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# SEM IN SELECTED TOPICS MA 501

UK

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This 20 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 501 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/228158/ma-501-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15

Geometry 3 Before Tuesday September 4 Read Dunham Chapter 2 and read Notes on Geometry Sections 51754 avail able from the course website httpwwwmsukyedu leema501fa07ma501fa07html Go to the Forum Axiomatics and make at least one substantive contribution by 11 pm Tuesday September 4 and at least one substantive response to others7 postings before class on Thursday September 6 Respond to my musings on my levels in Section 53 by writing about the following On which level do you place yourself On what level shouldcan students be in various points in the K716 timeline What does this have to do with the van Hiele levels Before Thursday September 6 Read Dunham Chapter 2 and read Notes on Geometry Sections 54754 available from the course website httpwwwmsukyedu leema501fa07ma501fa07html Familiarize yourself with the website aleph0clarkuedu djoycejavaelementselementshtml As you read think about the following questions for discussion H For axiom systems what is the meaning of consistent independent and complete What are models and how can they help determine whether axiomatic systems are consistent independent andor complete E0 Lewis Carroll the author of Alice s Adventures in Wonderland and Through the Looking Glass also wrote a book called Euclid and His Modern Rivals What is this book about and why did Lewis Carroll write it 3 Why mention I Kings 723 on page 30 of Dunham 4 Study some of the statements of Euclid7s Elements on the website httpaleph0clarkuedu djoycejavaelementselementshtml thinking about the wording and meaning and studying the diagram Does having a dynamic draggable diagram help in your understanding 5 Look at some of the proofs of the propositions not covered in Dunham 6 Try to carry out some of Euclid7s constructions with Geogebra or other dynamic ge ometry software Then try dragging around some of the points 5 00 p H CH H 5 Try to nd a website with which you can illustrate some of the statements of hyperbolic geometryifor example7 one in which you can sketch a triangle and measure its angles Find a list of Hilbert7s axioms for geometry Find a list of the SMSG axioms for geometry these are the ones my teacher used in my 10th grade geometry course Find a list of axioms from some other geometry text Compare these various lists with with those of Euclid Look at the material related to the Pythagorean Theorem on the National Library of Virtual Manipulatives7 httpnlvmusueduennavvlibraryhtml Take a look at the website httpwwwcut the knotorgpythagorasindexshtml Study Euclid7s proposition Vl317 and relate this to your past homework What is the connection of the Pythagorean Theorem to the distance formula ls this a standard connection typically made in High School Which basic trigonometric identity is actually a statement of the Pythagorean Theo rem There is a song in the Gilbert and Sullivan comic opera The Pirates of Penzance mentioning the Pythagorean Theorem Find these lyrics When the Scarecrow discovered he had a brain at the end of the Wizard of Oz7 he quoted the Pythagorean Theorem Or did he Yes7 even this can be found on the web The compasses commonly used in school hold a xed opening7 allowing lengths to be transferred from place to place A Euclidean compass did not do that as soon as it was picked up7 the size of the span was lost However7 Propositions 2 and 3 of Book I prove that lengths can be transferred7 thereby legitimiZing modern compasses Study these propositions to see how Euclid does this Find a list of axioms for a Set theory Zermelo Fraenkel b The natural numbers Peano c d e Vector spaces The real numbers and why dont the rational numbers satisfy all of them Groups Name some college level mathematics courses that typically begin with a set of axioms for some collection of objects Thursday September 6 779 pm Attend the Adobe Connect session to discuss the readings discussion questions fo rum and comments and questions on the assigned homework due on Sunday Before Sunday September 9 11 pm Homework problems due Sunday September 9 11 pm uploaded to the Moodle site as a single le less than 2 MB or else emailed to the address leemsukyedu Please use Word or pdf les only H E0 00 r U 03 Notes on Geometry Exercise 521 Notes on Geometry Exercise 541 Notes on Geometry Exercise 543 I dont remember the answer to this one Notes on Geometry Exercise 544 The ratio of the circumference of a circle to its diameter is not 7139 in non Euclidean geometry In fact it is not even a constant lllustrate this by considering a sphere of radius 1 marked with lines of latitude and longitude as on a standard globe a Look at the circle A of 45 degrees north latitude 45 degrees up from the equator Regard the north pole N as its center and a great circle are from N to the cir cumference 0f the circle A as its radius Calculate the ratio of the circumference to the diameter giving an exact answer and then also evaluating it to several decimal places b Repeat with the circle B that is 30 degrees north latitude c Repeat with the circle G that is 0 degrees north latitude ie the equator d Try to develop a formula for the ratio of the circumference to the diameter of a circle of z degrees N latitude How does this number compare to 7T7 Extra Credit Notes on Geometry Exercise 547 Isometries 1 Identifying Isometries 1 Modeling isometries as dynamic maps 2 GeoGebra les isoguesslggb7 isoguess2ggb7 isoguess3ggb7 isoguess4ggb 3 Guessing isometries 4 What can you construct or trace to nd the de ning elements l D 00 4 0 Three Re ection Theorem A point is uniquely determined by its distances from three non collinear points 7 model with GeoGebra Therefore an isometry is uniquely determined by its action on three noncollinear points Three Re ection Theorem lf T and T are congruent triangles7 then T can be mapped onto T using at most three re ections GeoGebra le threere ectionsggb Therefore every isometry is the composition of zero7 one7 tvvo7 or three re ections Similarly7 there is a Four Re ections Theorem for isometries in space OON t 4 0 Ch Compositions of Isometries Zero re ections is the identity isonietry One re ection is a re ection The composition of two re ections in parallel lines is a trans lation perpendicular to the lines by a distance equal to twice the distance between the two lines GeoGebra le twore ec tionsggb The composition of two re ections in intersecting lines is a rota tion around the point of intersection by an angle twice that of the angle of intersection GeoGebra le twore ectionsggb The composition of three re ections is either a re ection or a glide re ection The composition of two rotations is either a rotation or a trans lation GeoGebra le tworotations Express each rotation as a double re ection to determine the center and angle of rotation Dgt t 00 Kaleidoscopes Perform repeated reflection7 alternating between two given lines What angles of intersection between the two lines yield only nitely many images There are three dimensional variants of this involving re ections in three intersecting planes l D 00 4 Formulas for Isometries Translation by the amount p q Rotation by 6 about the point p q c 5 pcqsp s c ps qcq O O 1 Where c cos 6 and 5 sin 6 Both of these matrices are of the form c su scv 001 Where 62 82 1 Further7 given any matrix of the above form7 one can solve for 67 p and q7 so any such matrix is an isometry Both of these matrices have determinant equal to one These are the direct isometries 0 Ch 1 00 Every isometry is in fact one of the above forms Re ection across the line With equation pqyT O7 assuming mm i p2lq2 2pq 2pr n r qQ Qm 0 0 1 Glide re ection by re ecting across the line With equation pa qy 7 0 followed by translation by the amount tq tp7 assuming that p2 q2 1 p2 q2 2pq 2pr tq 2pq p2q2 2qrtp 0 0 1 Both of these matrices are of the form csu 5 011 001 Where 62 82 1 Further7 given any matrix of the above form7 one can solve for t p and q7 so any such matrix is an isometry Both of these matrices have determinant equal to negative one These are the indirect isometries See Exer cise 812 in Geometry for Middle School Teachers l OTHgtOOD Examples with Maxima Maxima is a free computer algebra system Creating isonietry matrices Inverses Compositions Solving for isornetries l 00M C OTHgt PointLine Incidence via Isometries Let P be a point and Hp be the isometry that is a rotation about P by 180 degrees Let L be a line and BL be the isometry that is a re ection in L Then P is incident to L if and only if RPORLRLORPa equivalently7 if and only if RPRLRPRL I Veri cation geometrically Veri cation algebraically This idea can be expanded to give an axiomatic system for Eu clidean geometry or other geometries in terms of groups 8 l D gtJgtoo Interlude Mathematical and Physical Re ections Why does the image in a mirror appear to be in the location de ned by a mathematical re ection If the human eye detects a set of light rays that7 when traced backward7 appear to emanate from a common point7 then the brain makes the interpretation that that common point is the origin of the light rays GeoGebra le physicalre ectionggb Because the atmosphere can bend the path of light rays7 The same phenomenon accounts for the apparent presence of the sun just above the horizon after it has actually set below the horizon l D 00 Applying Isometries to Figures De ned by Equa tions We are accustomed to applying an isometry to a drawn gure in the plane7 or to individual points via formulas7 but what about to gures described by equations Example Apply the translation by the amount p q to the circle whose equation is 2 12 100 The isometry is given by f p7yyq Thuszc m p7 y y q Substituting yields T p2 3 q2 100 for the equation of the translated circle ln general7 if we translate the graph of a function described by y by the amount p q then we get the new graph de scribed by g q p Thinking of things this way helps eliminate the necessity of memorizing rules for how to shift up or down or right or left 4 0 Completing the Square and the Quadratic Formula are merely using translations to simplify equations of parabolasl Example Consider the parabola described by y 23c 12m 13 Lets nd a translation of the form f 33 so that the equation of the translated parabola has no 3 term Using 13 m p we get y m p2 12f p13 which simpli es to y 2m 4p 12 2p2 12p 13 We want to choose p so that 4p 12 is zero7 so choose p 3 ln other words7 translating 3 units to the left results in a parabola centered on the y axis Then the equation further simpli es to y 2f2 5 Finding the two roots as intercepts is now easy 5 i i quotT 2 Translating the parabola and its roots back to the original posi 5 mm pBig 11 tion gives 6 Lets do this in general Consider the parabola described by y a2bc Using the translation f 33 and substituting a f p this becomes yam p2bm pc which simpli es to y 2 b 2apm ap2 bp c We want I 261 to be zero7 so choosing p results in a parabola centered on the y axis With this choice of p the equation then simpli es to 72 b2 4610 y am 4a The roots are simple to calculate b2 4610 2a 39 Translating the parabola and its roots back to its original posi fl tion gives the roots i b xb2 4ac p ij 2a 2a Which is the quadratic formula 7 Of course7 we dont have to use only translations to simplify 00 things ln general7 if we have a gure described by an equation of the form f 13 y O we might consider applying an isometry with some rule mapping 3631 to my and then substituting into the equation7 choosing our isometry wisely so that after wards things look simpler Hold on tightil am now going to do an example of this that l was taught in my public high school Baltimore County7 Maryland eleventh grade course in TrigonometryAnalytic Geometry by Mr Laferty 9 Problem lolentify the curve given by the equation 73 52y 723w 290 2801 325 0 We will try a rotation about the origin so that after the rotation there will be no my term A rotation about the origin by an angle 6 with sin 6 5 and cos6 c is given by f 66 51 y 53 0y Solving for 13 and y is equivalent to rotating by 67 hence 6 cf 53 y 5T 03 ll we substitute for 13 and y in the original equation7 the coef cient of W can be calculated to be 4205 7202 52 We want this to equal zero7 so we want 72 i 12 i cs 42 i 7 i c2 5239 But this latter expression7 by the double angle forniulas7 equals 1 E sin 26 i 1 7 it 2 cos 26 2 an 6 So tan 26 24 7 Drawing a right triangle with legs 24 and 77 and using the Pythagorean Theorem to determine that the 14 hypotenuse is 257 we calculate that cos 26 725 Hence7 using the half angle forrnula7 6 1 cos26 3 5sin A 2 5 so 0 cos6 V1 sin26 45 Using these values for c and 5 in the desired rotation7 and sub stituting into the original equation7 this simpli es to 100m2 400 25y2 503 325 0 or 4y 16f32 2y13 0 Completing squares 4T2 16f16322y113161 4f 22y12 4 So nally we have 7 2 2 i 1 2 ltsc gt lty gt 1 4 which is the equation of an ellipse centered at the point 2 1 We could further translate it by the amount 2 1 to obtain the ellipse 1 D D 1 am New centered at the origin Now it is possible to sketch the curve in its new position By using the inverses of the translation and the rotation in that order we can sketch the curve in its original position 10 One nal commentiwe can use the above technique to verify 1 l that certain gures have certain symmetries For example7 if we consider the gure described by the equation 33y 100 and apply the isometry m C y y we see that the equation is unchanged W 100 This this gure is symmetric under 180 degree rotation about the origin Also7 if we apply the isometry f 11 13 we again see that the equation is unchanged Thus this gure is symmetric under the action of re ection across the line y as Of course7 all of these ideas can be extended into three and higher dimensions

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