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# Tpc Geomtry Sec MATH 409

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Notes on transformational geometry Judith Roitman Jeremy Martin April 22 2009 1 The intuition When we talk about transformations like re ection or rotation informally we think of moving an object in unmoving space For example in the following diagram when we say that the shaded triangle is the re ection of the unshaded triangle about the line Z we think about physically picking up the unshaded triangle and re ecting it about the liner This is not how mathematicians think of transformations To a mathematician it is space itself 2D or 3D or that is being transformed The shapes just go for the rider To understand how this works letls focus on the following basic transformations of the plane translations along a vector re ections about a line rotations by an angle about a point To help us consider these as transformations of the plane itself youlve been given two transparencies You ll keep one xed on your desk Youlll move the other one The one that moves represents what happens when you move the entire planet The one that stays xed tells you where you started fromi Project 1 Draw a dot on your paper Take a transparency sheet put it over your paper and trace the dot What can you do to the top transparency iiei plane so that the dots will still coincide lief which translations re ections and rotations leave the dot xed Now place two dots on the bottom sheet and trace them on the transparency The dots should be two different colors For the sake of discussion let s assume one is red and one is blue What can you do to the transparency so the dots still coincide red on red blue on blue lief which translations re ections and rotations leave the two dots xed Which translations re ections and rotations put the blue dot on top of the red dot and the red dot on top of the blue dot Now try this with three dots in of course three different colors which are not collineari Which translations re ections and rotations leave the three dots xed Now try this with a straight line Pretend that it s in nite Which translations re ections and rotations leave the line xed Which translations re ections and rotations don7t leave the line xed but still leave it lying on top of itself 2 The basics 21 Transformations Let7s formally de ne what a transformation is De nition 1 A transformation of a space S is a map 45 from S to itself which is 11 and onto Notation 45 S A S Remember that 11 means that if p q are different points then f that is there7s no more than one way to get to any given point in 5 via while onto means that for every point 4 there is some point p such that L that is there7s at least one way to get to any given point via Here are some examples of transformations of R2 the plane 1 Re ecting the plane across a line 2 Rotating the plane about a point by a given angle 3 Translating a plane by a given vector 4 Contracting or expanding the plane about a point by a constant factor All of these kinds of transformations can be applied to R3 3 space as well with some modi cation For example re ection in R3 takes place across a plane not across a line and rotation occurs around a line not a point Question for those who have had some linear algebra or vector calculus How do these various transformations behave in R Here are some functions that are not transformations 1 The function taking all points zy E R2 to the point z E R It s not 11 and the space you start with isn7t the space you end up with E0 The map taking all points z E R to the point 10 6 R2 Its not onto and the space you start with isn7t the space you end up with even though R is geometrically isomorphic to its image CA3 Folding a plane across a line Z this is 21 rather than 11 off Z and it isn7t onto the whole plane he The function f R A R de ned by 12 It s neither 11 nor onto On the other hand the function 91 13 is a transformation An important note When we talk about transformations we only care about the map not how it is described For example the following three descriptions all describe the same transformation 1This is still an interesting map geometrically even though it isn7t a transformation Its an example of projection in this case projecting a plane onto a ne o rotate the plane by 90 about the origin 0 rotate the plane by 72700 about the origin 0 re ect the plane across the zaxis then re ect across the line y I To be precise we consider two transformations 45 S A S and 1 S A S to be the same iff for all points p in 5 Re ections rotations and translations have a special property they don7t change the distance between any pair of points That is these transformations are isometries2 Welll come to back this idea later 22 Groups Transformational geometry has two aspects it is the study of transformations of geometric spaces and it studies geometry using transformations The rst thing people realized when they started to get interested in transformations in their own right in the 19th century was that there was an algebra associated with them Because of this the development of the study of transformations was closely bound up with the development of abstract algebra In particular people realized that transformations behaved a lot like numbers in the following ways Closure Since transformations are functions you can compose any two transformations to get another transformation Speci cally if 45 and 1 are transformations of a space S then so is ab 0 Remember this means rst do 1b then do 415 ie an Mp we 0 Existence of an inverse Recall the de nition of an inverse function q5 1p q if 11 Since every transformation of S is 11 and onto it has an inverse which is also a transformation of S emember that inverting a function switches its domain and range but in this case both domain and range are just 0 Existence of an identity element The identity transformation denoted id is the transformation that leaves everything alone idp p for all points p E S o Associativity If 45 1 and w are three transformations of a space then 45 o o w o o w These four properties show up together in a lot of places For instance consider the set R of real numbers and the operation of addition If you add two real numbers you get a real number Every real number has an additive inverse namely its negative Therels an additive identity namely 0 And addition is associative ab 0 a bc One way to think about associativity is that it doesn7t matter how you parenthesize an expression like a b 0 These properties together 7 we can compose two transformations to get a new transformation there is an identity transformation every transformation has an inverse and composition is associative 7 say that the transformations of a given space form an algebraic structure called a group Analogously the real numbers form a group because we can add two real numbers to get a real number there is an additive identity every real number has an additive inverse and addition is associative One big difference between the group of real numbers and the group of transformations is that addition is commutative but composition of transformations is not That is if 7 s are real numbers then T s 8 7 2From Greek iso same metry distance but if 4511 are transformations then it is rarely the case that ab 0 1b 1b a Thatls okay 7 the operation that makes a set into a group doesn7t have to be commutative but it does have to be associative The idea of a group is absolutely fundamental in mathematics3 As we7ll see later on groups come up all the time in geometry In some sense a lot of modern geometry is about groups just as much as it is about things like points and lines 23 Notation for transformations Here are the major types of transformations of the plane that we7ll study Transformation Notation Re ection across line Z w Rotation about point p by angle 9 pp Translation by vector 17 73917 Glide re ection rst re ect across line Z then translate by vector 17 71 Dilation about point p with constant factor k 617 In some sense these are the most interesting77 kinds of transformations though certainly not all possible transformations This notation makes it easier to describe relations between transformations For example the fact that re ecting about a line twice ends up doing nothing can be expressed by the following equation 77 o w id Instead of saying Rotating counterclockwise about a point p by angle 9 is the inverse transformation of rotating clockwise about p by 9 7 which is true but extremely awkward 7 we can write a simple equation pp971 prg 3 Transformations and geometry In the previous section we looked at transformations by themselves Now we look at the interaction between transformations and sets of points First one piece of notation If 45 S 7 S is a transformation of S and A is a subset of S then we7ll write A for the image of A under 5 That is MA 4510 l P E A If we read this notation symbol for symbol it says A is the set of all points 4510 where p is any point in A For example here7s the rst gure again 3To learn more about groups take Math 558 If A is the unshaded triangle and B is the shaded triangle then we can Write TAA B and TAB Al De nition 2 A transformation 45 xes a point p iff p lt xes a set A iff for all p E A p It is a symmetry of A iff 5 Example 1 Consider the following picture What happens to each of these lines under the re ection 7 1 0 First of all 7 1 xes every point on E itself So certainly TM Zr 0 Second TAa a On the other hand 77 does Lot x most of the points on a except for P it ips them across Z to other points that are also on a So w is a symmetry of a but does not x it 0 Third TAb c and TAG 12 So 77 is not a symmetry of b or of cl Some more brief examples to think about 1 pp xes p no matter What 9 is 2 If p E Z then p is a symmetry of E but does not leave it xed 171300 3 If E is perpendicular to m then 7 1 is a symmetry of m but does not leave it xed On the other hand 7 1 does leave Z itself xed g lf17 0 then 73917 does not have any xed points On the other hand if E is parallel to 17 then 73917 is a symmetry of Using these terms we can rephrase the questions asked in Project 1 Which transformations x a single point two points three points a line Which transformations are symmetries of two points of a line First here7s an easy observation about the symmetries of all objects Fact 1 If 45 xes A then 45 is a symmetry of A In particular the identity transformation id of S xes every point in 5 so it is a symmetry of every subset of S 4 Special kinds of transformations isometries similarities and af ne maps The next step is to categorize transformations according to how much geometric structure they preserve For example consider the transformation of the plane that takes the point zy to the point Lyg letls call this transformation Note that 45 is 11 and onto as required On the other hand 45 is not very nice from a geometric standpoint For instance 45 takes the line y z and turns it into the curve y 13 So it doesn7t preserve straight lines And this means it doesn7t preserve the angle 180 so it doesn7t preserve angles lt doesn7t preserve distances either for example the points 11 and 12 are at distance 1 from each other but 45 sends them to 11 and 18 which are at distance 7 45 is an example of the kind of transformation we are not interested in De nition 3 Suppose we have a transformation l 45 is an isometry iff it preserves distances that is if A and B are any two points then AB A B where A and B 2 45 is a similarity iff it preserves angles that is if A B C are any three points then ABC 2 AA B C where A B and Ch 3 45 is an af ne map iff it preserves straight lines that is if X is a line then so is X and if is a line then so is X We7ve already observed that if 45 is any rotation re ection or translation then it is an isometry Therefore 45 is also a similarity and an af ne map Dilations are similarities and are af ne maps but not isometries In fact the ideas of isometry similarity and af ne map are successively more and more general Theorem 1 Every isometry is a similarity but not every similarity is an isometry Every similarity is a ne but not every a ne map is a similarity Proof Every isometry is a similarity by SSS Speci cally suppose 45 is an isometry lf A B and C then the triangles ABC and A B C are congruent by 555 so ABC 2 AA B C so 45 is a similarity On the other hand dilations are similarities but not isometriesi If 45 is a similarity then it preserves angles so in particular it preserves the angle 1800 That is it preserves straight lines On the other hand we7ve just seen in the exercise above a map that is af ne but not a similarity D Theorem 2 The isometries form a group the similarities form a larger group and the a he maps form a still larger group Proof Well just consider the case of isometries 7 the proofs that the other two sets are groups work exactly the same way To prove that the set of isometries forms a group we show that it satis es the four conditions listed in Section 22 1 Closure We need to show that the composition of two isometries is an isometry ie that if 45 and 1 preserve distance then so does iloqbi Let A B be any two points and let A 45A B B A A B i 1 B Then AB A B because 45 is an isometry and A B A B because 1 is an isometry but that means that AB A B and A cab A and B o Therefore 1 0415 is an isometry by de nition 2 Inverses Suppose that 45 is an isometryi In particular 45 is a transformation so it has an inverse transformation 4V1 which we want to show is af ne So let be any two points and let A 1A Bquot 1Bi Then A and Bi Since 45 is an isometry AB ABi Thatls exactly what we need to show that qb l is an isometryi These were the hard parts 3 Identity element The identity transformation is an isometry because clearly AB idA idBi 4i Associativity lsometries are functions so their composition satis es the associative lawi D Not every class of transformations forms a group For example consider the set of all re ections 7 1 l E is some line This set only satis es some of these properties so it does not form a group For example the composition of two re ections is not a re ection Another example For every point A the set of transformations 45 such that A forms a group However if A and B are different points then the set of transformations such that B does Lot form a group 5 The structure of isometries In this section we focus on isometriesi There are three major theorems about isometries Two of their proofs are fairly complicated so we won t give themi But we will give applications The ThreePoint Theorem Every isometry of the plane is determined by what it does to any three non collinear pointsi That is 4511 are isometries and A B C are noncollinear points such that A MB MEL and MC Wok th lt15 11A The proof of this theorem is rather technical but you7ve already seen the idea behind it 7 think about the threedot example in Project 1 The ThreePoint Theorem is useful for checking whether two isometries are equal all you have to do is check that they agree on each of three noncollinear pointsi Of course you may have to use some ingenuity in choosing those points appropriately Example 2 Suppose that mn are perpendicular lines that meet at a point A see gure below We will prove that Tm o Tn pAJgOOi We need to nd three noncollinear points and describe what each of these two isometries 7 the composition of re ections Tm o Tn and the rotation pAJgOO 7 does to themi The point A is a clear choice for one of the three points For the others let s draw a square BCDE centered at A with its diagonals parallel to m and n shown in blue below Why Because all the transformations we7ve described are symmetries of this square so its easy to see what they do to its verticesi We see that MA A MD D M0 0 MA A MD D we E and therefore em 0 TnA A m o mug D m 0 90 E On the other hand pA180 A A7 pA180 B D7 pA180 C E So the three noncollinear points AB C are mapped to the same pointsinamely ADE respectivelyiby Tm on and DA 1800 Therefore by the ThreePoint Theorem Tm or which is what we were trying to prove n pmgoet The ThreeRe ection Theorem Every isometry is the composition of at most three re ections If you believe the ThreePoint Theorem then you can prove the ThreeRe ection Theorem constructively and in fact you will do so as a homework problem That is if 1 is any isometry then the ThreePoint Theorem says that 1 is de ned by what it does to any three noncollinear points ABCi So to prove the ThreeRe ection Theorem it is sufficient to show that if AB CABC are six points such that AABC AA B C then there is some way of transforming AABC to AABC using three or fewer re ectionsi One application of the ThreeRe ection Theorem is the following theorem 7 which in case you thought everything was about the number 3 is about the number 4 The Isometry Classi cation Theorem There are only four different kinds of isometries re ection rotation translation and glide re ectioni What about the identity it can be described as either translation by the zero vector or as rotation about any point by 00 There are many ways to prove this theorem all of them tedious so we won t give a proof But all of the proofs rely to some extent on the ThreeRe ection Theoremi On the other hand the lsometry Classification Theorem has a nice corollaryi Theorem 3 Let 45 be an lsometry Either 45 is a symmetry of same line or it xes a point Proof By the lsometry Classification Theorem there are only four cases to consider o if 45 is a re ection 7 1 then 45 fixes every point on E so it certainly is a symmetry of Zr o if 45 is a rotation pp 9 then it xes the point p o if 45 is a translation 7395 then it is a symmetry of any line parallel to the translation vector 17 o if 45 is a glide re ection Wm then it is a symmetry of the line if D How many symmetries does a regular tetrahedron have How about a cube Or an icosahedron 6 Symmetries of bounded gures What can we say about the set of symmetries of a figure4 First of all let s agree that when we talk about a symmetry of a figure we restrict ourselves to isometriesi This captures our intuitioni There are many transformations that look like isometries in a small region of space but then do strange things outside it and it complicates our discussion too much to talk about those Call the figure yr The symmetries of y are closed under composition the identity transformation of the plane is a symmetry of y and each symmetry of g has an inverse which is also a symmetry of yr So they form a group which we7ll call Sym i 61 An example Suppose that y is an equilateral triangle AABCi in addition to the identity the group SymAABC contains two notrivial rotations 7 namely pZ 120 and pZ 240 where Z is the center of the triangle 7 and three re ections 7 1 Tm and Tn where Lm n are the bisectors of the three sides of the triangle 4Here 17m using the Euclidean de nition of gure an object built out of curves and line segments So triangles pentagons and circles are gures but not for example a lledein circle That is SymAABC id palm pzjm 7 1 Tm Tni Here s how we know that this is the complete list of symmetries Every symmetry 45 of AABC takes vertices to vertices that is 45 is a symmetry of the set ABCi For example Tm xes B and swaps A with C while pZ 120 maps A to C B to A and C to B The identity of course xes each of the three verticesi On the other hand by the ThreePoint Theorem any isometry is determined by what it does to A B and Cl So there are only 3 6 possibilities which means that we7ve listed them all Since the symmetries form a group we can ask how they behave under compositioni That is if 4511 are transformations in SymAABC then which element of SymAABC equals 45 0 1b This question is really a set of thirtysix questions eigi What is palm o Tm What is Tn o T73 whose answers can be collected in a table The easiest way to calculate a single composition is to see what it does to AB C For example pz120 MUD pz120A C TmA WTmA TAO B Pzgm Ab pz120WB pz1200 B TmB WTmB TAB C Pz240B7 pmww pmw A we wltrmltcgtgt MA A who so palm o w Tm and 7 1 0 Tm r2124 In the following table the rows and columns are labeled by the elements of SymAABC and the entry in column 45 and row 1 is ab 0 id pz120 pz240 W Tm T id id pz120 pz240 W Tm T pz120 pz120 pz240 id Tm T W pz240 pz240 id pz120 T W Tm T2 T2 Tm Tn id P2420 1240 Tm Tm M W 1240 id P2420 Tn Tn W Tm pz120 pz240 id Notice the following things o It is not always true that ab 0 1 9 1 o For example if r1 0 rm pZ 240 but rm 0 7 1 pZ 120i 0 The composition of two rotations or of two re ections is a rotation while the composition of a rotation and a re ection in either order is a re ection This is analogous to the sign of the product of two real numbers the product of two positive numbers or of two negative numbers is positive while the product of a positive number with a negative number is negative 62 De ning gures by their symmetry groups To a modern geometer ie any geometer since the late 19th century what characterizes a geometric gure isn7t the number or characteristics of its sides andor angles but its symmetry group For example suppose that y is a convex quadrilateralr Most of the time g has only one symmetry namely the identity transformation But we can identify several special kinds of quadrilateralsr D lel isosceles square rectangle rhombus parallelogram trapezoid kite What makes these special quadrilaterals special is exactly that they have a lot of symmetriesr In fact we can de ne them in terms of their symmetry groups A parallelogram is a convex quadrilateral whose only nontrivial5 symmetry is palm for some point C which we call the center of the parallelogram it s where the diagonals meetl A rectangle is a convex quadrilateral with one nontrivial rotational symmetry and two re ection symmetriesr So is a rhombusl An isosceles trapezoid has just the identity and one re ection symmetryr A square of course has the most symmetries of any quadrilateral four rotational symmetries including the identity and four re ection symmetriesr By the way we could de ne a circle as follows Let p be a point A circle with center p is a gure y such that every rotation around p is a symmetry of y and every re ection across a line containing p is a symmetry of a This seems like a roundabout way to de ne a circle but if you think about it its correct 7 every circle certainly has these symmetries and any gure with these symmetries has to be a circle A gure is called bounded if it ts inside some circlel For example a triangle is bounded a line isnltr Not all isometries occur as symmetries of bounded gures Theorem 4 If y is a bounded gure in the plane then every symmetry of y is either the identity a rotation or a re ection Equivalently no translation or glide re ection can possibly be a symmetry of y The equivalence of these two statements comes from the lsometry Classi cation Theoremr Here s a really slick proof that depends on the de nition of circle 5By nontrivial we mean other than the identity Proof If y is bounded then there is some unique smallest circle 9 that y ts inside Every symmetry of y must be a symmetry of if and by the de nition of circle must be either a re ection or a rotation D In particular this means that every symmetry of y xes at least one point namely the center c of 9 We could call this point the center of y For example if y is a triangle then if is the circumscribed circle so c is the circumcenter of L that is the intersection of the perpendicular bisectors of the sides 6 3 Regular polygons Most gures don7t have any nontrivial symmetries For example if you draw a random triangle then it will almost certainly be scalene and the only isometry that xes it will be the identity However there are special gures with more symmetries The equilateral triangle of Section 61 is an example of this More gener y De nition 4 A polygon P is regular if all of its sides are congruent and all of its angles are congruent Suppose p is a regular nsided polygon or ngon How big is the set SymP First of all let c be the center of P If 45 E SymP then c Also 45 takes vertices to vertices and it preserves adjacency among vertices if v and w are vertices of P that are adjacent to each other then so are and But the points c 1110 are noncollinear 7 but by the ThreePoint Theorem 45 is determined by what it does to all of them There are clearly n possibilities for namely all the vertices of P and once we know 451 there are 2 possibilities for namely the vertices adjacent to 451 so we conclude that P has exactly 2n symmetries In fact it is not too hard to say what the symmetries are Theorem 5 Let P be a regular polygon with n sides The nontrivial symmetries of P are as follows 0 all reflections across its angle bisectors 0 all reflections across the perpendicular bisectors of its sides and 0 all rotations about its center by 360hn for 0 lt h lt n Proof All of these transformations are certainly symmetries of P On the other hand if 45 E SymP then 45c c where c is the center of P as de ned above so 45 must either be a rotation about p or a re ection across a line containing c and any rotation or not re ection that is not one of those listed above does not take vertices to vertices D 64 Other polygons What about polygons that are not regular but have lots of symmetries nevertheless For example what does the group of symmetries of a rectangle look like Remember we said that a rectangle is a convex quadrilateral with one nontrivial rotational symmetry and two re ection symmetries across the perpendicular bisectors of each pair of opposite sides Of course so is a rhombus 7 although in this case the lines of re ection symmetry are the diagonals Here s the multiplication table for the symmetry group of a rectangle id pX180 T1 Tm id id prgo 77 Tm pX180 pX180 id Tm W m w Tm id prgo Tm Tm W pX180 id And here7s the multiplication table for the symmetry group of a rhombus id py 180 Tk Tn id id py 180 Tk Tn pY180 pY180 id Tn T16 m m Tn id py 180 Tn Tn m py 180 id These two multiplication tables are essentially the same if you take the rst table and replace X with Y7 E with k and m with n7 you get the second tablei Algebraically7 we say that the symmetry groups of the rectangle and the rhombus are isomorphic Therels a good reason for this the two gures can be superimposed so that their symmetry groups consist of exactly the same sets of transformations This is an example of how modern mathematics uses groups to study geometric objects The fact that the symmetry groups of the rhombus and rectangle are the same indicates that there s some relationship between the two gures Of course you don7t need groups to realize that you can form a rhombus by joining the midpoints of a rectangle but the same technique can be applied to more complicated gures 7 Counting symmetries If we know the symmetry group of an object that is if we know its multiplication table then of course we know how many symmetries there are But it is often possible to count the symmetries without having to work out the full symmetry group Example 3 Let 73 ABCDE be a regular pentagoni Every symmetry 45 of 73 permutes its ve ver tices that is it is a symmetry of the vepoint set ABCDE and by the ThreePoint Theorem is completely determined by what it does to any three of the ve But actually 45 is determined by even less information For instance if we know what and are then we know 45 completely by the ThreePoint Theorem again 7 because 0 where O is the center of 73 and the points A B O are noncollineari The point can be any of the 5 vertices of 73 and once we know 45A we know that must be one of the 2 vertices sharing a side with whatever that is Therefore the number of symmetries of 73 is 5 2 10 We can describe a symmetry of by its permutation wordi That is write the ve letters A B C D E in the order 45A 4MB i i i Here are the permutation words for all ten symmetries of the regular pentagon ABCDE AEDCB BAEDC BCDEA CBAED CDEAB DCBAE DEABC EABCD EDCBAA For example p0gt144o correponds to the permutation word DEABC because p0144o A D p0144o B E etc and Tb corresponds to the permutation word CBAED because TbC A TbB B etc The permutation word ABCDE corresponds to the identity transformation Notice that each of the 5 possible rst letters occurs twice in the table once with each of its neighbors next to it This corresponds exactly to our earlier observationi Generalizing this argument we can see that every regular polygon with n sides has exactly 2n symmetriesi Of course we already knew that from Theorem 5 but its nice to con rm it another way This method of counting symmetries doesn7t tell us explicitly what the symmetries are but on the other hand it is applicable to lots and lots of geometric objects 7 not just in the plane but also in threedimensional space and even in four and higherdimensional spacesl Example 4 Let R WYXZ be a rectangle that is not a square as shown below and let 45 be a symmetry of R Then can be any of the four vertices but once we know 45X there s only one possibility for 45 because must be the vertex adjacent to by one of the short sides of Qi HQ were a square then there would be two choices for instead of one So Sym 4 which con rms what we found earlier The permutation words for the four symmetries are WXYZ XWZY YZWX ZYXW Similarly let H ABCDEF be the hexagon you studied in homework problem TG 15 and let 1 be a symmetry of Hi Again can be any of the six vertices but once you choose A you immediately know what 1 does to the other ve vertices of Hi Therefore SymR 6i There7s nothing special about A we could just have well argued that 1 is determined by which can be any of the six verticesi What about higherdimensional objects Example 5 Let T be a regular tetrahedron ie a triangular pyramid in which every side is an equilateral triangle Call the vertices AB C Di How many symmetries does T have Equivalently what are all the permutation words of symmetries of T If 45 is a symmetry then clearly can be any of 45A 4MB or Four choices there 15 Having chosen 4514 there are three choices for any of the other three verticesi Having chosen and 4MB there are two choices for Then once we choose 41507 there is only one possibility left for of the other three verticesi ln total7 there are 4321 4 24 symmetries of Ti ln fact7 every rearrangement of the letters A7 B7 C7 D is a permutation word of a symmetry What these symmetries look like geometrically For example7 you can draw a line connecting a vertex with the center of the opposite triangle and rotate T by 120 or 240 around this line7 as in the following gure There are four ways to choose that vertextriangle pair7 so we get a total of eight rotations this way Here are the permutation wordsi Vertex Opposite triangle Permutation words A B C D BOD ACDB7 ADBC CBDA7 DBAC BDCA7 DACB BCAD7 CABD Another way to construct a rotation line is to connect the midpoints of two opposite edges of T as in the following gure ltls probably easiest to visualize if you dangle T from one of its edges 7 the righthand gure is an attempt at illustrating this There are three such pairs of opposite edges AB and CD AC and BD and AD and BCi This gives three more permutation words7 respectively BACD7 CDle7 and DCBAi We7ve accounted for twelve symmetries so far the identity and 3 8 11 nontrivial rotationsi The other twelve are re ections for example7 re ecting across the plane containing edge AC and the midpoint of edge BD or compositions of re ections and rotationsi CHAPTER 6 Planar Symmetries As has been mentioned before one of the most serious deficiencies in Euclid s axiomatic development of geometry was his failure to provide an explicit discussion of rigid motions despite the fact that they play an important role in several of his proofs beginning with that of Proposition 4 of Book I These transformations are not mentioned in Hilbert s axiomatization either where they are replaced by several congruence axioms Other axiom systems notably that of Mario Pieri 1860 1925 do refer to such motions explicitly The 19th century also witnessed the creation of many alternative geometries each with its own collection or group of rigid motions This proliferation of geometries called for their classification and in 1872 Felix Klein 1849 1925 promulgated his Erlanger Program in which he suggested that they be classified by their groups of rigid motions This chapter is devoted primarily to the classification of the rigid motions of the Euclidean plane and the allied topic of planar symmetry Some information is also obtained about the rigid motions of the hyperbolic plane 61 61 TRANSLATIONS ROTATIONS AND FIXED POINTS 1 Translations Rotations and Fixed points Informally speaking a rigid motion of the plane is a transformation that does not alter the distances between the points More formally a rigid motion is a function f of the plane into itself such that for any two points P and Q PQ P Q where P fP and Q f Q The prototypical rigid motion is the translation that slides the plane on itself so that all straight lines remain parallel to their original positions More precisely given any two points A and B the translation that carries A onto B is denoted by TAB and if P is any point then TABU Q where Q is the unique point such that AB PQ AB PQ and the segments AB and PQ are similarly directed If P does not lie on AB then this by Virtue of Proposition 317 is tantamount to saying that the quadrilateral ABQP is a parallelogram In Figure 61 TABPl Q for all i 1 2 3 4 Note that in this figure P1P is both parallel and equal to QiQJ whenever if j and hence TAB is indeed a rigid motion 62 61 TRANSLATIONS ROTATIONS AND FIXED POINTS Q P2 P1 P3 1 4 Figure 61 A translation The same translation can be represented in many different ways Thus the translatlon TAB of F1gure 61 can also be denoted by 1P Q TPZQZ and so on Two rigid motions f and g are said to be equal provided that fP gP for all points P in the plane In other words if the rigid motion is Visualized as a physical movement of the plane then the intermediary stages of the motion are immaterial all that matters are the final positions of the points This chapter s goal is the classification of all the rigid motions of the plane and the most important tool in this text s approach is the composition of rigid motions The reader is reminded that if f and g are functions of any set into itself then their composition g o f is a function of the same set into itself such that g ofP gfP The identity transformation Id is defined by the equation 63 61 TRANSLATIONS ROTATIONS AND FIXED POINTS IdP P for every point P and has the property that for any rigid motion f fold Idofzf The operation of composition is associative in the sense that for any three such functions f g and h f g h f g h We begin with the composition of translations PROPOSITION 611 If A B C are any points of the plane then TBC TAB TAC39 PROOF Let P be any point of the plane and set see Fig 62 P TABP P TBCP TBCO 1ABP It is necessary to show that P TACP However as was noted above ABP P and 64 61 TRANSLATIONS ROTATIONS AND FIXED POINTS B c P P Figure 62 The composition of translations BCP P are both parallelograms It follows from Proposition 318 that AP and BP are equal to and parallel to BP and CPH respectively Hence by Proposition 317 ACP P is a parallelogram and so P TACP QED It follows from this proposition that the composition of any two translations is itself a translation For if f and g are any translations and P is any point then we could set P fP P gP and conclude that g f TP39Pquot TPP39 TPPquot 1 The inverse f of the rigid motion f is a rigid motion such that 71 71 f f f f 1d 1 It 1s clear that for any two pomts A and B TAB TBA Another type of rigid motion is the rotation If C is any point of the plane and a is some directed angle then the rotation RC 1 is the rotation that moves the general 65 61 TRANSLATIONS ROTATIONS AND FIXED POINTS point P to the point P RC aP where CP 2 CP and LP CP a Fig 63 Exercise Figure 63 The rotation RC a 1 calls for the formal proof of the rigidity of rotations The point C is the pivot point of the rotation RC a The angle a of the rotation is understood to be oriented in the sense that it can be either positive or negative and the rotation accordingly proceeds either counterclockwise or clockwise Moreover if n is any integer and a n360 then RCHB R Consequently in describing any rotation R the angle will generally be chosen so that 0 s a lt 360 Note that R400 2 Rafa The composition of the rotations RC0 and Ra is clearly RC OH but what about the composition of RC a with RD where C and D are distinct points In order to answer this natural question it is first necessary to deal with the issue of identifying rigid motions in general The following sequence of propositions aims to answer the question of How much information is it necessary to have about a rigid motion before we can say that it is known It will soon be seen that surprisingly little is needed 66 61 TRANSLATIONS ROTATIONS AND FIXED POINTS PROPOSITION 612 Every rigid motion transform straight lines into straight lines PROOF Let f be a rigid motion let m be a straight line with two distinct points A and B on it and set A fA and B fB Fig 64 If P is any point of m between A Figure 64 A PP B 2 AP PB 2 AB A B and B and P fP then it follows from Proposition 2325 that P is on the line segment AB A similar argument Exercise 5 demonstrates that as long as P is on m 9 then P is on the line A39By even when P is not between A and B 9 Conversely let P be any point of A39By that lies on the line segment A B Since A B 2 AB there is a unique point P of m such that AP A P and BP 2 B P If P fP then AP 2 AP A P and B PH 2 BP 2 B P 67 61 TRANSLATIONS ROTATIONS AND FIXED POINTS 9 so that P and P fP must be identical The same holds even when P is on AVE 9 but not between A and B This means that every point of A39By is covered by some 9 point of m In other words fm A39B39 QED PROPOSITION 613 If two rigid motions agree on two distinct points then they agree at every point of the straight line joining them PROOF Let f and g be two rigid motions and A and B two distinct points such that 1 gA A and BgB 32 If P is any point of AB then by Proposition 612 fP and gP are both points of A B whose distances from A and B are respectively equal It follows that fP gP QED THEOREM 614 If two rigid motions agree at three noncollinear points then they agree everywhere PROOF Let f and g be two rigid motions that agree at the three noncollinear points A B C By Proposition 613 f and g agree at every point on the straight lines EC and KC If P is any point of the plane then there clearly exists a straight line through P that intersects the union of these three straight lines in some two distinct points X and Y Since f and g agree at X and Y it follows from Proposition 613 that they must also agree at P QED 68 61 TRANSLATIONS ROTATIONS AND FIXED POINTS Thus in order to pin down a rigid transformation it suffices to know how it affects some triple of noncollinear points A fixed point of the transformation f is a point P such that fPP It is clear that the point C is a fixed point of the rotation Rae and is in fact the only fixed point of that rotation It is equally clear that with the exception of the identity translations have no fixed points whatsoever On the other hand every point is a fixed point of the identity The following corollary is an immediate consequence of Theorem 614 COROLLARY 615 If a rigid motion xes three noncollinear points then it must be the identity 11 EXERCISES 61 1 Prove that every rotation is a rigid motion 2 Prove that every rigid motion transforms circles into circles 3 Prove that if A B C are any three points then TCA 0 TBC 0 1A3 Id 4 Let AABC be a clockwise triangle with oriented interior angles a 3 y at A B C respectively Prove that RC2 0 RBYZ 0 RAYZO Id 5 Complete the proof of Proposition 612 by providing the details for the case where P is on the infinite line AB but outside the segment AB 6 Let Aa1 a2 and 31 b2 be two points Explain why the transformation fP Q that takes the point Px y to the point Qx y where x xb17a1 y yb27a2 69 61 TRANSLATIONS ROTATIONS AND FIXED POINTS is in fact the translation TAB 7 Let a be an angle Explain Why the transformation fP Q that maps the point Px y to the point Qx y Where r X xcosaiysrna I y xsrnaycosa is in fact the rotation R0 a Where O is the origin 8 Prove that rigid motions preserve angles In other words show that if f is a rigid motion and m and n are straight lines that form an angle of measure a then m and fn are also straight lines that form an angle of measure a 2 Reflections Given a straight line m the re ection pm is the transformation that fixes every pornt of m and associates to each point P not on m the unique point P such that m is the perpendicular bisector of PP see Fig 65 and Exercise 28 It follows directly from the definition that pm 0 pm 2 Id and hence pm 1 2 pm This text s classification of the Q Figure 65 rigid motions is based on the fact that these re ections are the building blocks of all the rigid motions in the sense that every rigid motion can be expressed as the composition of 610 62 REFLECTIONS some re ections The next two propositions show that such is indeed the case for translations and rotations PROPOSITION 621 Let m and n be twopdrdllel straight lines Let AB be a directed line segment that first intersects m and then n and whose length is twice the distance between A and B Then a pnopmztAB b pnOTAszm c TABopmzpn39 PROOF Let P be any point outside the infinite strip bounded by m and n such that the Figure 66 distance from P to m is less than the distance between m and n Fig 66 Set P pmP and PH 2 pnP 611 62 REFLECTIONS It is clear that P P and PH are collinear and that PP 2 PP P39P 2XP39 2P39Y 2XY twice the distance between m and 11 Hence pn pmP pP P TPPJP 1 Since it is easy to find three noncollinear positions of P that satisfy the constraints specified in the beginning of this proof it follows that Equation 1 holds for three noncollinear points and hence by Theorem 614 pm p TPP This completes the proof of part a Parts b and c follow immediately since pnOTAB pn pn m pnopn pm dopm pm TAgopm pnopmhpm pnommopm pn Id pn QED Conversely given any translation TAB there clearly exist two parallel straight lines that are perpendicular to AB and whose distance from each other equals half of AB By the above proposition either TAB 2 pm 0 p or TAB p 0 pm and in either case the arbitrary translation TAB has been expressed as the composition of two reflections This expression is of course not unique since m can be any line that is perpendicular to AB 612 62 REFLECTIONS PROPOSITION 622 Let m and n be two straight lines that intersect at apoint A and let a be the counterclockwise angle from m to n at A Then pn 0 pm RA2a39 PROOF Let P be a point outside 4 BAC a Fig 67 but close enough to m so that P pmP is inside the angle Set PH 2 pnP p o pmP Then m bisects A PAP and n bisects LP AP Consequently LPAP ZLBAP ZLP AC 2a Hence RA2aP P p 0 pmP 2 Since it is easy to find three noncollinear positions of P that satisfy the constraints specified in the beginning of this proof it follows that Equation 2 holds for three noncollinear points and hence by Theorem 614 p 0 pm 2 R A 2a QED 613 62 REFLECTIONS Figure 67 It was noted above that the composition of rotations that share their pivot points is a rotation about the same point but that the nature of the composition of rotations with distinct pivot points was unclear We are now ready to dispose of this and other similar issues PROPOSITION 623 Let A and B be twopoints and let a and B be two oriented angles Then the composition RB o R A a is a a translation if a B is a multiple of 3600 b a rotation R a if a B is nota multiple of 3600 PROOF This is obvious if A and B are identical points as well as when either a or B is zero It is therefore assumed that A and B are distinct and neither a nor is zero Let m ltAgtB let k be the line through A such that the oriented angle from k to m is 12 let n be the line through B such that the oriented angle from m to n is Z Fig 68 Then by the above proposition RM oRA pnopmhmmopk pnommopm opk pnopk 614 62 REFLECTIONS which is either a translation or a rotation depending on whether the lines k and n are parallel or not However these lines are parallel if and only if 052 2 is a multiple of 1800 which is of course equivalent to a it being a multiple of 3600 Hence by Proposition 621 the composition is a translation if a f is a multiple of 3600 When a f is not such a multiple then by Proposition 622 the composition is the rotation RXa3 QED a2 32 Figure 68 EXAMPLE 624 Given any twopaints A and B identify R3600 o RA 600 It follows from Proposition 623 that this composition is a rotation RCY1200 The pivot point C is located as follows Set see Fig 69 A RC1200A R3450quot RA60quotA RB600A Then C is that unique point such that A ACA is isosceles with vertex angle A ACA 1200 In other words C is the center of the equilateral A AA B 615 62 REFLECTIONS A Figure 69 PROPOSITION 625 Let R be a rotation which is not the identity and let 7 be a translation Then both R 7 and 7 R are rotations with the same angle as R PROOF Suppose R RA a A 1A and let B be the midpoint of the segment AA Figure 610 Let k and m be the lines through B and A respectively that are Figure 610 perpendicular to AA and let n be the line through A such that the oriented angle from n to m is equal to 12 Then HR pkopmhmmopn pkommopmhpn pkopn 62 REFLECTIONS which is a rotation by angle a because k and 11 when extended intersect in an angle of 12 The proof that R 7 is also a rotation is relegated to Exercise 25 QED EXAMPLE 626 For the two given points A and B of Figure 611 identify both TAB 0 RA Qoa and RA Qoa 0 TAB Y A A X B Figure 611 By Proposition 625 TAB o R A 900 is a 900 rotation such that TAB o RA 900A TABA B It follows that the pivot point of TAB o R A 900 is that point X such that A ABX is an isosceles right triangle In other words TAB o R A 900 RX900 Similarly R A 900 o TAB is a 900 rotation such that RAygoo o TABA RAngdB A It follows that the pivot point of R A 900 o TAB is the point Y where A AA Y is an isosceles right triangle In other words R A 900 o TAB RYgoo Note that the two 617 62 REFLECTIONS compositions TAB o R Aygoo and R Aygoo o TAB are not equal In general rigid motions do not commute EXERCISES 62 Identi i the compositions of Exercises 1718 where ABCD is the square of Figure 612 1 4 7 10 16 RA90 RB90 2 RB90 RA90 3 RC180 RA90 RA90 TBC 5 RA90 TCA 6 TCA A90 TBC TBA 8 TBC TAD 9 IDA TBC RA270 RC90 11 RA180 RD180 12 RA45 RC135 RA45 RB45 14 RA60 RB120 15 RD120 RC120 RA90 RB30 17 TAB A60 18 RB60 TAB 1 C A a Figure 612 Let AABC be a clockwise triangle with oriented interior angles a 3 y at A B C respectively Use Proposition 622 to prove that RC2 0 RBYZ 0 RAYZO Id Let A1 A2 An be the clockwise successive vertices of a polygon with n sides If the interior A12a139 Let A1 A2 A be the midpoints of the successive sides of a polygon with n sides Identify the n composition RA uaRA nRA if n39 239 139 angle at Ai is ai Identify the composition RA 2a 0IRA Za 0R n n a n 3 b n 4 c n is an arbitrary positive integer Let n be an even integer and let A1 A2 An be the successive vertices of a regular nisided polygon and let mi be the bisector of the interior angle at Ai Identify the composition pm 0 0 n p op m2 quot 1 Let P be any point on the straight line In and let 6 be any angle Prove that both RP 9 0 pm and pm 0 RP 9 are reflections What are their axes Prove that if P is apoint on the straight line m then pm 0 RPYe 0 pm 2 RP 79 618 62 REFLECTIONS 25 Complete the proof of Proposition 625 by showing that R 1 is also a rotation with the same angle as a 26 Let ABCD be a cyclic quadrilateral Identify the composition pDA 0 pCD 0 pBC 0 pAB 27 Let a be an angle and let m be the straight line through the origin with inclination a to the positive X axis Explain why the transformation fP Q that maps the point Px y to the point Qx y39 where I x xcos2ays1n2a y xsin2a7y cos 2a is in fact the re ection pm 28 Prove that every reflection is a rigid motion 3 Glidere ections So far re ections have been used merely in order to explain how translations and rotations interact under compositions We now examine how these two types interact with re ections A special case of this issue was resolved by parts b and c of Proposition 621 wherein it was proved that the composition of a re ection with a translation whose direction is perpendicular to the direction of the translation is another re ection with an axis parallel to that of the given re ection The composition of a re ection with a rotation whose pivot point lies on the re ection s axis is also a re ection Exercise 6223 However in general the composition of either a translation or a rotation with a re ection forms a new kind of rigid motion Let A and B be two distinct points The composition pAB o TAB is called a glidere ection and is denoted by yAB It is easily seen that the reverse composition TAB o pAB also equals yAB and that the inverse of yAB is yBA Fig 613 In order to simplify the statements of some of the subsequent propositions re ections will be considered as special cases of glide re ections The line AB is called the axis of the 619 63 GLIDEREFLECTIONS glide re ection MB and it is easily seen that for any point P not on the axis A the line segment joining P to yABP is bisected by AB see Fig 613 and Exercise 25 7 P A s P yAB Q Figure 613 PROPOSITION 631 gt 7 be any translation and y any glidere ection Then y o 7 and 7 o y are both glidere ections PROOF Suppose 7 TAB If yszB thenclearly yo 7 To y 2 MB If y 2 pm where m H AB then there exist points A B on m such that 7 TAB TAB Consequently by the previous argument T o 7 7 pA39B39 TA39B39 VA39B39 TA39B39 pA39B39 pm where m J AB then this proposition follows from Proposition If 621bc If y 2 pm where m is skew to AB let C be a point such that AC H m and 6 20 63 GLIDEREFLECTIONS Figure 614 BC J m Fig 614 By Proposition 611 7 pm TAB pm TCB TAC pm TCB TAc By Proposition 621b there is a line 11 J BC such that pm 0 ICE 2 p and hence 7 7 pn TAO which since 11 M AC is known to be a glide re ection The proof that 7 o y is also a glide re ection is relegated to Exercise 23 Finally let y be an arbitrary glide re ection If y yCD pCD 0 1CD then by Proposition 611 TIOCD TCD TABIOCD TCD TABIOCD T for some translation 7 This however is known to be a glide re ection The proof that 7 o y is also a glide re ection is relegated to Exercise 23 QED EXAMPLE 632 Identify the compositions VAD o TAB and TAB o VAD where ABCD is the square of Figure 615 621 63 GLIDEREFLECTIONS 1quot Q 339 l 393 M P A 539 Figure 615 By the previous proposition these compositions are both glide re ections Moreover since YAD TABM YADUB B39 it follows that the axis of yAD TAB must contain the midpoint M of the segment AB In addition MD 0 TABM mm P It follows that MP is the axis of yAD o TAB and in fact yADo TAB VMPV Again TAB o yADA TABD C and hence the axis of TAB o yAD contains the midpoint P of AC In addition 1A3 0 VADP rABaD Q and hence TAB o yAD yPQ 6 22 63 GLIDEREFLECTIONS PROPOSITION 633 Let R be any rotation and y any glidere ection Then both y o R and R o y are glidere ections PROOF Let R R A a and suppose first that y 2 pk Let m be the straight line through A that is parallel to k and let n be the straight line through A such that the oriented angle from n to m is 12 Figure 616 Then Figure 616 Y R pkoRAa pkommopn pkopm pn Since k H m it follows from Proposition 621 that pk 0 pm is a translation and hence by Proposition 631 y o R 2 pk 0 pm 0 p is a glide reflection If y is the arbitrary glide reflection 1CD o pCD then 1 R TCD pCD RAa TCD pCD RAa39 By the first part of the proof pCD o R A a is a glide reflection and hence it follows from Proposition 631 that y o R 2 1CD o pCD o RA a is also a glide reflection The proof that R o y is a glide reflection is relegated to Exercise 24 QED EXAMPLE 634 Identify the composition VAD o R A Qoo and R A Qoo o VAD where ABCD is the square of Figure 617 6 23 63 GLIDEREFLECTIONS 1 Y C W X Z I X l A B Figure 617 By the previous proposition both of these compositions are glide re ections Moreover YAD RA900A YADM D so that the axis of this composition contains the midpoint X of AD Since VAD RA90quotX YADX39 Y it follows that yAD o R A Qoa yXY Similarly RA90quot YAD A RA900D Z so that the axis of this composition contains the midpoint X of AZ Since RA900 YAD XI RA90quotY W it follows that RA Qoa o VAD YXW 624 63 GLIDEREFLECTIONS PROPOSITION 635 Let yAB and yCD be two glidere ections The composition yAB o yCD is a a translation if AB CD 17 a rotation of angle 2a otherwise where a is the oriented angle from CD to AB PROOF Note that 7A3 7CD TAB pAB pCD 7CD TAB 0013 pCD 7CD TAB o f 0 1CD where by Propositions 621 2 f is a translation if AB II CD and a rotation by angle 2a otherwise The desired results now follow from Proposition 611 in the first case and from Proposition 625 in the second case QED EXAMPLE 636 Identify the compositions VAD o yAB and yAB o yCD where ABCD is the square of Figure 618 Bf l C M A B D Figure 618 By Proposition 635b VAD o yAB is a 1800 rotation Since 6 25 63 GLIDEREFLECTIONS YAD 74304 YADB B it follows that the pivot point of this rotation is the midpoint M of AB Hence VAD oyAB RM1800 By Proposition 635a yAB o yCD is atranslation Since 7A3 7CDC YABD DI it follows that yAB o yCD 2 1CD EXERCISES 63 Identiy the compositions of Exercises 1718 where ABCD is the square ofFigure 612 l 4 7 RD90 VDC 2 VDC RD90 3 RP180 VAB VAB RP180 5 TAB VDC 6 VCD TAB TAB VBC 8 VBC TBA 9 VAD VBC VAD VCB 11 VBA VBC 12 VCB VBA VAC VBD 14 PAD TAB 15 PAD RC90 VCD S VBC VAB 17 VCD TBC VAB 18 7CD VBC TAB If k m n are the perpendicular bisectors of the sides AB BC CA of AABC respectively show that pk 0 pm 0 p is a re ection What is the axis of this re ection Let A and B be any two distinct points Prove that the composition RBY1800 0 pAB 0 RAY1800 is a glideire ection and find its axis Show that the composition of the re ections in the three angle bisectors of a triangle is a re ection whose axis is perpendicular to one of the triangle s sides Let n be an odd integer and let A1 A2 An be the successive vertices of a regular nisided polygon and let mi be the bisector of the interior angle at Al Identify the composition pm 0 0 sz Pml Complete the proof of Proposition 631 Complete the proof of Proposition 633 Prove that if yAB is a glideire ection and yABP P then the axis AB contains the midpoint of PP39 626 63 GLIDEREFLECTIONS 26 Show that pk 0 pm 0 p p 0 pm 0 pk whenever the lines k m n are either concurrent or parallel 27 Show that the composition of an even number of glideire ections is either a rotation or a translation 28 Show that the composition of an odd number of glideire ections is a glide reflection 4 The Main Theorems Enough tools are now available to demonstrate that there are no Euclidean rigid motions above and beyond those described above PROPOSITION 641 Suppose A ABC 5 A DEF Then there exists a sequence of no more than three re ections such that the composition of these re ections maps the points A B C onto the points D E F respectively PROOF Suppose the two given triangles are identical then the composition of two identical re ections will clearly accomplish the required task If the two triangles share exactly two vertices then it may be assumed that their relative position is described by Figure 619 In that case p AB itself constitutes the Figure 619 required sequence of re ections 627 64 The Main Thenrems If the two triangles share exactly one vertex say A D let M be the midpoint of the segment BE Then the re ection pAM transforms AABC into A DEF that shares at least two vertices with A DEF It follows from the previous argument that at most one more re ection will be required to transform A DEF into A DEF Finally if the two triangles share no vertices let M be the midpoint of the segment AD The re ection pm then transforms A ABC into A DE F that shares at least one vertex with A DEF By the above argument at most two more re ections will transform A DE F into A DEF It follows that at most three re ections are required to transform AABC into ADEF QED The following is this chapter s main theorem THEOREM 642 Every rigid motion is the composition of at most three re ections PROOF Let f be a rigid motion let A B C be three noncollinear points and set A A B fB C fC Since A ABC 2 A A B C it follows from Proposition 641 that there exist at most three re ections whose composition say g also transforms A B C onto A B C respectively It follows from Theorem 614 that f g QED The following classification theorem is a consequence of the above THEOREM 643 Every rigid motion is either a translation a rotation or a glide re ection PROOF The composition of no re ections is the identity which can be viewed is either a rotation R A 00 or a translation 1AA The composition of one re ection is a glide 6 28 64 The Main Thenrems re ection The composition of two re ections is by Propositions 621 2 either a translation or a rotation It follows that the composition of three re ections is also the composition of a re ection with either a translation or a rotation which by Propositions 631 and 633 is a glide re ection QED EXAMPLE 644 Let f and g be two rigid motions Prove that g is a re ection if and only if f o g of1 is a re ection Suppose first that g is a re ection By Theorem 643 f is either a translation a rotation or a glide re ection In the first two cases it follows from Proposition 631 and 633 that f o g o f1 is also a glide re ection The same conclusion can be drawn in the third case if f is a glide re ection but this time Proposition 635 is also needed In order to show that f o g o f1 is a re ection it suffices to show that it has a fixed point Let P be any fixed point of the re ection g and set P fP Then f g f1P39 f g f1 P f 3P fP P39 so that P is the requisite fixed point of f o g 0 f1 Conversely suppose f o g o f1 is a re ection It then follows from the above that g is also a re ection because 1 10 f g f1gt ltf1391f1 f goflofg In conclusion we point out that the definitions of rotations and re ections as well as the proofs of Propositions 614 641 2 are all neutral and hence they also hold for the hyperbolic plane In particular Every rigid motion of the hyperbolic plane is the composition of at most three hyperbolic re ections These hyperbolic re ections will be described in the next chapter 629 64 The Main Thenrems EXERCISES 64 In the exercises below f and g denote two rigid motions 1 pmaomeww 11 Prove that g is a glideire ection if and only if f 0 g 0 f1 is a glideire ection Prove that g is a rotation if and only if f 0 g 0 f1 is a rotation Prove that g is a translation if and only if f 0 g 0 f1 is a translation Is it true that g is a re ection if and only if f 0 g 0 f1 is a re ection Justify your answer Prove that f 0 g is a glideire ection if and only if g 0 f is a glideire ection Prove that f 0 g is a rotation if and only if g 0 f is a rotation Prove that f 0 g is a translation if and only if g 0 f is a translation Is it true thatf 0 g is a re ection if and only if g 0 f is a reflection Justify your answer Is it true thatf 0 g is a translation if and only if both f and g are translations Justify your answer Is it true thatf 0 g is a rotation if and only if both f and g are rotations Justify your answer Is it true thatf 0 g is a re ection if and only if bothfand g are reflections Justify your answer 5 Symmetries of Polygons A mathematical symmetry of a figure I is a rigid motion f such that d q Thus the square of Figure 620 possesses the symmetries p d p 8 pm and p 630 65 SYMMETRIES OF POLYGONS Figure 620 Some symmetries of the square This is the mathematical formalization of the more intuitive observation that the square is symmetrical about its diagonals and about the lines joining the midpoints of its opposite sides However the mathematical definition of symmetry is broader than the common usage of the term If C denotes the geometrical center of the square then the rotations R0900 RCY1800 RCY2700 see Fig 621 all rotate the square back onto itself and so they too constitute mathematical symmetries even though they wouldn t be recognized as symmetries by the proverbial person in the street The identity rigid motion Id is another x Figure 621 More symmetries of the square such symmetry The set of all the symmetries of a figure is called its symmetry group or just group Thus the symmetry group of the square is Id pd pe pm pn Rawquot RCJSOO RC270quot By definition every plane figure I has a symmetry group that contains at least the identity motion Id The isosceles triangle of Figure 622 has 1d pv as its symmetry group whereas that of the equilateral triangle of Figure 623 is 1d pd pg pf RC 1200 RC240quot 631 65 SYMMETRIES OF POLYGONS Figure 622 The symmetries of an Figure 623 The reflections of isosceles triangle an equilateral triangle While figures of finite extent cannot have either translations or glide reflections as their symmetries infinitely extended figures do admit such symmetries and a variety of interesting examples will be discussed in the next two sections This section however is concerned with finite figures only and for the symmetries of such polygons there is a useful algebraic description that is obtained by restricting attention to the action of the symmetry on the polygon39s vertices This action is described by means of the positions occupied by the vertices Thus if the positions occupied by the four vertices of the square are labeled 1 2 3 4 respectively Fig 624 then any symmetry f of the square can be thought of as a 1 2 Figure 624 The symmetries of the square function f 1 2 3 4 gt 1 2 3 4 632 65 SYMMETRIES OF POLYGONS where for each i 1 2 3 4 fi denotes the new position of the vertex that was prior to the execution of f in position i Accordingly see Appendix F the eight symmetries of the square have the following permutation representations M 1234 pd 1 324 pg 12 43 pm 2 1 23 4 n 1 4x2 3 12000 1 2 3 4 RC180 1 3X2 4 RCY2700 1 4 3 2 Note that this involves some abuse of notation as the same symbol f is being used to denote both the symmetry as it acts on the whole plane and its restriction to the vertices alone This will lead to no difficulties and obviates the need for a new notation Mathematicians and physical scientists have a great interest in groups of symmetries of solids in spaces of an arbitrary number of dimensions and their classifications The composition operation plays an important role in the classification of both the rigid motions of the plane and the symmetry groups The advantage of the permutation representations is that they allow for an algebraic representation of composition Thus since pg 2 12 43 and pm 2 1 32 4 it follows that P3010 12 43o1 32 4 1 324 Pd and since R0900 1 2 3 4 it follows that pngCBOo 12 4301 2 3 4 1 42 3 p Similarly the composition of the reflection p14 2 12 63 54 and the rotation R0600 1 6 5 4 3 2 both symmetries of the regular hexagon of Figure 625 is 633 65 SYMMETRIES OF POLYGONS p14RCF600 2 ma 6X3 54o1 6 5 43 2 1 23 64 5 2 pm 4 5 Figure 625 Symmetries of the regular hexagon EXERCISES 65 1 Write down the symmetry groups of the following figures a the rectangle with unequal sides b the regular pentagon c the regular hexagon d the regular heptagon e the regular octagon Identify the following compositions of the symmetries of the square of Figure 623 Describe them both geometrically and with permutation representations a pmape b RC9000pm C pnape d Pea d e PmaRCJsoquot f pnaRCJ 80quot g RC18000pe h RC9000RC1800 i peaRCJSOO Identify the following compositions of the symmetries of the regular hexagon of Figure 624 Describe them both 39 and with1 A a pm0p14 b RCY6000pm 0 p140p36 d P250Pm e PmaRCJsoquot f PmaRCJzoquot g RC24000P25 h RC76000RC1800 i P140RC1800 6 Frieze Patterns 634 66 FRIEZE PATTERNS A frieze pattern is a one dimensional repeating figure More formally a frieze pattern is generated by a finite figure lt17 called a block and a translation 7 The pattern itself consists of the union of all the figures 2015 1015 q 14 3015 where 7 denotes 11 applications of 7 and 7 denotes 11 applications of 771 These frieze patterns are the mathematical idealization of such decorative designs as borders used to accent wallpapers and trim sewn or printed around a cloth Fig 626 However unlike their physical manifestations frieze patterns are understood to extend indefinitely in both directions just like a straight line The frieze pattern created by the repetition of a block I is denoted by p01 and it inherits some of the symmetries of I see Exercises 1 3 This observation however does not account for all the symmetries of the frieze pattern 017 By definition every such pattern possesses its generating translation 7 as a symmetry since this translation shifts the infinitely extended pattern onto itself In the case of the block I 1 of Figure 627 the frieze pattern has no other symmetries and so its symmetry group is denoted by F1 2 lt 7gt 635 66 FRIEZE PATTERNS W ji gi aingQkka c Figure 626 Chinese ornamental fn39ezes RepIinted from The Grammar of Chinese Ornament by Owen Jones With the permission of Studio Editions 66 FRIEZE PATTERNS 91 V 031 W Figure 627 A frieze pattern with symmetry group F1 2 lt 7gt Block DZ of Figure 628 possesses the symmetry pk h for horizontal which is of course also a symmetry of its frieze pattern In addition this pattern also necessarily 12 Figure 628 A frieze pattern with symmetry group F lt 7 pk y gt possesses the composite glide reflection y phn as a symmetry This frieze39s symmetry group is denoted by F2 2 lt 7 pk y gt The symmetry pv of D3 of Figure 629 results in a multitude of symmetries of the frieze p 173 which are all essentially identical It gt3 ltlgt 4b b b Figure 629 A frieze pattern with symmetry group F3 2 lt 7 0V gt 637 66 FRIEZE PATTERNS should be noted however that this frieze pattern possesses an additional symmetry namely the re ection pv which has no counterpart in the generating block D3 Because of its similarity to pv the symmetry pv is not listed in the symmetry group F3 2 lt 7 0V gt of this pattern Such an additional re ection could not have appeared with the horizontal re ection of 172 but similar quotaccidentalquot symmetries can arise in other cases as will be seen below The symmetry RC 1800 of the block D4 of Figure 630 results 14 I7CA pm l 6 CI 0 6 4 l 01 UV C1 1 Figure 630 A frieze pattern with symmetry group F lt 7 R gt in the symmetries R 2 RC 1800 of the frieze pattern Once again the frieze pattern has the additional symmetry RC 1800 This frieze pattern s symmetry group is F4 2 lt 7 R gt The block PS of Figure 631 possesses all three of the above symmetries as does the generated frieze pattern Its symmetry group is F5 2 lt 7 pk pv R V gt Just like F 1 5 mp5 W Figure 631 A frieze pattern with symmetry group F5 2 lt 7phpvRgt 638 m M Ma pamms at Flgnle 5 32 and Flgnle 5 as have symmzh39y yullps lt 139 ygt and r7 lt x y x pv gt ha camam ghdz n zconns Hawevex mum h ghdzrle ecunnaf r um n and r7 anmmemwcampmmmmn amn zconnmthz ymlp W no1 r mmz A 1122 pamm wnh symmmygmp r z lt x y magma A 1122 pamm wnh symmmwunp r lt x y x m gt m mumquot thzalem mum m m was ma 7 195a sums mm m fungamg s a camplzm Im at a m mm types at symmuy ymlps um mm pamms can passes PROPOSI39HON 551 9 Niggx 1925 vafnnzpanzm m axymmrvygmwp mundane Womofwgmupx r1 r ltw3Vgt gum r rlt1Rgt r z mphway r my r awnv EXERCISES 66 66 FRIEZE PATTERNS If the block P s frieze pattern p p has a horizontal symmetry must 1 necessarily have a horizontal symmetry If the block P s frieze pattern pm has a vertical symmetry must vertical symmetry If the block P s frieze pattern p p has a rotational symmetry must rotational symmetry Identi i the groups ofthe following frieze patterns 10 14 18 coco QXQXQXQX 96969696 00000000 W gtltgtltgtltgtlt 7 ll 13 15 17 t1 necessarily have a 1 necessarily have a ggggggg 69696969 CgtCgtCgtCgt OOOOOOOO K X W BgtBgtBgtBgtl 19 640 BdB BdB 66 FRIEZE PATTERNS CCCCCCCC OOOOOOOO AAAAAAAA NNNNNNNN 7 Wallpaper Designs Wallpaper designs are the two dimensional analogs of frieze patterns More technically let p be the frieze pattern generated by a block I and a translation 7 If 1 is another translation whose direction is not parallel to that of 7 then the union of the figures zwx lw p Two My is the wallpaper design W017 generated by I 7 and 77 see Figures 634 35 As their name implies such designs are the mathematical patterns that underlie the repeating decorative artworks illustrated in Figure 636 Unlike the carpets and walls that carry these artworks the mathematical wallpaper designs extend ad infinitum in all the directions of the plane It is clear that both 7 and 1 are symmetries of the wallpaper design they generate As was the case for frieze patterns the generated design WltIgt may possess 641 67 WALLPAPER DESIGNS further symmetries that are not present in D In contrast with the seven different groups of symmetries of frieze patterns there are seventeen different possibilities for the symmetry groups of wallpaper designs These together with their labels are exhibited below The presence of reflectional and glide reflectional symmetries is denoted by a dashed line with a label of either p for reflection or y for glide reflection The centers of rotational symmetries are denoted by the symbols 0 1800 A 1200 h 900 C 600 The following table lists the salient symmetry characteristics of each design A rotation through an angle of 360011 is said to have order n A glide reflection is said to be nontrivial if its component translation and reflection are not symmetries of the pattern The symbols p1 pgg p31m are used to denote both a type of wallpaper design and its symmetry group This is known as the crystallographic notation for the symmetry groups If the second character in this symbol is an integer it is the highest order of all the rotations in that group The significance of the other characters is too complicated to explain here 642 67 WALLPAPER DESIGNS 39639 DE DD 0 501 0 DD 0 Figure 634 A wallpaper design 643 674 WltIgt 67 WALLPAPER DESIGNS gtDlt DltgtC gtDlt DltgtC gtDlt DltgtC gtDlt y JVUVUVUVK gtDltgtDltgtDltgtDlt DltgtOltgtOltgtOltgtC gtDltgtDltgtDltgtDlt DltgtOltgtOltgtOltgtC gtDltgtDltgtDltgtDlt DltgtOltgtOltgtOltgtC gtDltgtDltgtDltgtDlt A A A Figure 635 644 67 WALLPAPER DESIGNS The following is the two dimensional analog of Proposition 661 PROPOSITION 671 There are exactly seventeen wallpaper symmetry groups The characteristics of the wallpaper designs that correspond to these groups are displayed in the table below The table itself is then followed by Figures 637 638 639 which display one illustration for each of these groups Proposition 671 was first discovered in 1891 by EVgraf Stepanovich Fedorov 1853 1919 thirty five years before Niggli stated and proved its 1 dimensional analog on frieze groups Proposition 661 Curiously this work had been preceded by Fedorov s and Arthur Scho nflies s 1853 1928 independent classifications of the 230 crystallographic groups these being the 3 dimensional analogs of the wallpaper groups It has since then been established that there are exactly 4783 classes of such groups in 4 dimensional space For spaces of more than four dimensions it is only known that the number of such symmetry groups is finite 646 E p p2 pm Pg pmm ng p3 p3m1 p31 m 176 Hi ghest order of rotation 1 2 67 WALLPAPER DESIGNS Recognition Chart for Wallpaper Symmetry Groups From Doris Schattschneider s article Glide Re ections Re ections no no no no yes no no yes yes yes yes no yes yes no yes yes yes no no yes yes yes yes no no yes yes yes yes no no yes yes Non Trivial 647 Helpful Distinguishing Properties parallel re ection axes perpendicular re ection axes 4 fold centers on re ection axes 4 fold centers not on re ection axes all 3 fold centers on re ection axes not all 3 fold centers on re ection axes 4444 9444 4444 9444 gt444 9444 4444 4444 4444 4444 444 gt444 x w 0000 Figure 638 xxxx d a 9 m M m xxxxxxxxx x X artrx S A 6 67 WALLPAPER DESIGNS EXERCISES 67 Determine the crystallographic symbol of each of the wallpaper designs of Exercises 134 In each case a display a rotation of the highest order b denote the presence of a glideire ection by drawing its aXis with an accompanying y c denote the presence of a re ection by drawing its aXis with an accompanying p d avoid redundancy by only drawing only one aXis in any direction e in case you have to choose between a p and a y display the y J J J J 9 Q 9 Q MHMHMHMH 1 A 4 A A Zgt ltgt O O lt1 K K JV HRH r1 r1 1 M L 90 ltgt 0 lt1 L Ll N L il b 0 0 ltgt lt1 14 qu me w 1 o o o 2 WWW 3 K 39 K 39lt V V W 33 131 WW W pt pl v V v v 4 5 6 All 31 Ni Lirail V V V V L vr v vrlr 333 XXXX 7rv7rvjr A A A A V V V V VWVVWVV V V V V A A A A 7 1 l 1 8 9 I x A x A A x WM i K A A 10 11 u A u A u 12 I 67 WALLPAPER DESIGNS 21 LLLIV JGOCOCL z xz yl yl lt15 H 13 20 m39 N 47474747474 47474747474 47474747474 47474747474 27 6 52 67 WALLPAPER DESIGNS u u u ugugu ugugu ugugu 28 29 M M M 30 31 32 33 34 CHAPTER REVIEW EXERCISES 1 If f is any rigid motion and 1 is any translation identify 1 of o 171 2 If f is any rigid motion and R is any rotation identify R o f 0 R71 3 If f is any rigid motion and p is any reflection identify p of o p7 4 Identify the symmetry groups of the frieze patterns in Figure 626 5 Identify the symmetry groups of the wallpaper designs in Figure 636 6 Are the following statements true or false in Euclidean geometry Justify your answers a The composition of two rigid motions is a rigid motion b The composition of two translations is a translation c The composition of two rotations is a rotation d The composition of two reflections is a reflection 6 53 CHAPTER REVIEW The composition of two glideireflections is never a glideireflection The only rigid motion that fixes three distinct points is the identity Every geometrical figure has at least one symmetry There exist only seven frieze patterns The frieze patterns have only seven pairwise distinct frieze pattern symmetry groups There exist only seventeen wallpaper designs There exist only seventeen distinct wallpaper symmetry groups Every frieze pattern has a translation in its symmetry group For each n l 2 3 4 6 there is a wallpaper design which has a rotation of order n and a nontrivial glideireflectional symmetry For each n l 2 3 4 6 there is a wallpaper design which has a rotation of order n and no nonitrivial glideireflectional symmetries The inverse of every rigid motion is a rigid motion of the same type 654 CHAPTER 4 Circles and Regular Polygons Circles and regular polygons are the subject of Books 111 and IV of The Elements Euclid39s abstract exposition of the interrelation of chords arcs and tangents lines is augmented with the computation of the circle s circumference and area 1 The Neutral Geometry of the Circle Equal circles are circles that have equal radii A chord of a circle is a line segment that joins two of its points A diameter is a chord that contains the center of the circle An arc of a circle is a portion of the circle that joins two of its points Every chord determines two arcs of the circle Consequently it takes at least three letters to denote an arc unambiguously and the two arcs of the circle of Figure 41 with endpoints A and B should be denoted properly speaking by arcAEB and arcAFB Nevertheless it is customary to label both of these arcs arcAB and to rely on the context for clarification A segment of a circle is the portion between a chord and either of its arcs A sector of a circle is the portion between two radii The arcs determined by a diameter are each called a semicircle That the two semicircles determined by a diameter are equal in length is a 41 proposition that Euclid mentions in Definition 17 Chapter 2 This observation is proved as part of Proposition 411 below Figure 4 1 A central angle of a circle is one both of whose sides are radii Every arc subtends a central angle that is either greater or less than 1800 according as the arc is greater or less than a semicircle Every chord subtends a central angle that is at most 180 The following four propositions of Euclid39s are established here with a single unified proof PROPOSITION 411III26 In equal circles equal central angles stand on equal arcs PROPOSITION 411III27 In equal circles central angles standing on equal arcs are equal to one another PROPOSITION 411III28 In equal circles equal chords cut off equal arcs the greater equal to the greater and the less to the less PROPOSITION 411III29 In equal circles equal arcs are subtended by equal chords 42 41 THE NEUTRAL GEOMETRY OF THE CIRCLE GIVEN Equal circles centered at E and E39 respectively Points A B on the first circle and points A B39 on the second Fig 42 T0 PROVE The following are equivalent 1 arcAB arcA B 2 AB A39B39 3 LAEB LA39E39B39 AT ll O B A AB Figure42 PROOF 1 gt 2 Since the given circles are equal it is possible to apply the circle centered at E to that centered at E39 so that E and A fall on E39 and A respectively and arc AB falls along arc A39B39 The two arcs having equal lengths B falls on B39 It follows from PT 1 that the chord AB falls on the chord AB and hence by CN 4 AB A39B39 2 gt 3 AAEB s AA E B by SSS because AB A B Given AE A 39E39 Given BE 2 B E Given 4 AEB A A E B 43 41 THE NEUTRAL GEOMETRY OF THE CIRCLE 3 gt 1 Since A AEB A A39E B it is possible to apply the first circle to the second so that E falls on E 39 and these angles coincide Since the circles have equal radii it follows that arc AB falls on arc A39B39 Consequently these arcs have equal lengths QED COROLLARY 412 In a circle all the semicircles are equal to each other See Exercise 1 PROPOSITION 4131113 In a circle a radius bisects a chord not through the center if and only ifthe radius and the chord are perpendicular to each other See Exercise 2 EXERCISES 41A amp 9quot Prove Corollary 412 Prove Proposition 413 Prove that in a circle a diameter is greater than any chord which is not a diameter Prove that two chords of a circle are equal if and only if they are at equal distances from its center Exercise 23N2 can be used to produce a neutral proof Prove that a circle cannot contain three collinear points 1112 Prove that in a circle the radius perpendicular to a chord bisects that chord39s central angle and arc Prove that in a circle two equal intersecting chords cut each other into respectively equal segments Prove that of two unequal chords in a circle the greater one is closer to the center This is Proposition 11115 1t can be easily proved on the basis of the Theorem of Pythagoras but such a proof is would not be neutral Euclid39s neutral proof is based on Proposition 124 Exercise 23Q5 44 41 THE NEUTRAL GEOMETRY OF THE CIRCLE Let A be a given point and p a circle centered at C If the point P moves along the circle p prove that the midpoint of AP describes a circle centered at the midpoint of CA See Exercises 31D778 It is necessary to consider three cases depending on the relative positions of A and 1 Construct the midpoint of a given arc on a given circle Given an arc of a circle construct the center of the circle Given points A B C D construct a circle through A and B Whose center is equidistant from C and D Given a point A inside a circle construct a chord that is bisected by A Prove that this chord is the shortest of all the chords through A Given an angle a and a line segment a construct a circle Whose center is on one side of a and which cuts a segment equal to a on the other side Given a circle 17 and a point A outside it construct a straight line through A which cuts the circle so that the segment from A to the circle equals the segment in the circle See Exercises 31D778 Comment on Proposition 411 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 413 in the context of the following geometries d maxi a spherical b hyperbolic c taxicab An infinitely extended straight line is said to be tangent to a circle if they have exactly one point in common and that point is called their point of contact PROPOSITION 414III16 18 If a straight line intersects a circle then they are tangent if and only if the straight line is perpendicular to the radius through the point of contact GIVEN Circle C CP straight line PT Fig 43 T0 PROVE PT is tangent to C CP if and only if CPJ PT 45 41 THE NEUTRAL GEOMETRY OF THE CIRCLE P Q Figure 43 PROOF Suppose first that PT is tangent to C CP Hence if Q is any point of PT that is distinct from P it must lie outside the circle so that CP lt CQ Consequently CP J PT PN 2324 Conversely suppose that CPJ PT Then by Proposition 2324 for any point Q of PT that is distinct from P CP lt CQ Consequently no such point Q can lie on the circle C CP and hence PT is tangent to it QED EXERCISES 41B 1 Suppose S and T are the contact points of the tangents to a circle from a point P outside it Prove that PS 2 PT 2 Prove that in a circle the contact points of two parallel tangents are the endpoints of a diameter 3 Prove that the straight line that joins the center of a circle to the intersection of two of its tangents bisects the angle between these tangents 4 Prove that for each side of the triangle there is a circle that is tangent to that side at one of its interior points and tangent to the other two sides at points on their extensions Construct these circles Use Exercise 3 above Two circles are said to be tangent if they intersect in exactly one point If one circle lies inside the other the tangency is said to be internal otherwise it is external 5 Prove that if two circles are externally tangent then the line segment joining their centers contains the point of contact Hint Proceed by contradiction and examine the triangle formed by the centers and the contact point 46 41 THE NEUTRAL GEOMETRY OF THE CIRCLE Prove that if two circles are internally tangent then the line joining their centers contains the point of contact Prove that if two circles are tangent to each other then they have a common tangent line at their point of contact Prove that if two circles lie outside each other then they have four different common tangent lines Let m and n be common tangents to unequal circles such that both circles lie inside one of the angles formed by these tangents Prove that the line joining the centers of the circles bisects this angle Let m and n be common tangents to unequal circles such that the circles lie in vertically opposite angles formed by these tangents Prove that the line joining the centers of the circles bisects these angles Given two circles with the same center and unequal radii prove that all the chords of the larger circle that are tangent to the smaller circle have the same length Construct a circle with a given radius tangent to a given line Construct a circle with a given radius tangent to a given line and containing a given point Construct a circle containing a given point and tangent to a given straight line at a given point on the line Construct a circle that is tangent to two given parallel straight lines Construct a circle that is tangent to two given parallel straight lines and contains a given point between them How many solutions are there Construct a circle that is tangent to two intersecting straight lines Construct a circle that is tangent to two given intersecting straight lines and contains a given point Construct a circle that is tangent to two given parallel straight lines as well as to a given third line that intersects them Given a circle 17 and a point A construct a straight line containing A such that its segment inside 17 has a given length Hint See Exercise 11 Construct a point such that the lengths of the tangents from it to two given circles are given Comment on Proposition 414 in the context of the following geometries d maxi a spherical b hyperbolic c taxicab The following proposition was proved by Euclid in its entirety The proof offered in this text is incomplete in two ways In the first place the argument is restricted to rational values of the ratios in question Moreover given an angle 4 ABC and a positive 47 41 THE NEUTRAL GEOMETRY OF THE CIRCLE integer m this argument makes use of the angle LABC n even though it has not been demonstrated that such an angle can be constructed within Euclid39s system PROPOSITION 415VI33 In equal circles central angles are proportional to the arcs on which they stand GIVEN Equal circles with centers G and H respectively Fig 44 LBGL arcBL TO PROVE LEHN arcEN B B4 E Figure 44 SUPPORTING ARGUMENT The argument is limited to the case where the given ratios are rational In other words it is assumed that there exist positive integers m and n such that LBGL a LBGL LEHN LEHN n m n Let a be an angle such that a LBGL LEHN m n 48 41 THE NEUTRAL GEOMETRY OF THE CIRCLE It follows that there exist points B1 BZ Bmi1 on arcBL and points E1 E2 En1 on arcENsuch that LBGBI 431G132 2 LB mil GL 2 LEHE1 LEIHEZ 2 LE n71 Hence by Proposition 411 arcBBl arcB 1B2 arcBm1L arcEEl arcE1E2 arcEm1L If the common length of these arcs is denoted by it then arcBL m LBGL arcEN n n LEHN QED Proposition 415 was used by Eratosthenes ca 275 194 BC director of the Alexandrian library to obtain a remarkably accurate estimate of the circumference of the earth He knew that on the summer solstice the sun shone down at mid day directly into a well in the city of Syene whereas in Alexandria 5000 stadia to the north the shadows indicated that the sun formed an angle of 150 of 3600 720 with the vertical Assuming that the sun is so far away that its rays can be considered to be parallel when they reach the earth Fig 45 he then used Proposition 415 to obtain the equation 49 41 THE NEUTRAL GEOMETRY OF THE CIRCLE circumference of earth 3600 distance from Alexandria to Syene 720 50 SSW Syene Figure 45 from which he concluded that the circumference is 505000 2 250000 stadia In order to make his answer divisible by 60 probably because of the in uence of the Babylonian sexagesimal number system he adjusted this result to 252000 stadia The standard stade of the time had a length of 1786 meters which converts his rounded estimate to 45007 km an overestimate of 123 since the circumference of the earth is actually 40075 km EXERCISES 41C 1 A circle has circumference 10 ft Find the lengths of the arcs that subtend the following angles at the center of the circle 0 o 0 a 10 b 30 c 90 0 0 0 d 110 e 120 f 180 2 A location on earth has latitude 250 N Find its distance from the equator and from the North Pole 3 A location on earth has latitude 700 N Find its distance from the equator and from the North Pole 410 41 THE NEUTRAL GEOMETRY OF THE CIRCLE 4 A location on earth has latitude 700 S Find its distance from the equator and from the North Pole 5 A location on earth has latitude 700 S Find its distance from the equator and from the North Pole A location on earth lies 2000 km north of the equator Find its latitude A location on earth lies 1234 km north of the equator Find its latitude A location on earth lies 1000 km south of the equator Find its latitude 5000 0 A location on earth lies 617 km south of the equator Find its latitude 10 Comment on Proposition 415 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 2 The NonNeutral Geometry of the Circle The next proposition is one of the most surprising in The Elements Unlike those appearing the previous section its implications are quite unexpected PROPOSITION 421III20 In a circle the angle at the center is double of the angle at the circumference when the angles have the same arc as base GIVEN Points A B C on the circumference of a circle centered at E Fig 46 T0 PROVE A BEC 24 BAC A V B Case 1 Case 2 411 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROOF It is necessary to distinguish three cases Case 1 One of the sides of A BAC contains the center E A BEC A BAC A ECA Exterior PN 316 A BAC A ECA AE 2 EC PN 235 A BEC 2 A BAC Case 2 The center E lies in the interior of A BAC Let F be the other intersection of XE with the circumference of the given circle Then A 2 2 A 1 Case 1 A 4 2 A 3 Case 1 A BEC 2 A BAC CN 2 Case 3 The center E lies outside of A BAC Let F be the other intersection of AE with the circumference of the given circle Then A 2 2 A 1 Case 1 2 4 2 2 3 Case 1 A BEC 2 A BAC CN 3 QED Proposition 421 has several corollaries whose proofs are relegated to the exercises PROPOSITION 422III21 In a circle the angles in the same segment are equal to one another GIVEN Points A B C D on the circumference of a circle such that A and D lie on the same side of BC Fig47 TO PROVE A BAC A BDC 412 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE A Figure 47 PROOF See Exercise 1 This proposition is somewhat counterintuitive Suppose the points A and B in Figure 48 are fixed whereas P slides clockwise around the circle occupying positions P1 P2 P5 successively Proposition 422 implies that as long as the point P remains in the interior of the upper or longer arcAB the angle APB retains a constant acute value When P passes through A or B APB is no longer an angle Finally when P is in the interior of the shorter or lower arcAB the angle APB P2 P4 Figure 48 A discontinuous function assumes a different obtuse value In other words even though the point P moves in a continuous manner A APB varies as a discontinuous function of the position of P 413 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 423III31 In a circle the angle subtended by a diameter from any point on the circumference is a right angle See Exercise 2 PROPOSITION 424III32 Let AB be a chord ofa circle and let ltAgtT be any straight line at A Then the line ET is tangent to the circle ifand only if A TAB is equal to the angle at the circumference subtended by the intercepted arc GIVEN Circle p with chord AB straight line KT arcAB Fig 49 T0 PROVE ET is tangent to 17 if and only if A TAB equals the angle at the circumference of p subtended by arcAB PROOF Let AD be the diameter of the circle containing A and join BD Figure 4 9 Figure 49 By Proposition 423 A ABD 900 Hence the following statements are all equivalent to each other TT is tangent to the circle 4 DAT 900 4 1 900 4 2 4143 414 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 425III36 37 Let P be apoint outside a given circle p let T be apoint on p and let PAB be a secant line with chord AB Then PT is tangent to p ifandonlyif PA PB 2 PT2 Figure 410 GIVEN Point P outside circle p straight lines PT and PAB that intersect p in T A B Fig 410 TO PROVE PTis tangent to p ifand only if PA PB 2 PT2 PROOF The following statements are all equivalent to each other The line PT is tangent to p A 1 A 2 PN 424 A TPA and A BPT are similar to each other PN 357 2 PT B PA PB 2 PT2 QED A polygon is said to be cyclic if all of its vertices lie on a circle 415 PROPOSITION 426III22 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE The opposite angles of a cyclic quadrilateral are equal to two right angles See Exercise 3 EXERCISES 42A MP9 Prove Proposition 422 Prove Proposition 423 Prove Proposition 426 Prove that in a circle parallel chords enclose equal arcs Prove that if the quadrilateral ABCD is cyclic then the exterior angle at A equals the interior angle at C In a circle the extensions of the chords AB and KL intersect in a point P outside the circle Prove that LAKP LLBP and LBKP A LAP In a circle the extensions of the chords AB and CD intersect in a point P outside the circle Prove that PAPB PCPD Proposition 11135 In a circle the chords AB and CD intersect in a point P inside the circle Prove that PAPB PCPD Proposition 11136 Prove that two equal and parallel chords in a circle constitute the opposite sides of a rectangle Prove that if the hexagon ABCDEF is cyclic and the interior angles at A and D are equal then BC ll EF 1n the cyclic quadrilateral ABCD AD 2 BC Prove that the interior angles at A and B are equal to each other as are those at C and D Prove that the sum of the interior angles at A C and E in the cyclic hexagon ABCDEF is four right angles Prove that if the perpendicular chords AB and CD of a circle intersect at the point M inside the circle then the straight line through M that is perpendicular to AD bisects the chord BC Prove that every cyclic rhornbus is a square Prove that if A and B are two distinct points and D is any other point on AB then the locus of all the points P in the plane such that IE B 2 1 is a circle This is the circle of Apollonius State and prove the converse of Proposition 426 416 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE 17 Given a line segment AB construct the circle which consists of all the points from which AB subtends an angle of 900 18 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle of 600 19 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle of 1200 20 Given a line segment AB construct the arc which consists of all the points from which AB subtends an angle equal to a given angle a 21 Construct a triangle given the data a ahbhc b ahaa c amaa d abchaa 22 Construct a parallelogram given its two diagonals and one of its angles 23 Given line segment AB and CD and angles a and constructapoint P such that A APB a and 2 CPD 24 In a given AABC construct apoint P such that A APB A BPC A CPA 25 Comment on Proposition 421 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 26 Comment on Proposition 422 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 27 Comment on Proposition 423 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 28 Comment on Proposition 426 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 29C Use a computer application to verify the following propositions a 421 b 422 c 423b 1 424 Three or more straight lines are said to be concurrent if they all contain the same point PROPOSITION 427 The three perpendicular bisectors 0f the sides of a triangle are concurrent 417 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE GIVEN A ABC DD 39 EE39 FF 39 are the perpendicular bisectors of AB AC and BC respectively Fig 411 TO PROVE DD 39 EE F F 39 are concurrent B F C B Figure 411 PROOF Exercise 31A4 guarantees that DD39 and EE39 intersect in some point M Draw AM BM CM Then AM 2 BM PN 2312 AM 2 CM PN 2312 BM 2 CM CN 1 M is on the perpendicular bisector to BC PN 2313 QED A circle is said to circumscribe a triangle if all of the triangle s vertices are on the circle Its center and radius are respectively the triangle s circumcenter and circumradius PROPOSITION 428IV5 About a given triangle to circumscribe a circle See Exercise 1 418 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE PROPOSITION 429 The bisectors 0f the three interior angles of a triangle are concurrent GIVEN A ABC AA BB CC are the bisectors of A BAC A ACB 4 ABC respectively Fig 412 TO PROVE AA BB CC are concurrent A G E AMA B C B F C Figure 412 PROOF Since 1 1 0 41 43 24 lt 5180 PN2321 it follows from Postulate 5 that 3339 and CC intersect in some point D Let EF and G be those points on AB BC CA respectively such that DE J AB DF J BC and DG J AC Then DE DF DG PN 2331 DE DG CN 1 DA bisects 4 BAC PN 2332 QED 419 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE A circle that lies in the interior of a triangle and is tangent to all of its sides is said to be inscribed in the triangle Its center and radius are respectively the triangle s incenter and inradius PROPOSITION 4210IV4 In a given triangle to inscribe a circle See Exercise 2 EXERCISES 42B MPWP Prove Proposition 428 Prove that similar triangles have circurnradii that are proportional to their sides Prove Proposition 4210 Prove that similar triangles have inradii that are proportional to their sides Prove that the circurncenter of a right triangle is the midpoint of its hypotenuse Prove that if the circurncenter of A ABC lies inside the triangle then the triangle is acute if the center is on a side the triangle is right and if the center is outside the triangle then the triangle is obtuse Prove that the circurncenter of an acute triangle lies inside the triangle Prove that the circurncenter of an obtuse triangle lies outside it Prove that the area of the triangle equals the product of half its perimeter with the im adius Hint Examine the three triangles formed by the center of the circle with the triangle39s three sides In a given circle inscribe a triangle similar to a given triangle Prove that the altitudes of the triangle are concurrent Hint Through each vertex of the triangle draw a line parallel to the opposite side Then show that the altitudes in question are the perpendicular bisectors of the triangle formed by these parallels Comment on Proposition 427 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 428 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 429 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 4210 in the context of the following geometiies a spherical b hyperbolic c taxicab d maxi 4 20 42 THE NONNEUTRAL GEOMETRY OF THE CIRCLE 16C Use a computer application to verify the following propositions a 425 b 427 3 Regular Polygons A polygon is regular if all of its sides and all of its interior angles are equal The equilateral triangles are the regular triangles and they are the subject matter of Proposition 1 of Book I Squares are the regular quadrilaterals and they are constructed in Proposition 331 Book IV of The Elements is mostly concerned with the constructibility of other regular polygons and their inscription in circles Regular hexagons are also easily constructed PROPOSITION 431IV 15 In a given circle to inscribe a regular hexagon GIVEN Circle p 0 r Fig 413 TO CONSTRUCT Points A B C D E F on p such that ABCDEF is a regular hexagon Figure 413 CONSTRUCTION Let A be an arbitrary point on the circle p Let B be the intersection of an arc of radius r and center A with p Let C be the intersection of an arc of radius r and center B with p and let D E F be constructed in a similar manner Then ABCDEF is a regular hexagon 421 43 REGULAR POLYGONS PROOF By construction A AOB A BOC A COD A DOE and A EOF are all equilateral so that A AOB A BOC A COD A DOE A EOF 600 It follows that A F 0A 2 3600 5600 2 600 and hence the isosceles A F 0A is also equilateral Thus FA 2 0A and so each of the sides of ABCDEF has length r It also follows that each of the interior angles of ABCDEF equals 1200 Thus ABCDEF is a regular hexagon QED A slight variation on the construction of the regular hexagon yields the flower like configuration of Figure 414 Figure 414 The construction of the regular pentagon is a considerably more difficult matter Some of the technically demanding details are isolated in the following lemma Others were listed as exercises above PROPOSITION 432IV10 To construct an isosceles triangle having each of the angles at the base equal to double of the remaining one TO CONSTRUCT A ABC such that 4 ABC 2 A ACB 2L BAC Figure 415 CONSTRUCTION Let AB be an arbitrary line segment and let D be a point such that AB BD 2 AD2 PN 341 Then the required A ABC is that triangle such that AC 2 AB and BC 2 AD PN 2327 4 22 43 REGULAR POLYGONS Figure 415 PROOF By construction BC2 2 AD2 2 ABBD It therefore follows that from Proposition 425 that BC is tangent to the circle p that circurnscribes AACD Hence 41 42 PN424 43 44 4244 45 46 DC BC AD 1 l l 41 442243 246 245 QED If the smallest of the angles of the triangle of Proposition 432 is denoted by x then the other two angles are each 226 and so by Proposition 316 180 x2x2x 5x from which it follows that x 360 Hence the following corollaries hold PROPOSITION 433 To construct angles of 360 and 720 4 23 43 REGULAR POLYGONS We are now ready to construct the regular pentagon PROPOSITION 434IV11 In a given circle to inscribe a regular pentagon GIVEN Circle 17 0 r Fig 416 TO CONSTRUCT Points A B C D E on p such that ABCDE is a regular pentagon CONSTRUCTION At the center 0 of the circle construct five non overlapping central angles of 720 PN 433 Label the successive intersections of their sides with the circle A B C D E Then ABCDE is the required pentagon Figure 416 PROOF The five constructed isosceles triangles are all congruent by SAS It follows that the five sides AB BC CD DE and EA are all equal Moreover the base angles of 0 1 these triangles are 5 1800 720 54 each and hence all of the pentagon39s interior angles are equal to 1080 each QED Euclid took the trouble to prove that the regular 15 sided polygon is constructible Exercise 3 It is reasonable to suppose that this was his way of pointing out that there is 4 24 43 REGULAR POLYGONS an interesting question to be pondered here Namely for which integers n can the regular n sided polygon be constructed It has already been shown above that this is possible for n 3 4 5 6 Some more such 11 can be easily produced by simply doubling the number of sides of any constructible regular polygon see Exercises 1 2 4 However this does not answer the question for such numbers as 7 9 11 13 14 17 The surprising intricacy of the construction of the pentagon indicates that such polygons might pose an even greater challenge This problem continued to excite the interest of mathematicians after Euclid but no progress was made for over 2000 years until the young Gauss demonstrated the constructibility of the 17 sided polygon in 1796 Actually he did much more Using the newly emergent theory of complex numbers Gauss proved that a regular p sided polygon can be constructed for every prime integer p 2quot that has the form 2 1 for some nonnegat1ve integer n These include the values 0 1 21 211 3 21 221 5 3 221 241 17 221 281 257 24 16 1 2 1 2 65537 25 32 Curiously the next number in this sequence namely 2 1 2 1 4294967297 fails to be a prime since it can be factored as 6416700417 a fact that had already been noted by Euler over fifty years earlier The same is true for all the numbers of this form for n 6 7 16 and several other values including 11 1945 In fact it is not known whether there are any more primes p that can be expressed in this form above and beyond the five listed above Gauss completely resolved the issue of the constructibility of regular polygons as follows 4 25 43 REGULAR POLYGONS PROPOSITION 435 It is possible to construct in the sense de ned by Euclid a regular ggon g 2 3 if and only if the factorization of g into primes has the form k g 2 p1p2pm Yl where m 2 0 and p1p2 pm are distinctprimes each of which has theform 22 1 Thus the regular 2040 gon is constructible because 2040 233517 whereas the regular 28 sided and 100 sided polygons are not constructible because 28 227 and 100 2252 EXERCISES 43 PW Fquot Prove that the regular octagon is constructible Prove that the regular decagon is constructible Prove that the regular 15 sided polygon is constructible Proposition IV16 Let g be a positive integer Prove that if the regular gisided polygon is constructible so is the regular 2gisided polygon Let p and q be two prime integers Prove that if the regular pisided and qisided polygons are constructible so is the regular pqisided polygon Let g and h be relatively prime integers such that the regular gisided and hisided polygons can be constructed Prove that the regular ghisided polygon can be constructed Let g h gt 1 be integers such that h is an integer multiple of g Prove that if the regular hisided polygon is constructible so is the regular gisided polygon Use a calculating device to prove that 226 1 is not a prime integer For which n 3 4 100 is the regular nigon constructible For which n 101 102 200 is the regular nigon constructible Comment on Proposition 431 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 432 in the context of the following geometries 4 26 15C 43 REGULAR POLYGONS a spherical b hyperbolic c taxicab d maxi Comment on Proposition 434 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Show that in taxicab geometry equilateral triangles are not necessarily equiangular Can all three of the angles of an equilateral taxicab triangle be distinct Perform the construction of Proposition 431 using a computer application 4 Circle Circumference and Area The fundamental observation that the circumference of a circle is proportional to its diameter and hence also its radius goes back several millennia BC Surprisingly Euclid says nothing on this topic in The Elements PROPOSITION 441 Circumferences of circles are proportional to their radii GIVEN Circles p1 01 r1 and p2 02 r2 of circumferences c1 and 02 respectively Fig 417 01 1 TO PROVE a a SUPPORTING ARGUMENT inscribe a regular hexagon in each of the given circles Figure 417 It follows from Proposition 431 that it is possible to By repeatedly bisecting the 4 27 44 CIRCLE CIRCUMFERENCE AND AREA central angles subtended by the sides of the polygons it is possible to inscribe in each circle pi i 1 2 a regular n sided polygon of side say am where n is an integer of the form 32 It is clear that 4 1310101 4 320202 3600n Since A 013101 and A 02B2C2 are isosceles they must be equiangular PN 316 and hence they are similar PN 357 In other words the sides of the inscribed polygons are proportional to the radii Making the reasonable assumption that for large n the difference between the circumferences of each circle and that of its inscribed polygon is negligible it follows that quot91 91 C2 quot92 2 r2 QED An alternate supporting argument that makes use of calculus is described in Exercise 1 It follows from the above proposition that if c and r denote the circumference and radius of an arbitrary circle then the ratio has a constant value say a This constant number can be used to restate the above proposition in the following form PROPOSITION 442 There isa number 0L such thatif c and r are respectively the circumference and radius ofany circle then c 2 ar 4 28 44 CIRCLE CIRCUMFERENCE AND AREA The numerical value of a is of course of interest and will be estimated at the end of this section Next the area of the circle is examined The following proposition was proved by Euclid using the Method of Exhaustion which was the Greeks version of the integral calculus This method was developed by Euclid39s predecessor Eudoxus whom Archimedes 287 212 BC credits with this and other similar propositions PROPOSITION 443XII2 The areas of circles are proportional to the squares of their radii GIVEN Circles p1 01 r1 and p2 02 r2 of areas A and A2 respectively Fig 417 A1 r3 TO PROVE A 7 2 r 2 SUPPORTING ARGUMENT It follows from Proposition VI19 Exercise 35E10 that the areas of A OlBlC1 and A 0232C2 are proportional to the squares of the radii r1 and r2 Making the reasonable assumption that for large n the difference between the areas of each circle and that of its inscribed polygon is negligible it follows that 2 nA 013101 A 01310 r1 A2 nA 023202 A 023202 r QED An alternate supporting argument makes use of calculus see Exercise 2 It follows from the above proposition that if A and r denote the area and radius of an arbitrary circle then the ratio NDgt 4 29 44 CIRCLE CIRCUMFERENCE AND AREA has a constant value say 7 This number can be used to restate the above proposition in the following form PROPOSITION 444 There is a number if such that if A and r are the area and radius respectively of any circle then A 2 m2 The numerical value of n is of course of interest but the relationship between a and 7139 needs to be addressed first The discovery of this relationship is attributed by Proclus to Archimedes PROPOSITION 445 The proportionality constants of the circumference and area of a circle are related by the equation a 2n SUPPORTING ARGUMENT Suppose the circle p is divided into n equal sectors Figure 418 each of which has a central angle of 3600n and let OBD be a typical sector Fig 418 If n is large it may be assumed that OBD is a triangle with altitude 0C 2 r Applying Proposition 325 it follows that this triangle has area rarcBD2 Hence the circle p has area 430 44 CIRCLE CIRCUMFERENCE AND AREA r r r 12 Aniarc BD ci ar iir 2 2 2 2 I a It now follows from Proposition 444 that n E or a 2n COROLLARY 446 The circumference of a circle of radius r is 2m H 2 O Figure 419 An alternate supporting argument for Proposition 445 can be based on Figure 419 Imagine that the circle of radius r on the left is filled with circular strands Cut the circle along the vertical dashed radius and straighten out all the strands as indicated until they form an isosceles triangle that its sides are straight follows from Proposition 442 It follows from Proposition 325 that the area of this triangle and hence also the area of the circle is A circumferencer2 cr2 arr2 a2r2 and the rest of the argument proceeds as before Appealing as this argument is it is fraught with logical perils which are discussed in Exercise 21 431 44 CIRCLE CIRCUMFERENCE AND AREA Yet another alternative argument in support Proposition 445 calls for slicing the circle into an even number of equal sectors and rearranging these to form the near parallelogram at the top of Figure 420 As the number of sectors increases to infinity the Halfcircumference Halfcircumference Half circumference Half circumference Figure 420 near parallelogram converges to the rectangle below it whose area is clearly A 2 half circumferencer cr2 arr2 a2r2 The procedure of successive approximations used in Proposition 441 also yields a method for obtaining numerical estimates of the constant of proportion 71 This was first carried out by Archimedes and constitutes the first of a long and still ongoing series of scientific estimations of 71 Let an denote the length of the chord AB which is one side of the regular n gon inscribed in a circle of radius 1 Fig 421 If C is the midpoint of the corresponding arcAB then the chord AC has length a2 Two 432 44 CIRCLE CIRCUMFERENCE AND AREA Figure 421 applications of the Theorem of Pythagoras then yield 2 a a2 AC2 AD2 DCZ 7 2 0C 0D2 a 2 7 2 1 OAZADZ a 2 7 2 1 V1 an2f anz anz an2 712 17 17 2 4 ai Hence PROPOSITION 447 If an denotes the length of the regular polygon with n sides that is inscribed in a circle of radius 1 then 433 44 CIRCLE CIRCUMFERENCE AND AREA 2 a2 2 4 an Since the length of the side of the inscribed regular hexagon equals the radius it follows that a6 1 and so a12 l2 l41 lZVg 5176380902 24 2 H 2610523844 6148 2 4 2 23 2 2 2 13 1308062585 6196 2 2 2 2 V3 0654381656 Since the length of any arc exceeds that of its chord another one of those reasonable assumptions the circumference of the circle exceeds that of any inscribed polygon As the circle of radius 1 has circumference 271 it follows that for each positive integer n and in particular 71 gt 485496 2 314103195089 434 44 CIRCLE CIRCUMFERENCE AND AREA To obtain upper bounds for the value of n Archimedes examined regular n gons whose sides were tangent to the circle Suppose a regular n gon is inscribed in a circle of radius 1 and at each of the vertices a straight line tangent to the circle is constructed It is easily verified that the resulting polygon surrounding the circle is also a regular n gon of bu 7 Figure 422 Comparing a circumscribed polygon with an inscribed one side say I Fig 422 The side 17 can be estimated by showing that see Exercise 9 i 2 41 2 b6 2 E and b2 2 b 1 n Alternately the figure above can be used to show that 2a This gives us 1796 0654732208 and hence 961796 71 lt 2 31427145996 435 44 CIRCLE CIRCUMFERENCE AND AREA Archimedes did not have the decimal number system at his disposal and he had to work within the much more cumbersome systems that were then current Using some of the quite complicated methods for the estimation of square roots by means of fractions that were then known he was able to show that 6336 10 48d96 gt 1 gt 3i 2 31408 2017 and 14688 1 481796 lt 1 lt 37 31428 4673 Thus Archimedes proved that PROPOSITION 448 3 lt n lt 3 This section and chapter conclude with a discussion of some paradoxes and problems regarding the areas of circles Consider Figure 423 where A ABC is both right and B a C Figure 423 isosceles with legs of length a and hypotenuse of length The outside arc is a semicircle with diameter AB and radius m5 2 whereas the inside arc is a quarter 436 44 CIRCLE CIRCUMFERENCE AND AREA circle centered at C with radius 61 Regions bounded between two such arcs are called lunes but these are different from the spherical lunes Note that the area of this lune is the difference between the entire figure and the quartercircle centered at C Thus area of lune 2 area of semicircle area of AABC area of quartercircle 2 2 2 2 2 2 2 i i 2 22394 42394 2 2 area of A ABC This surprising equation is due to Hippocrates of Chios A similar equation appears in Exercise 10 There are two unexpected aspects to this equation First the area of the lune whose boundary consists of circular arcs turns out to have an expression that is free of 71 Second this curvilinear figure has the same area as a triangle This leads naturally to the question of whether it is possible to construct a triangle or a square for that matter whose area equals that of a given circle the simplest of all the curvilinear regions The operative word here is construct It is clear that any circle of radius r has the same area as a square of side The difficulty lies in constructing r7 I within the framework of Euclidean geometry This problem drew the attention of many mathematicians and non mathematicians both in classical times and during the subsequent two and a half millennia Although many individuals dedicated their lives to the solution of this problem and some even deluded themselves into believing they had discovered the construction all their efforts were in fact wasted In 1882 the German mathematician C L Ferdinand von Lindemann 1852 1939 proved a theorem which had the following corollary amongst many others 437 44 CIRCLE CIRCUMFERENCE AND AREA Given a line segment a it is impossible to construct in the sense of The Elements a square whose area equals that of the circle of radius a EXERCISES 44A b f 2 The length of the graph of the function y fx for a s x s b is I f39 dx Use this a formula to prove Proposition 441 Use calculus to prove Proposition 443 Using 314 for the value of n in a circle of radius 10quot find the length of the arc and the area of a sector determined by a central angle of a 60 b 20 c 90 d 100 e 180 f 230 Compute the radical expressions for 1192 and 12192 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 1384 and 22384 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 0768 and 12768 and use them and a calculator to obtain decimal bounds of the value of J17 Compute the radical expressions for 11536 and 121536 and use them and a calculator to obtain decimal bounds of the value of J17 Prove Equations 1 Prove that of two circular arcs joining two given points the one with the longer radius has shorter length Semicircles are constructed on the sides of a right A ABC Fig 424 Prove that the sum of the areas of the two lunes l and II equals the area of AABC Figure 424 438 44 CIRCLE CIRCUMFERENCE AND AREA Show that the circumference of a circle of spherical radius r on a sphere of radius R is 2nR sin rR Show that the area of a spherical circle of spherical radius r on a sphere of radius R is 21R2l 7 cos rR Comment on Proposition 441 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 442 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 443 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 444 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 445 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 446 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 447 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 448 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Explain the following paradox Suppose the method that was used to convert a circle into a triangle see paragraph following Proposition 446 is applied to the same square ABCD in two different manners 7 first by slicing from a corner to the center and second by slicing from the middle of a side to the center Fig 425 The two triangles so obtained have their bases equal to the perimeter of the square but their altitudes are clearly different Why are two triangles of different areas obtained 439 44 CIRCLE CIRCUMFERENCE AND AREA Figure 425 45 Impossible Constructions Mostly Part of the legacy that the Greek mathematicians passed on to their successors was a collection of construction problems they could not resolve by ruler and compass alone While most of these problems have already been discussed Sections 23 and 43 it might be a good idea to reexamine this topic here in order to provide a better perspective on its outcome We begin by listing the specific construction problems in question I To divide a given angle into three equal parts 2 To construct a regular n gon for each integer n 2 3 3 To construct a square whose area equals that of a given circle 4 To construct a cube whose volume is double that of a given cube The reader will recall that Cartesian coordinates were invented for the purpose of expressing geometrical problems in the language of algebra Since construction problems are geometrical this applies to them as well Some of this relation between geometry and algebra has already been pointed out If a and b are the lengths of two given line segments then it is possible to construct line segments of lengths a 17 Exercise 23A3 and a 17 Proposition 233 Assuming a to be a unit length Exercise 35B13 shows how given segments of lengths b and c it is possible to construct a segment of length bc Assuming c to be a unit length the same exercise can be used to 440 IMPOSSIBLE CONSTRUCTIONS construct for any given segments of lengths a and b a segment of length ab Finally assuming 17 to have unit length Exercise 35B 14 can be used to construct for any given segment of length a a line segment of length a Thus the four arithmetical operations as well as the taking of square roots can be mimicked by ruler and compass constructions The Cartesian coordinate system can be used to argue that the power of ruler and compass constructions cannot be extended beyond these five algebraic operations To do this it is first necessary to formalize some notions A configuration is a set of points straight lines and circles An elementary ruler and compass construction is any of the following five operations 1 Draw the line joining two given points 2 Draw a circle with a given center and radius 3 Find the intersection of two given straight lines 4 Find the intersection of a given circle and a given straight line 5 Find the intersection of two given circles A configuration T is said to be constructible from configuration S provided every element of T can be obtained from the elements of S by a succession of elementary ruler and compass constructions In particular note that every construction problem stipulates a given configuration and aims at the derivation of a desired configuration Assume now that a Cartesian coordinate system has been chosen to which all the configurations below are referred The numerical aspects of the point x y are x and 441 IMPOSSIBLE CONSTRUCTIONS y The numerical aspects of the straight line with equation ax by c 0 are a b and c The numerical aspects of the circle with equation x2 y2 ax by c 0 are a b and c A real number r is said to be a Hippasian function of the set S provided it is obtainable from the elements of S and the rational numbers by rational operations and extractions of real square roots this terminology honors Hippasus of Metapontum the discoverer of the irrationality of V2 For example the numbers below are all Hippasian functions of the set S 71 3V2 e 12 n3e 3wg 4 5 e 22 3 IE The Hippasian numbers are those that are obtainable from the rational numbers alone by the rational operations and the extractions of real square roots In fact this is tantamount to saying that they are obtainable from the number 1 by the said operations The numbers below are all Hippasian numbers 3 11ME Lg linHm Jim The following theorem formalizes the intuitively plausible connection between constructibility and Hippasian functions 4 42 IMPOSSIBLE CONSTRUCTIONS THEOREM 451 If configuration T is constructible from configuration S then the numerical aspects of T are Hippasianjunctions of the numerical aspects of S OUTLINE OF PROOF Suppose configuration T is obtained from configuration S by the elementary construction i where i 1 2 3 4 5 If i 5 for example let the two given circles have equations x2y2axbyc0 and x2y2a cb yc 0 By Exercise 5 the intersection point of these two circles if it exists has coordinates which are Hippasian functions of a b c a b c The proof of the cases i 1 2 3 4 is similar see Exercises 1 4 and we conclude that if T is constructible from S by any ruler and compass operations then the numerical aspects of T are obtainable from those of S in the desired manner QED We note in passing that the converse of Theorem 451 is also valid albeit somewhat harder to prove As this converse is not needed for the proof of the impossibilities below it is relegated to Exercise 6 The strategy for demonstrating the non feasibility of a ruler and compass construction calls for demonstrating that the numerical aspects of the desired configuration are not Hippasian functions of those of the given configuration Matters can and will be simplified below by setting things up so that the numerical aspects of the given data are either integers or Hippasian numbers and hence it will suffice to show that the desired configuration has a non Hippasian number as one of its numerical aspects 4 43 IMPOSSIBLE CONSTRUCTIONS The following unproven proposition provides an easily applied criterion for recognizing non Hippasian numbers It is found in a more general form in many undergraduate modern algebra texts PROPOSITION 452 Let x be a real solution of the equation ax3bx2cxd0 1 where a b c d are integers Then x is Hippasian if and only if this equation has a rational solution u Unlike the above proposition the next one is found in many precalculus texts and is easily proven Exercise 7 PROPOSITION 453 The Rational Zeros Theorem Let Px anx an71x 1 alx a0 where all the ai s are integers If pq is a rational number in lowest terms such that Ppq 0 then p is afactor of a0 and q is afactor of an u 444 IMPOSSIBLE CONSTRUCTIONS The four construction problems of antiquity are now reexamined one at a time The simplest of these turns out to be the doubling of the cube DOUBLING THE CUBE If it were possible to double a given cube by ruler and compass constructions then it would certainly be possible to construct a cube of volume 2 The length of the side of such a cube would be 3V2 and it would have to be a Hippasian number However 3V 2 is clearly a solution of the equation x3 220 2 and hence by Proposition 452 this equation would have to have a rational solution say 1761 in lowest terms By Proposition 453 17 must be a factor of 2 and q a factor of 1 Hence pq must be one of the numbers 1 1 t 21 none of which by Exercise 9 is a solution of the Equation 2 Thus the supposed feasibility of a ruler and compass doubling of the cube has lead to a contradiction and we conclude that The cube cannot be doubled by ruler and compass alone ANGLE TRISECTION We next argue that there is no method for trisecting angles by ruler and compass alone Suppose to the contrary that such a method exists and is used to trisect the 60 angle of an equilateral triangle whose side has unit length Here it may be supposed that the given configuration consists of the three points 00 0 A0 1 445 IMPOSSIBLE CONSTRUCTIONS B12 32 Fig 426 all of whose numerical aspects are Hippasian numbers The hypothetical construction yields an angle of 20 that may be placed at the origin with one side on the x axis Fig 426 The constructible intersection P of this angle s other side with the circle 0 1 has coordinates cos 20 sin 20 and hence it follows from Theorem 451 that cos 20 is a Hippasian function of Hippasian numbers Consequently cos 20 is a Hippasian number We now go on to obtain an analog of Equation 2 If x cos 20 then by Exercise 8 12 cos 60 2 4 cos3 20 3 cos 20 2 4x3 43x and hence 8x3 6x 120 3 By Proposition 452 this equation has a rational solution say pq where p and q are integers By Proposition 453 pq must be one of the fractions 1 12 14 18 none of which by Exercise 10 is a root of Equation 3 Thus the assumption of the feasibility of a ruler and compass method for trisecting angles has resulted in a contradiction and hence 446 IMPOSSIBLE CONSTRUCTIONS There is no method for trisecting angles by ruler and compass alone Figure 426 REGULAR nGONS Whether or not a regular ngon is constructible by ruler and compass turns out to depend on the value of n In general such a polygon can be constructed if and only if an angle of 360011 can be constructed When this angle is placed at the origin with one side on the x axis the other side intersects the circle 0 1 in the point cos 360011 sin 360011 By Theorem 451 and Exercise 18 this general ruler and compass construction is feasible if and only if cos 360011 is a Hippasian number We now show that there is no ruler and compass construction for the regular 7 gon Set A 2 36007 and x cos 36007 If such a method existed then 6 would be a Hippasian number However by Exercises 8 and 12 cos 3A 4cos3A 3cosA 4x3 3x cos4A 8cos4A 8cos2A1 8a4 8x21 4 47 IMPOSSIBLE CONSTRUCTIONS Since 3A 4A 3600 it follows that cos 3A 2 cos 4A and hence 4263 326 2 8 4 8x21 S 4x3 8x23x1 0 x 18x34x2 4x 1 0 However cos 36007 1 and hence 8x34x2 4x 10 It follows from Proposition 453 that the only possible rational solutions of this equation are again 1 12 14 18 Since by Exercise 10 none of these is a solution it follows from Proposition 452 that cos 36007 is not a Hippasian number and hence no such method for constructing regular 7 gons can exist According to Exercise 11 cos 36005 is a Hippasian number and so the regular pentagon is indeed constructible as demonstrated in Proposition 434 As was mentioned above the regular 17 gon is also constructible and hence cos 3600 17 must also be a Hippasian number In fact it is known to equal 1161 16116134 21 448 IMPOSSIBLE CONSTRUCTIONS ugh17 31 J34 2W 2134 2 SQUARING THE CIRCLE To square a circle of unit radius is tantamount to constructing a line segment of length a V75 As it happens neither 313 nor J39l3 are the solutions of any equation of form 1 In fact it was proven by Lindemann that there exists no polynomial Px with integer coefficients of any degree such that either if or 171 are solutions of Px 0 Since every Hippasian number is known to be the solution of such an equation it follows that V7 is not a Hippasian number and hence It is impossible to square a circle by ruler and compass alone EXERCISES 45 1 Show that the straight line joining the points a b and a b has equation b 7bx a7a y a b7ab 0 2 Show that the circle with center a b and radius r has equation x2 y2 72116 72by a2 b2 7 r2 0 3 Show that if the two lines with equations ax by c 0 and 1 by c 0 intersect then their point of intersection has coordinates bc 7 b cab 7 a b a c 7 ac ab 7 ab 4 Show that if the line with equation ax by c 0 and the circle with equation x2 y2 1 by c 0 intersect then their points of intersection have coordinates x 7B 1 Bz 7 4AC2A and y 7ax7cb where A a2 b2 B 2m a b27abb and C 027 bb c c bz 5 Show that if the two circles with equations x2 y2 ax by c 0 andx2 y2 a c by c 0 intersect then their points of intersection have coordinates x 7B 1 Bz 7 4AC2A and 4 49 IMPOSSIBLE CONSTRUCTIONS y ia cicOb Where A a 2 b a B 2a c a b 2 1a b b C 0 41 b b c c b Z and a aia b bib c 070 State and prove the converse of Theorem 451 Prove Proposition 453 Prove that cos 3A 4 cos3A7 3 cosA Verify that none of the numbers 111 121 is a solution of the equation x3 1 2 0 Verify that none of the numbers 111 112 114 118 is a solution of the equation 813 1 6x 11 0 Prove that if x cos 72quot then 4162 2x7 1 0 Prove that cos 4A 8 cos4A 7 8 coszA 1 Prove that it is impossible to construct a regular 97gon by ruler and compass alone Prove that it is impossible to triple a cube by ruler and compass alone Prove that it is impossible to halve a cube by ruler and compass alone Prove that the following numbers are not Hippasian a 35 b 23 5 c 12737 17 Is it possible to construct an angle of 1 18 Let A be any number Explain Why sinA is a Hippasian number if and only if cos A is a Hippasian number 4 50 IMPOSSIBLE CONSTRUCTIONS CHAPTER REVIEW EXERCISES 15 Circle 17 intersects two concentric circles Prove that the arcs of 17 cut off by the two circles are equal Prove that if each of the sides of a square is extended in both directions by the length of the radius of the circle that circumscribes the square one obtains the vertices of a regular octagon Two circles centered at C and D respectively intersect at apoint A Prove that if PAQ is the lldouble chordll that is parallel to CD then PQ 2CD The area of the annular region bounded by two concentric circles equals that of a circle whose diameter is a chord of the greater circle that is tangent to the smalller one In a circle a diameter bisects the angle formed by two intersecting chords Prove that the chords are equal Prove that every equiangular polygon all of whose sides are tangent to the same circle is regular Through the center of a circle passes a second circle of greater radius and their common tangents are drawn Prove that the chord joining the contact points of the greater circle is tangent to the smaller circle Prove that every cyclic equilateral pentagon is regular Three circles through the point 0 and of radius r intersect pairwise in the additional points A B C Prove that the circle circumscribed about A ABC also has radius r Prove that in the regular hexagon ABCDEF the diagonals AC and AE cut the diagonal BF into three equal segments Each of the sides of a cyclic quadrilateral is the chord of a new circle Prove that the other four intersection points of these new circles also form a cyclic quadrilateral A circle of radius r is inscribed in A ABC in which A ACB is a right angle Prove that a b c 2r The chord AB of a circle of radius 1 has the property that if the circle is folded along AB so as to bring AB39s arc into the circle then the arc passes through the center of the circle Compute the lengths of the chord AB and its arc In a given AABC construct a point whose distances from the sides of the triangle are proportional to three given line segments Through the midpoint M of a chord PQ of a circle any other chords AB and CD are drawn chords AD and BC meet PQ at points X and Y Prove that M is the midpoint of XY Thisis the notorious Butter y Problem 451 IMPOSSIBLE CONSTRUCTIONS The points of intersection of the adjacent trisectors of any triangle are the vertices of an equilateral triangle This is known as Morley39s Theorem Are the following statements true or false Justify your answers The Greeks believed that the world is flat The area of a plane Euclidean figure whose perimeter is composed of circular arcs must involve J17 in its eXpression The Greeks knew that J17 314 If two sectors have equal angles then their arcs are proportional to their radii Given a line segment a it is impossible to construct in the sense of The Elements a square whose area equals that of the circle of radius 1 Given a line segment a it is impossible to construct in the sense of The Elements an equilateral triangle whose area equals that of the circle of diameter 1 Of two equal chords in unequal circles the one in the larger circle lies further from the center The diameter is the circle s longest chord It is possible to construct a regular 3407sided polygon in the sense of The Elements It is possible to construct a regular 1407sided polygon in the sense of The Elements In a circle all the angles subtended by a chord are equal to each other In a circle arcs are proportional to their chords Every circle has only one center Every circle has only one tangent line If two chords of a circle bisect each other then they are both diameters 4 52 PERMUTATIONS APPENDIX F Permutations If S is a finite set then a permutation of S is a function fS gtS that has the following two properties 1 if a and b are distinct elements of S then u and fb are also distinct elements of S 2 for every element y of S there is an element x of S such that y fx It is customary to display permutations as a collection of cycles A cycle of a permutation f is a cyclic sequence a1 a2 ak where di1 di for i12k 1 and a1 fak PERMUTATIONS EXAMPLE F1 If S 1 2 3 4 5 6 7 and f1 6f2 5f3 7f4 4 f5 3 f6 1f7 2 then 1 6 5 3 7 2 and 4 are cycles of f as are 6 1 and 3 7 2 5 However since cycles are by their definition cyclically ordered it follows that 1661 and 5372372572532537 Hence 1 6 5 3 7 2 and 4 are the complete set of cycles of f and we write f 1 65 3 7 24 The order of the cycles is immaterial Thus f1653 7 24 53 7 2416 41653 7 2 3 7 2 546 1 EXAMPLE F2 If S123 4 5 6 7 8 9 a b c and f1af21f3 c f4 8 f5 9 f6 7 f7 3 f8 419 6 fa 2 jib b fc 5 then f 21a73 c 5 9 64 8b If f and g are permutations of the same set S then the composition f g is also a permutation of S such that F2 PERMUTATIONS fgx gx for all x in 5 EXAMPLE F3 If f1 653724 and g17263 54 then gxl g1 f7 2 gxz g2 16 1 16 8X3 g3 f5 3 fg4 g4 f1 6 16 8X5 g5 f4 4 fg6 g6 13 7 fg7 11530 12 5 Consequently fg 1 234 6 7 5 Similarly gf1 NH g6 3 ltgfgt2 gm g5 4 gf3 NS g7 2 gf4 gf4 g4 1 gf5 gf5 g3 5 gf6 gf6 g1 7 g 7 NO g2 6 Consequently gf 1 3 2 456 7 F3 PERMUTATIONS EXERCISES F Rewrite the functions of exercises 1 5 in terms of their cycles 1 2 3 f1 62 53 7f4 25 3 61f7 4 f1 62 53 7f4 85 3 61f7 2 f8 4 f1 92 53 7f4 85 3 61f7 2 f8 49 6 f0 910 513 74 85 3y 61f7 a 18 49 6 u 2 f0 910 513 III4 85 3y 61f7 a 18 49 6 fa 2 jib 7 Suppose f 1 2 3 4 5 6 7 8 9 g 4 3 2 159 8 7X6 h 1 2X3 45 67 8X9 Display the following compositions in terms of their cycles a f g b gf c fh d hf e gh f hg g ff h gg i hh F4 CHAPTER 3 NonNeutral Geometry This chapter s propositions differ from those of the previous one in that they depend on Postulate 5 for their validity Their proofs are therefore not valid in the context of non Euclidean geometry 1 Parallelism The first of the non neutral propositions is the converse of Proposition 2335 the last proposition of the previous chapter PROPOSITION 311I29 A straight line falling on parallel straight lines makes the alternate angles equal to one another the corresponding angles equal to one another and the interior angles on the same side equal to two right angles A B 1 3 2 C D H 4 F Figure 31 31 31 PARALLELISM GIVEN Straight lines AB II CD straight line EF intersecting AB and CD at G and H respectively Fig 31 TOPROVE41L2L3L4 L2L3 2rightangles PROOF By contradiction Suppose A 1 and A 2 are unequal Then it may be assumed without loss of generality that A 1 gt A 2 A 1 A 3 gt A 2 A 3 but A 1 A 3 2 right angles PN 2317 2 right angles gt A 2 43 AB and CD intersect PT 5 This however contradicts the fact that AB II CD and so A 1 A 2 Moreover 41 43 42 44 2 right angles PN 2317 43 44 CN 3 and also A 2 43 2 right angles QED The following proposition provides an alternative and intuitively appealing characterization of parallel lines PROPOSITION 312 The locus of all points on one side of a straight line that are equidistant from it is a straight line See Exercise 11 32 31 PARALLELISM EXERCISES 31A Prove that if two parallel straight lines are cut by a third line then the two bisectors of a pair of alternate interior angles are parallel to each other Prove that if a straight line is perpendicular to one of two parallel straight lines then it is also perpendicular to the other one Suppose AB J KL and CD J MN are all straight lines such that KL H MN Prove that AB H CD Suppose AB J KL and CD J MN are all straight lines such that KL Vl MN Prove that AB Vl CD Prove that two angles whose sides are respectively parallel are either equal or supplementary Prove that two angles whose sides are respectively perpendicular are either equal or supplementary In AABC AD is the bisector of A BAC and E is a point on AC such that DE H AB Prove that AE DE For a given A ABC AD H BC and AD AB Prove that BD bisects either the interior angle or the exterior angle at B Prove that if the points A B are on the same side of the straight line In and at the same distance from m then AB Hm Prove that if the points A B are such that AB H m then they are at the same distance from m Use the above two exercises to prove Proposition 312 Prove that the internal bisectors of each pair of angles of a triangle intersect Given two distinct parallel lines construct a straight line that is parallel to both and also equidistant from both Comment on Proposition 311 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 312 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi PROPOSITION 313I30 Distinct Straight lines parallel to the same straight line are also parallel to one another GIVEN Distinct straight lines AB EF CD EF Fig 32 T0 PROVE AB II CD 33 31 PARALLELISM A D I C 1 B J 3 E F Figure 32 PROOF By contradiction Suppose AB and CD intersect in some point I Join I to any point J of EF Then 41 43 PN311 AB IIEF 42 43 PN311 CDIIEF 41 A 2 CN 1 but this is impossible since the straight lines AB and CD are distinct Hence AB II CD QED Euclid begins his proof of this proposition by drawing a straight line PQ that intersects all the three given lines While intuitively plausible the existence of such a line calls for a justification and Euclid39s proof is therefore incomplete The need for such a justification is demonstrated by Figure 33 which exhibits three pairwise parallel hyperbolic geodesics such that no single geodesic intersects all three Ann Figure 33 Three hyperbolic parallel straight lines that are not intersected by the same hyperbolic straight line 34 31 PARALLELISM EXERCISES 3 IB 1 If a straight line intersects one of two parallel straight lines in only one point then it also intersects the other one 2 Comment on Proposition 313 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi PROPOSITION 314I31 Through a given point to draw a straight line parallel to a given straight line GIVEN Straight line BC point A not on FC Fig 34 TO CONSTRUCT A straight line AE such that AE BC Figure 34 CONSTRUCTION Let D be any point on BC and draw AD Construct A DAE 4 ADC PN 2328 Then AE is the required straight line PROOF A EAD 4 CDA Construction AE BC PN 2335 QED The following proposition has supplanted Euclid39s Postulate 5 in many texts where it is known as Playfair39s Postulate Although this will not be demonstrated here the two are in fact logically equivalent 35 31 PARALLELISM PROPOSITION 315Playfair s Postulate Through a point not on a given straight line there exists exactly one straight line that is parallel to the given line See Exercises 1 and 2 Just like Postulate 5 Playfair s postulate does not hold in hyperbolic geometry Figure 35 exhibits three distinct geodesics p q r all of which contain the same point P and all of which are parallel to the same geodesic m V Figure 35 A hyperbolic counterexample to Playfair s Postulate It is now possible to give a more precise definition of hyperbolic geometry This calls for negating Playfair s postulate which is equivalent to Postulate 5 In View of Proposition 237 the following postulate is the proper negation of Playfair s postulate H Hyperbolic There exists a straight line that is parallel to two intersecting distinct straight lines Hyperbolic geometry is the geometry based on Euclid s Postulates 1 2 3 4 A S and Postulate H 36 31 PARALLELISM PROPOSITION 316032 In any triangle if one of the sides be produced the exterior angle is equal to the two interior and opposite angles and the three interior angles of the triangle are equal to two right angles GIVEN AABC side BC extended to D Fig 36 TO PROVE 4ACD 4ABC 4 CAB 4ABC 4 BCA 4 CAB 2 right angles E D Figure36 PROOF Draw CE IIAB PN 314 Then 45 42 PN311 46 41 PN311 44 2 42 41 CN2 4443 414243 CN2 2 right angles 2 41 42 43 PN 2317 QED Recall that according to Chapter 1 the sum of the angles of every spherical triangle is greater than 1800 PN 115 whereas the sum of the angles of every hyperbolic triangle is less than 1800 PN 126 EXERCISES 31C 37 31 PARALLELISM 1 Prove Proposition 315 2 Prove that in AABC the bisector of the exterior angle at A is parallel to BC if and only if AB AC 3 Prove that a straight line that is parallel to one side of an isosceles triangle cuts off another isosceles triangle Note There are two distinct cases to be considered here 4 A straight line cuts off an isosceles triangle from a given isosceles triangle Prove that the straight line is parallel to one of the sides of the given isosceles triangle 5 In an isosceles AABC a line perpendicular to the base BC intersects AB and AC in the points D and E respectively Prove that AADE is also isosceles 6 In AABC 4 BAC 900 and 4 ACB 30 Prove that BC 2AB 7 In AABC 4 ABC 600 and BC 2AB Prove that AABC is a right triangle 8 Prove that in a right triangle the angle between the altitude to the hypotenuse and one of the legs equals the angle opposite that leg 9 Let D be that point on side BC of A ABC such that AD is the bisector of A BAC Prove that 4 ADC is half the surn of the interior angle at B and the eXterior angle at C 10 Prove that in AABC the bisectors of the interior angle at B and the exterior angle at A form an angle that is half the interior angle at C 11 Prove that in A ABC the angle bisector and the altitude at A form an angle that is half the difference between the interior angles at B and C 12 Prove that in a right A ABC the bisector of 4 ABC the altitude to the hypotenuse BC and the side AC form an isosceles triangle 13 The point D on the hypotenuse BC of the right isosceles A ABC is such that BD AB Prove that 4 BAD 6750 14 Prove that if the diagonals of quadrilateral ABCD are equal and the sides AB CD then AD H BC 15 Prove that the surn of the interior angles of a quadrilateral is 3600 16 Prove that in quadrilateral ABCD the bisectors of the interior angles at A and B form an angle that is half the surn of the interior angles at C and D and if the bisectors of the interior angles at A and C intersect they form an angle that is half the difference between the angles at Band D A polygon is said to be convex if all of its diagonals fall in its interior 17 Prove that the surn of the interior angles of a conveX nisided polygon is n 7 21800 18 Prove that the surn of the interior angles of an arbitrary nisided polygon is n 7 21800 Go ahead and use the difficult to prove fact that every polygon has a diagonal that lies completely 38 31 PARALLELISM inside it 19 Prove that the number of acute interior angles of a convex polygon cannot exceed 3 20 Prove that the sum of the exterior angles of a convex polygon is 3600 Try to prove this without making use of Exercise 17 above 21 Construct AABC given the data a A B 22 Construct an isosceles triangle given one of its angles and one of its sides Construct AABC given the data a b c A Construct AABC given the data b c A B Comment on Proposition 314 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 315 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 316 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Explain why there are no rectangles in spherical geometry Explain why there are no rectangles in hyperbolic geometry Are there rectangles in taxicab geometry Are there rectangles in maxi geometry The following method for trisecting an arbitrary angle is credited to Archimedes If that attribution is correct he must have been aware of its shortcomings as a construction in the sense of Euclid Let a be a given angle with vertex A Fig 37 Draw a circle of radius AB 2 AC On a ruler mark two points D and E such that DE 2 AB 2 AC and place the ruler on the page so that the point E falls on the extension of AB the point D falls on the circle A AB and the ruler also passes through the point C Prove the following assertions a LADCLACD7 b L AED L EAD y c a3y or y 13 Figure 37 An angle quotttisectionquot 39 35C 36C 31 PARALLELISM Explain why this quottrisectionquot of or does not meet Euclid39s standards for a construction Criticize the following quotneutral proofll of Playfair39s Postulate offered by Proclus 4107485 quotI say that if any straight line cuts one of two parallels it will cut the other also For let AB CD be parallel and let EFG cut AB at F with G between AB and CD I say that it will cut CD also For since BF FG are two straight lines from one point F they have when produced indefinitely a distance greater than any magnitude so that it will be greater than the interval between the parallels Whenever therefore they are at a distance from one another greater than the distance between the parallels FG will cut CDll Criticize the following quotproofquot of the fact that the surn of the interior angles of a triangle is 1800 Let ABC be a given triangle let d be a line segment that lies on the straight line AB with its center at A Slide d along AB until its center falls on B and then rotate it through the exterior of the triangle about B as a pivot until it falls along side BC Next slide d along BC until its center reaches C and rotate it about C as a pivot through the exterior of the triangle until it falls along CA Finally slide d along CA until its center reaches A and rotate it about A as pivot so that it comes into its initial position If the triangle39s interior angles are a 3 y then the segment d has been rotated successively by the angles 1800 7 1800 7 y and 1800 7 a before it returned to its original position Consequently 1800 7 8 1800 7 y 1800 7 a 3600 from which it follows that a 3 y1800 Perform the construction of Proposition 314 using a computer application Use a computer application to verify Proposition 316 Euclid39s statement of the following proposition is awkward and so it appears here in a paraphrased forrn PROPOSITION 317I33 A quadrilateral in which two opposite sides are both equal and parallel to each other is a parallelogram GIVEN Straight line segments AB II CD AB 2 CD Fig 38 T0 PROVE AC M ED AC 2 BD 310 PROOF and 31 PARALLELISM B A 9 Figure 38 Draw BD BC AC Then A ABC 5 A DCB by SAS because AB 2 DC Given 4 1 A 2 Alternating angles AB II CD PN 311 BC 2 CB AC 2 DB 4 4 A 3 AC BD Equal alternating angles PN 2334 QED PROPOSITION 318I34 If both pairs of opposite sides of a quadrilateral are parallel to one another then they as well as the opposite angles are equal to one another and the diameter bisects the area GIVEN Quadrilateral ACDB AB II CD AC BD TO PROVE AB 2 CD AC 2 BD 4 CAB A BDC A ABD A DCA AABCADCB ABDC 311 31 PARALLELISM A B I or 61 I C D Figure 39 PROOF A ABC 5 A DCB by ASA because A 5 A 6 Alternating angles AB II CD PN 311 BC 2 CB 4 7 A 8 Alternating angles AC DB PN 311 AB 2 CD AC 2 BD 43 44 l AABC ADCB E 39ABDC Also 41 42 CN 2 QED EXERCISES 31D 1 Both pairs of opposite sides of a quadrilateral are equal to each other Prove that the quadrilateral is a parallelogram 2 Both pairs of opposite angles of a quadrilateral are equal to each other Prove that the quadrilateral is a parallelogram Prove that the diagonals of a parallelogram bisect each other Prove that if the diagonals of a quadrilateral bisect each other then it is a parallelogram Prove that a parallelogram is a rectangle if and only if its diagonals are equal to each other Prove that a parallelogram is a rhombus if and only if its diagonals are perpendicular to each other SQMPW Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equals half its length 312 ll 12 31 PARALLELISM The midpoint of side AB of AABC is D and E is a point of AC such that DE H BC Prove that AE EC and DE BC2 Prove that the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram Prove that each of two medians of a triangle is divided by their intersections into two segments one of which is double the other Prove that the three medians of a triangle all pass through one point Point E in the interior of square ABCD is such that A ABE A BAE 150 Prove that A CDE is equilateral In AABC AB 2 AC and D E F are points on the interiors of sides BC AB AC respectively such that DE J AB and DF J AC Prove that the value of DE DF is independent of the location of D Prove that the three segments joining the midpoints of the three sides of a triangle divide it into four congruent triangles Prove that three parallel straight lines that cut off equal line segments on one straight lines also cut off equal line segments on every straight line that intersects them Hint Through the middle intersection point on one straight line draw a line parallel to the other straight line A trapezoid is a quadrilateral two of whose sides are parallel Prove that the line segment joining the midpoints of the noniparallel sides of a trapezoid is parallel to the other two sides and equals half the sum of their lengths Construct angles of the following magnitudes a 60 b 30 c 1200 d 75 Through a given point construct a straight line such that its portion between two given parallel straight lines is equal to a given line segment Let A be a point in the interior of an angle Construct a straight line whose segment between the sides of the angle has A as its midpoint A pair of parallel straight lines is intersected by another pair of parallel straight lines Through a given point constr39uct another straight line on which the two given pairs cut off equal line segments Given an angle determine the locus of all the points the sums of whose distances from the sides of the angle equals a given magnitude In a given AABC constr39uct points M on AB and N on BC such that BM NC MN and VIN H BC Construct AABC given the data a 1 ha I3 b 1 ha hb c a hb a d hb he a e hb me a f a he b c g abc y h abc ha 313 31 PARALLELISM Construct a parallelogram given a two adjacent sides and the included angle b two adjacent sides and a diagonal c two adjacent sides and the distance between two opposite sides d a side and the two diagonals e the diagonals and the angles between them Construct a rectangle given one side and the diagonal Construct a rhombus given a its side and one of its angles b its side and one diagonal c both diagonals Construct a square given a its side b its diagonals Construct AABC given the data a womb b ahamb Comment on Proposition 317 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 318 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 2 Area Euclid defined the concept of area by means of axioms that he called Common Notions This axiomatic approach is customary today as well although the specific axioms are different from those used by Euclid The modern approach to area stipulates that a certain unit of length called unit has been chosen The square the length of whose side is unit 314 32 AREA Figure 310 1 unit Fig 310 is denoted by unit square or unit2 and serves as the unit for measuring areas It is then assumed that area is a measurement of figures that satisfies the three properties or axioms listed below UNIT The unit square has area 1 unit2 ADDITIVITY If a figure is divided by a line into two subfigures then the area of the figure equals the sum of the areas of the subfigures INVARIAN CE Congruent figures have equal areas Loosely speaking the additivity and invariance axioms were stated by Euclid as Common Notions 2 and 4 respectively The unit axiom however has no analog in Euclid39s system As a consequence Euclid39s Elements contains no proposition that computes areas explicitly Instead Euclid made comparative statements such as parallelograms on equal bases and between the same parallels are equal and if a parallelogram have the same base with a triangle and be in the same parallels the parallelogram is double the triangle This has the theoretical advantage of dispensing with units and the practical disadvantage of not answering the reasonable question of what is the area of a rectangle of dimensions 315 32 AREA 3 and 5 stadia Greek mathematicians did of course make use of units and could resolve such questions with ease It is just that Euclid for reasons that can only be guessed at and that in the author s opinion were probably esthetic decided to develop his geometry without any units whatsoever Propositions 321 and 322 below are the explicit modern day analogs of Euclid39s 135 PN 323 They give explicit formulas for the areas of rectangles and parallelograms Their complete proofs unfortunately contain elements that are beyond the scope of this text Specifically one runs into the difficulty inherent in proving propositions regarding line segments with irrational non fractional lengths These difficulties were first encountered by the Greeks in the sixth century BC and eventually surmounted by Eudoxus two hundred years later Euclid39s book did incorporate Eudoxus39s treatment of irrational numbers but it would be impractical to expound this theory here Instead a mere supporting argument for the fact that the area of a rectangle is given by the product of the lengths of its sides is offered It is customary in today39s high school geometry textbooks to circumvent these difficulties by stating this formula as yet another axiom the Rectangle Axiom In the author s opinion this is a misguided solution to a pedagogical problem since it opens up the possibility of stating many other interesting and non trivial geometrical facts as axioms even when elementary and convincing albeit logically incomplete arguments are available PROPOSITION 321 Ifa rectangle has dimensions a units and 17 units then it has area ab unitZ GIVEN ABCD with sides a 17 TO PROVE Area of ABCD 2 ab unit2 SUPPORTING ARGUMENT If a and b are positive integers then a rectangle of dimensions a and b can be divided into ab unit squares by means of straight lines that 316 32 AREA are parallel to its sides Fig 311 Consequently by the Additivity Property the given 2 rectangle has area ab un1t 1 13 Areaunitlunit2 D1 Area ofrectangle 15 unit2 Figure 311 Similarly if a rectangle has dimensions a 1m and b 111 for some positive integers m and n then the unit square can be divided into mn copies of the given rectangle all of which by the Invariance Property have the same area Fig 312 Hence the given rectangle has area 2 1 unit i l 2 2 m m n unit 2 db un1t Area of rectangle 112 unit2 lunit2 Figure 312 317 32 AREA Next if a rectangle has dimensions a mn and b pq where m n p q are all positive integers then it can be decomposed into mp rectangles each of which has l l d1mens1ons n and F1g 313 Since each of these latter rectangles is now known to have area ml unit2 it follows from the Additivity Property that the given rectangle has area L 2 2 2 quot1179161 11nlt un1t ab un1t 14 El Area 124 unitz Area of rectangle 3524 Figure 313 This verifies the proposition for all rectangles with fractional dimensions As was mentioned above the extension of this formula to rectangles with arbitrary real dimensions lies beyond the scope of this text QED An altitude of a parallelogram is any line segment cut off by two opposite sides from a straight line that is perpendicular to both of them It follows from Proposition 312 that all the altitudes joining the same pair of opposite sides of a parallelogram have equal length PROPOSITION 322 The area of a parallelogram with base 17 units and altitude h units is bh unitZ 318 32 AREA GIVEN ABCD with base 17 and altitude h Fig 314 TO PROVE Area of 39ABCD bh unit2 Figure 314 PROOF In the given parallelogram draw AE 39 CD and BF 39CD with E F on CD Thus ABFE is a rectangle with area bh unit2 Since A ADE s A BCF it follows that they have the same area and hence by the Additivity Property 39ABCD ABFE bh unit2 QED Euclid39s version of Propositions 321 and 322 is now stated together with his proof as well as another proof that is more consistent with modern pedagogy PROPOSITION 323I35 Parallelograms which are on the same base and in the same parallels are equal to one another GIVEN ABCD and 39 EBCF such that A D E F are collinear Fig 315 TO PROVE 39ABCD 2 EBCF F 319 32 AREA Figure 315 PROOF Euclid Since AD 2 BC 2 EF PN 34 it follows from CN 2 that AE DF Then ABAE s A CDF by SAS because AE DF See above A 1 A 2 Corresp angles AB DC PN 311 AB 2 DC Parallelogram ABCD PN 318 A EAB A FDC ABGD EGCF Subtract A DGE CN 3 39ABCD 2 EBCF Add A GBC CN 2 QED Euclid39s proof of Proposition 323 is incomplete albeit easily fixed because it depends on the relative position of the points A D E F on their common line Exercise 12 PROOF modern B J C Figure 316 Draw HJ perpendicular to AD and BC Fig 316 It then follows from PN 322 that NABCD BCHJ NEBCF QED 320 32 AREA PROPOSITION 324036 Parallelograms which are on equal bases and in the same parallels are equal to one another See Exercise 1 The area of the triangle will be given the same dual treatment as that of the rectangle First the modern formula is offered PROPOSITION 325 The area of a triangle with base 17 units and altitude h unitsis bhZ unitZ A b C Figure 317 PROOF Through the vertices B and C of AABC draw straight lines parallel to AC and AB respectively and let their intersection be D Fig 317 It is clear that ACDB is a parallelogram and hence by Proposition 318 1 bk 2 AABC 5 ACDB 7mm QED Next comes Euclid39s version 321 32 AREA PROPOSITION 326037 Triangles which are on the same base and in the same parallels are equal to one another GIVEN AABC A DBC AD II BC Fig 318 TO PROVE A ABC 2 A DBC C Figure 318 PROOF Let E be the intersection of AD with the straight line through B parallel to AC and let F be the intersection of TD with the straight line through C parallel to BD PN 314 Then AEBC 39DBCF PN 322 A ABC 2 39AEBC PN 318 A DEC 2 DBCF PN 318 A ABC 2 A DBC QED PROPOSITION 327038 Triangles which are on equal bases and in the same parallels are equal to one another See Exercise 2 PROPOSITION 328039 Equal triangles which are on the same base and on the same side are also in the same parallels 322 32 AREA GIVEN A ABC 2 A DBC A and D are on the same side of BC TO PROVE AD BC Figure 3 19 PROOF By contradiction Suppose AD and BC are not parallel and let E be the intersection of BD with the straight line through A parallel to BC Then A ABC 2 A EBC PN 326 A EBC lt A DBC CN 5 AABC lt A DBC This however contradicts the give equality of the two triangles Hence AD BC QED PROPOSITION 329I40 Equal triangles which are on equal bases and on the same side are also in the same parallels According to Heath Proposition 329 is an interpolation into The Elements by a later georneter Exercise 2 PROPOSITION 3210I41 If a parallelogram have the same base with a triangle and be in the same parallels the parallelogram is double of the triangle See Exercise 2 323 32 AREA There is no analog of Proposition 325 for the area of a general quadrilateral In practice any such quadrilateral can be divided into triangles by means of a diagonal and then the area of each of the parts can be evaluated by means of Proposition 325 A similar procedure can be used to dissect any polygon regardless of the number of its sides into triangles Neither Euclid39s nor the modern approach to areas are applicable to spherical geometry Both of these approaches rely heavily on the notion of parallelism and the sphere has no parallel geodesics Thus another approach is required in order to develop a theory of spherical areas As spherical polygons can also be dissected into spherical triangles it suffices to provide a formula for the latter It is clear that any two lunes of the same angle a on the same sphere can be made congruent by a series of rotations of that sphere Consequently every two such lunes have the same area This in turn implies that the area of a lune is proportional to its angle Since the lune of angle 2n radians has area 471R2 the sphere s total surface area the following lemma is obtained LEMMA 3211 On a sphere of radius R the area of a lune of angle a radians is 2 2 20cR unit The following theorem was first discovered by the Flemish mathematician Albert Girard 1595 1632 The proof presented here is due to Euler PROPOSITION 3212 On a sphere of radius R the area of the spherical triangle ABC with angles of radian measures a it y is a it y nR2 unit2 GIVEN Spherical A ABC with interior angles a y measured in radians TO PROVE A ABC a f y nR2 unitZ 324 32 AREA Figure 320 PROOF Let A B C be the respective antipodes of A B C Fig 320 Draw the great circles that contain the geodesic segments AB BC and CA The hemisphere in front of the great circle BCB C is thereby divided into four spherical triangles ABC AB C AB C ABC whose areas are denoted respectively by T1 T2 T3 T4 From the construction it follows that the spherical A ABC is congruent to the A AB C of area T2 Hence T1 T2 lunea Similarly T1 T3 lune and 325 32 AREA T1 T4 lune y Consequently 2T1 lunealune luney T1T2T3T4 2a 23 2 2nR2 unit2 and the statement of the theorem now follows immediately QED EXERCISES 32A 1 Prove Proposition 324 2 Use Proposition 325 to prove a Proposition 326 b Proposition 327 0 Proposition 328 d Proposition 329 e Proposition 3210 3 One of the triangle39s sides is divided into 11 equal segments and the division points are joined to the opposite vertex Prove that the triangle is divided into 11 equal parts 4 Prove that the area of the trapezoid equals the product of half the sum of its parallel sides with the distance between them 5 Prove that the diagonals of a parallelogram divide it into four equal triangles 6 Prove that the line segmentjoining the midpoints of two sides of a triangle cuts off a triangle that is equal to one fourth of the original triangle 7 Prove that the parallelogram formed by the midpoints of the sides of a quadrilateral equals one half of that quadrilateral 8 Prove that the triangle s medians divide it into siX equal triangles 9 The diagonals of a quadrilateral divide it into four equal triangles Prove that the quadrilateral is a parallelogram 326 32 AREA 10 Prove that if the point P lies in the interior of ABCD then the parallelogram equals twice the sum of AABP and A CDP 11 Each of the sides AB BC CA of an equilateral triangle is extended by their common length to points D E F respectively all in the same sense Prove that A DEF 7A ABC 12 Complete Euclid39s proof of Proposition 323 Both the taxicab and maxi areas of a gure are de ned to equal its Euclidean area 13 Comment on Propositions 321 322 and 325 in the context of taxicab geometry 14 Comment on Proposition 323 324 and 32610 in the context of taxicab geometry 15 Comment on Propositions 321 322 and 325 in the context of maxi geometry 16 Comment on Proposition 323 324 and 32610 in the context of maxi geometry Euclid s Propositions 142 45 are of limited interest They are included here only in order to facilitate the later discussion of the Golden Ratio Proposition 341 PROPOSITION 3213I42 To construct in a given rectilineal angle a parallelogram equal to a given triangle See Exercise 1 The above proposition is an example of a conversion which consists of the construction of a polygon IT of some prespecified nature that is equal to a given polygon H PROPOSITION 3214I43 In any parallelogram the complements of the parallelograms about the diagonal are equal to one another GIVEN 39ABCD K is a point on the diagonal AC 39BGKE 39 KFDH Fig 321 TO PROVE 39BGKE KFDH 327 32 AREA A H D E V B G C Figure 321 PROOF See Exercise 2 PROPOSITION 3215I44 To a given straight line to apply in a given rectilineal angle a parallelogram equal to a given triangle See Exercise 3 PROPOSITION 3216I45 To construct in a given rectilineal angle a parallelogram equal to a given rectilineal figure See Exercise 4 EXERCISES 32B H Prove Proposition 3213 Prove Proposition 3214 Prove Proposition 3215 Prove Proposition 3216 Convert a given parallelogram into a rectangle with the same base Convert a given parallelogram into a rhombus with the same base Convert a given parallelogram into another parallelogram with the same base and a given angle Convert a given parallelogram into a triangle with the same base and a given angle WPO QMPWN Convert a given triangle into a right triangle with the same base H 0 Convert a given triangle into an isosceles triangle with the same base H H Convert a given triangle into another Uiangle with the same base and a given angle 328 32 AREA 12 Bisect the area of a parallelogram by means of a straight line that is parallel to a given straight line 13 Given AABC construct apoint O in its interior such that the triangles AOB BOC COA all have equal areas 3 The Theorem of Pythagoras The Theorem of Pythagoras was discovered independently by several cultures and has been given more different proofs than any other theorem It is considered by many mathematicians to be the most important of all theorems and has the dubious distinction of being misquoted in the classic movie The Wizard of Oz and of being the subject of popular jokes It will be presented following an easy lemma PROPOSITION 331I46 On a given straight line to describe a square GIVEN Line segment AB Fig 322 TO CONSTRUCT ABCD E C D J2 4K 1 3 f A E Figure 322 CONSTRUCTION Draw EA JAB PN 2311 and let D on AE be such that AD 2 AB Let C be the intersection of the straight lines through B and D that are parallel 329 33 THE THEOREM OF PYTHAGORAS to AD and AB respectively PN 314 Then quadrilateral ABCD is the required square PROOF By construction ABCD is a parallelogram Since AB 2 AD it follows that AB 2 AD 2 DC 2 CB It remains to show that all of the angles of ABCD are right angles However 41 42 2 right angles PN311 A 2 right angle A 1 is a right angle A 3 A 4 right angle PN 318 QED PROPOSITION 332I47 The Theorem of Pythagoras In rightangled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle GIVEN AABC A BAC right angle quotABFG quotACKH quot BCED Fig 323 TO PROVE quotBCED quot39ABFG quotACKH H G K A F B J M C D L E Figure 323 330 33 THE THEOREM OF PYTHAGORAS PROOF Let L M be the respective intersections of the straight lines DE and BC with the straight line through A parallel to BD and CE PN 314 Note that the points G A C are collinear A GAB LBAC 2 right angles PN 2318 the points B A H are collinear A HAC LBAC 2 right angles PN 2318 A ABD s A FBC by SAS because BD 2 BC Sides of the same square 4 ABD A FBC Both equal 4 ABC right angle AB 2 FB Sides of the same square A ABD A F BC 39 BDLM quot ABF G Doubles of equal triangles PN 3210 A similar argument yields the equation 39 CELM quotACKH and hence quotBCED 7 BDLM quotCELM quotABFG TACKH Two other proofs of this theorem are now sketched out If a and b are the legs and c is the hypotenuse of a right triangle then the square of side a b can be dissected in the two ways depicted in Figure 324 The dissection of I calls for no explication That of 11 requires a proof that the interior quadrilateral labeled as 02 is indeed a square see Exercise 8 However once these dissections are granted it is clear from Figure 324 that a2 b2 02 This proof is attributed to the Chinese mathematician Chou pei Suan ching who lived circa 250 BC 331 33 THE THEOREM OF PYTHAGORAS Figure 324 The next proof is due to the Indian mathematician Bhaskara 1114 1185 The square of side 0 can be dissected in the manner depicted in Figure 325 It then follows that ab 0 47ab2 2aba22abb2 a2b2 ab2 Figure 325 Yet another proof of the Theorem of Pythagoras in indicated in Exercise 17 This one is due to president James Garfield 183 1 1881 The Theorem of Pythagoras has a converse 332 33 THE THEOREM OF PYTHAGORAS PROPOSITION 333I48 If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle the angle contained by the remaining two sides is right See Exercise 9 2 2 2 Since 3 4 5 it follows that any tr1angle whose s1des have lengths 3 units 4 units and 5 units is necessarily a right triangle So is the triangle whose sides have lengths 5 12 13 Triples ofintegers a b c such that are known as Pythagorean triples but the interest in such triples precedes Pythagoras by over a thousand year The Babylonian tablet PLIMPTON 322 dated between 1900 and 1600 BC contains fifteen Pythagorean triples the largest of which consists of 12709 13500 and 18541 Although it is highly unlikely that the Babylonians found these numbers by trial and error it is not known what method they used to generate these triples Not surprisingly the earliest method for generating Pythagorean triples appears in Euclid39s The Elements Lemma 1 to Proposition 29 of Book X states that if m gt n are any positive integers then 2 2 2 2 2 2 2 2mn mn m n 2 2 2 2 Exercise 13 so that 2mn m n m n form a Pythagorean tr1ple For example m 5 and n 4 yield the triple 2542 52 422 52422 or 333 33 THE THEOREM OF PYTHAGORAS 2 2 2 40 9 41 Pierre Fermat 1601 1665 took it for granted that Euclid s method can be used to generate all the Pythagorean triples and this fact was proven by Euler a hundred years later Specifically Euler proved that if a b c are numbers whose only common divisor is 1 and which constitute a Pythagorean triple then there exists a pair of relatively prime integers m n such that a b c 2 2mm m2 n2 m2 n2 All other Pythagorean triple of course proportional to these EXERCISES 33A 1 Which of the following triples of numbers are the lengths of the sides of a right triangle a 71015 b 51213 c 203750 364056 417194 d 57302 491714 650463 a2 3 2 Show that an equilateral triangle of side a has area 4 3 An isosceles right triangle has a hypotenuse of length 0 Compute its other sides and its area 4 A right triangle has an angle of 300 and a hypotenuse of length 1 Compute its other sides and its area 5 Compute the area of a rhombus whose sides equal 13 and one of whose diagonals has length 10 6 Compute the area of a parallelogram whose sides have lengths 11 and 8 and one of whose angles is 450 7 The diagonals and one side of a parallelogram have lengths 30 16 17 respectively Prove that it is a rhombus and compute its area 8 Show that the interior quadrilateral in Dissection II of Figure 324 is indeed a square of area 02 9 Prove Proposition 333 10 Find the error in the following quotproofquot of the quotpropositionquot that every triangle is isosceles GIVEN AABC 334 33 THE THEOREM OF PYTHAGORAS TO PROVE AB AC M Figure 326 PROOF Let N be the intersection of the bisector of 4 ABC and the perpendicular bisector of side BC M the midpoint of BC and ND39AB NE AC Then ND NE PN 2333 AD 2 AE Pythagoras Also EN CN PN 2314 BD 2 CE Pythagoras AB AC CN 2 QED Given a square of side a construct a square of double its area Given a square of side a and a positive integer n construct a square Whose area equals 11 times that of the given square Construct a square Whose area equals the sum of three given squares Use algebra to prove that Euclid39s method does indeed generate Pythagorean triples Assume that a line segment of length 1 inch is given Prove that line segments of the following lengths can be constructed a 5 inch b 13 inch c d IE inch Where n is any positive integer e 5 inch WE inch Find two nonicongruent isosceles triangles Whose sides have integer lengths and Whose perimeters and areas are equal Prove the Theorem of Pythagoras by applying Exercise 32A4 to Figure 327 335 33 THE THEOREM OF PYTHAGORAS b a Figure 327 18C Perform the construction of Proposition 331 using a computer application 19C Use a computer application to verify the Theorem of Pythagoras Both spherical and hyperbolic geometry have their own versions of the Theorem of Pythagoras Their proofs follow directly from the appropriate trigonometries Exercises 1 3 PROPOSITION 334The spherical Theorem of Pythagoras If the spherical A ABC has a right angle at C then cos c cos a cos b PROPOSITION 335The hyperbolic Theorem of Pythagoras If the hyperbolic A ABC has a right angle at C then cosh c cosh a cosh b EXERCISES 33D l Derive the spherical Theorem of Pythagoras from Proposition 112 336 33 THE THEOREM OF PYTHAGORAS 2 Find the length of the hypotenuses of the three spherical isosceles right Triangles Whose legs have lengths l l 01 respectively Compare the answers to the lengths of the hypotenuses of the three Euclidean isosceles right triangles both of Whose legs have lengths l l 01 respectively 3 Derive the hyperbolic Theorem of Pythagoras frorn Propositions 122 4 Find the length of the hypotenuses of the three hyperbolic right triangles Whose legs have lengths l l 01 respectively Compare the answers to the lengths of the hypotenuses of the three Euclidean triangles both of Whose legs have lengths l l 01 respectively 5 Is there a taxicab version of the Theorem of Pythagoras 6 Is there a maxi version of the Theorem of Pythagoras 4 Consequences of the Theorem of Pythagoras optional Book II of Euclid39s Elements contains a variety of consequences of the Theorem of Pythagoras of which only a sample are presented here The first of these is tantamount to a construction of the Golden Ratio This proposition will be used later in the construction of the regular pentagon PROPOSITION 341II6 To cut a given line segment so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment GIVEN Line segment AB Fig 328 TO CONSTRUCT A point C on AB such that AB BC 2 AC2 D C Figure 328 337 34 CONSEQUENCES OF THE THEOREM OF PYTHAGORAS 1 CONSTRUCTION At B construct BD AB and BD 5 AB Join AD let E be the point on AD such that DE 2 DB and let C be the point on AB such that AC 2 AE Then C is the required point PROOF Set AB 2a Then DE 2 BD 2 a and AC AE AD DE 2a2a2 a 03 1a sothat 2 2 2 2 2 AC 2 IE 1a 5 2 1a 6 23a and ABBC ABABAC ABAB AE ABAB AD DE 2a2a IEa an 23 13a2 AC2 QED In the context of the above proposition the common value 7 of the ratios 2 0618 E E ADDE 3 1a AC AB AB 2a 2 is called the Golden Ratio While of demonstrated mathematical interest this quantity has also been the subject of much nonsensical speculation Typical of this latter variety is an article that reports that the average ratio of the height of a man39s navel off the ground to his height equals the Golden ratio The following two propositions constitute Euclid39s analog of the modern day Law of Cosines Their proofs are relegated to the exercises PROPOSITION 342II12 In obtuseangled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse 338

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