STRUCTURE OF MODERN GEOMETRY
STRUCTURE OF MODERN GEOMETRY M 333L
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Chapter 2 EUCLIDEAN PARALLEL POSTULATE 21 INTRODUCTION There is a welldeveloped theory for a geometry based solely on the five Common Notions and first four Postulates of Euclid In other words there is a geometry in which neither the Fifth Postulate nor any of its altematives is taken as an axiom This geometry is calledAbsolute Geometry and an account of it can be found in several textbooks in Coxeter s book Introduction to Geometry for instance or in many college textbooks where the focus is on developing geometry within an axiomatic system Because nothing is assumed about the existence or multiplicity of parallel lines however Absolute Geometry is not very interesting or rich A geometry becomes a lot more interesting when some Parallel Postulate is added as an axiom In this chapter we shall add the Euclidean Parallel Postulate to the five Common Notions and first four Postulates of Euclid and so build on the geometry of the Euclidean plane taught in high school It is more instructive to begin with an axiom different from the Fi h Postulate 211 Playfair s Axiom Through a given point not on a given line exactly one line can be drawn parallel to the given line Playfair s Axiom is equivalent to the Fifth Postulate in the sense that it can be deduced from Euclid s five postulates and common notions while conversely the Fifth Postulate can deduced from Playfair s Axiom together with the common notions and first four postulates 212 Theorem Euclid s five Postulates and common notions imply Playfair s Axiom Proof First it has to be shown that if P is a given point not on a given line I then there is at least one line throughP that is parallel to Z By Euclid s Proposition I 12 it is possible to draw a line tthrough P perpendicular to Z In the figure below let D be the intersection of l with t E By Euclid s Proposition I 11 we can construct a line m throughP perpendicular to t Thus by construction I is a transversal to l and m such that the interior angles on the same side atP andD are both right angles Thus m is parallel to because the sum of the interior angles is 180 Note Although we used the Fifth Postulate in the last statement of this proof we could have used instead Euclid s Propositions I 27 and I 28 Since Euclid was able to prove the rst 28 propositions without using his Fifth Postulate it follows that the existence of at least one line through P that is parallel to I can be deduced from the rst four postulates For a complete list of Euclid s propositions see College Geometry by H Eves Appendix B To complete the proof of 2 l 2 we have to show that m is the only line through P that is parallel to I So let 7 be a line through P with m at n and let E at P be a point on n Since m at n AEPD cannot be a right angle If szPD lt 90 as shown in the drawing then szPD m APDA is less than 180 Hence by Euclid s fifth postulate the line 71 must intersect l on the same side of transversal tas E and so 71 is not parallel to Z If szPD gt 90 then a similar argument shows that n and must intersect on the side of l opposite E Thus m is the one and only line throughP that is parallel to l QED A proof that Playfair s axiom implies Euclid s fth postulate can be found in most geometry texts On page 219 of his College Geometry book Eves lists eight axioms other than Playfair s axiom each of which is logically equivalent to Euclid s fth Postulate ie to the Euclidean Parallel Postulate A geometry based on the Common Notions the rst four Postulates and the Euclidean Parallel Postulate will thus be called Euclidean plane geometry In the next chapter Hyperbolic plane geometry will be developed substituting Altemative B for the Euclidean Parallel Postulate see text following Axiom 122 22 SUM OF ANGLES One consequence of the Euclidean Parallel Postulate is the well known fact that the sum of the interior angles of a triangle in Euclidean geometry is constant whatever the shape of the triangle 221 Theorem In Euclidean geometry the sum of the interior angles of any triangle is always 180 Proof Let AABC be any triangle and construct the unique line I throughA parallel to the sidelY39 as shown in the gure Thenm AEAC mzACB and mADAB mzABC by the alternate angles property of parallel lines found in most geometry textbooks Thus mzACB mzABC mABAC 180 QED Equipped with Theorem 221 we can now try to determine the sum of the interior angles of gures in the Euclidean plane that are composed of a nite number of line segments not just three line segments as in the case of atriangle Recall that a gollggon is a gure in the Euclidean plane consisting of points P P Pquot called vertices together with line segments PliP2 I 1 2 called edges or sides More generally a gure consisting of the union of a nite number of nonoverlapping polygons will be said to be a Qiecewise linear figure Thus G are piecewise linear gures as is the example of nested polygons below This example is a particularly interesting one because we can think of it as a gure containing a hole But is it clear what is meant by the interior angles of such gures For such a polygon as the following we obViously mean the angles indicated But what about a piecewise linear gure containing holes For the example above of nested polygons we shall mean the angles indicated below This makes sense because we are really thinking of the two polygons as enclosing a region so that interior angle then refers to the angle lying between two adjacent sides and inside the enclosed region What this suggests is that for piecewise linear gures we will also need to specify what is meant by its interior The likely formula for the sum of the interior angles of piecewise linear gures can be obtained from Theorem 221 in conjunction with Sketchpad In the case of polygons this was probably done in high school For instance the sum of the angles of any quadrilateral ie any foursided gure is 360 To see this draw any diagonal of the quadrilateral thereby diViding the quadrilateral into two triangles The sum of the angles of the quadrilateral is the sum of the angles of each of the two triangles and thus totals 360 If the polygon has 7 sides then it can be diVided into n2 triangles and the sum of the angles of the polygon is equal to the sum of the angles of the 712 triangles This proves the following result 222 Theorem The sum ofthe interior angles ofan nsided polygon n 2 3 is n 2 180 222a Demonstration We can use a similar method to determine the sum of the angles of the more complicated piecewise linear gures One such gure is a polygon haVing holes that is a polygon haVing other nonoverlapping polygons the holes contained totally within its interior Open a new sketch and draw a gure such as An interesting computer graphics problem is to color in the piecewise linear gure between the two polygons Unfortunately computer graphics programs will only ll polygons and the interior of the gure is not a polygon Furthermore Sketchpad measures angles greater than 1800 by using directed measurements Thus Sketchpad would give a measurement of 900 for a 2700 angle To overcome the problem we use the same strategy as in the case of a polygon join enough of the vertices of the outer polygon to vertices on the inner polygon so that the region is sub divided into polygons Continue joining vertices until all of the polygons are triangles as in the gure below Color each of these triangles in a different color so that you can distinguish them easily We call this a triangular tiling of the gure Now use Theorem 222 to compute the total sum of the angles of all these new polygons Construct a different triangular tiling of the same gure and compute the total sum of angles again Do you get the same value Hence complete the following result 223 Theorem When an nsided piecewise linear gure consists of a polygon with one polygonal hole inside it then the sum of its interior angles is Note Here n equals the number of sides of the outer polygon plus the number of sides of the polygonal hole End of Demonstration 222a Try to prove Theorem 223 algebraically using Theorem 222 The case of a polygon containing h polygonal holes is discussed in Exercise 251 23 SIMILARITY AND THE PYTHAGOREAN THEOREM Of the many important applications of similarity there are two that we shall need on many occasions in the future The rst is perhaps the best known of all results in Euclidean plane geometry namely Pythagoras theorem This is frequently stated in purely algebraic terms as a2 b2 cl whereas in more geometrically descriptive terms it can be interpreted as saying that in area the square built upon the hypotenuse of a right angled triangle is equal to the sum of the squares built upon the other two sides There are many proofs of Pythagoras theorem some synthetic some algebraic and some visual as well as many combinations of these Here you will discover an algebraicsynthetic proof based on the notion of similarity Applications of Pythagoras theorem and of its isosceles triangle version to decorative tilings of the plane will be made later in this chapter 234 Theorem The Pythagorean Theorem In any triangle containing a right angle the square of the length of the side opposite to the right angle is equal to the sum of the squares of the lengths of the sides containing the right angle In other words if the length of the hypotenuse is c and the lengths of the other two sides area andb then a2 b2 0 Proof Let AABC be a rightangled triangle with right angle at C and let CT be the perpendicular from C to the hypotenuse A B as shown in the diagram below A B o ShowACAB is similar to ADAC o ShowACAB is similar to ADCB 0 Now let H have length x so that E has length 0 x By similar triangles x a c x and a c b 0 Now eliminate x from the two equations to show a2 b2 0 There is an important converse to the Pythagorean theorem that is often used 235 Theorem Pythagorean Converse Let AABC be a triangle such that a2 b2 0 Then AABC is rightangled with AACB a right angle 235a Demonstration Pythagorean Theorem with Areas You may be familiar with the geometric interpretation of Pythagoras theorem Ifwe build squares on each side of AABC then Pythagoras theorem relates the area of the squares 0 Open a new sketch and draw a rightangled triangle AABC Using the Square By Edge tool construct an outward square on each edge of the triangle haVing the same edge length as the side of the triangle on which it is drawn 0 Measure the areas of these 3 squares to do this select the vertices of a square and then construct its interior using Construct Polygon Interior tool Now compute the area of each of these squares and then use the calculator to check that Pythagoras theorem is valid for the rightangled triangle you have drawn End of Demonstration 235a This suggests a problem for further study because the squares on the three sides can be thought of as similar copies of the same piecewise linear gure with the lengths of the sides determining the edge length of each copy So what does Pythagoras theorem become when the squares on each side are replaced by say equilateral triangles or regular pentagons In order to investigate we will need tools to construct other regular polygons given one edge If you haven t already done so move the document called Polygons gsp into the Tool Folder and restart Sketchpad or simply open the document to make its tools available 235b Demonstration Generalization of Pythagorean Theorem 0 Draw a new rightangled triangle AABC and use the SPentagon By Edge script to construct an outward regular pentagon on each side having the same edge length as the side of the triangle on which it is drawn As before measure the area of each pentagon What do you notice about these areas Repeat these constructions for an octagon instead of a pentagon Note You can create an Octagon By Edge script from your construction for Exercise 135b What do you notice about the areas in this case Now complete the statement of Theorem 236 below for regular n gons End of Demonstration 235b 236 Theorem Generalization of Pythagoras theorem When similar copies of a regular ngon n 2 3 are constructed on the sides of a rightangled triangle each ngon having the same edge length as the side of the triangle on which it sits then The gure below illustrates the case of regular pentagons 237 Demonstration Reformulate the result corresponding to Theorem 236 when the regular ngons constructed on each side of a rightangled triangle are replaced by similar triangles This demonstration presents an opportunity to explain another feature of Custom Tools called AutoMatching We will be using this feature in Chapter 3 when we use Sketchpad to explore the Poincare Disk model of the hyperbolic plane In this problem we can construct the first isosceles triangle and then we would like to construct two other similar copies of the original one Here we will construct a similar triangle script based on the AA criteria for similarity Tool Composition using Auto Matching Open a new sketch and construct AABC with the vertices labeled Next construct the line not a segment DE Select the vertices BA C in order and choose quotMark Angle BACquot from the Transform Menu Click the mouse to deselect those points and then select the point D Choose Mark Center D from the Transform Menu Deselect the point and then select the lineDE Choose Rotate from the Transform Menu and then rotate by Angle BA C Select the vertices ABC in order and choose Mark Angle ABC from the Transform Menu Click the mouse to deselect those points and then select the point E Choose Mark Center E from the Transform Menu Deselect the point and then select the line DE Choose Rotate from the Transform Menu and rotate by Angle AB C Construct the point of intersection between the two rotated lines and label it F ADEF is similar to AABC Hide the three lines connecting the points D E and F and replace them with line segments Now from the Custom Tools menu choose Create New Tool and in the dialogue box name your tool and check Show Script View In the Script View double click on the Given Point A and a dialogue box will appear Check the box labeled Automatically Match Sketch Object Repeat the process for points B and C In the future to use your tool you need to have three points labeledA B and C already constructed in your sketch where you want to construct the similar triangle Then you only 10 need to click on or construct the points corresponding to D and E each time you want to use the script Your script will automatically match the points labeled A B and C in your sketch with those that it needs to run the script Notice in the Script View that the objects which are automatically matched are now listed under Assuming rather than under Given Objects If there are no objects in the sketch with labels that match those in the Assuming section then Sketchpad will require you to match those objects manually as if they were Given Objects 0 Now open a new sketch and construct atriangle with vertices labeled A B and C o In the same sketch construct a right triangle Use the similar triangle tool to build triangles similar to AABC on each side of the right triangle For each similar triangle select the three vertices and then in the Construct menu choose construct polygon interior Measure the areas of the similar triangles and see how they are related End of Demonstration 237 24 INSCRIBED ANGLE THEOREM One of the most useful properties of a circle is related to an angle that is inscribed in the circle and the corresponding subtended arc In the gure below LABC is inscribed in the circle andArc ADC is the subtended arc We will say that zAOC is a central angle of the circle because the vertex is located at the center 0 The measure of Arc ADC is de ned to be the angle measure of the central angle zAOC 240 Demonstration Investigate the relationship between an angle inscribed in a circle and the arc it intercepts subtends on the circle 0 Open a new script in Sketchpad and draw a circle labeling the center of the circle by O mZBCA 53 szDA 53 D mZBOA 106 0 Select an arbitrary pair of points AB on the circle These points specify two possible arcs let s choose the shorter one in the gure above that is the arc which is subtended by a central angle of measure less than 180 Now select another pair of points CD on the circle and draw line segments to form ABCl and ABDA Measure these angles What do you observe o If you drag C or D what do you observe about the angle measures Now nd the angle measure of 430A What do you observe about its value 0 Drag B until the line segment ZB passes through the center of the circle What do you now observe about the three angle measures you have found Use your observations to complete the following statements proving them will be part of later exercises 241 Theorem Inscribed Angle Theorem The measure of an inscribed angle of a circle equals that of its intercepted or subtended arc 242 Corollary A diameter of a circle always inscribes at any point on the circumference of the circle 243 Corollary Given a line segment 1E3 the locus of a pointP such that LAPB 90 is a circle having 1 as diameter End of Demonstration 240 The result you have discovered in Corollary 242 is a very useful one especially in constructions since it gives another way of constructing rightangled triangles Exercises 254 and 255 below are good illustrations of this The Inscribed Angle Theorem can also be used to prove the following theorem which is useful for proving more advanced theorems 244 Theorem A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary A quadrilateral that is inscribed in a circle is called a cyclic quadrilateral 25 Exercises Exercise 251 Consider a piecewise linear figure consisting of a polygon containing h holes nonoverlapping polygons in the interior of the outer polygon has a total of n edges where 71 includes both the interior and the exterior edges Express the sum of the interior angles as a function of n and h Prove your result is true Exercise 252 Prove that if a quadrilateral is cyclic then the opposite angles of the quadrilateral are supplementary ie the sum of opposite angles is 180 This will provide half ofthe proof of Theorem 244 Exercise 253 Give a synthetic proof of the Inscribed Angle Theorem 241 using the properties of isosceles triangles in Theorem 146 Hint there are three cases to consider here w is the angle subtended by the arc and 8 is the angle subtended at the center of the circle The problem is to relate w to 9 Case 1 The center of the circle lies on the subtended angle Case 2 The center of the circle lies within the interior of the inscribed circle 13 Case 3 The center of the circle lies in the exterior of the inscribed angle 7a End of Exercise 253 For Exercises 254 255 and 256 recall that any line tangent to a circle at a particular point must be perpendicular to the line connecting the center and that same point For all three of these exercises the Inscribed Angle Theorem is useful Exercise 254 Use the Inscribed Angle Theorem to deVise a Sketchpad construction that will construct the tangents to a given circle from a given pointP outside the circle Carry out your construction Hint Remember Corollary 242 Exercise 255 In the following gure the line segmentsEZl and TB are the tangents to a circle centered at 0 from a pointP outside the circle Prove that TA and TB are congruent Exercise 256 Letl and m be lines intersecting at some pointP and let Q be a point on I Use the result of Exercise 255 to deVise a Sketchpad construction that constructs a circle tangential to l and m that passes through Q Carry out your construction For Exercises 257 and 258 we consider regular polygons again that is polygons with all sides congruent and all interior angles congruent If a regular polygon has 7 sides we shall say it is a regular ngon For instance the following gure is a regular octagon above ie a regular 8gon By Theorem 222 the interior angle of a regular ngon is 180 2 o 360 The measure of any central angle is In the gure ADEF is an interior angle and n LABC is a central angle Exercise 257 Prove that the vertices of a regular polygon always lie on a circumscribing circle Be care Jl Don t assume that your polygon has a center you must prove that there is a point equidistant from all the vertices of the regular polygon Exercise 258 Now suppose that the edge length of a regular ngon is l and letR be the radius of the circumscribing circle for the ngon TheAQothem of the ngon is the perpendicular distance from the center of the circumscribing circle to a side of the ngon The Apothem a With this notation and terminology and using some trigonometry complete the following Rl lR ApothemR Use this to deduce 2 l b area ofngon 7 HR2 sinki c perimeter ofngon 271R sin 2 n n d Then use the wellknown fact from calculus that lim 1 e gt o 9 to derive the formulas for the area of a circle of radius R as well as the circumference of such a circle Exercise 259 Use Exercise 258 together with the usual version of Pythagoras theorem to give an algebraic proof of the generalized Pythagorean Theorem Theorem 236 Exercise 2510 Prove the converse to the Pythagorean Theorem stated in Theorem 235 26 RESULTS REVISITED In this section we will see how the Inscribed Angle Theorem can be used to prove results involving the Simson Line the Miquel Point and the Euler Line Recall that we discovered the Simson Line in Section 18 while exploring Pedal Triangles 261 Theorem Simson Line If P lies on the circumcircle of AABC then the perpendiculars from P to the three sides of the triangle intersect the sides in three collinear points Proof Use the notation in the gure below 0 Why do PDA andE all lie on the same circle Why do P A C andB all lie on another circle Why do PDB andF all lie on a third circle Verify all three of these statements using Sketchpad o In circle PDAE mzPDE E mzPAE mzPAC Why 0 In circle PACB mzPAC E mzPBC mzPBF Why 0 In circle PDBF mzPBF E mzPDF Why Since mzPDE E mzPDF points DE andqust be collinear QED In Exercise 194 the Miquel Points of atriangle were constructed 262 Theorem Miquel Point If three points are chosen one on each side of atriangle then the three circles determined by a vertex and the two points on the adjacent sides meet at a point called the Miquel Point Proof Refer to the notation in the gure below LetDE andF be arbitrary points on the sides of AABC Construct the three circumcircles Suppose the circumcircles for AAFD and ABDE intersect at point G We need to show the third circumcircle also passes through G Now Gmay lie inside AABC on AABC or outside AABC We prove the theorem here in the case that G lies inside AABC and leave the other two cases for you see Exercise 281 0 AF GD and ADAF are supplementary Why 0 AEGD and ADBE are supplementary Why Notice sz GD szGE szGF 360 Combining these facts we see the following 180 mzA 180 sz szGF 360 So szGF 180 mzC or AC and AEGF are supplementary Thus C E G andF all lie on a circle and the third circumcircle must pass through G QED The proof of Theorem 263 below uses two results on the geometry of triangles which were proven in Chapter 1 The first result states that the line segment between the midpoints of two sides of a triangle is parallel to the third side of the triangle and it is half the length of the third side see Corollary 154 The second results states that the point which is 23 the distance from a vertex along a median to the midpoint of the opposite side is the centroid of the triangle see Theorem 156 263 Theorem Euler Line For any triangle the centroid the orthocenter and the circumcenter are collinear and the centriod trisects the segment joining the orthocenter and the circumcenter The line containing the centroid orthocenter and circumcenter of atriangle is called the Euler Line Proof In the diagram below A39 is the midpoint of the side opposite toA and O G andH are the circumcenter centroid and orthocenter respectively Since A G and A39are collinear we can show that O G andH are also collinear by showing that LAGH E AA39GO To do this it suf ces to show that AAH G N AA39OG If we also show that the ratio of similiarity is 2 1 then we will also prove that G trisects 0 H A 39A The proof that AAH G N AA39OG with ratio 21 proceeds as follows Let I be the point where the ray CO intersects the circumcircle of AABC Then why It follows that ABC N AA39CO with ratio 21 why It is also true that AIBH is a parallelogram why and hence AH 13 2OA39 Since G is the median we know thatAG 2GA Thus we have two corresponding sides proportional The included angles are congruent because they are 19 alternate interior angles formed by the parallel lines 743 and O7 and the transversal 717p Why are 217 and O7quot parallel Thus MHG N AA39OG with ratio 21 by SAS Of course as we noted in Chapter 1 we must be careful not to rely too much on a picture when proving a theorem Use Sketchpad to find examples of triangles for which our proof breaks down ie triangles in which we can t form the triangles AAH G and AA39OG What sorts of triangles arise You should nd two special cases Finish the proof of Theorem 263 by proving the result for each of these cases see Exercise 282 27 THE NINE POINT CIRCLE Another surprising triangle property is the socalled Nine Point Circle sometimes credited to KW Feuerbach 1822 Sketchpad is particularly well adapted to its study The following Demonstration will lead you to its discovery 270 Demonstration Investigate the nine points on the Nine Point Circle The First set of Three points 0 In a new sketch construct AABC Construct the midpoints of each of its sides label these midpoints D E and F 0 Construct the circle that passes through DE and F You know how to do this 0 This circle is called the Nine Point Circle Complete the statement The ninepoint circle passes through The Second set of Three points In general the ninepoint circle will intersect AABC in three more points If yours does not drag one of the vertices around until the circle does intersect AABC in three other points Label these points J K and L 0 Construct the line segment joining and the vertex opposite J Change the color of this segment to red What is the relationship between the red segment and the side of the triangle containing J What is an appropriate name for the red segment Construct the corresponding segment joining K and the vertex opposite K and the segment joining L to the vertex opposite L Color each segment red What can you say about the three red segments Place a point where the red segments meet label this pointM and complete the following statement The ninepoint circle also passes through 20 The Third set of Three points The red segments intersect the circle at their respective endpoints J K orL For each segment there exists a second point where the segment intersects the circle Label them N O and P 0 To describe these points measure the distance betweenM and each of AB and C Measure also the distance betweenM and each of N O andP What do you observe Con rm your observation by dragging the vertices of AABC 0 Complete the following statement The ninepoint circle also passes through You should create a Nine Point Circle tool from this sketch and save it for future use End of Demonstration 270 To understand the proof of Theorem 271 below it is helpful to recall some results discussed earlier As in the proof of the existence of the Euler Line it is necessary to use the fact that the segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle Also we recall that a quadrilateral can be inscribed in a circle if and only if the opposite angles in the quadrilateral are supplementary It is not difficult to show that an isosceles trapezoid has this property Finally recall that a triangle can be inscribed in a circle with a side of the triangle coinciding with a diameter of the circle if and only if the triangle is a right triangle 271 Theorem The Nine point Circle The midpoints of the sides of atriangle the points of intersection of the altitudes and the sides and the midpoints of the segments joining the orthocenter and the vertices all lie on a circle called the ninepoint circle Your final figure should be similar to 21 Proof 7 Figure 1 See Figure 1 In AABC label the midpoints of BC 54 and 1E3 byA 39 B39 and C39 respectively There is a circle containingA 39 B39 and C 39 In addition we know A 39C39AB39 is a parallelogram and so A 39C39 AB 39 See Figure 2 Construct the altitude fromA intersecting E atD As W is parallelto 37C and ID is perpendicular to 137639 then E must be perpendicular to Denote the intersection of 1 and E by P ThenzAPB E ADPB g E 1 and APEDP See Figure 3 Consequently AAPB EADPB by SAS So AB39 B39D By transitiVity withA39C39 AB39 we haveB39D A39C39 Thus A 39C39B39D is an isosceles u n u A u u A Figure 3 22 trapezoid Hence by the remarks preceding this theorem A39 C 39 B39 andD are points which lie on one circle See Figure 4 By a similar argument the feet of the other two altitudes belong to this circle See Figure 5 Let denote the midpoint of the segment joining vertexl and the orthocenter H Then again by the connection of midpoints of the sides of a triangle E1 is parallelto See Figure 6 Now W l lTC andTCE but llC J Hence C A39La See Figure 7 Therefore C 39 lies on a circle with diameter A39 J 23 Figure 4 Figure 7 A similar argument shows that 339 lies on the circle with diameter A39 J and hence J lies on the circle determined by A 39 B39 and C 39 Likewise the other two midpoints of the segments joining the vertices with the orthocenter lie on the same circle QED 28 Exercises In this exercise set Exercise 283 7 287 are related to the nine point circle Exercise 281 Using Sketchpad illustrate a case where the Miquel Point lies outside the triangle Prove Theorem 262 in this case Exercise 282 Prove Theorem 263 for the two special cases a The triangle is isosceles b The triangle is a right triangle Exercise 283 For special triangles some points of the ninepoint circle coincide Open a new sketch and draw an arbitraryAABC Explore the various possibilities by dragging the vertices of AABC Describe the type of triangle if it exists for which the nine points of the ninepoint circle reduce to 4 points 5 points 6 points 7 points 8 points Exercise 284 Open a new sketch and draw an arbitrary triangle AABC 0 Construct the circumcenter O the centroid G the orthocenter H and the center of the nine point circle N for this triangle What do you notice 0 Measure the length of 5v 1 1 and What results for a general triangle do your calculations suggest 0 Measure the radius of the ninepoint circle of AABC Measure the radius of the circumcircle of AABC What results for a general triangle do your calculations suggest Drag the vertices of the triangle around Do your conjectures still remain valid 24 Exercise 285 Open a new sketch and draw an arbitrary AABC LetH be the orthocenter and 0 be the circumcenter of AABC Construct the ninepoint circles for AOHA AOHB and AOH C Use sketchpad to show that these ninepoint circles have two points in common Can you identify these points Check your observation by dragging the vertices A B and C around If one starts with given vertices AB and C then the locations of the midpoints P Q and R of the sides of AABC are uniquely determined Similarly the locations of the feet of the altitudes DE andF will be determined once AB and C are given The remaining two problems in this exercise set use the geometric properties we have developed so far to reverse this process ie we construct the vertices AB and C knowing the midpoints or the feet of the altitudes Use the notation from the following figure Exercise 286 a Prove the line segment IQ is parallel to side b Given points P Q and R show how to construct points AB and C so that P Q andR are the midpoints of the sides of AABC c Formulate a conjecture concerning the relation between the centroid G of AABC and the centroid of APQR Exercise 287 25 a Assume AABC is acute to ensure the feet of the altitudes lie on the sides of the triangle Prove thatPC PB PE PF and thatP lies on the perpendicular bisector of the line segment 5 b Given points DE and F show how to construct points AB and C so thatDE andF are the feet of the altitudes from the vertices of AABC to the opposite sides Hint remember the ninepoint circle 29 THE POWER OF A POINT AND SYNTHESIZING APOLLONIUS Another application of similarity will be to a set of ideas involving what is o en called the power of a point with respect to a circle The principal result will be decidedly useful later in connection with the theory of inversion and its relation to hyperbolic geometry Demonstration 290 Discover the formula for the power of pointP with respect to a given circle 0 Open a new sketch and draw a circle Select any point P outside the circle and let A B be the points of intersection on the circle of a line through P Compute the lengths PA PB of T4 andP B respectively then compute the product PA PB of PA and PB Drag lwhile keepingP fixed What do you observe Investigate lrther by considering the case when l is tangential to the given circle Use this to explain your previous observation What happens to the product PA PB whenP is taken as a point on the circle Now letP be a point inside the circle I a line through P and A B its points of intersection with the circle Again compute the product PA PB of PA and PB Now vary 1 Investigate lrther by considering the case when 1 passes through the center of the given circle Can you reconcile the three values of the product PA PB for P outside on and inside the 2 where O is the center of the given given circle Hint consider the value of OP 2 7 r circle and r is its radius End of Demonstration 290 The value of PA PB in Demonstration 290 is often called the power of P with respect to the given circle Now complete the following statement 26 291 Theorem LetP be a given point 2 a given circle and l a line throughP intersecting 2 at A and B Then 1 the product PA PB of the distances from P to A and B is whenever P is outside whenever it is inside or when it is on E 2 the value of the product PAPB is equal to where O is the center of Z and r is the radius of Z The proof of part 2 of Theorem 291 in the case whenP is outside the given circle is an interesting use of similarity and the inscribed angle theorem In the diagram below let C be a point on the circle such that P C is a tangent to the circle By the Pythagorean Theorem OP2 r2 PC2 so it suffices to show that PAPB PC2 292 Theorem Given a circle 2 and a pointP outside 2 let I be a ray throughP intersecting 2 at pointsA and B If C is a point on the circle such that P C is a tangent to 2 at C then PAPB PC2 Proof The equation PAPB PC2suggests use of similar triangles but which ones Let C7 be a diameter of the circle By the Inscribed Angle Theorem mzPAC szDC and ACBD is a right angle Thus szDC szCB 90 and aslTC is tangent to the circle szCB mzPCB 90 Therefore mzPAC mzPCB By AA similarity APAC is similar to APCB proving PA 3C PCPB or PAPB PC2 QED 27 Theorem 293 Given a circle 2 and a point P inside 2 let I be a line through P intersecting 2 at pointsA and B Let C D be the chord perpendicular to the segment O P Then the value of the product PA PB is equal to r2 OP2 PC 2 where O is the center of Z and r is the radius of Z Proof By AA similarity AACP is similar to ADBP so thatPA 3C PDPB Thus PA PBPC PD By HL ACPO is congruent to ADPO so thatPCPD By the Pythagorean Theorem PD2 OP2 ODZ Rearranging and substituting we obtainPC PD r2 0P2 Therefore PA PB r2 OP2 as desired QED There is a converse to theorem 292 that also will be useful later You will be asked to provide the proof in Exercise 2 l 11 below 294 Theorem Given a circle 2 and a pointP outside 2 let I be a ray throughP intersecting 2 at pointsA and B If C is a point on 2 such thatPAPB PC2 then P C is atangent to 2 at C In Chapter 1 we used Sketchpad to discover that when a pointP moves so that the distance from P to two fixed points AB satisfies the condition PA 2PB then the path traced out by P is a circle In fact the locus of a pointP such that PA mPB is always a circle when m is any positive constant not equal to one From restorations of Apollonius work Plane Loci we infer that he considered this locus problem now called the Circle of Apollonius However 28 this is a misnomer since Aristotle who had used it to give a mathematical justi cation of the semicircular form of the rainbow had already known the result That this locus is a circle was con rmed algebraically using coordinate geometry in Chapter 1 However it can be also be proven by synthetic methods and the synthetic proof exploits properties of similar triangles and properties of circles Since the synthetic proof will suggest how we can construct the Circle oprollom39us with respect to xed points AB through an arbitrary point P we shall go through the proof now The proof requires several lemmas which we consider below 295 Lemma Given AABC letD be onA B andE on 1 such thatD E is parallel to 39 Then ADAE ABAC an DB EC DB EC Proof LetF be the intersection of KC with the line parallel tm passing through E Then AD AE AAED N AECF by AA similarity and E7 E The quadrilateral EFBD is a parallelogram AD AE AB AC therefore EFDB and i i A s1m11ar argument showsi i QED DB EC DB EC 295a Lemma Converse of Lemma 295 Given AABC letD be onZB andE on AiC AD AE AB such that i AC 7 7 i or i i see gure below thenDE is parallel toBC DB EC DB EC 29 AB AC Proof Assume i i The line through DB EC D parallel to BiC intersects E at pointF with 17 parallel toB C By Lemma 295 AB AC AB AC u alsoso DB FC DB EC ACAC which implies thatF E Thus FC EC D E IT is paralleltolY39 AD AE If the proof is similar QED DB EC 296 Theorem The bisector of the internal angle AABC of AABC divides the opposite side 1 in the ratio of the adjacent sides E4 and E39 In other words 2 DC BC Proof Suppose E bisects zABC in AABC AtC construct a line parallel to E intersecting A B atE producing the gure below E But then zABD E ACBD and ABEC E zABD since they are corresponding angles of parallel lines In addition ABCE E ACBD since they are altemate interior angles of parallel lines Hence ACBE is isosceles andBE BC By the preVious lemma Q BE DC39 B But BE BC so AB BC DC39 A D C This completes the proof QED 297 Exercise The converse to Theorem 296 states that if AB AD BC DC then 17 bisects zABC in the gure above Prove this converse You may use the converse to Lemma 295 proven in Lemma 295a 298 Theorem The bisector of an external angle of AABC cuts the extended opposite side at a point determined by the ratio of the adjacent sides That is to say if 1E3 is extended and intersects the line containing the bisector of the exterior angle of C at E then ACAE E E 39 Proof There are two cases to consider Either mABAC lt mzABC or szAC gt mzABC IfszAC mzABC then the bisector of the exterior angle at C is parallel to A B Assume that mABAC lt mZABC Then as shown in the gure the bisector of ABCG will intersect the extension of 1E3 atE andAE gtAB AtB construct a line parallel to CTE intersecting 1E atF Then ABFC E AECG G since they are corresponding angles F of parallel lines And AECG E ABCE since CTE bisects LBCG and ABCE E ACBF since they are alternate interior angles of parallel lines AC AE Hence ABFC is isosceles andFC BC Now by aprevious lemma But FC BE FCBC so BC BE This proves the assertion for the case when mLBAC lt mZABC If mZBAC gt mZABC then the line containing the bisector of ABCG intersects the extension of 1E3 at pointE on the other side of A withA betweenE and B A similar argument proves the assertion for this case as well and the theorem is proved QED 299 Exercise The converse to Theorem 298 states that if ACAE EZBE in the gure above then CTE bisects the external angle of AABC at C Prove this conjecture We are now able to complete the proof of the main theorem 2910 Theorem Circle of Apollonius The set of all points P such that the ratio of the distances to two xed points A andB that is 133 is constant but not equal to l is a circle Proof Assume the notation above and that PA mPB where m gt 1 is a constant There are two points on 213 indicated by C and D in the gure with the desired ratio By the converse to Theorem 296 and the converse to Theorem 298 P C and 17 are the internal and external angle bisectors of the angle at P Thus they are perpendicular why so ACPD is a right angle This means thatP lies on a circle with diameter QED 32 In the prsvmus pmufwhat happens mLhe case where m lt17 Alsu see Exercisellll 1 mm or Flu y underlying symmemes Twu guudxllustxatmns nuns arethe gunned Demls and Angels39 C Escher aw n x 3 V 1 gm 1 vote A 2m A um a I E7411 sz Ile 1 3 NE 4 r Example 1 The above example shows atypical Arabic design This was drawn starting from a regular hexagon inscribed in a circle Demonstration 2100 Construct the design in Example 1 using Sketchpad 0 First draw a regular hexagon and its circumscribing circle Now construct a regular 12 sided regular polygon having the same circumscribing circle to give a gure like the one below 0 To construct a second 12sided regular polygon having one side adjacent to the rst regular hexagon re ect your gure in one of the sides of the rst regular hexagon Now complete the construction of the previous Arabic design End of Demonstration 2100 2101 Exercise If the radius of the circumscribing circle of the initial regular hexagon is R determine algebraically the area of the sixpointed star inside one of the circles Continuing this example inde nitely will produce a covering of the plane by congruent copies of three polygons a square a rhombus and a sixpointed star Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges ie the polygons do not overlap The second example Example 2 if continued inde nitely also will provide a covering of the plane by congruent copies of two regular polygons two squares in fact Again adjacent polygons do not overlap but now the individual tiles do not meet along full edges The next example Example 3 is one very familiar one from oor coverings or ceiling tiles when continued inde nitely it provides a covering of the plane by congruent copies of a single regular polygon a square But now adjacent polygons meet along the full extent of their edges Finally notice that continuations of the fourth example Example 4 produce a covering of the plane by congruent copies of two regular polygons one a square the other an octagon again the covering is edgetoedge To describe all these possibilities at once what we want is a general de nition of coverings of the plane by polygons without overlaps Specializations of this de nition can then be made when the polygons have special features such as the ones in the rst four examples 2102 De nition A tiling or tessellation of the Euclidean plane is a collection T1 T2 Tquot of polygons and their interiors such that o no two of the tiles have any interior points in common 0 the collection of tiles completely covers the plane When all the tiles in a plane tiling are congruent to a single polygon the tiling is said to have order one and the single region is called the fundamental region of the tiling If each tile is congruent to one of n dilTerent tiles also called fundamental regions the tiling is said to have order n Now we can add in special conditions on the polygons For instance when the polygons are all regular we say that the tiling is a regular tiling Both the second third and fourth examples above are regular tilings but the rst is not regular since neither the sixpointed polygon nor the rhombus is regular To distinguish the second example from the others we shall make a crucial distinction 2103 De nition A tessellation is said to be edge to edge if two tiles intersect along a full common edge only at a common vertex or not at all Thus examples one three and four are edgetoedge whereas example two is not edgeto edge The point of this edgetoedge condition is that it reduces the study of regular tilings to combinatorial problems for the interior angles of the regular polygons meeting at a vertex It is in this way that the Euclidean plane geometry of this chapter particularly the sums of angles of polygons comes into play So from now on atiling will always mean an edgetoedge tiling unless it is explicitly stated otherwise A major problem in the theory is to determine whether a given polygon can serve as fundamental region for a tiling of order one or if a collection of n polygons can serve as fundamental regions for a tiling of order n The case of a square is wellknown from oor coverings and was given already in example 3 above 2104 Demonstration Investigate which regular polygons could be used to create an edgeto edge regular tiling of order one Use the 3Triangle By Edge script to show that an equilateral triangle can tile the plane meaning that it can serve as fundamental region for a regular tiling of order one Try the same with a regular hexagon using the 6Hexagon By Edge script what in nature does your picture remind you of Now use the SPentagon By Edge to check if a regular pentagon can be used a fundamental region for a regular tiling of order one Experiment to see what pattems you make One example is given below can you find others End of Demonstration 2104 Can you tile the plane with a regular pentagon To see why the answer is no we prove the following result 2105 Theorem The only regular polygons that tile the plane are equilateral triangles squares and regular hexagons In particular a regular pentagon does not tile the plane Proof Suppose a regular psided polygon tiles the plane with q tiles meeting at each vertex Since the interior angle of a regular psided polygon has measure 18pr it follows that q180l 2 p 360 But then ie p 2q 2 4 The only integer solutions of this last equation that make geometric sense are the pairs P79 36 44 of 63 These correspond to the case of equilateral triangles meeting 3 at each vertex squares meeting 4 at each vertex and regular hexagons meeting 3 at each vertex QED Tilings of the plane by congruent copies of a regular polygon does not make a very attractive design unless some pattern is superimposed on each polygon that s a design problem we shall return to later What we shall do rst is try to make the tiling more attractive by using more than one regular polygon or by using polygons that need not be regular Let s look rst at the case of an equilateral triangle and a square each having the same edge length Demonstration 2105a Construct a regular tiling of order 2 where the order of the polygons is preserved at each vertex 0 Open a new sketch and draw a square not too big since this is the starting point and draw an equilateral triangle on one of its sides so that the side lengths of the triangle and the square are congruent Use the scripts to see if these two regular polygons can serve as the fundamental regions of a regular tiling of order 2 where the order of the polygons is preserved at each vertex Here s one such example AA 39 AAA U U E J 4 Notice that the use of colors can bring out a pattern to the ordering of the polygons at each vertex As we move in counterclockwise order around each vertex we go from Sgreen gtSyellow gt Twhite gt Tblue gt Twhite and then back to Sgreen where S square and T equilateral triangle This is one example of an edgetoedge regular tiling of order two Consider how many are there End of Demonstration 2105a 2106 Theorem Up to similarity there are exactly eight edgeedge regular tilings of order at least 2 where the cyclic order of the polygons is preserved at each vertex Keeping the order S gtS gt T gt T gt T of squares and triangles produced one such tiling Convince yourself that S gt T gt T gtS gt T produces a different tiling Why are these the only two possible orderings for two squares and three triangles How many permutations are possible for the letters S S T T and T What are the other six tilings Algebraic conditions limit drastically the possible patterns so long as the tiling is edgetoedge and that the order of the polygons is the same at each vertex Using the angle sum formulas for regular polygons one can easily see that you need at least three polygons around a vertex but can have no more than six In the case of a pgon a qgon and an rgon at each vertex you get the equation 2 2 2 180 LJ 180 LJ 180r J 360 p q r You can check that 488 4612 and 31212 are solutions There are afew other solutions as well but they will not make geometric sense Thus S gt O gt O S gtH gtD and T gtD gtD all produce tilings where 0 stands for Octagon H for hexagon andD for Dodecagon We are 40 still missing three tilings but you can have fun looking for them See Exercise 2113 Now we will take a look at some less regular tilings It is surprising how much of geometry can be related to tilings of the plane Let s consider two instances of this the second being Pythagoras theorem The rst instance is atheorem known familiarly as Napoleon s theorem after the famous French general though there is no evidence that he actually had anything to do with the theorem bearing his name Recall that earlier we proved the form of Pythagoras theorem saying that the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides On the other hand Napoleon s theorem says that the centers of these three equilateral triangles themselves form an equilateral triangle as we saw in Exercise 185 The gure below makes this result clearer Here DE andF are the centers of the three equilateral triangles where by center is meant the common circumcenter centroid and orthocenter of an equilateral triangle Napoleon s theorem says that ADEF is equilateral it certainly looks as if its sides are congruent and measuring them on Sketchpad will establish congruence You will provide a proof of the result in Exercise 2115 The question we consider here is how all this relates to tilings of the plane Notice now that we have labeled the interior angles of the triangle because we are going to allow polygons which are not necessarily regular Since the interior can then be different the particular interior angle of polygons that appears at a vertex is going to be just as important as 41 which polygon appears Now we will see how we can continue the gure above inde nitely and thus tile the plane One should notice that the edgetoedge condition imposes severe restrictions on the angles that can occur at a vertex Label the angles in the original gure as follows Of course the angles of the equilateral triangles are all the same but we have used dilTerent letters to indicate that they are the interior angles of equilateral triangles of different size Since a b c d e f 360 three copies ofthe rightangled triangle and one copy ofeach of the three different sizes of equilateral triangle will t around a vertex with no gaps or overlaps The gure can thus be constructed inde nitely by maintaining the same counterclockwise order a gt e gt c gt f gt b gt d at each vertex Now draw the gure for yourself It may be instructive to use a different color for each equilateral triangle to highlight the fact that the equilateral triangles are not necessarily congruent 2106a Demonstration 0 Open a new sketch and in the top lefthand comer of the screen draw a rightangled triangle as shown in the gure above Make sure that your construction is dynamic in the sense that the triangle remains rightangled whenever any one of the vertices is dragged Use the Circle By Center Radius construction to construct a congruent copy of your triangle in the center of the screen Draw an outwardly pointing equilateral triangle on each side of this rightangled triangle 42 0 Continue adding congruent copies of the rightangled triangle and the equilateral triangles to the sides of the triangles already in your gure One way to add congruent copies of the 0 right triangle is to use your AutoMatching similar triangle script Just label your original right triangle appropriately 0 Experiment a little to see what gures can be produced Check that your construction is dynamic by dragging the vertices of the rst rightangled triangle you drew End of Demonstration 2106a Here s one that looks as if it might tile the plane if continued inde nitely B C A Napoleon Tiling The gure above of the Napoleon Tiling has an overlay of hexagons over it To see where it came from apply Napoleon s Theorem to the tiling That is around each right triangle connect the centers of the equilateral triangle to create a new equilateral triangle Six of those new equilateral triangles make up each hexagon above Thus Napoleon s theorem brings out an underlying symmetry in the design because it showed that a regular tiling of the plane by regular hexagons could be overlaid on the gure The same design could have been obtained by 43 putting a design on each regular hexagon and then tiling the plane with these pattemed regular hexagons This brings out a crucial connection between lings and the sort of designs that are used for covering walls oors ceilings or any at surface A design is said to be wallpaper design if a polygonal portion of it provides a tiling of the plane by translations in two different directions Thus all the examples obtained in this section are wallpaper designs It is very clear that the Islamic design in problem 2101 is a wallpaper design because the portion of the design inside the initial regular hexagon will tile the plane as the figure below clearly shows i 2107 Exercise Find a square portion of Example 4 in Seection 210 that tiles the plane In other words show that that example is a wallpaper design Example 2 is sometimes called the Pythagorean Tiling It is created by atranslation of two adjacent noncongruent squares This tiling occurs o en in architectural and decorative designs as seen in this sidewalk tiling To see why this tiling might be called a Pythagorean Tiling open a new sketch and draw the tiling as it appears in example 2 using two squares of different sizes Construct an overlaying of this design by a tiling which consists of congruent copies of a single square What is the area of this square Use Pythagoras theorem to relate this area to the area of the two original squares you used to construct your pattern 44 211 Exercises Exercise 2111 Prove Theorem 294 Given a circle 2 and a pointP outside 2 let I be a ray throughP intersecting 2 at pointsA and B If C is a point on 2 such that PAPB PC2 then P C is tangent to 2 at C Exercise 2112 Given points AB andP use Sketchpad to construct the Circle of Apollonius passing through P In other words construct the set of points Q such that QA mQB where PAPB m Exercise 2113 Produce two different orderpreserving edgetoedge regular tilings of order 2 just using triangles and hexagons Produce an orderpreserving edgetoedge regular tiling of order 3 using triangles squares and hexagons We now have the eight tilings mentioned in Theorem 2105 Exercise 2114 Using Sketchpad construct the Napoleon Tiling Choose a regular hexagon in your gure and describe its area in terms of the original triangle and the three equilateral triangles constructed on its sides Now choose a different larger or smaller area regular hexagon having a different area and describe the area of this hexagon in terms of the original triangle and the three equilateral triangles Exercise 2115 While the tiling above makes a very convincing case for the truth of Napoleon s theorem it doesn t prove it in the usual meaning of proof Here is a coordinate geometry proof based on the figure on the following page and on the notation in that figure a The points DE andF are the centers of the equilateral triangles constructed on the sides of the rightangled triangle AABC Show that length7 cJg Determine also the lengths of A D and B E b If AABC 6 and ACAB write the values of sine cos6 sin and cos in terms of a b and c c Write down the addition formulas for sine and cosine cosu v sinu v 45 d Let the lengths of F E 177 and TE be x y and 2 respectively Use the Law of Cosines to show that 22 sz 122 2abcos30 Determine corresponding values for x and y Deduce that x y z Use all the previous results to nish off a coordinate geometry proof of Napoleon s theorem Exercise 2116 Instead of starting with a rightangled triangle start with an arbitrary AABC and draw equilateral triangles on each of its sides and repeat the previous construction 0 Open a new sketch and draw a small triangle near the top comer of the screen label the vertices A B and C By using the Circle By CenterRadius tool you can construct congruent copies of this triangle 0 Draw one congruent copy of AABC in the center of the screen Draw an equilateral triangle on each of its sides 46 Continue this construction preserving cyclic order at each vertex to obtain atiling of the plane The following gure is one such example Construct the centers of all the equilateral triangles and draw hexagons as in the case of rightangled triangles Do you think Napoleon s theorem remains valid for any triangle not just rightangled triangles Exercise 2117 Can the plane be tiled by copies of the diagram for Yaglom s Theorem given below as in the manner of the tiling corresponding to Napoleon s Theorem If so produce the tiling using Sketchpad Recall that Yaglom s Theorem said if we place squares on the sides of a parallelogram the centers of the squares also form a square 47 212 One nal Exercise Exercise 2121 To the le in the gure below are two triangles one obtuse the other right angled The interior angles of the two triangles have been labeled Since the sum of these siX angles is 360 there should be a tiling of the plane by congruent copies of these two triangles in which the cyclic order of the angles at each vertex is the same as the one shown in the gure to the right 0 Open a new sketch and continue this construction to provide a tiling of the plane Unlike the preVious tilings the triangles in this tiling are not congruent Explain why this tiling is more like a Nautilus Shell 48 0 Construct the circumcenters of the three outwardly pointing obtuse triangles on the sides of one of the rightangled triangles and join these circumcenters by line segments What if any is the relation of the triangle haVing these three circumcenters as vertices and the original obtuse triangle Is there any relation with the original rightangled triangle Use Sketchpad if necessary to check any conjecture you make Don t forget to drag Investigate what happens if you construct instead the three circumcenters of the rightangled triangles on the sides of one of the obtuse triangles Draw the triangle haVing these circumcenters as vertices What if any is the relation between the original rightangled triangle and the triangle having the three circumcenters as vertices Is there any relation with the original obtuse triangle Again use Sketchpad if necessary to check Visually any conjecture you make Don t forget to drag 49 Chapter 5 INVERSION The notion of inversion has occurred several times already especially in connection with Hyperbolic Geometry Inversion is a transformation different from those of Euclidean Geometry that also has some useful applications Also we can delve further into hyperbolic geometry once we have developed some of the theory of inversion This will lead us to the description of isometries of the Poincare Disk and to constructions via Sketchpad of tilings of the Poincare disk just like the famous Devils and Angels picture of Escher 51 DYNAMIC INVESTIGATION One very instructive way to investigate the basic properties of inversion is to construct inversion via a custom tool in Sketchpad One way of doing this was described following Theorems 353 and 354 in Chapter 3 but in this section we ll describe an alternative construction based more closely on the definition of inversion Recall the definition of inversion given in section 5 of chapter 3 511 De nition Fix a point 0 and a circle C centered at O of radius r For a point P P at O the inverse of P is the unique point P39 on the ray starting from O and passing through P suchthat OPOP39r2 The point 0 is called the center of inversion and circle C is called the circle of inversion whiler is called the radius of inversion OP 051 inches OP39 108 inches r 074 inches OPOP39 055 inches2 r2 055 inches2 To create atool that constructs the inverse of a point P given the circle of inversion and its center we can proceed as follows using the dilation transformation Open a new sketch and draw a circle by center and point Label the center by O and label the point on the circle by R Construct a point P not on the circle Construct the ray from the center of the circle passing through P Construct the point of intersection between the circle and the ray label it D Mark the center of the circle this will be the center of dilation Then select the center of the circle the point P and then the point of intersection of the ray and the circleGo to Mark Ratio under the Transform menu This de nes the ratio of the dilation Now select the point of intersection of the ray and the circle and dilate by the marked ratio The dilated point is the inverse point to P Label the dilated point P39 Select 0 R P andP Under the Custom Tools Menu choose Create New Tool and check Show Script View You may wish to use AutoMatching for O andR as we are about to use our inversion script to explore many examples Under the Givens List for your script double click on O andR and check the box Automatically Match Sketch Object To make use of the AutoMatching you need to start with a circle that has center labeled by O and a point on the circle labeled by R 0 Save your script Use your tool to investigate the following 512 Exercise Where is the inverse of P if 0 P is outside the circle of inversion o P is inside the circle of inversion o P is on the circle of inversion o P is the center of the circle of inversion Using our tool we can investigate how inversion transforms various figures in the plane by using the construct Locus property in the Construct menu Or by using the trace feature For instance let s investigate what inversion does to a straight line 0 Construct a circle of inversion Draw a straight line and construct a free point on the line Label this free point by P 0 Use your tool to construct the inverse point P39 to P 0 Select the points P39 and P Then select Locus in the Construct menu Altematively one could trace the point P39 while dragging the pointP 513 Exercise What is the image of a straight line under inversion By considering the various possibilities for the line describe the locus of the inversion points Be as detailed as you c A line which is tangent to the circle of inversion can A line which passes through the circle of inversion Image Image 40 P I I A line which passes A line which doesn t through the center of the circle of inversion intersect the circle of inversion Image Image 514 Exercise What is the image of a circle under inversion By considering the various possibilities for the line describe the locus of the inversion points Be as detailed as you can A circle which is tangent to the circle of inversion Image A circle which passes through the center of the circle of inversion Image A circle which intersects the circle of inversion in two points Image A circle passing through the center of the circle of inversion also internally tangent Image A circle which is orthogonal to the circle of inversion Image You should have noticed that some circles are transformed into another circle under the inversion transformation Did you notice what happens to the center of the circle under inversion in these cases Try it now End of Exercise 514 You can easily construct the inverse image of polygonal gures by doing the following Construct your figure and its interior Next hide the boundary lines and points of your figure so that only the interior is visible Next select the interior and choose Point on Object from the Construct Menu Now construct the inverse of that point and then apply the locus construction Here is an example 515 Exercise What is the image of other gures under inversion By considering the various possibilities for the line describe the locus of the inversion points Be as detailed as you can A triangle external to the circle of inversion A triangle with one vertex as the center of the circle of inversion Image Image A triangle internal to the circle of inversion Image 52 PROPERTIES OF INVERSION Circular inversion is not atransformation of the Euclidean plane since the center of inversion does not get mapped to a point in the plane However if we include the point at in nity we would have a transformation of the Euclidean Plane and this point at in nity Also worth noting is that if we apply inversion twice we obtain the identity transformation With these observations in mind we are now ready to work through some of the basic properties of inversion Let C be the circle of inversion with center 0 and radius r Also when we say line we mean the line including the point at in nity The rst theorem is easily veri ed by observation 521 Theorem Points inside C map to points outside of C points outside map to points inside and each point on C maps to itself The center 0 of inversion maps to 00 522 Theorem The inverse of a line through 0 is the line itself Again this should be immediate from the de nition of inversion however note that the line is not pointwise invariant with the exception of the points on the circle of inversion Perhaps more surprising is the next theorem 523 Theorem The inverse image of a line not passing through 0 is a circle passing through 0 Proof LetP be the foot of the perpendicular from O to the line Let Q be any other point on the line Then P39 and Q39 are the respective inverse points By the de nition of inverse points OP OP39 0Q OQ39 We can use this to show that AOPQ is similar to AOQ39P39 Thus the image of any Q on the line is the vertex of a right angle inscribed in a circle with diameter OP39 The proof of the converse to the previous theorem just involves reversing the steps The converse states the inverse image of a circle passing through 0 is a line not passing through 0 Notice that inversion is dilTerent from the previous transformations that we have studied in that lines do not necessarily get mapped to lines We have seen that there is a connection between lines and circles 524 Theorem The inverse image of a circle not passing through 0 is a circle not passing through 0 Proof Construct any line through the center of inversion which intersects the circle in two points P and Q Let P39 and Q39 be the inverse points to P and Q We know that OP OP39 OQ OQ39 r2 Also by Theorem 292 Power ofa Point OPOQ 012 k OPOP39 O O39 2 OP39 039 2 Thus 7 M L or Q L In other words everything reduces to a OPOQ OPOQ k OQ OP k dilation 525 Theorem Inversion preserves the angle measure between any two curves in the plane That is inversion is conformal Proof It suf ces to look at the case of an angle between a line through the center of inversion and a curve In the gure below P and Q are two points on the given curve and P39 andQ39 are the corresponding points on the inverse curve We need to show that szPB szP39D The sketchpad activity below will lead us to the desired result mZOPB 6967 mZEP39D 6967 0 Open a new sketch and construct the circle of inversion with center 0 and radius r Construct an arc by 3 points inside the circle and label two of the points as P and Q Next construct the inverse of the arc by using the locus construction and label the points P39 and Q39 Finally construct the line OP it will be its own inverse Next construct tangents to each curve through P and P39 respectively Notice thatP Q P39 and Q39 all lie on a circle Why Thus AQPP39 and AP39Q39Q are supplementary Inscribed Angle Theorem Thus szPQ mzP39Q39O Check this by measuring the angles Next drag Q towards the point P What are the limiting position of the angles AOPQ and AP39Q39O What result does this suggest 526 Theorem Under inversion the image of a circle orthogonal to C is the same circle setwise not pointwise Proof See Exercise 531 dPM dQN dPN dQM 527 Theorem Inversion preserves the generalized cross ratio of any four distinct points P QM and N in the plane
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