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# College Geometry I MATH 461

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This 78 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 461 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/205298/math-461-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Coordinate Geometry JWR Tuesday September 6 2005 Contents 1 Introduction 3 2 Some Fallacies 4 21 Every Angle is a Right Angle 5 22 Every Triangle is lsoscelesl 6 23 Every Triangle is lsoscelesl ll 7 3 A ine Geometry 8 31 Lines 8 32 Af ne Transformations 12 33 Directed Distance 19 34 Points and Vectors 20 35 Area 20 36 Parallelograms 23 37 Menelaus and Ceva 24 38 The Medians and the Centroid 26 4 Euclidean Geometry 30 41 Orthogonal Matrices 30 42 Euclidean Transformations 31 43 Congruence 32 44 Similarity Transformations 33 45 Rotations 34 46 Review 36 47 Angles 37 51 US 48 Addition of Angles More Euclidean Geometry 51 Circles 52 The Circumcircle and the Circumcenter 53 The Altitudes and the Orthocenter 54 Angle Bisectors 55 The lncircle and the lncenter 56 The Euler Line 57 The Nine Point Circle 58 A Coordinate Proof 59 Simson7s Theorem 510 The Butter y 511 Morley7s Theorem 512 Bramagupta and Heron 513 Napoleon7s Theorem 514 The Fermat Point Projective Geometry 61 Homogeneous coordinates 62 Projective Transformations 63 Desargues and Pappus 64 Duality 65 The Projective Line 66 Cross Ratio 67 A Geometric Computer Inversive Geometry 71 The complex projective line 72 Feuerbach7s theorem Klein7s View of geometry 81 The elliptic plane 82 The hyperbolic plane 83 Special relativity Matrix Notation Determinants 69 69 69 70 70 70 70 71 73 C Sets and Transformations 75 1 Introduction These are notes to Math 461 a course in plane geometry I sometimes teach at the University of Wisconsin Students who take this course have com pleted the calculus sequence and have thus seen a certain amount of analytic geometry Many have taken or take concurrently the rst course in linear algebra To make the course accessible to those not familiar with linear al gebra there are three appendices explaining matrix notation determinants and the language of sets and transformations My object is to explain that classical plane geometry is really a subset of algebra ie every theorem in plane geometry can be formulated as a theorem which says that the solutions of one system of polynomial equations satisfy another system of polynomial equations The upside of this is that the criteria for the correctness of proofs become clearer and less reliant on pictures The downside is evident algebra especially complicated but elementary algebra is not nearly so beautiful and compelling as geometry Even the weakest students can appreciate geometric arguments and prove beautiful theorems on their own For this reason the course also includes synthetic arguments as well I have not reproduced these here but instead refer to the excellent texts of lsaacs 4 and Coxeter amp Greitzer 3 as needed It is my hope that the course as a whole conveys the fact that the foundations of geometry can be based on algebra but that it is not always desirable to replace traditional synthetic forms of argument by algebraic arguments The following quote of a quote which I got from page 31 of 3 should serve as a warning The following anecdote was related by ET Bell 1 page 48 Young Princess Elisabeth had successfully attacked a problem in elementary geometry using coordinates As Bell states it The problem is a ne specimen of the sort that are not adapted to the crude brute force of elementary Cartesian geometry77 Her teacher Rene Descartes who invented the coordinate method said that he would not undertake to carry out her solution in a month77 The reduction of geometry to algebra requires the notion of a transfor mation group The transformation group supplies two essential ingredients First it is used to de ne the notion of equivalence in the geometry in question For example7 in Euclidean geometry7 two triangles are congruent iff there is distance preserving transformation carrying one to the other and they are similar iff there is a similarity transformation carrying one to the other Sec ondly7 in each kind of geometry there are normalform theorems which can be used to simplify coordinate proofs For example7 in af ne geometry every tri angle is equivalent to the triangle whose vertices are A0 07 07 B0 17 07 00 01 see Theorem 313 and in Euclidean geometry every triangle is congruent to the triangle whose vertices are of form A 107 B b07 C 00 see Corollary 414 This semester the of cial text is In past semesters 1 have used 4 and many of the exercises and some of the proofs in these notes have been taken from that source 2 Some Fallacies Pictures sometimes lead to erroneous reasoning7 especially if they are not carefully drawn The three examples in this chapter illustrate this I got them from See if you can nd the mistakes Usually the mistake is a kind of sign error resulting from the fact that some point is drawn on the wrong side of some line 21 Every Angle is a Right Angle D P C 0 Figure 1 Every Angle is a Right Anglel Let ABCD be a square and E be a point with BC BE We will show that ZABE is a right angle Take R to be the midpoint of DE7 P to be the midpoint of DC7 Q to be the midpoint of AB7 and O to be the point where the lines PQ and the perpendicular bisector of DE intersect See Figure 21 The triangles AQO and BQO are congruent since 0Q is the perpendicular bisector of AB it follows that AO B0 The triangles DRO and ERO are congruent since R0 is the perpendicular bisector of DE it follows that DO EO Now DA BE as ABCD is a square and E is a point with BC BE Hence the triangles OAD and OBE are congruent because the corresponding sides are equal It follows that ZABE LOBE 7 ZABO LOAD 7 BAO LBAD 22 Every Triangle is Isoscelesl B D Figure 2 An lsosceles Trianglel Let ABO be a triangle we will prove that AB AO Let O be the point where the perpendicular bisector of BC and the angle bisector at A intersect7 D be the midpoint of BO7 and R and Q be the feet of the perpendiculars from O to AB and AC respectively see Figure 22 The right triangles ODB and ODO are congruent since OD OD and DB DO Hence OB 00 Also the right triangles AOR and AOQ are congruent since ZRAO ZQAO A0 is the angle bisector and ZAOR ZAOQ the angles of a triangle sum to 180 degrees and A0 is a common side Hence OR OQ The right triangles BOB and COO are congruent since we have proved OB OO and OR OQ Hence RB QO Now AR AQ as AOR and AOQ are congruent and RB QO as BOB and COO are congruent so AB AR RB AQ QO AO as claimed 23 Every Triangle is Isoscelesl II A B X Figure 3 AX bisects ZBAC In a triangle ABC7 let X be the point at which the angle bisector of the angle at A meets the segment BC By Exercise 22 below we have XB i XC 1 AB AO39 Now ZAXB ZACX ZCAX LC ZA since the angles of a triangle sum to 180 degrees By the Law of Sines Exercise 21 below applied to triangle AXB we have E 7 sin ZBAX i sin ZA 2 AB sin AXB sinZC zA Similarly ZAXC ZABX ZBAX LB ZA so XC i sin 4A 3 AC 7 sinZB ZAf From 1 3 we get sinZC ZA sinZB ZA so LC ZA LB iZA so LC LB so AB AC so ABC is isosceles Exercise 21 The law of sines asserts that for any triangle ABC we have sin LA 7 sin LB 7 sin LC BC CA AB Prove this by computing the area of ABC in three ways Does the argument work for an obtuse triangle What is the sign of the sine Exercise 22 Prove Hint Compute the ratio of the area of ABX to the area of ACX in two different ways 3 Af ne Geometry 31 Lines 31 Throughout R denotes the set of real numbers and R2 denotes the set of pairs of real numbers Thus a point of P E R2 is an ordered pair P Ly of real numbers De nition 32 A line in R2 is a set of form Zzy 6R2azbyc0 where abc E R and either a 31 0 or b 31 0 or both Three or more points are called collinear iff there is a line Z which contains them all Three or more lines are called concurrent iff they have a common point Two lines are said to be parallel iff they do not intersect 33 The two most fundamental axioms of plane geometry are Axiom 1 Two distinct nonparallel lines intersect in a unique point Axiom 2 Two distinct points determine a line Axiom 1 says that two equations a1b1ycl0 a2zb2y020 for lines have a unique common solution the usual case7 no common solution this means that the lines are parallel7 or else de ne the same line which is case if and only if the equations are nonzero multiples of one another The latter two cases are characterized by the condition 1le 7 agbl 0 and in the rst case the intersection point is C152 7 C251 77 1le 7 agbl y 1le 7 agbl 1102 7 1201 Axiom 2 says that for any two distinct points P1 17 yl and P2 27 yg there is a unique line Zxy azbyc0 containing both Remark 35 below gives a formula for this line Theorem 34 I Three points Pl are collinear if arid orily if II Three distinct liries ll am big cl 0 are concurrent or parallel if arid orily if det 12 b2 02 as 53 03 Proof A determinant is unchanged if one row is subtracted from another Hence 1 11 1 1 3 11 733 0 det x2 y2 1 det x2 7 x3 y2 7 yg 0 3 733 1 3 113 1 Evaluating the determinant on the right gives 1 11 1 det 2 732 1 1 3yz ya 2 3 11 93 3 733 1 Dividing by 1 7 3x2 7 x3 shows that the determinant vanishes if and only if 91 i 93 7 12 13 him i zi st This last equation asserts that the slope of the line P1133 equals the slope of the line P2133 Since P3 lies on both lines7 this occurs if and only if the lines are the same7 ie if and only if the points Pl7 P2 P3 are collinear The above proof assumes that 172 31 3 a special argument is required in the contrary case We give another proof which handles both cases at the same time The matrix equation 1 yl 1 a 0 2 yg 1 b 0 x3 y3 1 c 0 says that the points Pl lie on the line ax by c 0 Any nonzero solution ab7 c of this equation must have either a 31 0 or b 31 0 or both Hence the 9 three points P1 are collinear if and only if this matrix equation viewed as a system of three homogeneous linear equations in three unknowns abc has a nonzero solution Part I thus follows from the following Key Fact A homogeneous system ofn linear equations in n unknowns has a nonzero solution if and only if the matrico of eoe eients has determinant zero Part ll is similar7 but there are several cases The matrix equation a1 b1 01 0 0 12 2 02 yo 0 13 3 Cg 1 0 says that the point 010 lies on each of the three lines aiz big cl 0 The three lines are parallel and not vertical if and only if they have the same slope7 ie if and only if ialbl 12b2 13b3 This happens if and only if the matrix equation 11 1 C1 1 0 a2 2 02 m 0 2 13 3 Cg 0 0 has a solution in The lines are vertical and hence parallel if and only if 1 b2 b3 0 This happens if and only if the matrix equation 11 1 C1 0 0 12 2 02 1 0 13 3 Cg 0 0 holds This and the above Key Fact proves only if For if77 assume that the matrix equation 11 1 C1 u 0 12 2 02 1 0 13 3 Cg w 0 has a nonzero solution 1111112 If U 71 07 then x0 11117 10 ow satis es If U 0 and u 71 07 then m 1111 satis es If U u 07 then 1 71 0 so 1 holds 1 Figure 4 P tP1 1 itP0 Remark 35 The point P Ly lies on the line joining the distinct points P1 zhyl and P2 27y2 if and only if the points P177 P27 P are collinear Thus Theorem 34 implies that an equation for this line is 1 711 det x2 121 0 m y 1 It has form ax by c 0 where 091 927 b217 C192291 The points P1 and P2 satisfy this equation since a determinant vanishes if two of its rows are the same Theorem 36 The line connecting the two distinct points P0 x07y0 and P1 xhyl is given by ie a point P Ly lies on Z if and only if ztx117tz07 yty117ty0 for some t E R See Figure 4 Proof These are the parametric equations for the line as taught in Math 222 The formula 0 90 1 0 110 1 0 110 1 det 1 y1 1 tdet 1 y1 1 1 7 t det 1 y1 1 0 9 1 1 11 1 0 110 1 11 shows that any point P of form P tP117t P0 lies on the line Conversely in P lies on the line7 choose t to satisfy one of the two parametric equations and then the equation for the line in the form ax by c 0 shows that the other parametric equation holds as well 1 De nition 37 The line segment connecting points P0 and P1 is the set We say that a point P on the line joining P0 and P1 lies between P0 and P1 iff it lies in the segment P0P1 We call P0 and P1 the end points of the line segment The ray emanating from P0 in the direction of P1 is the set tP1 1 7 tP0 t 2 0 We call Po the initial point of the ray De nition 38 Some general terminology An ordered sequence P17P2 7P of n distinct points no three of which are collinear is called a polygon or an n gon The points are called the vertices of the polygon Two consecutive vertices in the list are said to be adjacent and also the vertices Pn and P1 are adjacent The lines and line segments joining adjacent vertices are called the sides of the polygon and the other lines and line segments joining vertices are called diagonals The term extended side is employed if we want to emphasize that the line and not the line segment is intended for emphasis but the adjective is often omitted A 3 gon is also called a triangle7 a 4 gon is also called a quadrangle7 a 5 gon is also called a pentagon7 a 6 gon is also called a hexagon7 etc For triangles the two notations ABC and AABC are synonymous the latter is more common 32 Af ne Transformations De nition 39 A transformation T R2 a R2 is called a ine iff it has the form by 10 T ltgt 95 m w w ymwy where ad 7 be 31 0 310 It is convenient to use matrix notation to deal with af ne transforma tions To facilitate this we will we not distinguish between points and column vectors ie we write both Pxy and P Then in matrix notation an af ne transformation takes the form TP MP V and detM 31 0 Theorem 311 The set of all a ne transformations is a group ie 1 the identity transformation P P is a ne 2 the composition T1 0T2 of two a ne transformations T1 and T2 is a ne and 3 the inverse T 1 of an a ne transformation T is a ne Proof The identity transformation I has the requisite form with a d 1 and b c p q 0 Suppose T1 and T2 are af ne say T1P M1P V1 and T2P MZP V2 Then T1 0 T2P T1T2P M1M2PV2 V1 MPV where M M1M2 and V M1VV2 Since detM1M2 detM1 detM2 31 0 this shows that T1 0 T2 is af ne To compute the inverse transformation we solve the equation P TP for P weget P MPV ltgt MPP 7V ltgt PM 1P M 1V In other words T 1P M P V where 1 d 7b 1 dp bq M 7 V adibe6 al adibe6paq This shows that the inverse T 1 of the af ne transformation T is itself an af ne transformation 1 Theorem 312 An a he transformation maps mes ohto lines the seg ments onto the segments and rays ohto rays Proof Let Z be a line and T be an af ne transformation the theorem asserts that the image TZ TP P E Z is again a line Fix two distinct points P0P1 E Z Let M and V be the matrices which de ne T7 ie TP MP V Choose P E Z Then P 1 itP0 tP1 for some t E R Hence Tp M1itP0tP1V 1itMPoVtMP1l 11TP0tTP1 which shows that TP lies on the line Z connecting TP0 and TP1 The same argument reading T 1 for T shows that if P E Z then T 1P E Z Hence TZ Z as claimed Reading 0 S t S 1 for t E R proves the theorem for line segments Reading t 2 0 for t E R proves the theorem for rays 1 Theorem 313 For my two triangles AABC and AA B C there is a unique a he transformattohT such that TAABC AA B C ie TA A TB 3 and TO 0 Proof Let A 114127 B b17b27 C 0102 De ne To by T0xy ny where lfl Hllxl lcll y 02 i 02 52 i 02 y 02 Then T0ltA0gt A T0ltB0gt B7 T000 C where A0 1707 B0 0717 CO 07 0 As in the proof of Theorem 34 we have 01 02 1 det b1 b1 1 detalcl 2101 Cl 621 02 02 2 02 and this is nonzero since A7 B7 0 are not collinear Hence To is an af ne transformation Similarly there is an af ne transformation T1 such that T1A0 A T1B0 B moo 0 By Theorem 311 T T1 0 T1 14 is an af ne transformation It satis es the conclusion of the theorem For example7 TA T2T1 1A T2A0 A To prove uniqueness let T be another af ne transformation such that T A A 7 T B B7 and T O Cquot Let I T24 oT oTl By Theorem 311 I is af ne so IP MP V for some 2 gtlt 2 matrix M and 2 gtlt 1 matrix V Also ICO CO IA0 A0 and IB0 B0 From V MOO V ICO CO 0 it follows that V is the zero matrix and from MAO A0 and MBO B0 it follows that M is the identity matrix Hence I is the identity transformation Thus TTZon1 TgoIon1 TgoTziloToTlon1 T as required 1 Remark 314 Often it is possible to nd a coordinate proof of a theorem which is both straight forward and uncomplicated by using af ne transfor mations For example7 imagine a theorem involving ve points A7 B7 0 D7 and three lines a7 b7 c7 and suppose the hypothesis includes the condition that A7 B7 0 are the vertices of a triangle7 ie they are not collinear We can the prove the theorem by arguing as follows Choose an af ne transformation T with TA0 A7 TB0 B7 TOO 0 Let D0 T lD7 a0 Tquotla7 b0 Tquotlb7 c0 T 1c Prove the theorem for A0 B07 00 D07 a07 bo7 c0 Check that the hypotheses and conclusion are preserved by af ne trans formations This will be true if the hypotheses and conclusion involve only assertions about lines see Theorem 3127 ratios of collinear distances see Theorem 3257 and ratios of areas see Theorem 332 For example7 if the lines a and b are parallel7 then so are a0 and be We can now conclude that the theorem holds for A7 B7 0 D7 a7 b7 c We will often signal this kind of proof by saying Choose a ne coordinates my so that The following theorem illustrates this technique Theorem 315 Parallel Pappus7 Theorem Assume that the three points ABC are collinear and that the three points ACEC are collinear Let the lines joining them in pairs intersect as follows X BO m 30 Y OA m O A z AB m A B See Figure 5 If the lines ABC and ABC are parallel then the points XY Z are collinear This theorem is a special case of Theorem 619 below 15 A B C Figure 5 Parallel Pappus7 Theorem Proof Choose af ne coordinates Ly so that the line ABC has equation y 0 and the line A B C has equation y 1 Then A a 0 B b0 C c0 A a 1 B b 1 C c 1 The equation of the line AB is a 01 0 b 1 1 7b 7aya z y 1 and similarly the equation of the line AB is z b 7 a y 7 b 0 To nd the intersection point Z 2122 we solve these two equations The result 1S 7 a 7 b 7 aa 7 bb Ta7ba 7b 7 21Ta7ba 7bquot Now calculate the coordinates of X 12 and Y y1y2 by cyclically permuting the symbols and then use Theorem 34 Note that the columns of the resulting matrix sum to zero D 22 Exercise 316 Do the calculations required to complete the proof of The orem 315 Theorem 317 Parallel Desargues Theorem Let the corresponding sides of two triangles AABC and AA B C intersect in X BO m B O Y CA m O A z AB m A B See Figure 6 If the lines AA BB CC are parallel then the points XY Z are collinear This theorem is a special case of Theorem 616 below 16 B Figure 6 Parallel Desargues Theorem Proof Choose coordinates Ly so that the lines AA BB CC are the vertical lines z a z b x 0 Then A 11 A ap B bq B bq C QT C cr The line AB has equation a p 1 b q 1 0 z y 1 ie p 7 q b 7 ay Cu 7 bp 0 Similarly the equation of line AB is p 7 q x b 7 ay aq 7 bp 0 The two lines intersect in the solution of the matrix equation q 7 p a 7 b x 7 cu 7 bp d7p a7b y 7 047M so the intersection is Z 2122 where M7WW7W W7M 444 47V 444f 1079710qu piqianq Similarly X 12 where cq7brbr 7cq qr 7q r 1 q7r7q r 7 2 q7r7q r and Y y1y2 where ar7cpcp 7ar rp 7r p 7 7 92 yli r7p7r p Tr7p7r p 17 At this point we could complete the proof using Theorem 34 see Re mark 318 below but here is a trick which nishes the proof more easily The proof uses the following two assertions I An af ne transformation of form TWW 960 96110 WC y transforms each vertical line z k to another vertical line II Given any line Z y ms y0 there is an af ne transforma tion T as in part I such that TM is the x axis Using these two facts we may suppose wlog that the line XY is the axes ie that 2 yg 0 From the above formulas it follows that qr 19 and rp r p Multiplying these two equations and dividing by rr gives qp q p ie 22 0 Hence Z lies on the s axis as well ie the points X Y Z are collinear as required Remark 318 To show that X Y Z in Theorem 317 are collinear we could show that the determinant 1 21 1 cq7brbr 7cq qr 7q r q7r7q 7r y1 y2 1 7 ar7cpcp 7ar rp 7r p r7p7r 7p 21 22 1 m bp7aqaq 7bp pq 7p q p7q7p q vanishes Here m q 7 r 7 q r r 7 p 7 r p p 7 q 7 p q A three by three determinant has six terms each of which has three factors one from the rst column one from the second and one from the third In the case at hand the rst and third factors have four terms each and the second has two Thus the fully expanded determinant has 6 gtlt 4 gtlt 2 gtlt 4 192 terms 1 evaluated it using a computer program Maple which does symbolic calculation and all the terms cancel leaving zero If you believe in computers this is an alternative proof Exercise 319 Complete the proof of Theorem 317 by proving l and II in the proof Exercise 320 The proof of Theorem 317 assumes that m 31 0 ie that the three numbers p 7 p q 7 q r 7 r are distinct What if this is false Exercise 321 The last step in the proof of Theorem 317 assumes that rr 31 0 What if this is false Hint lf rr 0 then either 7 0 or r 0 There are three cases 7 r 0 r 31 r 0 r 31 7 0 and the last two are treated the same way 33 Directed Distance Theorem 322 Let P 1 7 tP0 tP1 and Q 1 7 sP0 5P1 be two points on the line P0P1 Then the distance1 lPQl between P and Q is given by PQ 15 7 tl P0P1 Proof P7Qs7tP07P1 D De nition 323 The directed distance PQ from P to Q in the direction from P0 to P1 is de ned by PQ 8 i t P0P1 324 Distances are always nonnegative However7 the directed distance can be negative For example7 this is the case if the points appear on the line in the order P07P17Q7P7 ie if P1 is between P0 and Q and Q is between P1 and P See De nition 37 lnterchanging P0 and P1 reverses the sign of the directed distance and hence leaves a ratio of directed distances un changed Most af ne transformations do not preserve distances those which do preserve distance are called Euclidean transformations and will be stud ied in Section 4 However7 af ne transformations preserve ratios of collinear distances ln fact7 Theorem 325 A ne transformations preserve ratios of collinear directed distances Proof As in the proof of Theorem 3127 TP 17tTP0tTP1 Hence we 7 TQ 7 s 7 tgtltTltPogt 7 TltP1gtgt so PQl 787 7 PQ P6131 T T P0131 for P5 TP0 Pf TP1 P TP Q TQ 1 Corollary 326 A ne transformations preserve midpoints of segments Proof The midpoint of the segment AB is the unique point M on the line AB which is equidistant from A and B It is given by MAB Read A P07 B P17 M P and t 12 in the parametric equation P 17tP0 tP for a line 1 1See De nition 45 below 34 Points and Vectors De nition 327 The difference W P1 7 P0 between two points P0 and P1 is called the vector from P0 to P1 Remark 328 An af ne transformation TP MP where V 0 is called a linear transformation These are studied in the rst course in linear algebra An af ne transformation T is a linear transformation if and only if it xes the origin ie if and only if T0 0 When points undergo an af ne transformation7 the corresponding vectors undergo a linear transformation This means the following If TP MP V is an af ne transformation7 and P6 TP07 Pf TP1 are the images of points P07 P1 under T7 then the vectors W P1 7 P0 and W P 7 P6 are related by the formula W MW Contrast this with the formula P MP V for P TP The set of all linear transformations form a group the fact that the composition of two linear transformations is again linear follows from the associative law for matrix multiplication ie M2M1W M2M1W Remark 329 An af ne transformation TP P V where M is the identity matrix is called a translation The set of all translations forms a group the identity transformation is the translation with V 07 the composition of two translations T1P P V1 and T2P P V2 is T20T1P PV1V2 and the inverse transformation of the translation TP P V is T 1P P i V 35 Area In Math 222 you learn how to compute the area of a parallelogram using determinants cross products well take this as the de nition of area The calculations in this section are easy if you are familiar with matrix algebra The neophyte can skip the proofs in this section 2The reader is cautioned that some authors use the term linear transformation for what we call a ne transformation 20 De nition 330 The oriented area of a triangle AABC is half the deter minant ABC detlB 7 A C 7 A of the 2 gtlt 2 matrix whose columns are the edge vectors B 7 A and C 7 A from A Thus if A 11702 B bpbg7 C 0102 then b 7 a c 7 a 7 1 1 1 1 1 ABC 7 2 det b2 7 a2 CZ 7 W The area is the absolute value of the oriented area Remark 331 The oriented area of AABC is positive and hence equal to the area when the points A7 B7 0 occur in counter clockwise order Theorem 332 An a ne transformation preserves ratios of oriented areas ie it changes all oriented areas by the same factor Proof Suppose that TP MP V Then TB 7 TA MB 7A and TC 7 TA MC 7 A Hence we have an equality of 2 gtlt 2 matrices TB i TA TO i TA M B i A O i A Taking the determinant of both sides and using the fact that the determinant of a product is the product of the determinants shows that the oriented area of TAABC is the determinant of M times the oriented area of AABC D Theorem 333 The oriented areas of the four triangles obtained by deleting a vertem of a quadrangle satisfy ABD DEC ABC CDA Proof See Figure 7 The determinant is linear in its columns7 reverses sign if the columns are interchanged7 and vanishes if two of the columns are the same From this we see that ABC detA B detlB C deth A Combining this with the corresponding formulas for the other three triangles in the quadrangle gives the result 21 Figure 7 Two ways to compute the area of a quadrangle De nition 334 The oriented area of the convex quadrangle ABC7 D is the sum of the oriented areas of the two triangles AABC and AC DA7 ie by de nition it is the sum of the oriented areas of the two triangles with common side AC By Theorems 333 this is the same as the sum of the oriented areas the two triangles with common side DB See Figure 7 In other words the oriented area of a quadrangle is independent of the choice of the diagonal used to compute it Figure 8 shows that this might not work for areas as opposed to oriented areas because one of the terms in the formula can have the wrong sign De nition 335 Two points P1 zhyl and P2 27y2 are said to be separated by the line Z ax by c 0 if axl byl c and an by2 e have opposite signs This gives a precise meaning to the phrase P1 and P2 lie on opposite sides of Z A polygon is called convex iff no edge separates any two of the remaining vertices For example7 the quadrangle on the right in Figure 8 is not convex because 0 and D lie on opposite sides of the edge AB Figure 8 Convex and not convex 22 Theorem 336 If the quadrangle A7 B7 0 D is COTZUCCE then all four terms in the formula in Theorem 333 have the same sign Hence Theorem 333 holds for COTZUCCE quadrangles and area instead of oriented area Proof By Theorems 313 and 332 and Exercise 3397 we may assume wlog that A 07 07 B 10 and C 01 Now use Exercise 340 1 Remark 337 One could can carry out the theory of this section for any polygon and prove algebraically that the oriented area of any polygon may be de ned by breaking it up triangles and that this area is independent of how the polygon is thus broken up Similarly for areas and convex polygons Exercise 338 Show that ABC BOA CAB 7BAO 7AOB 7OBA Exercise 339 Let T be an af ne transformation7 Z be a line7 and A and B be points not on Z Show that Z separates A and B if and only if TZ separates TA and TB Hint Given a7 b7 0 there are numbers a b 7 e such that z y Tdy gt ac by e ad by 0 If Z ax by e 07 then TZ sz a x b y e 0 Exercise 340 Let A 107 B 007 C 017 and D d17d2 Calculate the four areas ABD7 DB07 ABC7 CDA and verify the formula in Theorem 333 Use the de nition to show that the quadrangle A7 B7 0 D is convex if and only if d1 gt 07 d2 gt 07 and d1 d2 gt 1 and that in this case all four oriented areas are positive 36 Parallelograms De nition 341 A parallelogram is a quadrangle A7 B7 0 D such that the lines AB and CD are parallel and the lines AD and B0 are parallel Theorem 342 A quadrangle A7 B70 D is a parallelogram if and only if B 7 A O 7 D Proof Let A 11702 B b17b27 C 0102 D d17d2 Two lines are parallel if and only if they have the same slope Hence A7 B7 0 D is a parallelogram if and only if b2702 027d2 d2702 027b2 d b1 7 11 Cl 7 d1 an d1 7 11 Cl 7 b1 23 The rst equation clearly holds if B 7 A C 7 D since the nurnerators and denominators are equal But B 7 A C 7 D gt D 7 A C 7 B so the second equation holds as well Conversely assume that ABC7 D is a parallelogram and let in and n be the slopes of the sides Then bg71277 lbl7117 027b271017b17 02 7 d2 77101 7 d17 d2 7 12 71d1 7 11 Subtracting the two equations on the left and subtracting the two on the right gives two equations 1 mu and i nu where i b2 7 a2 7 02 7 d2 and u b1 7a1 7 cL 7d1 Since in 31 n else the points would be collinear we conclude u i 0 so B 7 A C 7 D as required Exercise 343 Addition in A ne Geometry Let 0 007 A 107 B b07 C 00 be four points on the n axis and OA C B be a parallelograrn ie the lines OA and BC are parallel and the lines OB and AC are parallel such that the lines AA 7 BB 7 CC are parallel Show that c a b Exercise 344 Subtraction in A ne Geometry Let 0 07 07 A 107 B b7 0 be three points on the n axis and OA 7 B 7 0 be a parallelograrn such that the lines OO 7 AA 7 BB are parallel7 ie the lines OA and B O are parallel and the lines OB and A O are parallel Show that b 7a Exercise 345 Multiplication in A ne Geometry Let 0 007 I 107 A 107 B b07 C 00 be ve points on the x axis7 and 0 7 1 7 B be three points such that a the lines OO 7 II 7 BB are parallel7 b the lines IB and I B are parallel7 c the points 0 7 1 7 A are collinear7 and d the points O 7 B7 0 are collinear Show that c ab Exercise 346 Division in A ne Geometry Let 0 007 I 107 A 107 B b0 be four points on the x axis7 and 0 7 1 7 B be three points such that a the lines OO BB 7 II are parallel7 b the lines IB and FE are parallel7 c the points 0 7 1 7 A are collinear7 and d the points O 7 I7 B are collinear Show that b 1a 37 Menelaus and Ceva 347 Let P7Q7R be points on extended sides BC7 CA7 AB of triangle AABC Thus there are numbers p7q7r with PpB17pC QqO17qA RrA17rB 24 The six distances BP PC 0Q QA AR BB in Theorems 348 and 349 below are directed ie XY 7YX The sign convention is such that the signs of BP PC 0Q QA AR RB are the same as the signs of 1 7 p p 1 7 q q 1 7 rr respectively Theorem 348 Menelaus The points P Q and R are collinear if and only if E w E PC Q4 RB Proof See also 4 page 146 and 3 page 66 By Theorem 312 the condition that the lines AP BQ and CR are concurrent is preserved by af ne trans formations By Theorem 325 the ratios and hence product of the ratios is preserved by af ne transformations Hence by Corollary 414 we may assume that A 10 B 01 and C 00 so P 010 Q 17q0 and R r 1 7 r By Theorem 34 the points P QR are collinear if and only if 0 p 1 det 17q 0 1 0 r 17r1 The condition on the ratios is 17p17q 17 p70 070 2070 Both conditions simplify to 1 7 p 7 q 7 r pq qr rp 0 1 Theorem 349 Ceva The lines AP BQ and CR are concurrent if and only if 71 BP 0Q AR PO Q4 RB Proof See also 4 page 126 and 3 page 4 As in Theorem 348 we may that A 10 B 01 o 00 P 02 Q 17 10 R r 1 7 r The lines AP BQ CR have equations MHz19 1Qy17q T1ry0 By Theorem 34 these lines are concurrent if and only if p 1 7p det 1 1 7 q q 7 1 0 r 7 1 r 0 25 The condition on the ratios is 22 14 p70 Q70 r70139 Both conditions simplify to 1 7 p 7 q 7 r pq qr rp 2pqr 1 Remark 350 Ceva7s Theorem has the following six corollaries H D 00 7 CT CT 38 The medians of a triangle are concurrent The common point is called the centroid See 38 The altitudes of a triangle are concurrent The common point is called the orthocenter See 53 The perpendicular bisectors of a triangle are concurrent The common point is called the circumcenter See 52 The angle bisectors of a triangle are concurrent The common point is called the incenter See 55 The lines connecting each vertex of a triangle to the opposite point of tangency of the inscribed circle are concurrent The common point is called the Gergonne point See 55 The lines connecting each vertex of a triangle to the point of tangency of the opposite exscribed circle are concurrent The common point is called the Nagel point See 55 The Medians and the Centroid De nition 351 The medians of a triangle are the lines connecting the vertices to the midpoints ofthe opposite sides The triangle formed by joining these midpoints is called the medial triangle Theorem 352 The medians Ufa triangle AABC are concurrent The com mon point G is called the centroid and is given by G71ABO g 26 MB MA A MC E Figure 9 The Centroid Proof The three ratios in Ceva7s Theorem are all one To check the for mula for G let MA7 MB7 MC be the midpoints of the sides BC7 CA7 AB respectively Then as in the proof of Corollary 326 MA BO7 MB OA Mo AB G A MA B MB o MC In other words G lies on each median two thirds of the way from the vertex to the opposite midpoint Remark 353 In Math 221 one learns that the center of mass of a collec tion of 71 point masses 77117712 7771 located at point P171327 7P is the weighted average p Li 2139 mi An analogous formula 15 f Pdm f dm 27 holds for continuous mass distributions The centroid of a triangle is both the center of mass of three equal mass points at its vertices and also the center of mass of a uniform mass distribution spread over its area See 4 page 58 Somewhat surprisingly the center of mass of a triangle made from three uniform rods is the incenter not the centroid See 4 page 59 As explained in 4 on page 129 the point of concurrency in Ceva7s theorem can also be viewed as the center of mass of three unequal point masses Exercise 354 Show that the medians of a triangle divide it into six trian gles of equal area Hint It is enough to prove this for an equilateral triangle Exercise 355 Show that the centroid of a triangle divides each median into two segments one of which is twice as long as the other Exercise 356 Points P and Q are se A lected on two sides of AABC as shown P and segments AQ and BP are drawn Then QX PY are drawn parallel to BP and AQ respectively Show that XYHAB B Q Y C Exercise 357 In the gure vertices B A and C of AABC are joined to points P and Q on the opposite sides and lines BP and Q p CQ meet at point X Suppose that BX 23BP and OX 23CQ Prove that BP and CQ are medians of AABC B C Exercise 358 Show that there is no point P inside AABC such that every line through P cuts the triangle into two pieces of equal area Hint Show that if there were such a point it would have to lie on each median of the triangle Exercise 359 In the gure the side B0 of AABC is trisected by points R and S A Similarly T and U trisect side AC and V and W trisect side AB Each vertex V U of AABC is joined to the two trisection points on the opposite side and the inter sections of these trisecting lines determine AXYZ as shown Prove that the sides of B R S C AXYZ are parallel to the sides of AABC 28 Exercise 360 Varignon s Theorem Points W X Y and Z are the midpoints of the sides of quadrangle ABCD as shown and P is the intersection of WY with XZ Two of the four small quadrangles are shaded Show that P is the midpoint of both WY and XZ and that the shaded area is exactly half of the area of quad rangle ABCD Hint For the second part decompose the whole area into four trian gles so that exactly half the area of each triangle is shaded Exercise 361 Points P and Q are chosen on two sides of AABC as shown and lines BP and QC meet at X Show that X lies on the median from vertex A if and only if QPHBC Exercise 362 Given AABC39 let A be the point 13 of the way from B to C39 as shown Similarly B is the point 13 of the way from C39 to A and Cquot lies 13 of the way from A to B ln this way we have constructed a new triangle AA B Cquot start ing with an arbitrary triangle Now apply the same procedure to AA B Cquot thereby creating AA B C39 Show that the sides of AA B C39 are parallel to the appropri ate sides of AABC39 What fraction of the area of AABC39 is the area of AA B C39 7 Exercise 363 lf we draw two medians of a triangle we see that the interior of the triangle is divided into four pieces three triangles and a quadrilateral Prove that two of these small triangles have equal areas and show that the other small triangle has the same area as the quadrilateral 4 Euclidean Geometry 41 Orthogonal Matrices De nition 41 A square matrix M is said to be orthogonal iff its transpose M is it inverse Theorem 42 A 2 gtlt 2 matn39r M is orthogonal if and only if it has one of the two forms a b a b Miiba or Mibia where a2 b2 1 Proof Let abcd be the entries of M so that M and M are given by 7 a b i 7 a 0 Mile dlv M lb all Then M 1 M if and only if MM MM I the 2 gtlt 2 identity matrix7 ie ac ab 7 a202 abcdi 10 bd cdibad0b2d27017 ab acia2b2acbd 10710 Cd bdicadbczd2 bd70139 This happens if and only if a202b2d2a2b202d21and abcdacbd0 From the former we get 12d27 b202 sodia andcib The additional equation abcd0shows that ifda7 0 thenb7candifd7a7 0thenbc 1 and Remark 43 The two forms in Theorem 42 are distinguished by their determinants detM a2b2 1 for the rst and detM iazibz 71 for the second Remark 44 Each of the equations MM I and MM I implies the other More generally7 if M and N are square matrices and NM I7 then N M l See paragraph A5 Hence to show that a 2 gtlt 2 matrix M is orthogonal it is enough to verify one of the two equations MM I and MM I 30 42 Euclidean Transformations De nition 45 The distance between the two points A a17a2 and B b17b2 is de ned by the Pythogorean formula b1 7 112 b2 7 122 ln elementary geometry it is customary to we AB instead of lABl We also write lWl xu2 o2 for the distance between the origin 07 0 and the point W um De nition 46 A Euclidean transformation is an af ne transformation T which preserves distance7 ie lA B l lABl whenever A TA and B TB Euclidean transformations are called isometries in The term rigid motion is also commonly used Remark 47 The Euclidean transformations form a group This is obvious and has nothing to do with distance The set of invertible transforma tions which preserve any function contains the identity and is closed under composition and inverses Theorem 48 An a ne transformation TP MP V is Euclidean if and only if the matrics M is orthogonal Proof Given points A a17a2 and B b17b2 let W B 7 A denote the vector from A to B7 let A TA7 B TB be the images of A and B under T7 and let W B 7 A Then W MW see Remark 328 Thus the theorem asserts that lW l for all W if and only if M is orthogonal lf W um then W an omen do and lWlZ u2 o2 lWl2 a2 02u2 2ab cduo b2 d2o2 If M is orthogonal7 then a2 02 b2 d2 1 and ab Cd 0 see the proof of Theorem 42 so lW lZ Conversely7 assume that lW lZ lWlZ for all W Taking W 10 and W 01 gives a2 02 b2 d2 1 ab Cd 0 These equations say MM I It follows that MM I See Remark 44 Hence M is orthogonal 1 31 43 Congruence De nition 49 Two gures in the plane are said to be congruent iff there is a Euclidean transformation carrying one onto the other In particular two triangles AABC and AA B C are congruent iff there is a Euclidean transformation T such that TA A 7 TB B 7 and TC C We write F E F as an abbreviation for the sentence F is congruent to F Remark 410 We use the word gure rather than the word set because the order of the points is important The triangles AABC and ACBA are different7 and usually there is no Euclidean transformation T such that TA O TB B7 and TO A Remark 411 Congruence is an equivalence relation7 ie for any gures 1 F g F 2 ifF g F and F g F 7 the F F 3 ifF 2 F7 then F 2 F This is an immediate consequence of the fact that the Euclidean transforma tions form a group 412 Any two points are congruent ln fact7 for any two points A and B there is a unique translation TP P B 7 A such that TA B Two directed line segments AB and AR are congruent if and only if they have the same length7 ie lABl lA B l This is an immediate consequence of the following Theorem 413 Let b lABl be the distance between distinct points A and B Then there are emactly two Euclidean transformations T such that TA 00 and TB b70 Proof Using a translation we may assume wlog that A 00 Let B b17b2 and de ne a7 b7 and M by b1 b2 a b J a b M 7b a 32 Then M is orthogonal and MB b0 Now suppose that T1A T2 00 and T1B T2B b0 Then T0 0 00 and Tb0 b 0 where T T1 0 T24 As T is an Euclidean transformation xing the origin Theorem 42 says that it has form Ty ow By x i ay where 042 62 1 From Tb 0 b 0 we conclude that B 0 and Oz 1 We conclude that there are two choices for T namely T I the identity or T S where Sxy x 7y is re ection in the x aXis Hence either T1 T2 or T1 S 0 T2 D Corollary 414 Every triangle is congruent to a triangle AABC where A a0 B b 0 and C 0 0 Proof By Theorem 413 We may assume that the triangle has vertices A1 0 0 B1 b1 0 01 c1c2 Apply the translation TP P7 V where V Cl D Theorem 415 SSS Two triangles AABC and AA B C are congruent if and only if the corresponding sides are equal AABcgnABc a lABl lA B l lBOl lB O l lOAl lO A l Proof Only if77 is immediate since Euclidean transformations preserve dis tance Hence assume that the corresponding sides are equal By Theo rem 413 we may assume that A A 00 and B B c 0 Let a lBCl and b lACl then both C and C lie on the intersection of the two circles 2y2b27 ic2y2a239 But these two circles intersect in the two points amigo Speci cally 0 c2 7 a22c and go xbz 7 Thus either 0 C or the re ection Sxy x 7y carries 0 to C Either way AABC E AA B C 1 44 Similarity Transformations De nition 416 A similarity transformation is an af ne transformation T which preserves ratios of distances ie there is a positive constant a such that lA B l ulABl whenever A TA and B TB 33 De nition 417 Two gures in the plane are said to be similar iff there is a similarity transformation carrying one onto the other In particular two triangles AABC and AA B C are similar iff there is a Euclidean transfor mation T such that TA A 7 TB B 7 and TC C Theorem 418 The similarity transformations form a group and similarity is an equivalence relation Proof As for Euclidean transformations and congruence D Theorem 419 An a ne transformation TP MP V is a similarity transformation if and only M has one of the two forms mm or My bl where a2 b2 gt 0 Proof If TP MP V is a similarity transformation7 then M 1TP fi lMP n lV is Euclidean D 420 We summarize De nitions 46 and 416 and Theorem 325 o Euclidean transformations preserve distance 0 Similarity transformations preserve ratios of distances 0 Af ne transformations preserve ratios of collinear distances 45 Rotations De nition 421 An af ne transformation TP MP V is called orien tation preserving iff detM gt 0 and orientation reversing iff detM lt 0 Remark 422 A Euclidean transformation has form TP MP V where M has one of the two forms in Theorem 42 The rst is orientation preserving and the second is orientation reversing De nition 423 A xed point of a transformation T is a point P which is not moved by T7 ie TP P 34 Theorem 424 An orientation preserving Euclidean transformation which is not a translation has a unique cced point 0 such a transformation is called a rotation about 0 Proof Assume that T is an orientation preserving Euclidean Transformation which is not a translation Then TP MP V where Ml a ll W 7b a q and a2b2 1 a lt1 To nd the xed point 0 we solve MOV O for O These equations are 1 0 by107 by1ayq The determinant of the matrix of coef cients is 1 7 a2 b2 21 7 a gt 0 so there is a unique solution B 425 Recall from Math 222 that rotation through angle 6 is described by the orthogonal matrix cos0 isind sin0 cos0 39 Rel The rotation R9 carries the point 10 to the point cos 67sin 0 The ma trices R9 satisfy the identities R0 1 Bang RM R R10 The second identity is an immediate consequences of the trigonometric ad dition formulas cosoz cosa cos Bisina sin 7 sincv sinozcos cosoz sin De nition 426 A matrix of form RC 78 c2521 s c rotation matrix the corresponding Euclidean transformation is then a ro tation about the origin 00 35 46 Review Next we review how angles are treated in Math 222 We use calculus notation For the proofs see any calculus textbook We will not use the material in this section in the formal development7 but it motivates the de nitions De nition 427 The dot product of two vectors u and V is de ned by u V cos 6lul lvl where lul is the length of the vector u and 6 is the angle between the vectors u and V Theorem 428 The dot product between u u1i uzj and V v1i Ugj is given by u V ulvl u2v2 De nition 429 The cross product of two vectors u and V is de ned by u gtlt V sin6lul lvlw where w is the unit vector normal to the plane of u and V and 6 is the angle between the vectors u and V There are two unit vectors perpendicular to a plane the choice of w is determined by the condition that uV7 W forms a right hand frame like ijk and not a left hand frame like ij7 7k Theorem 430 The cross product between u u1i uzj and V v1i vzj is given by u gtlt V iile 7 u1v2k Corollary 431 Let 6 be the angle from the vector u u1i uzj to the vector V v1i vzj Then U101 UZUZ U102 211 cos6 and s1n6 ui u i v5 ui u i v 432 In the sequel we will write U uhuz instead of u u1i uzj and U A 7 O for the vector from O to A instead of the notation u if used in Math 222 36 47 Angles De nition 433 An angle is a pair p17 p2 of rays having a common initial point The common initial point of p1 and p2 is called the vertex of the angle and the rays p1 and p2 are called the arms of the angle Given three distinct points 07 P7 Q the notation LPOQ denotes the angle p17 p2 where p1 is the ray with initial point 0 through P and p2 is the ray with initial point 0 through Q De nition 434 The trigonometric measure of an angle phpg is the pair c7 5 de ned by C i U101 U202 8 i U102 U201 T 2 2 2 27 T 2 2 2 27 Vu1u2VU1U2 Vu1u2VU1U2 where O is the vertex of the angle phpg and AB are points on phpg distinct from O7 and 1117112 UA707 11702 VB70 The de nition is independent of the choice of A and B as follows If A 7 B are other points on phpg distinct from O and U A 7 07 V B 7 07 then U MU and V VV where n V gt 0 The numbers n and V factor out in the above expressions for c and 5 Exercise 435 Show that c2 52 1 Theorem 436 Let c7 5 and c 7 5 be the trigonometric measures of angles phpg and paw2 respectively Then 1 There is an orientation preserving Euclidean transformation T such that Tp1 pi and Tp2 p z if and only if c 75 c7 5 2 There is an orientation reversing Euclidean transformation T such that Tp1 pi and Tp2 p z if and only if c 75 c7 75 Proof 1 De nition 437 For three points A 11702 O 01027 B bhbl of R2 with A 31 0 and B 31 0 de ne cos ZAOB c7 sin ZAOB s 37 00 Figure 10 An Angle as in Theorem 436 with ui al 7 01 111 bl 7 oi In other words7 the pair cos ZAOB7 sin ZAOB is the trigonometric measure of the p17 p2 where p1 is the ray A and p2 is the ray OB In these notes the nota tion ZAOB ZA O B shall henceforth be considered as an abbreviation for the two equations cos ZAOB cos ZA O B and sin ZAOB sin ZA O B Thus cos ZAOB cos ZA O B ZAOB ZAOB ltgt an sin ZAOB sin ZA O B 438 In elementary trigonometry the radian measure of an angle is de ned as follows Slide the angle in the plane so that the vertex is at the origin and the rst ray p1 is the positive x axis Then the rst ray p1 intersects the unit circle 2 y2 1 in the point 17 0 and the second ray p2 intersects the unit circle a point a7b The radian measure 6 of the angle is then the length of the arc of the unit circle traced out by a point moving counter clockwise from the point 17 0 to the point 07 s moreover7 the sine and cosine of the angle are given by e cos 0 s sin0 ln coordinate calculations we generally trigonometric measure rather than radian measure A careful treatment of radian measure requires calculus Remark 439 It is an immediate consequence of De nitions 433 and 330 that the oriented area of AAOB is given by AOB lAOl isol sin AOB 38 Remark 440 The reader is cautioned that in most elementary books no distinction is made between ZAOB and ZBOA7 but with the de nitions presented here sin ZBOA 7 sin ZAOB7 ie reversing the order of the rays in an angle reverses the sign of the sine With our de nitions cos ZBOA cos ZAOB7 and in elementary texts the condition that two angles are equal can usually be replaced by the condition that they have the same cosine When a diagram is present ZAOB is often denoted by 0 especially when the sign of the sine not important Corollary 441 Orientation preserving similarity transformations preserve trigonometric measures of angles Orientation reversing similarity transfor mations preserve cosines of angles Corollary 442 Vertical angles are equal ie if A1 0 A2 are collinear with 0 between A1 and A2 and B1 0 B2 are collinear with 0 between B1 and B2 then ZAlOBl Corollary 443 Corresponding angles determined by parallel lines are equal ie if lines A1B1 and flng are parallel and a point C is on the line A1A2 is such that C is between A1 and A2 then ZBlAlC ZBZAZC Proof The translation TP P A2 7 A1 sends the ray A1B1 onto the ray flng and the ray A10 onto the ray A20 1 48 Addition of Angles Theorem 444 Let p17 pg 0 A B c s be as in De nitions 433 and434 and TP RP 7 O 0 where Then T is the unique orientation preserving Euclidean transformation such that Tp1 pg 39 Proof The transformation T is orientation preserving and Euclidean the condition 02 52 1 is Exercise 435 It follows from the identity U101 U202 112 112002 Clearly TO O Letu1u2 U A 7 O and 01112 V B 7 O as in De nition 434 The equations U101 U202U1 U102 Uz iwz 3 17 U102 11210111 U101 UZUZWZ 3 show that RU is a positive multiple of V ie TA 7 O is a positive multiple of B 7 O and hence that T sends p1 to oz The uniqueness of T is a consequence of the uniqueness in Theorem 413 D De nition 445 Let oz y be angles Then the notation va is understood to be an abbreviation for the assertion that there exist four distinct points 0417 B7 0 such that 74400 a4AOB 54300 lf oz ZAOB7 the notation oz 180 means A7 07 B are collinear with 0 between A and B the notation B 7oz means 6 LBOA the notation oz 90 means cos ZAOB 0 and sin ZAOB 1 the notation oz 45 means ozoz 90 etc We also say that lines 0A and OB are perpendicular the word orthogonal is used in some textbooks when ZAOB i90 Angles which sum to 90 are called complementary angles which sum to 180 are called supplementary Theorem 446 Let Ra R R7 be the rotation matrices corresponding to the angles oz 6 y as in Theorem 444 Then rya ltgt R7RaR Proof Let p17p27p3 be three rays through the origin lf Rap1 p2 R p2 p3 and then R Rap1 pg 1 Theorem 447 The angles of a triangle sum to 180 40 Proof The theory presented thus far justi es the usual proof draw a parallel to a side through the third vertex D Theorem 448 SSSSASASA For triangles AABC and AA B C the following conditions are equivalent i the triangles AABC and AA B C are congruent ii lABl lA B l lBOl lB O l and lOAl lO A l iii lABl lA B l lBCl lB C l and cos LB cos ZB iv cos LA cos LAC cos LB cos LB and lABl lA B l Proof The assertion ltgt ii is Theorem 415 The other assertions are similar 1 Corollary 449 Pons Asinorum The base angles of an isosceles triangle are equal Proof If lABl lCBl Then AABC g ACBA by ASA7 so cos LA cos 0 since Euclidean transformations preserve cosines of angles D Remark 450 Strictly speaking7 the correct conclusion of Pons Asinorum is cos ZBAC cos LBC A7 sin ZBAC isin ZBCA See Remark 440 Theorem 451 AAASSSSAS For triangles AABC and AA B C the following conditions are equivalent i the triangles AABC and AA B C are similar ii cos LA cos LAC cos LB cos LB and cos 0 cos LC AB B C C A E BO CA A B O A AB OA39 Proof Two triangles are similar if and only if there is a number n such that the af ne transformation TP MP carries one of them to a triangle congruent to the other Apply Theorem 448 1 iii the sides are proportional ie iv cos LA cos LA and 41 Exercise 452 Let A 00 B 10 0 0812 A 23 B 2436 Then AB A B 1 There are two points C such that AABC and AA B C are congruent Find them and for each one nd the Euclidean transformation T such that TA A TB B TC C For which ones is T orientation preserving 42 5 More Euclidean Geometry 51 Circles De nition 51 The locus of all points P at a xed distance r from a given point 0 is called the circle of radius r and center or centered at 0 lf 0 a7 b circle is the set of all points P Ly satisfying the equation 96702y7b2 rz Any line segment 0P where P lies on the circle is called a radius of the circle Theorem 52 Though any point P0 0n the circle there is a unique line Z intersecting the circle at the single point P0 This line is called the tangent line to the circle at P0 The radius 0P0 is perpendicular to the line tangent line Z through P0 Proof The line through P0 x07y0 has an equation of form iz 7 x0 7uy 7 0 Du is the slope and hence parametric equations zz0tu yy0ti where u2 112 1 lnserting this into and gives 0 7 a2 2x0 7 atu tzu2 yo 7 b2 2y0 7 bti t212 r2 But P0 lies on the circle7 ie 0 7 a2 yo 7 b2 r27 so the equation simpli es to 2x0 7 atu 2y0 7 bti t2 0 There are two distinct solutions for t unless 0 7 au yo 7 bi 07 in which case the line has equation 950 a95 r950 yo WZI 90 0 Note that this equation says that the vector TPO from the center 0 of the circle to a point P0 on its circumference is perpendicular to vector 1W3 from P0 to a point P on the tangent line at P0 This is in agreement with the method used in Math 222 1 43 Theorem 53 The locus of all points P equidistant from two given points A and B is a straight line perpendicular to line AB and passing through the midpoint M of the segment A B It is called the perpendicular bisector of the segment A B Proof Exercise D 52 The Circumcircle and the Circumcenter Theorem 54 For any triangle AABC there is a unique circle containing the three points A B C This circle is called the circumcircle short for circumscribed circle of AABC and its center 0 is called the circum center The circumcenter is the intersection of the perpendicular bisectors of the sides Proof Let 0 be the intersection of the perpendicular bisectors of A B and BC Then OA OB and OB OC soOA OC 0 lies on the perpendicular bisector of A C and is equidistant from the vertices 1 Remark 55 In high school algebra you learned that equation of form x2y272ax72byic0 determines a circle a point or the empty set you determine which by com pleting the square and that is also how you nd the center of the circle Three non collinear points phyl x2y2 zgy3 determine a multiple of an equation of this form namely z2y2 z y l 2 det 1yi 1 11 1 0 1 Since each of the three points zy satis es this equation because a determinant with two identical rows vanishes this must be the equation of the circumcircle 53 The Altitudes and the Orthocenter Theorem 56 Let t be a line and P be a point The there is a unique line through P perpendicular to l The point H on t where the perpendicular 44 HB HA A HO E Figure 11 The Othocenter through P to Z intersects Z is called the foot of the perpendicular yo Z from P Exercise 57 The foot H of the perpendicular to Z from P is the point on Z closest to P7 ie the distance from P to H is less than the distance from P to any other point of Z The distance lPHl is called the distance from the point P to the line Z De nition 58 The altitudes of a triangle are the lines through the vertices perpendicular the opposite sides Theorem 59 The altitudes of a triangle are concurrent The point of concurrency is called the orthocenter 0f the triangle Proof Let AABC be a triangle and HA7 HB7 HO be the feet of the altitudes through A7 B7 0 respectively The lengths in Ceva7s Theorern are BHA ABcosZB7 HAG CAcos 0 CH3 BC cosZC7 HBO ABcosZA7 AHC CAcos LA HOB BCcos AB The corresponding product or ratios is one so the altitudes are concurrent 1 45 Remark 510 Another proof can be based on the fact that circumcenter of a triangle is the orthocenter of its medial triangle De nition 511 The triangle AHAHBHC formed by joining the feet of the altitudes of AABC is called the orthic triangle3 of AABC 54 Angle Bisectors De nition 512 Let A7 07 B be distinct points Then the locus of all points P such that either P O or ZAOP LPOB is called the angle bisector of the angle AOB Theorem 513 Every point on the angle bisector of the angle AOB is equidistant from the lines A0 and B0 Proof Drop perpendiculars from P to 0A and OB and use ASA D 55 The Incircle and the Incenter Theorem 514 The angle bisectors of a triangle are concurrent The point of concurrency is equidistant from the sides of the triangle and is therefore the center of a circle tangent to the three sides of the triangle This circle is called the incircle short for inscribed circle of the triangle and its center is called the incenter Proof The bisector of an angle is the locus of all points equidistant from the arms of the angle Hence the point of intersection of two angle bisectors of AABC is equidistant from all thee sides and hence must lie on the third angle bisector 1 Remark 515 One can give a proof using Ceva7s Theorem See 4 page 21 Theorem 112 and 3 page 10 Theorem 134 Theorem 516 Let X on B0 Y on CA Z on AB be the points of tangency of the inscribed circle of triangle AABC Then the lines AX BY CZ are concurrent Their common intersection is called the Gergonne point of AABO 3Some authors call it the pedal triangle 46 Proof Ceva7s Theorem plus the fact that the two tangents to a circle from a point are equal D Theorem 517 Let X on B0 Y on CA Z on AB be the points of tangency of the emscribed circles of triangle AABG Then the lines AX BY GZ are concurrent Their common intersection is called the Nagel point ofAABG Proof The line joining two center of excircle tangent at X and the center of the excircle tangent at Z passes through the corresponding vertex B The ratio of BX to BZ is the same as the ratio of the radii of these two excircles Now use Ceva 56 The Euler Line Theorem 518 Let AABG be a triangle G be its centroid 0 be its cir cumcenter and H be its orthocenter Then H G O are collinear and G g0 H The line OGH is called the Euler line Proof Consider the similarity transformation TP G7 P i G This transformation satis es TG G TA MA TB MB TG MC where MA M3 M0 are the midpoints of B G GA AB It carries the orthocenter H to the circumcenter 0 As its shrinks distances by a factor of we conclude that lOGl The transformation rotates through 180 about G so H G and O TG are collinear D 57 The Nine Point Circle This section uses material not covered above See 3 page 20 Theorem 519 The medial triangle and the orthic triangle of a triangle have the same circumcircle This circumcircle is called the nine point circle4 of the original triangle It also contains the three midpoints called the Euler points of the line segments joining the orthocenter to the vertices 4 Sometimes called the Euler Circle or the Feuerbach Circle 47 Proof Denote the midpoints of the sides of AABC the feet of the altitudes and the Euler points by MA M3 M0 HA H3 H0 EA E3 E0 with the subscripts chosen so that the medians are AMA BMB 0M0 the altitudes are AHA BHB CH0 and the points EA E3 E0 lie on these altitudes respectively We claim that MAMBEBEA is a rectangle ie adjacent sides are per pendicular This is because 0 AMAMBC is similar to AABC by SAS so MAME is parallel to AB 0 AEAEBH is similar to AABH by SAS so EAEB is parallel to AB 0 AAMBEA is similar to AACHC by SAS so MBEA is parallel to CH0 and hence perpendicular to the parallel lines AB and MAME o ABMAEB is similar to ABCHC by SAS so MAEB is parallel to CH0 and hence perpendicular to the parallel lines AB and MAME The rectangle MAMBEBEA is inscribed in a circle with diameters MAEA and MBEB Reading B and C for A and B we obtain that the rectangle MBMCECEB is inscribed in a circle with diameters MBEB and MCEEC These two circles share a common diameter MBEB and so must be the same circle As this circle contains MA M3 M0 it is the circumcircle ofthe medial triangle ie the nine point circle The line EAHA is the altitude AHA and the line HAMA is the side BC opposite A Hence LEAHAMA is a right angle so the point HA and similarly HB and HO lies on the nine point circle D Theorem 520 The orthoeenter H of an acute angled triangle AABC is the ineenter of its orthie triangle AHAHBHC Proof We must show ZHAHCH ZHBHAH The same argument shows ZHCHBH ZHCHBH and ZHCHBH ZHAHCH Now the quadrangle HHCBHA is cyclic ie is inscribed in a circle since HHCB HHAB 90 Hence ZHHAHC ZHBHC But ZHBHC ZHBBA is the complement oz of ZCBA so HHAHC a The same argument reverse the roles of B and C shows that HHAHB oz as required 1 Remark 521 For an obtuse triangle the argument shows that H is the center of one of the exscribed circles not the center of the inscribed circle 48 Corollary 522 The triangle of least perimeter inscribed iri a giveri triangle AABC is the orthie triangle of AABC Proof For three points X Y Z on lines BC CA AB respectively let f fX Y Z XYYZ ZX This function is continuous as a function of three variables Each of X Y Z is constrained to a line When one or more of the points X Y Z are far from A B and C the function f is large so the minimum occurs either a point where the partial derivatives of f are zero or else at a point where the function f is not differentiable The function f is differentiable except where one of the three lengths is zero ie whenXYCorYZAorZXB D 58 A Coordinate Proof Here are some coordinate calculations which could be used to prove Theo rems 518 and 519 I used a Maple program to do these calculations and verify that the nine points of the nine point circle are equidistant from N the midpoint of the segment OH 523 By Corollary 414 we may assume that the vertices of the triangle are A 00 B 50 C 06 The midpoints of the sides are given by a b a e b e M 0 M77 M 77 A lt 2 B 2 2 C b 2 and the centroid is a b e G l A B O 7 3 lt 3 3 The slopes of the lines AB A0 B0 are given by e e mAB 0 mAC ii 77130 1 b The equations for these lines are respectively y 0 and exayae exbybe 49 The orthocenter H lies on the y axis say H 0 h The slopes of the altitudes AH and BH are h 1 i b i h i 1 a mAH7 7 WBH77 777 a 77130 C a MAC C H 022 C Equations for the altitudes CH AH BH are x 0 and 7 and solving for h gives bx7cyab a7cyab The feet of these altitudes are respectively HO 00 and ab2b02 b207abc ba2a02 a207abc HA 7 HB b202 b202 a202 a202 The Euler points EC C H EA A H EB B H are 02 7 ab 1 ab b ab E 0 E 7 77 E 7 77 c 26 A 2 26 B 2 2C The perpendicular bisectors of the sides AB A0 B0 have equations kw 12051 19951 2 92 c 2 92 c 2 The common point is the circumcenter ab 02ab O lt 2 7 20 59 Simson s Theorem See 3 page 40 524 Let ABC be a triangle The lines AB B0 A0 divide the plane into seven regions let the point P lie in the unbounded region containing the edge AG in its boundary see diagram Let X Y Z be the feet of the perpendiculars from P to the lines B0 A0 AB respectively See Figure 59 50 Figure 12 Sirnson7s Theorem Theorem 525 Sirnson7s Theorem The points X Y Z are collmcar if and only ifP lies on the circumcirclc of ABC Proof The point P lies on the circurncircle of ABC if and only if APO 1800 7 LB 1 Because the opposite angles at X and Z in the quadrangle BXPZ are right angles we have 1800 7 LB LZPX so condition 1 is equivalent to APO LZPX 2 and on subtracting ZAPX we see that 2 is equivalent to LXPO LZPA 3 A quadrilateral containing a pair of opposite right angles is cyclic7 ie its vertices lie on a circle in fact7 the other two vertices are the endpoints of a diameter of this circle Hence each of the quadrilaterals AYPZ7 BXPZ7 CXPY is cyclic Since the quadrangle CXPY is cyclic we have LXYO LXPO 4 Since the quadrangle AYPZ is cyclic we have ZYA ZPA 5 51 From 4 and 5 we conclude that 3 is equivalent to LXYO LZYA 6 But clearly 6 holds if and only if the points X7 Y7 Z are collinear 1 Alternate Proof Let 0 be the circumcenter of AABC By trigonometry BX BP cos LPBC7 XO PO cos LPC39B7 CY OP cos ZPC39A7 YA PB cos ZPAC7 AZ AP cos ZPAB7 ZB PA cos ZPBA Using the fact that the inscribed angle is half the subtended arc BX BPcos lPOC7 XO POcos lZPOB7 CY CPcos izPOA YA PBcos 741300 AZ AP cos EZPOB7 ZB PA cos EZPOA Now use Menalaus 1 526 Here is a computer assisted coordinate calculation which proves Sim son7s Theorem lt shouldnt be to dif cult to do by hand7 especially if we take 7 704 below to simplify the formulas We begin by loading the Maple package for doing linear algebra with LinearAlgebra To calculate the foot Z of the perpendicular from the point P to the line AB we use the formulas ZAtBiA PZLAB solve for t and plug back in to get Z Here is a Maple procedure to compute this footprocABP local t tB1A1P1A1B2A2P2A2 B1A1 2B2A2 2 A1tB1A1A2tB2A2 end proc 52 We choose A B C on the unit circle and P arbitrarily A cos alpha sinalpha B cos beta sinbeta C cos gamma singamma Pxy2 We use the procedure to calculate X YZ X footBCP YfootCAP ZfootABP We de ne the matrix whose determinant vanishes when X Y Z are collinear MMatrix X1 X2 1 Y1 Y2 1 Z1 Z2 1 J We compute its determinant W2Determinant M The determinant W vanishes exactly when X Y Z are collinear The fol lowing commands show that 2 y2 7 1 divides W and that the quotient m is independent of P msimplifyWX 2y 2 1 simplifyW mx 2y 2 1 The last command evaluates to 0 and proves that W m2 y2 71 Thus X Y Z are collinear if and only if 2 y2 1 ie if and only if P lies on the circumcircle of AABC The commands mm eXpandsinalpha beta sin beta gamma sin gamma alpha 4 simpl ify m mm produce an output of zero which shows that m sina 7 B sin 7 y sin y 7 oz 4 We have proved the following 53 Theorem 527 Algebraic form of Simsor s Theorem Let A COS047S1HOL B cos sin C cos ysin y be three points art the urttt circle x2 y2 1 let P my be art arbitrary point artd X 1727 Y 91712 Z Z1722 be the feet of the perpendiculars from P to the ltrtes BC CA AB respectively Thert 1 2 1 y1 y2 1 mx2 112 71 21 22 1 where m i s1na 7 B s1n 7 y s1n y 7 a 7 4 510 The Butter y See 3 page 45 511 Morley s Theorem See 3 page 47 512 Bramagupta and Heron See 3 page 56 513 Napoleon s Theorem See 3 page 63 514 The Fermat Point See 3 page 83 54 6 Proj ective Geometry Projective geometry developed historically at the same time artists learned to draw in perspective7 ie to draw pictures on a at canvas which look three dimensional See for example Figure 13 where a two dimensional theorem is illustrated by a three dimensional picture A line in space will be drawn as a line in the picture 7 but a circle in space will be drawn as an ellipse in the picture Parallel lines in space will if extended meet in the picture Look down State street from Bascom hill the sidewalks and roofs of the buildings all aim at a point behind the Capitol 61 Homogeneous coordinates 61 When Ly7 z is a point of R3 distinct from the origin 07 07 0 we denote the line connecting the origin and this point by mg2 Thus zyz tx7ty7tz tE R Note that x y z Kym if and only if x Hz y My 2 M2 for some non zero number p De nition 62 The projective plane is the set P2 of all lines through the origin in R3 We say that Luz are homogeneous coordinates for the point mg2 63 A point zyz E P2 is a line in R3 and7 if z 31 07 intersects the plane 2 1 in the unique point 2 17y2 1 17 ie z 31 0 gt zyz 24 yz l 1 The points ie lines of form zy0 do not intersect the plane 2 1 the set of these points is called the line at in nity Exercise 64 Find 2 if 1 2 3 2 4 Remark 65 Imagine an object in three dimensional space R3 above the plane 2 1 Place your eye at the origin 07 07 0 view the object through a transparent canvas lying on the plane 2 1 Each point on the object determines a line though your eye For each point of the object paint a dot on the point of intersection of this line with the plane 2 1 and you have 55 a portrait of the object If you have another object and a correspondence between the points of the objects such that each pair of corresponding points is collinear with the origin7 then the two portraits will be indistinguishable This is why two points7 Luz and uuyuz on the same line through the origin determine the same point my 2 in projective space 66 A plane through the origin in R3 has an equation of form abycz 0 where a7b7c 31 000 We denote by abc E P2 azbycz 0 the set of lines through the origin which lie in this plane If u 31 07 the equations Max uby ucz 0 and ax by c2 0 de ne the same plane Hence uaubuc abc We call abc a projective line Several points 1 y1 2117 x2yg227 are said to be collinear if there is a line abc which contains all of them several lines 11 b1 cj7 a2 Cbg c27 are said to be concurrent if there is a point xy2 in their intersection Theorem 67 Two distinct points in P2 lie in a unique projective line ii Two distinct projective lines intersect in a unique point Proof Two lines in R3 through the origin lie in a unique plane necessarily containing the origin ii Two distinct planes in R3 which pass through the origin intersect in a line which passes through the origin Exercise 68 Find the intersection of the two lines 123 and 321 Remark 69 Parallel lines in the af ne plane have equations abyc 0 and ax by d 0 where c 31 d The corresponding projective lines abc and abd intersect in the point 7ba0 This point lies in the line at in nity Hence parallel lines intersect at in nity ln af ne geometry two distinct points determine a unique line7 but dis tinct lines need not determine a point they might be parallel Because of this asymmetry the statement proof of part ll of Theorem 34 was more complicated then that of part ln projective geometry this asymmetry dis appears two distinct points determine a unique line and two distinct lines intersect in a unique point We recover af ne geometry by specifying a line at in nity77 where parallel lines meet 56 Theorem 610 I Three points 4142 are collinear if and only if 1 2 3 det 91 12 733 0 23 23 23 II Three lmes atbird are concurrent if and only if 11 b1 01 det 12 b2 C2 13 b3 C3 Proof As in Theorem 34 but without the special cases Note that in part I the determinant is zero if and only if 123 a b cy1y2 13 0 0 0 23 23 23 for some a b c 31 0 0 0 and in part II the determinant is nonzero if and only if 11 b1 01 i 0 02 52 02 y 0 13 b3 C3 2 0 for some z7y72 31 07 07 0 1 62 Projective Transformations De nition 611 Let flj be a 3 gtlt 3 matrix with nonzero determinant Then there is a unique transformation T P2 a P2 satisfying the condition 2 112 It129 132 y tglz tggy t232 gt z y z Tzy 2 312 13329 It332 The de nition is legal since matrix multiplication sends lines through the origin to lines through the origin A transformation of this form is called a projective transformation Note that multiplying the matrix flj by a nonzero number does not change the transformation T but it does change the matrix 57 612 As in af ne geometry7 the formulas de ning a projective transformation can be written using matrix notation 11 1312 tis 11 21 1322 1323 y 2 31 1332 tag 2 A matrix transformation preserves the plane 2 1 if and only if t31 tgg 0 and tgg 1 For such a transformation we may rewrite in the form z abp z 2ch y 1 0011 with a tn b tn 0 tgl d tgg p tm7 q tgg This is equivalent to x axby 10 and y Cdyq in agreement with De nition 39 In this way every af ne transformation is a projective transformation The power of the theory will become evident when we use projective transformations which are not af ne 613 We say that the triple z7y72 represents the point Thus if M 31 0 the triples pxwywz and Luz represent the same point It is convenient to think of Ly7 2 as a column matrix 157172 y 2 We also say that the row matrix a7b70 represents the line abc and the matrix tij represents the projective transformation T Theorem 614 A projective transformation maps lines to lines Proof If L a7b70 represents the line abc and X Luz represents the point mg2 then zyz lies on abc if and only if LX 0 But LX LT lTX so LX 0 ltgt LT 1TX 0 so the line represented by LT l is the image under T of the line abc 1 58 Theorem 615 Given four points A B C D no three ofwhieh are collinear there is a unique projective transformation T such that T100 A T010 B T001 C T111 D Compare Theorem 313 and Corollary 414 Proof Let A a1a2a3 B b1b2b3 C 010203 Then the projec tive transformation S represented by the matrix 11 b1 01 02 52 02 as 53 03 satis es the rst three of the four conditions namely S100 A S010 B S001 0 Let SD d1d2d3 and dfl 0 0 R 0 d1 0 0 0 dgl We have d1 31 0 since otherwise SB 80 SD would be collinear con tradicting the hypothesis that B C D are not collinear Similarly d2 31 0 and d3 31 0 Now RSA R100 df100 1 00 and similarly RSB 010 and RSC 001 Also RSD Rd1d2d2 1 1 1 The projective transformation T R o S satis es all four condi tions To prove uniqueness suppose that T is another projective transformation satisfying the four conditions Then T oT 1100 100 T oT 1010 010 T oT 1001 001 T oT 1111 111 The rst three conditions imply that T o T 1 is represented by a diagonal matrix and then the fourth condition says that the three diagonal entries must be equal This means that T o T 1 is the identity transformation of P2 so T T 1 59 Figure 13 Desargues7 Theorem 63 Desargues and Pappus Theorem 616 Desargues Given triangles AABC and AA B C let X Y Z be the intersections of the corresponding sides ofAABC and AA B C ie X BO m B O Y CA m O A Z AB m A B Then the lines AA BB CC are concurrent if and only if the points X Y Z are collinear See Figure 13 Proof Assume the lines AA BB CC are concurrent ie there is a point Q such that each triple QAA QBB Q 06 is collinear Choose an af ne transformation T which sends Q to the line at in nity Then the lines TATA TBTB TCTC are parallel By Exercise 317 the points TX TY TZ are collinear Apply T 1 to conclude that the points X Y Z are collinear The converse can be proved in the same way but it also follows from the principle of duality as explained in 623 below Remark 617 Sometimes Desargues7 Theorem is stated as follows Two triangles are perspective from a point if and only if they are perspective from a line 60 Remark 618 Think of Figure 13 as representing a three dimensional pic ture where Q is the apex of two tetrahedra one with base AABC and the other with base AA B C The plane A B C intersects the plane ABC in a line containing X Y Z Note that projecting the three dimensional gure to a plane either from a point your eye or by parallel lines gives the con gura tion in Desargues7 theorem if no triangle projects to a line This observation is undoubtedly how the theorem was discovered We give a proof based on it Synthetic proof of Desargues theorem Denote by H the plane containing the points in the statement of the theorem ie the points A B C A B C X Y Z and Q Let E the eye be a point not on H Each point P in H determines a line EP and each line L in H determines a plane EL In par ticular the points E Q A A are coplanar Let A be the intersection of the three planes EAB EAC and EQAA Similarly let 3 be the intersection of the three planes EEC EBA EQBB and Q be the intersection of the three planes EOA EOB EQOO The three planes EQAA A EQBB E EQCC Q intersect in a point A diagram illustrating the situation can be obtained from Figure 13 by replacing Q A B C by Q A E Q The original points Q A B C are hidden because they are on the line of sight with The points X n m EQBB B m EQoo o Y n m EQoo o m EQAA A Z n m EQAA A m EQBB B lie in the intersection of the plane of AEQ and H and are therefore collinear D Theorem 619 Pappus Assume that the three points A B C are collinear and that the three points A B C are collinear Let the lines joining them in pairs intersect as follows X 30 m B O Y OA m O A Z AB m A B Then the points XY Z are collinear See Figure 14 Proof As in the proof of Desargues7 theorem we may assume wlog that the lines ABC and ABC are parallel Then the theorem follows from Theorem 315 1 61 ii B C 39 Figure 14 Pappus7 Theorem Remark 620 Any theorem involving only points and lines in the projec tive plane can be viewed as a theorem about determinants For example Desargues7 theorem assumes we are given column matrices A B C A B C X Y Z satisfying detABZ detBOX detOAY detA B Z detB O X detO A Y 0 and concludes that 3Q detQAA detQBB detQOO 0 ltgt detXYZ 0 li appus7 theorem assumes we are given column matrices A B C A B C X Y Z satisfying WWZF4MBUXF WWW dam32 detB OX detO AY 0 and concludes that detABO detA B O 0 gt detXYZ 0 Exercise 621 Let the lines joining three points ABC to three other points ACEC in pairs intersect as follows X BO m B O Y OA m O A z AB m AB 62 Show that if A B C are collinear and X Y Z are collinear then A B C are collinear Hint Don7t work too hard 64 Duality 622 Let the column matrix yz represent the point Kym and the row matrix abz represent the line abc Then the equation abycz0 says that the point zyz lies on the line abc But it also says that the point abc lies on the line ln matrix theoretic terms taking the transpose converts points to lines and line to points If we have a theorem about matrices and replace every matrix in the statement by its transpose we get another theorem about matrices If we have a theorem in projective geometry and systematically replace points by lines and lines by points and phrases like the point P lies on the line Z by the line p passes through the line L then we get another theorem in projective geometry This is called the principle of duality We illustrate this using the principle of duality to complete the proof of Desargues7 Theorem Theorem 616 623 Consider AABC with opposite sides a b 0 Then aBC bCA CAB Ab c Bc a Ca b Similarly for AA B C we have a BO b OA C AB A b c B c a C a b De ne Xa a Yb b Zc c zAA yBB 200 Then Desargues7 Theorem is x y z are concurrent ltgt X Y Z are collinear We proved gt in our proof of Theorem 616 above This together with the principle of duality proves lt 63 Remark 624 The theorems of Menelaus and Ceva Theorems 348 and 349 above appear to be dual as they are usually stated When they are stated with directed distances instead of distances as above it becomes clear that they are not dual Menelaus has a minus one in the conclusion and a quadratic equation in the proof Ceva has a plus one in the conclusion and a cubic equation in the proof One cannot transform one to the other by taking transposes of matrices Exercise 625 State the dual of li appus7 theorem and draw a diagram illustrating it 65 The Projective Line 626 The projective line llD1 is the space of lines through the origin in R2 In analogy with the projective plane each point my 6 R2 with my 74 00 determines a point x y tLty E R2 t E R in the projective line P2 and pony x y for u 74 0 A transformation M LPquot1 a llD1 satisfying x ax by j i ywdy i 96 i MWy where ad 7 be 74 0 is called a projective transformation of the line These de nitions exactly parallel De nition 62 of the projective plane and De nition 611 of projective plane transformation Theorem 627 Given three distinct points A B C in lP l there is a unique projective transformation of the line M such that M10A7 M01B7 M11C Proof The same as Theorem 615 but easier D 628 If y 74 0 then my 21 where z This establishes a one one correspondence between lP l and and the space R U 00 obtained from the real line R by adjoining a point which we denote by 00 The real number 2 corresponds to the point z 1 on the projective line lP l and the point 00 64 corresponds to the point 1 0 With this identi cation llD1 and R U 00 a projective transformation takes the form 7 a2 b 7 02 d M 2 T where it is understood that 00 if 02 d 07 Moo ac if c 31 07 and Moo 00 if c 0 These conventions are consistent with the following formulas from Math 221 a2b a2bia zLIEoczd ZJEHOOcandiC if c 31 0 and7 for 20 idc a2bi i ioo7 lim w zazo CZ d 7 i ZHZoi CZ d 00 7 where the signs on the two ioo7s are opposite In calculus we usually think of adjoining two points7 00 and foo to the real numbers7 but here there is only one in nity A transformation of form is called a fractional linear transformation or a Mobius transformation 629 Let A x1y121 and B x2y222 be two distinct points in the projective plane P2 De ne b lPquotl a P2 by ts zyz where ztx1 sy1 y ty1sy2 z t21522 This de nition is legal since Myths pxwywz zyz for M 31 0 It is easy to see that b is a one one correspondence between the projective line lP l and the line AB in P2 We call b a projective parameterization of the line AB Note that b depends not just on the points A and B but also on the representatives chosen Exercise 630 Let A7 B7 0 be distinct collinear points in P2 Show that there is a unique projective parameterization of the common line such that 10 A 01 B7 and 1511 0 Exercise 631 Assume that b and 7 are projective parameterizations of the same line in P2 Show that there is a unique prpjective transformation M ll quotl a llD1 of the projective line such that g b o M 65 Exercise 632 Assume that o is a projective parameterization of a line Z in P2 and that T P2 a P2 is a projective transformation of the projective plane Show that T o o is a projective parameterization of a line TM 66 Cross Ratio De nition 633 The cross ratio of four distinct points A7 B7 0 D on the projective line is de ned by detA C detB D AB CD 7 detA D detB C In the formula the column vectors on the right are representatives respectively of the points by the same name It does not matter which representatives are used since the cross ratio is unchanged if any of the four column vectors is replaced by a nonzero scalar multiple of itself since the scalar factors out in the numerator and denominator and then cancels ln particular7 if A a17 B b17 C 017 D d17 then 7 a 7 c ABOD 7 a 7d Theorem 634 A projective transformation of the projective lirie preserves cross ratio Proof This is an immediate consequence of the fact that the determinant of the product is the product of the determinants7 ie detTA TB detT detA B where A7 B are 2 gtlt 1 column vectors and T is a 2 gtlt 2 matrix 1 A B O D Figure 15 Cross Ratio 66 Exercise 635 Let ABCD E R2 0 represent distinct points in lP l and O 00 denote the origin in R2 Then AOO BOD AOD BOO sin ZAOC sin ZBOD sin ZAOD sin ZBOC AC BD AD B0 AB CD In the rst formula AOB denotes the oriented area of AAOB and in the last formula it is assumed that the four points are collinear and AB denotes the directed distance along the common line The choice of direction doesnt matter since reversing the direction produces four sign reversals which cancel See Figure 66 Remark 636 Recall the relation between projective geometry and drawing in perspective from Remark 65 From a perspective drawing of three points A B C on a line in R3 we cannot conclude anything about the distances between them If A B C lie respectively on the line of sight from the origin 000 with A B C then the points A B C wll appear the same as the points A B C in the drawing although the ratios ABAC and A B A C can be different However in an accurate perspective drawing of four collinear points A B C D the cross ratio in the drawing will be the same as the cross ratio of the four points in space 67 A Geometric Computer In the following four exercises we will design a geometric computer It works better than the one in Exercises 343 346 because that one required us to draw parallel lines whereas this one requires only a straightedge to connect two points with a line Let 0 I Q A B C be six collinear points such that O I Q are distinct and a b 0 denote the three cross ratios 1 AQIO b BQJO c CQJO Exercise 637 Addition in Projective Geometry Assume given points M N P A B C satisfy the following conditions 67 a the points M N P Q are collinear b the points M O A are collinear c the points M B G are collinear d the points N O B are collinear e the points N A G are collinear f the points P A A are collinear g the points P B B are collinear and i the points P G G are collinear Draw a diagram illustrating this con guration and show that c a b Hint Take the common line of M N P Q to be the line at in nity and use Exercise 343 Exercise 638 Subtraction in Projectiye Geometry State and prove a theorem analogous to Exercise 344 in the same way that Exercise 637 is analogous to Exercise 343 The conclusion should be that b 7a Draw a diagram illustrating the con guration Exercise 639 Multiplication in Projectiye Geometry State and prove a theorem analogous to Exercise 345 in the same way that Exercise 637 is analogous to Exercise 343 The conclusion should be that c ab Draw a diagram illustrating the con guration Exercise 640 Division in Projectiye Geometry State and prove a theo rem analogous to Exercise 346 in the same way that Exercise 637 is anal ogous to Exercise 343 The conclusion should be that b 1a Draw a diagram illustrating the con guration 68 7 Inversive Geometry 71 The complex projective line 72 Feuerbach s theorem 69 8 Klein s View of geometry 81 The elliptic plane 82 The hyperbolic plane 83 Special relativity 70 A Matrix Notation Matrix notation is a handy way to describe messy calculations The matrix operations obey most of the rules of ordinary algebra the most important exception is the commutative law for multiplication The material presented in this section contains all the matrix algebra we shall need For more details consult any book in linear algebra A1 Fix positive integers m and n An m gtlt 71 matrix is an array an 012 aln 021 022 am A aml 1mg amn where each entry aij is a number There are m rows and 71 columns in an m gtlt 71 matrix We say that an m gtlt 71 matrix is a matrix of size m gtlt 71 when we want to call attention to the number of rows and columns A square matrix is one having the same number of rows as columns a column vector is an m gtlt 1 matrix a row vector is an 1 gtlt 71 matrix A2 To add or subtract matrices7 add or subtract the corresponding entries as in an 012 bll b12 an bll 012 b12 021 022 521 522 021 521 022 522 031 032 531 532 031 531 032 532 To multiply a matrix by a number7 multiply each entry by the number as in C an 012 013 7 Can 0012 0013 021 022 023 0021 0022 0023 The zero matrix of any size has all its entries zero and is usually denoted by 0 ThusA0A A3 If the number n of columns in A is the same as the number of rows in B the matrix product 0 AB is de ned by the rule that the number 017 in the 2th row and jth column of C is given by Cij E aikbkj k 71 For example an 012 511 512 011511a12521 allb12012b22 021 022 521 522 021511022521 021b1z022522 a b x 7 ax by c d y 7 0x dy 39 A4 The m gtlt m identity matrix I is the square matrix with is on the diagonal and 07s elsewhere For example if m 2 433 The identity matrix behaves like the number 1 multiplication by I leaves a matrix unchanged and A5 A square matrix A is called invertible there is a necessarily unique matrix A 1 called the inverse of A which satis es AA l A lA I where I is the identity matrix The following calculation shows that if A is invertible then the only matrix B satisfying BA I is B A l B BI BAA 1 BAA 1 IA l A l A square matrix is invertible if and only if its determinant is nonzero For example if a b A 7 l c d l 7 then detA ad 7 be and 1 d 7b A 1 ad 7 be 6 a A6 The transpose A of a matrix A is de ned by the rule that the entry of A in the 2th column and jth row is the same as the entry of A in the 2th row and jth column For example i an 012 7 an 021 021 022 012 022 72 and B Determinants ln elementary linear algebra Math 340 at UW one studies an operation which assigns to any square matrix A a number detA called the determinant of A In this appendix we state its key properties In the notes we only need determinants of 2 gtlt 2 and 3 gtlt 3 matrices these are often taught in high school algebra For these matrices one can check the properties using elementary algebra7 although the 3 gtlt 3 case is a bit hairy For the proofs in the general case see any book on elementary linear algebra My favorite is Theorem B1 There is a unique function called the determinant which assigns to each square matrid A a number detA satisfying the following properties Scale IfB results from A by multiplying some row by a number c then detB cdetA Swap IfB results from A by interchanging two rows then detB idetA Shear IfB results from A by adding a multiple of one row to another row then detB detA Identity The determinant of the identity matricc I is detI 1 Exercise B2 The determinant of a 2 gtlt 2 matrix A an 012 J 021 022 111022 7 112021 Verify the properties listed in Theorem B1 using this formula 73 Exercise B3 The determinant of a 3 gtlt 3 matrix 011 a12 a13 A 021 a22 023 031 032 033 detA 111022033 a12 l2 xan 013021032 011023032 a13022031 112021033 Verify the properties listed in Theorem Bl using this formula Remark B4 There is a general de nition for the determinant of an n gtlt n matrix as a sum detA Z iarauwz a 39an0n where the sum is over all n permutations of 17 27 7n The sign depends on the permutation o in a subtle way We do not need this formula in these notes and will not explain it Theorem B5 A square matrip is invertible if and only if its determinant is not zero Exercise B6 Find AB and BA where a b A l c d l B d b l 70 a Then show that A is invertible if and only if its determinant is not zero and give a formula for A l Theorem B7 The determinant of a matrip and its transpose are the same detA detA Hence the properties of Theorem Bl hold reading quotcol umn for row Exercise B8 Check this for 2 gtlt 2 matrices 74 Theorem B9 For a square matricc A the homogeneous system AX 0 has a nonzero solution X if and only if detA 0 Exercise B10 Check this for 2 gtlt 2 matrices Corollary B11 The determinant of a square matricc uanishes if and only if one of its columns is a linear combination of the others Theorem B12 The determinant function satis es the following 1 detI 1 2 detA 1 detA 1 3 detAB detA detB Exercise B13 Check this for 2 gtlt 2 matrices C Sets and Transformations C1 A set X divides the mathematical universe into two parts those objects x that belong to X and those that dont The notation z E X means z belongs to X The notation z X means that z does not belong to X In geometry the word locus is often used as a synonym for set as in the sentence The locus of all points P such that ZAPB is a given constant is an arc of a circle C2 If X is a set and Pz is a property that either holds or fails for each element x 6 X7 then we may form a new set S consisting of all z E X for which Pz is true This set S is denoted by S x E X 1 and called the set of all z E V such that Pz The set S is a subset of X meaning that every element of S is an element of X The notation used in equation 1 is called set builder notation Having de ned S by 17 we may assert that for all z zES ltgt aEXandPx and that for all z E X z e S ltgt Pz 75 Since the property Pz may be quite cumbersome to state7 the notation z E W is both shorter and easier to understand The symbol ltgt is an abbreviation for if and only if C3 Let X and Y be sets The notation f X a Y means that f is a func tion which assigns to each point z E X and element f E Y Mathemati cians have more words for this concept than any other calculus textbooks call f a function with domain X or de ned on X taking values in Y7 while textbooks on linear algebra or plane geometry call f a transformation from X to Y7 and in more advanced mathematics one says that f is a map or mapping from X to Y In these notes we use the word transformation and almost always we take X Y R2 the set of pairs of real numbers We will also use the letter T rather than f C4 For any set X there is a transformation IX X a X called the identity transformation of X and de ned by IXx z for z E X The composition gof X a Z of two transformations f X a Y and g Y a Z is de ned by 9 0 M96 90 for z E X A transformation f X a Y is called invertible iff there is a necessarily unique transformation f l Y a X such that f 1f96 96 and ff 1y y for z E X and y E Y The transformation f 1 is called the inverse trans formation to f Clearly hog of ho fog7 foIX f7 yof f7 and f 1 of IX and fof l 1y C5 When f X a Y and S is a subset of X we de ne the image of S by f to be the set of all points f as x ranges over S It is denoted by fS ln set builder notation this is written as fS fm 396 E S 2 The notation is often used to emphasize that the right hand side is the de nition of the left hand side7 so that no proof is required To prove that a point y lies in the set fX one must show there is an x E S with y 76 Example C6 ln calculus one learns that z cos 6 y sin6 3 are parametric equations for the unit circle In the notation introduced thus far this could be written xy 6R22y21cos6sin66ER 4 The left hand side uses the set builder notation of equation 1 while the right hand side uses the set builder notation of equation It is also true that xy E R2 2 y2 1 cos6sin6 0 S 6 lt 27139 C7 To prove that two sets are equal on must show that every element of one set is an element of the other set and vice versa For example to prove 4 we argue as follows If my lies in the set de ned on the right hand side of 4 then 3 holds for some number 6 Hence 2 y2 cos2 6 sin2 6 1 by the Pythagorean Theorem so my lies in the set on the left hand side of Conversely if 2 y2 1 then my cos 6 sin 6 where 6 tan 1yz if z gt 06 tan 1yz 7139 if z lt 06 7T2 if my 01 and 6 37r2 if zy 0 71 The case analysis is necessary because the equivalence m tan6 ltgt 6 tan 1m holds only if 77r2 lt 6 lt 7r2 77 References H M l3l E M E T Bell Men of Mathematics Simon amp Schuster 1937 L M Blumenthal A Modern View of Geometry W H Freeman 1961 H S M Coxeter amp S L Greitzer Geometry Revis ited Random House 1967 Reissued by the MAA see http wwwmaa orgpubsbooksnmllQ html A Classic written by a famous geometer l Martin lsaacs Geometry for College Students Brooks Coles 2001 A beautiful book with lots of problems written by a professor at UW H Levy Projectiye and Related Geometries Macmillan 1961 Professor Levy introduced me to the subject in a course at the University of Illinois in 1961 62 E A Maxwell Fallacies in Mathematics Cambridge University Press 1959 See httpbookscambridgeorg0521057000htm G E Martin Transformation Geometry An Introduction to Symmetry Springer Undergraduate Texts in Math 1982 MathWorld Website httpmathworldwolframcom A google search on any term from elementary geometry like Gergonne Point usually brings up this websiite J Robbin Matrir Algebra Using MINImal MATlab AK Peters 1995 78

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