suppose that one of A and C is an obtuse angle. Without loss of generality, suppose A is obtuse. Then we construct the altitude from vertex B to CA, and let X be the foot of the altitude, which is on CA but not on AC. Let h AX. Then AXB and CXB are both right triangles. By the right triangle trigonometry facts, sin A sin π A h c, sin C h a,

37 HIGHER GEOMETRY 37 so again we get Thus, in all cases we have h a sin C c sin A and sin A sin C. a c sin A a Repeating the arguments for the side BC, we get sin B b sin C. c sin C. c Theorem 7.. Law of Cosines et ABC be any triangle, with a, b, c the lengths of the sides opposite the angles A, B, C, respectively. Then c a + b ab cos C and two other formulas obtained by permuting the roles of a, b, c and the angles A, B, C. Proof. First, if C is a right angle, then cos C 0, and c a + b by the Pythagorean theorem. If A is a right angle, then cos C b and a b + c a, so b ab cos C ab b b b a b a + c, or c a + b ab cos C. Next, if A and C are both acute angles, then the altitude from B intersects AC at the point X, so that AXB and CXB are both right triangles with right angle at X. Then AX b CX and b CX + BX c, CX + BX a. Subtracting and using the fact that CX a cos C, so b CX CX c a b b CX + CX CX c a b ab cos C c a, c a + b ab cos C. Next, if A is obtuse and C is acute, then AX CX b and AX + BX CX b + BX c, CX + BX a.

38 38 KEN RICHARDSON Subtracting and using the fact that a cos C CX, CX b CX c a CX b CX + b CX c a b ab cos C c a, so c a + b ab cos C. Finally, the last case is when A is acute and C is obtuse. Then AX CX + b and AX + BX CX + b + BX c, CX + BX a. Subtracting and using the fact that a cos π C a cos C CX, CX + b CX c a CX + b CX + b CX c a b ab cos C c a, so c a + b ab cos C. Remark 7.3. Note that one can think of the Law of Cosines as a direct generalization of the Pythagorean theorem: the equation C π is true if and only if c a + b. The Law of Sines and the Law of Cosines can be used to find missing angles and side lengths in triangles. Example 7.4. A triangle has sides of length, 3, and 4. Using trigonometry and your calculator, find the angles of the triangle. Proof. Let A angle opposite the side; B angle opposite the 3 side, and C angle opposite the 4 side. By the Law of Cosines, cos C, so cos C 3, so C arccos π 4 sin B By the Law of Sines, sin C 3 4 B arcsin , so sin B , so Then A π 8. Vectors and vector operations In this section, we will study vectors in the coordinate plane R. As in analytic geometry, we will describe points of the plane using their coordinate locations x, y. Sometimes we will think of these points as vectors, meaning that we are considering them to be arrows, where the arrow starts at 0, 0 its tail and ends at x, y its head. In order to

39 HIGHER GEOMETRY 39 emphasize the directional nature of the vectors, we will sometimes use other notation: vector from 0, 0 to x, y x, y x ı + y j xı + yj x, y, but sometimes we still refer to the vector as vector from 0, 0 to x, y x, y. The most important things we need to remember about a vector is that it has two properties: length or norm and direction. A quantity that does not have a direction is called a scalar; for example, a real number is a scalar. The length of a vector is denoted various different ways and is given as usual by the Pythagorean theorem. length of x, y norm of x, y x, y x, y x, y x + y. We often will want to move vectors around from one starting location to another, but without changing length or direction. That is, we consider for example vector from, 3 to 4, 7 vector from 0, 0 to, 4, 4.

40 40 KEN RICHARDSON In general, vector from a, b to c, d vector from 0, 0 to c a, d b c a, d b. We often refer to vectors with a single letter, such as v or w. When we do that, we refer to the individual components using subscripts. For instance, if then w, 7., w, w 7. Vectors are added or subtracted by adding or subtracting the individual coordinates. That is, the definitions of addition and subtraction of vectors are as follows. Given v v, v and w w, w, v + w v + w, v + w v w v w, v w. In both cases, these could also be done in any number of dimensions in the same way: v + w v + w, v + w,... v w v w, v w,.... A scalar multiple of a vector v is a vector of the form cv cv, cv for some c R. The real number c is called a scalar, because it does not have a vector direction, just a magnitude. Again, this could easily be applied to vectors in more than two dimensions. We observe that these sum of vectors can be pictured as follows. The sum v + w can be obtained from the vectors v and w by first placing the vectors tail-to-head, say with the tail of w at the same

41 HIGHER GEOMETRY 4 point as the head of v; then the sum v + w is the vector from the tail of v to the head of w. Because of the commutative property, the vector v + w can also be pictured by reversing the roles of v and w, i.e. by first placing the tail of v at the same point as the head of w. The scalar multiple cv can be pictured as follows. Observe first that slope m v v does not change if v is multiplied by a nonzero scalar: cv cv v v, so the slope of the vector does not change. The length is changed: cv cv + cv c v + c v c v + v of a vector v c v + v c v. So the length of the vector v is multiplied by c. If c is a positive number, then cv points in the same direction as v. If c is a negative number, cv points in the opposite direction of v. Since we now understand scalar multiplication and addition, we can now understand subtraction of vectors. For any two vectors v, w R, v w v + w v + w. Note that w is the same vector as w but with the arrow pointing in the opposite direction. So one way of understanding v w in a picture is to first reverse the vector w and then to add the two vectors. But also one can see that since w + v w v, if we place the vectors v and w tail-to-tail, then the vector v w is the vector whose tail is at the head of w and whose head is at the head of v.

42 4 KEN RICHARDSON There is a special kind of multiplication of vectors called the dot product. The dot product v w of two vectors v and w is actual a scalar quantity and is defined by v w v w + v w. In higher dimensions, the formula is the same. For example, for x, y R 3, The dot product is sometimes denoted x y x y + x y + x 3 y 3. v w v, w Proposition 8.. Properties of addition, subtraction, scalar multiplication, dot products of vectors For any v, w, z R n, b, c R, we have: v w v + w Definition of vector subtraction v + w w + v Commutative property of vector addition v + w + z v + w + z Associative property of vector addition c v + w cv + cw Distributive property for scalar multiplication b + c v bv + bw Another distributive property for scalar multiplication c v w cv cw Another distributive property for scalar multiplication b c v bv cv Another distributive property for scalar multiplication b cv bc v Associative property of scalar multiplication v v v 0 and v v 0 only if v 0 Positivity of the dot product v w w v Symmetry of the dot product cv w c v w v cw Homogeneity of the dot product v w + z v w + v z Linearity of the dot product v + w z v z + w z Linearity of the dot product v w z v w v z Linearity of the dot product v w z v z w z Linearity of the dot product Proof. Left as exercises. To prove each equation, first compute the left side using the definitions, then compute the right side using the definitions, and show they yield the same result. Corollary 8.. A vector v is a unit vector of length if and only if v v. One important property of dot products is the following. Proposition 8.3. Geometric formula for the dot product. For any nonzero vectors v and w, let θ be the angle between the two vectors when the vectors are placed tail-to-tail. Then v w v w cos θ. [The formula also works for zero vectors, but the angle θ is not well-defined.]

43 HIGHER GEOMETRY 43 Proof. Consider any two nonzero vectors v and w, placed tail-to-tail. Letting the points at the heads of the two vectors and the tail be vertices of a triangle, the side lengths of the triangle are v, w, and v w. Using the Law of Cosines, Then v w v + w v w cos θ. Subtracting v v + w w, v w v w v v + w w v w cos θ v v w v v w + w w v v + w w v w cos θ v v + w w v w v v + w w v w cos θ v w v w cos θ, so that v w v w cos θ. We can now use this geometric formula to find the angle between two vectors. For the angle θ between v.5,.3 and w 6.3,.4 satisfies v w v w cos θ cos θ cos θ cos θ θ arccos Also, observe that θ 90 if and only if cos θ 0 if and only if v w 0. Thus: Corollary 8.4. Two vectors v and w are perpendicular if and only if v w 0. Remark 8.5. Two vectors are parallel if one vector is a scalar times the other. 9. Matrices, Linear and Affine transformations We will be using vector operations and matrices to exactly describe transformations of the coordinate plane R. We will be particularly interested in giving explicit formulas for isometries rigid motions. In this section, it will be most convenient to describe points and vectors as column vectors, rather the more customary row vectors. For example, the vector v, 3 can be described in all of these different ways: v, 3 ı 3j. 3 The simplest type of rigid motion is a translation, where every point of R is moved by a vector. Specifically, to move all points by the vector v, the function F : R R that does the job is x x v x + v F +. y y v y + v

44 44 KEN RICHARDSON For example, to move all points by the translation vector, 3, the translation function F is given by x x F +. y y 3 The picture below shows how sets of points are changed by the function F. The primes indicate the locations of the objects after translation by the vector, 3. To express reflections and rotations as functions, we need to use matrices, which are rectangular arrays of numbers. A matrix is an array of real or complex numbers, usually arranged in a rectangular grid. Examples include: x 3 3, π 0 Note that the entries may be variables. We will often label matrices by letters, often capitalized: 0 x 5 a, M π. 0 An n k matrix is one with n rows and k columns. The individual entries are label by the row,column position. For instance, the 3, entry of M above is 3.8, and the notation for this is M Similarly, a 5. Note that two n k matrices A and B are the same if and only if they have identical entries.

45 HIGHER GEOMETRY 45 Vectors are special cases of matrices those that have as one of the dimensions. For example v 3 is a column vector or vertical vector, and w is a row vector or horizontal vector. In these cases, we often use a single index to refer to entries or components of the vector. In the examples above, v 3 : v 3, and w 4 w 4. Addition and subtraction are most simple operations to define. Both of these operations are done entry-wise for matrices with the same dimensions. For example, 4 0 r 0 45 t 4 0 r 0 45 t t t r t r t Multiplication and related operations are slightly more complicated. However, one simple form of multiplication is called scalar multiplication. This concerns multiplying a matrix by a real or complex number called a scalar. The result is a new matrix of the same dimensions, where every entry is the scalar times the corresponding entry of the matrix. For example, x x In general, matrix multiplication is more complicated. First, one is permitted to multiply two matrices A and B if the middle dimensions agree that is, if A is an n k matrix and B is an r s matrix and k r. In this case, where A is an n k matrix and B is a k s matrix, the product AB is a new n s matrix whose entries are defined by AB ij A i B j + A i B j A ik B kj k A im B mj. m

46 46 KEN RICHARDSON For example, we can multiply the matrices P and Q because P has dimensions 3 and Q has dimensions 3 3 the 3 s match. The result is P Q P Q P Q P Q P Q Note that the dot product of vectors is a special case of matrix multiplication. That is, if v v, v,..., v n and w w,..., w n are two vectors of dimension n, the dot product v w is defined to be the scalar v w v w v n w n n k v kw k. Observe that this is the same as a matrix product of v as a row vector with w as a column vector, for the result is a matrix: v v... v n w w. w n n v k w k. k For example, , 6, 4, 5, 5,, 0. In general, if we multiply matrices P and Q, the i, j entry of P Q is the dot product of the i th row of P with the j th column of Q. This seems to be a counterintuitive definition of matrix multiplication; the only justification for its use is that it has many very nice applications that remind us of multiplication by numbers in algebra. One very special matrix I I n is called the identity n n matrix; this matrix satisfies the identity property for matrix multiplication. This matrix is defined by I ij δ ij the Kronecker delta symbol { if i j 0 otherwise,

47 so that HIGHER GEOMETRY I Notice that for any n m matrix A, I n A A. For example, Matrix operations satisfy the following properties. Let A, B, C, D be any matrices. First we discuss the five properties of matrix addition: Closure under Addition: If A + B is defined ie A and B have the same dimensions, then it is another matrix of the same size. Associative Property of Addition: A + B + C A + B + C if A, B, C have the same dimensions. Identity Property for Addition: 0+A A+0 A, if 0 is the matrix with the same dimensions as A but with every entry zero. Inverse Property for Addition: Let A A. Then A + A A + A 0, with 0 the corresponding matrix of zeros. Commutative Property of Addition: A + B B + A, whenever the matrix sum is defined. The following properties are satisfied for scalar multiplication. Let x, y be scalars. Closure under Scalar Multiplication: xa is another matrix of the same dimensions as A. Associative Property of Scalar Multiplication: x ya xy A Property of : A A Distributive Property #: x + y A xa + ya Distributive Property #: x A + B xa + xb The following properties are satisfied for matrix multiplication: Associative Property of Matrix Multiplication: AB C A BC if defined. Distributive Property of Matrices #: A + B C AC + BC if defined. Distributive Property of Matrices #: A B + C AB + AC if defined. Identity Property of Matrix Multiplication: AI k A and I n A A if A is an n k matrix. Note that some desirable properties do *not* hold for matrix multiplication: The commutative property does not hold in general for matrix multiplication, even for square n n matrices. Thus, the order matters a lot in matrix multiplication. Matrices do not in general have multiplicative inverses. In other words, if A is an n n matrix, then it may be impossible to find a matrix B such that AB I n or BA I n. Each one of the properties that are actually true can be proved using the definitions of the matrix operations. We will specialize to the use of matrices to describe functions from R to R. An affine function F : R R is a function of the form x a b x p F +, y c d y q

48 48 KEN RICHARDSON where a, b, c, d, p, q are fixed real numbers. Notice that the input of this function is a point or x x vector R y, and the output is a new point F R y. It is a combination of a matrix p p 0 multiplication and a translation addition of the vector. If the vector, then q q 0 this special case of an affine transformation is called a linear transformation. One important things to realize about linear and affine transformations: They map lines to lines. Let s consider an example. Let S : R R be defined by x S y x 3 y x + y 3. 0 Observe that this linear transformation maps the origin to itself always happens with linear 0 transformations but not in general with affine transformations. It maps to the first column of the matrix and maps to 0 the second column of the matrix. If we plot the original coordinate grid, we see that the matrix multiplication has the effect of transforming the coordinate grid to a new coordinate grid:

49 HIGHER GEOMETRY 49 Here is another geogebra file created by Jon Ingram where the transformation x F y x y is explored: you can change points in the xy-plane and see how the output changes: Note that in both cases, if we just examine the columns of the matrix, we are able to see what the matrix does. This also means that we are custom-design a linear transformation so that it does what we want it to. In particular, let s construct a linear transformation that will rotate points around the origin by an angle θ. We know that we want the point moved to 0 0 we want the point moved to cos θ sin θ. And also, cos θ + 90 sin θ + 90 cos θ cos 90 sin 90 sin θ sin θ cos 90 + sin 90 cos θ sin θ. cos θ Thus, the desired linear transformation RO θ : R R that rotates points by an angle θ is x cos θ sin θ x RO θ y sin θ cos θ y.

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