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# MODERN GEOMETRY MATH 532

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This 9 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 532 at University of South Carolina - Columbia taught by Ma Filaseta in Fall. Since its upload, it has received 25 views. For similar materials see /class/229525/math-532-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15

Or I U 4 L1 53 MATH 5327361 LECTURE 7 Homework Problem Sheet on Vector Notation Theorem 1 Let A and B be distinct points Then 0 is on E if and only if there is a real numbert such that O l 7 tA tB Basic Ideas of Proof 7 7 0 AC tAB o C 7 A t B 7 A Comment In Theorem 1 7 t 7 length of AC 1 7 t length of W7 where a plus sign occurs on the right if and only if C is between A and B and one denomi nator is 0 if and only if the other denominator is 0 Basic Idea of Proof Consider 3 cases depending on the position of 0 relative to A and B Theorem 2 If A B and O are collinear then there exist real numbers 2 y and 2 not all 0 such that xy20 and zAyB2CU Basic Ideas of Proof 0 HA Btakez Ly 7land2 0 0 Otherwise use Theorem 1 and take 2 l 7 t y t and 2 71 Theorem 3 If A B and O are points and there exist real numbers 2 y and 2 not all 0 such that xy20 and 2AyB2CU then A B and O are collinear Basic Ideas of Proof 0 Relabel so 2 7 0 o Deduce A 7yzB 72zO and l 7112 7 0 Take It 722 so that l 7 t 7112 0 Use Theorem 1 Theorem 4 If A B and O are not collinear and there exist real numbers 2 y and 2 such that xy20 and 2AyB2CU thenzy20 Basic Idea of Proof This is a rewording of Theorem 3 r L 4 EXAMPLES 0N TRANSLATIONS AND ROTATIONS Lecture Notes for Math 532 taught by Michael Filaseta Let AABC be given and let MB denote the midpoint of side E and let Mo gt denote the midpoint of side E Show that M 3 M0 is parallel to R and that the length of M 3M0 is onehalf of the length of m Solution A picture would help here From the theorem f RWgtMBR7rgtMC is a translation One checks that fB C Hence f is a translation which moves B to C Also fMc M6 where M6 is the result of rotating Mo about M3 by 7r This means that McMB and MBM6 have the same length and the three A pomts M0 M3 and M6 are collinear We get that McM BC and the result follows Let A B C and D be the vertices of an arbitrary quadrilateral Show that the midpoints of the sides of the quadrilateral form a parallelogram Solution Let M1 be the midpoint of E M2 the midpoint of W M3 the mid point of W and M4 the midpoint of m Then from example 1 M1M4 and MQMg each have the same direction as FD and half its length The desired con clusion follows Comment Instead one can let f I37BMAI3WgtM1 andg RWgtM2RWM3 Then f is a translation taking B to D and g is a translation taking D to B One then essentially repeats the argument in example 1 I In 2 consider instead midpoints of a 2n7gon Solution The problem is a bit vague but one can conclude the following us ing the argument in 2 Let M17 M2 7 M2 denote the midpoints along the edges moving counterclockwise beginning with some edge Then the segments M1M27 M3M47 7M2n71M2n can be translated without rotating them to form an nigon Similarly the segments MQnM17M2M37 7M2n72M27771 can be translated to form an nigon I Let AABC be given Draw an equilateral triangle exterior to AABC with one edge AB an equilateral triangle exterior to AABC with one edge W and an equilateral triangle exterior to AABC with one edge E Show that the centers of these 3 equilateral triangles form the vertices of an equilateral triangle Solution The argument is essentially the same as in the next problem This is worth going over separately but we do not do so here I 5 Generalize 4 as follows Let AABC and real numbers a B and 7 be given Let A B and C be points exterior to AABC such that ABAC a LABC and AAC B 7 Also suppose that the lengths of the sides BA and m are the same the lengths of the sides AB and B C are the same and the lengths of the sides AC and C B are the same Show that ifa B 7 2 then the interior angles of AA B C are a and 7 Solution Let f RaAR5BRYc Then f is atranslation since a 7 27f Since fB B we get that f is the identity translation Hence fC C Let C RABCV Then AC B C and the lengths ofm and B C are the same Also 0 fC RaARgBRcC RaARgBC Ra 0 This means that AC A C a and the lengths ofm and AC are the same One easily gets that the triangles AA B C and AA B C are congruent from which it follows that AC A B 12 and AC B A 52 It follows that AA C B 7r 7 12 7 52 72 giving the desired result I Observe that the following is a consequence of the problem Suppose that AABC is given and D E and F are points exterior to AABC such that ADB C AAEC and AABF are similar so that AD 4E and AF are the three angles associated with these similar triangles Let A B and C be the centers of the circumscribed circles for ADBC AAEC and AABF respectively Then AA B C is similar to ADBC and hence the other two exterior triangles as well 6 Let n be an odd positive integer and let P17 Pn be n not necessarily distinct points Let A A0 be an arbitrary point Forj E 17 7n de ne Aj as the point you get by rotating Aj1 about by 7r Forj E n 17 2n de ne Aj as the point you get by rotating Aj1 about Prn by 7r Prove that A2 Solution Write n 2k l where k is some nonnegative integer Let f denote the composition of the rst n rotations about P17 Pn each by 7r Then we want to show that ff A Note that every two rotations by 7r are equivalent to a translation and the composition of translations is a translation Hence we can view f as TabR7rp1 for some a7b where a b 0 if k 0 By a homework problem we can rewrite this as Rwa2b2 Rw00R7r2P1 RW2P1 Taking the product of the matrices as indicated by the parentheses above we get from the theorem that f is equivalent to a rotation about some point by 7r It is clear then that f f A completing the argument I Comment The situation when n is even is that A is translated since the composi tions of the rotations is a translation It is possible that the translation is the identity translation in which case A2 A 7 so Let n be an odd positive integer Suppose that we are given the n midpoints of the sides of an nigon Show how one can construct an nigon with these given midpoints along its edges Solution Let M1 Mn be the midpoints Consider the composition f ofthe ro tations about M1 Mn each by 7r By the solution to 6 we get that f is equiv alent to a rotation about some point say A by 7r We can construct A as follows Take any point B and apply f to it Note that this is done by successively taking the rotations about Mj by 7r with straightedge and compass for j 1 2 n We get some point C f Since B is obtained from C by a rotation about A by 7r we deduce that A must be the midpoint of W Since f is equivalent to a rotation about A we have that fA A Set A0 A and rotate it about M1 to obtain a new point Call the new point A1 and rotate it about M2 to obtain another point A2 Continue rotating Aj about Mj1 to obtain Aj1 forj 6 12 n 7 1 Since fA A we get that An A0 and the points A1 An are the vertices of an nigon as desired I Comments i There are many such nigons since the order of the Mj one chooses to do the above construction will affect the outcome ii There is another approach to the problem which may be worth discussing For example suppose n 5 though any odd n works here Call the given midpoints M1 M5 Let A0 A4 be the points we are trying to construct with Mj along edge AjL1Aj forj E 1 5 where A5 A0 Then the vertices A0 A1 A2 and A3 are the vertices of a quadrilateral and three of its midpoints M1 M2 and M3 are known Using the information from example 2 it is not dif cult to construct the midpoint M of AoAg Now we know the midpoints of the sides of triangle AA0A3A4 Using the information from example 1 we can construct the vertices A0 A3 and A4 One can modify this argument to obtain the other vertices or use the approach in the solution above Let P1 P2 P3 and P4 be 4 not necessarily distinct points Let A be an arbitrary point Beginning with A0 A forj E 1 2 3 4 de ne Aj as the point you get by rotating Aj1 about Pj by 7r Set Q1 P3 Q2 P4 Q3 P1 and Q4 P2 Beginning with B0 A forj 6 123 4 de ne Bj as the point you get by rotating B141 about Qj by 7r Prove that A4 B4 See Figure 1 Solution A rotation about P1 by 7r followed by a rotation about P2 by 7r is a translation say TR Similarly a rotation about P3 by 7r followed by a rotation about P4 by 7r is a translation say T5 The problem amounts to establishing that TSTR TRTS This is easy to establish but note that in general the product of two matrices does not commute I Let A B C and D be the vertices of a convex quadrilateral labelled counterclock wise Consider 4 squares exterior to the quadrilateral one square with an edge E one re with an edge B C one square with an edge 0 D and one square with an edge DA Let M1 be th mter of the square with edge AB let M2 be tlEenter of the square with edge B C let M 3 be the cent of the square with edge CD and let M4 be the center of the square with edge DA Show that the length of M1M3 m m 1s the same as the length of M2M4 and that M1M3 and M2M4 are perpend1cular Solution Let g Rw27M1R7r27M27 h Rw27M3R7r27M47 and f Then f A A and we get that f is the identity translation Also there are points P1 and P2 such that g is a rotation about P1 by 7r and h is a rotation about P2 by 7r It is easy to see draw a picture that since a rotation about P1 by 7r followed by a rotation about P2 by 7r is the identity we must have P1 P2 Next we observe that P1 is the only point such that gP1 P1 It follows that AM2P1M1 is a isosceles right triangle labelled counterclockwise since P1 so located is mapped to itself by 9 Similarly AM4 P2M3 is a isosceles right triangle labelled counter clockwise Since P1 P2 we get that AP1M2M4 is obtained from AP1M1M3 by a rotation about P1 by 7r 2 This implies that the length of M 1 M 3 is the same as the length of M2M4 Let Q1 be the pomt of 1ntersectlon of M1M3 and M2M4 and Q2 the p01nt of 1ntersectlon of M2M4 and M1P1 conV1nce yourself these eX lSt Then ZMlQQQl ZPngMQ and ZQngQl ZPlMQQQ and it follows a a that 1M1Q1Q2 1M2P1Q2 7r2 Thus the hnes M1M3 and M2M4 are perpendicular I Comment It can be shown that if the original quadrilateral is a parallelogram then M1 M2 M3 and M4 form the vertices ofa square 10 Let AABC be a given triangle with the angles AABC ABCA and ACAB all acute Let F be an altitude drawn from A so that P is on m Similarly let m be the altitude drawn from B and m the altitude drawn from C Show that AP QR is a triangle with minimum perimeter that can be inscribed in AAB C that is show that if AUVW is a triangle with U on W V on E and W on E then its perimeter is at least that of APQR Solution First we will show that APRB AQRA ARPB AQPC and APQC ARQA We establish one of these and the other two can be done in the same way Let D be the intersection of the altitudes Since ADRA DQA 7r2 the points A Q D and R all lie on a circle Hence ADRQ 7 ADAQ 7 g 7 ACP Since ADRB ADPB 7r2 the points B P D and R all lie on a circle Hence ADRP 7 ADBP 7 g 7 ABCQ 7 g 7 ACP 7 BBQ Since AARC ABRC we get that APRB AQRA As mentioned in a similar fashion one obtains ARPB AQPC and APQC ARQA If you haven t already started drawing pictures get out those crayons To match some of the discussion here you should label your points along the triangle as A B and C in a counterclockwise direction We begin with triangle AABC and re ect it about side W to get a new tringle AAlBC The 2 triangles are distinct congruent and share the edge W Next we re ect AAIBC about side AlC to get a new tringle AAlBlC Then we re ect AAlBlC about side A1B1 to get a new tringle AAlBlCl Next we re ect AA1B1C1 about side B1C1 to get a new tringle AAgBlCl Finally we re ect AAgBlCl about side A201 to get a new tringle Aflng C1 If your crayoning technique is mastered this is what should happen Leta ABAC and AABC Every point along segmentm is rst rotated about B by 2 7 25 then about A1 by 2 7 2a then about B1 by 25 and nally about A2 by 2a We can conclude that the segment E has been translated to 14ng This means that ill R is parallel to If we draw in APQR and its re ections the information from the rst paragraph implies that the points R P Q1 the rst re ection of Q R2 the next re ection of R there was already an R1 from the rst re ection P2 the next re ection of P Q3 the next re ection of Q and R4 the last re ection of R are all collinear Also the segment RR4 has length twice the perimeter of APQR If one similar re ects AUVW one nds that W is translated to some W4 Here WW4 and RR4 have the same length since R and W went through the same translation and the length of WW4 is g twice the perimeter of AUVW The desired conclusion follows I Figure 1 D U 4 MATH 5327361 LECTURE 6 Wrapping Things Up Recall the axioms of nite projective and af ne planes both of order n Axiom P1 Axiom P2 There exists at least 1 line with exactly 71 1 distinct points on it Axiom P3 Axiom P4 There exist at least 4 points no 3 of which are collinear Given 2 distinct points there is exactly 1 line that they both lie on Given 2 distinct lines there is at least 1 point on both of them Axiom A1 Axiom A2 Axiom A3 Axiom A4 Given any line Z and any point P not on Z there is exactly 1 line through P that does not intersect Z There exist at least 4 points no 3 of which are collinear There exists at least 1 line with exactly 71 distinct points on it Given 2 distinct points there is exactly 1 line that they both lie on Explain why the axioms for an af ne plane hold for a p gtlt p array of lines and points that we constructed Here p is a prime points are of the form a b where a and b are in the set say S 01 p 7 l and lines are of one of two forms z E u mod p or y E mz k mod p where u m and k are all in S Explain how to add 71 1 points and 1 line to an arbitrary nite af ne plane of order n to create a nite projective plane of order n This is done by adding new points at in nity and adding a line that passes through all of the new points Each new point is associated with lines of the form x E u mod p or lines with the same slope The new points are used to make the parallel lines in the af ne plane intersect at a new point at in nity Justify that the axioms for a projective plane of order 71 hold when the new points and line are added Explain how to create an af ne plane of order n from an arbitrary projective plane of order n by taking away any line and all the points on that line Justify in this case that the axioms for an af ne plane will hold wNH 4 L1 53 7 MATH 5327361 LECTURE 5 Finish Projective Plane Theorems from Homework 2 Homework Mention Modulo Arithmetic Homework De ne a modulo m Discuss and prove how congruences can be treated almost like equations see theorems below Theorem 1 Let a b c d and m be integers with m gt 0 Then the following are true 0 Ifa E b mod m and b E 0 mod then a E 0 mod o IfaEb mod m andCEd mod mthenaiCEbid mod o Ifa E b mod m and c E d mod then ac E bd mod o lfa E b mod then ak E bk mod m for every k E 0 l 2 Theorem 2 Let a and m be positive integers with no common prime divisors Then there is a positive integer b such that ab E 1 mod Examples 0 Is 25621235904 divisible by 3 by 4 by 5 by 9 by 11 o What is the last digit of 347223 o Is 5638287462039703 the sum of 2 squares o A Fermat number is a number of the form Fn 22 l The rst few Fermat numbers are F0 3 F1 5 F2 17 F3 257 F4 65537and F5 4294967297 The rst 5 Fermat numbers are primes Explain why F5 is divisible by 641 using that 641 5 x2712454 Homework 1 Test the number 1433456304672354 for divisibility by 2 3 4 5 9 and 11 2 Make up a test for divisibility by 8 and explain why it works 3 What is the last digit of 11704639 2170469 3170469 100170469 4 What is the last digit of 129459 5 Using an argument modulo 8 explain why 3462437654807 is not the sum of 3 squares

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