ࡱ> )+&'(tmaci y a r b 3 N!jbjb^^ xh<h<!l7i*7i*akaklll]p]p]p]p8pytT]pT IwB(ɐdW${,t lw|۬akakɐIw۬۬۬aklɐ۬1ln>akakakak۬b۬=k6lTl< v|Eϼ]p]pJy$TTH۬۬Chapter 2 EUCLIDEAN PARALLEL POSTULATE 2.1 INTRODUCTION. There is a well-developed 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 alternatives is taken as an axiom. This geometry is called Absolute Geometry, and an account of it can be found in several textbooks - in Coxeters 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 Fifth Postulate. 2.1.1 Playfairs Axiom. Through a given point, not on a given line, exactly one line can be drawn parallel to the given line. Playfairs Axiom is equivalent to the Fifth Postulate in the sense that it can be deduced from Euclids five postulates and common notions, while, conversely, the Fifth Postulate can deduced from Playfairs Axiom together with the common notions and first four postulates. 2.1.2 Theorem. Euclids five Postulates and common notions imply Playfairs Axiom. Proof. First it has to be shown that if P is a given point not on a given line l, then there is at least one line through P that is parallel to l. By Euclid's Proposition I 12, it is possible to draw a line t through P perpendicular to l. In the figure below let D be the intersection of l with t.  By Euclid's Proposition I 11, we can construct a line m through P perpendicular to t . Thus by construction t is a transversal to l and m such that the interior angles on the same side at P and D are both right angles. Thus m is parallel to l 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 first 28 propositions without using his Fifth Postulate, it follows that the existence of at least one line through P that is parallel to l, can be deduced from the first four postulates. For a complete list of Euclid's propositions, see College Geometry by H. Eves, Appendix B.) To complete the proof of 2.1.2, we have to show that m is the only line through P that is parallel to l. So let n be a line through P with m  EMBED Equation.2 n and let E  EMBED Equation.2  P be a point on n. Since m  EMBED Equation.2 n,  EMBED Equation.2 EPD cannot be a right angle. If m EMBED Equation.2 EPD < 90, as shown in the drawing, then m EMBED Equation.2 EPD + m EMBED Equation.2 PDA is less than 180. Hence by Euclids fifth postulate, the line n must intersect l on the same side of transversal t as E, and so n is not parallel to l. If m EMBED Equation.2 EPD > 90, then a similar argument shows that n and l must intersect on the side of l opposite E. Thus, m is the one and only line through P that is parallel to l. QED A proof that Playfairs axiom implies Euclids fifth postulate can be found in most geometry texts. On page 219 of his College Geometry book, Eves lists eight axioms other than Playfairs axiom each of which is logically equivalent to Euclids fifth Postulate, i.e., to the Euclidean Parallel Postulate. A geometry based on the Common Notions, the first 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 Alternative B for the Euclidean Parallel Postulate (see text following Axiom 1.2.2).. 2.2 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. 2.2.1 Theorem. In Euclidean geometry the sum of the interior angles of any triangle is always 180. Proof: Let  EMBED Equation.3  be any triangle and construct the unique line  EMBED Equation  through A, parallel to the side EMBED Equation.3 , as shown in the figure  Then mEAC = mACB and mDAB = mABC by the alternate angles property of parallel lines, found in most geometry textbooks. Thus mACB+mABC+mBAC=180. QED Equipped with Theorem 2.2.1 we can now try to determine the sum of the interior angles of figures in the Euclidean plane that are composed of a finite number of line segments, not just three line segments as in the case of a triangle. Recall that a polygon is a figure in the Euclidean plane consisting of points P1, P2,..., Pn, called vertices, together with line segments  EMBED Equation.2 ,  EMBED Equation.2 , ..., EMBED Equation.2 , called edges or sides. More generally, a figure consisting of the union of a finite number of non-overlapping polygons will be said to be a piecewise linear figure. Thus  are piecewise linear figures as is the example of nested polygons below.  This example is a particularly interesting one because we can think of it as a figure containing a hole. But is it clear what is meant by the interior angles of such figures? For such a polygon as the following:  we obviously mean the angles indicated. But what about a piecewise linear figure 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 figures 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 figures can be obtained from Theorem 2.2.1 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, i.e., any four-sided figure, 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 n sides, then it can be divided into n-2 triangles and the sum of the angles of the polygon is equal to the sum of the angles of the n-2 triangles. This proves the following result. 2.2.2 Theorem. The sum of the interior angles of an n-sided polygon,  EMBED Equation.2 , is  EMBED Equation.3 . 2.2.2a Demonstration. We can use a similar method to determine the sum of the angles of the more complicated piecewise linear figures. One such figure is a polygon having holes, that is, a polygon having other non-overlapping polygons (the holes) contained totally within its interior. Open a new sketch and draw a figure such as  An interesting computer graphics problem is to color in the piecewise linear figure between the two polygons. Unfortunately, computer graphics programs will only fill polygons and the interior of the figure is not a polygon. Furthermore, Sketchpad measures angles greater than 180 by using directed measurements. Thus Sketchpad would give a measurement of -90 for a 270 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 figure below. Color each of these triangles in a different color so that you can distinguish them easily.  SHAPE \* MERGEFORMAT  We call this a triangular tiling of the figure. Now use Theorem 2.2.2 to compute the total sum of the angles of all these new polygons. Construct a different triangular tiling of the same figure and compute the total sum of angles again. Do you get the same value? Hence complete the following result. 2.2.3 Theorem. When an n-sided piecewise linear figure 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 2.2.2a. Try to prove Theorem 2.2.3 algebraically using Theorem 2.2.2. The case of a polygon containing h polygonal holes is discussed in Exercise 2.5.1. 2.3 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 first is perhaps the best known of all results in Euclidean plane geometry, namely Pythagoras theorem. This is frequently stated in purely algebraic terms as  EMBED Equation.2 , 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 algebraic/synthetic 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. 2.3.4 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  EMBED Equation.2  and the lengths of the other two sides are EMBED Equation.2  and EMBED Equation.2 , then  EMBED Equation.2 . Proof: Let  EMBED Equation.3  be a right-angled triangle with right angle at C, and let  EMBED Equation.3  be the perpendicular from C to the hypotenuse  EMBED Equation.3  as shown in the diagram below.  Show EMBED Equation.3  is similar to  EMBED Equation.3 . Show EMBED Equation.3  is similar to  EMBED Equation.3 . Now let  EMBED Equation  have length  EMBED Equation.2 , so that  EMBED Equation  has length  EMBED Equation.2 . By similar triangles,  EMBED Equation.2  and  EMBED Equation.2  Now eliminate  EMBED Equation.2  from the two equations to show  EMBED Equation.2 . There is an important converse to the Pythagorean theorem that is often used. 2.3.5 Theorem. (Pythagorean Converse) Let  EMBED Equation.3  be a triangle such that  EMBED Equation.2 . Then  EMBED Equation.3  is right-angled with  EMBED Equation.2 ACB a right angle. 2.3.5a Demonstration (Pythagorean Theorem with Areas) You may be familiar with the geometric interpretation of Pythagoras theorem. If we build squares on each side of  EMBED Equation.3  then Pythagoras theorem relates the area of the squares. Open a new sketch and draw a right-angled triangle  EMBED Equation.3 . 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. 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 right-angled triangle you have drawn. End of Demonstration 2.3.5a. 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 figure 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 havent 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. 2.3.5b Demonstration (Generalization of Pythagorean Theorem) Draw a new right-angled triangle  EMBED Equation.3  and use the 5/Pentagon (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 1.3.5(b).) What do you notice about the areas in this case? Now complete the statement of Theorem 2.3.6 below for regular n-gons. End of Demonstration 2.3.5b. 2.3.6 Theorem. (Generalization of Pythagoras theorem) When similar copies of a regular n-gon,  EMBED Equation.2 , are constructed on the sides of a right-angled triangle, each n-gon having the same edge length as the side of the triangle on which it sits, then ____________________ ______________________________________________________. The figure below illustrates the case of regular pentagons.  2.3.7 Demonstration. Reformulate the result corresponding to Theorem 2.3.6 when the regular n-gons constructed on each side of a right-angled triangle are replaced by similar triangles. This demonstration presents an opportunity to explain another feature of Custom Tools called Auto-Matching. We will be using this feature in Chapter 3 when we use Sketchpad to explore the Poincar 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  EMBED Equation.3  with the vertices labeled. Next construct the line (not a segment)  EMBED Equation.3 . Select the vertices B-A-C in order and choose "Mark Angle B-A-C" 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 line EMBED Equation.3 . Choose Rotate from the Transform Menu and then rotate by Angle B-A-C. Select the vertices A-B-C in order and choose Mark Angle A-B-C 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  EMBED Equation.3 . Choose Rotate from the Transform Menu and rotate by Angle A-B-C. Construct the point of intersection between the two rotated lines and label it F.  EMBED Equation.3  is similar to  EMBED Equation.3 . 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 labeled A, B, and C already constructed in your sketch where you want to construct the similar triangle. Then you only 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. Now open a new sketch and construct a triangle with vertices labeled A, B, and C. In the same sketch, construct a right triangle. Use the similar triangle tool to build triangles similar to  EMBED Equation.3  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 2.3.7. 2.4 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 figure below,  EMBED Equation.3  is inscribed in the circle and Arc ADC is the subtended arc. We will say that  EMBED Equation.3  is a central angle of the circle because the vertex is located at the center O. The measure of Arc ADC is defined to be the angle measure of the central angle,  EMBED Equation.3 .  2.4.0 Demonstration. Investigate the relationship between an angle inscribed in a circle and the arc it intercepts (subtends) on the circle. Open a new script in Sketchpad and draw a circle, labeling the center of the circle by O.  Select an arbitrary pair of points A, B on the circle. These points specify two possible arcs - lets choose the shorter one in the figure above, that is, the arc which is subtended by a central angle of measure less than 180. Now select another pair of points C, D on the circle and draw line segments to form BCA and BDA. Measure these angles. What do you observe? If you drag C or D what do you observe about the angle measures? Now find the angle measure of BOA. What do you observe about its value? Drag B until the line segment  EMBED Equation.3  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. 2.4.1 Theorem. (Inscribed Angle Theorem): The measure of an inscribed angle of a circle equals _____________________ that of its intercepted (or subtended) arc. 2.4.2 Corollary. A diameter of a circle always inscribes _____________________ at any point on the circumference of the circle. 2.4.3 Corollary. Given a line segment  EMBED Equation.3 , the locus of a point P such that  EMBED Equation.3  is a circle having  EMBED Equation.3  as diameter. End of Demonstration 2.4.0. The result you have discovered in Corollary 2.4.2 is a very useful one, especially in constructions, since it gives another way of constructing right-angled triangles. Exercises 2.5.4 and 2.5.5 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. 2.4.4 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.) 2.5 Exercises Exercise 2.5.1. Consider a piecewise linear figure consisting of a polygon containing h holes (non-overlapping polygons in the interior of the outer polygon) has a total of n edges, where n 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 2.5.2. Prove that if a quadrilateral is cyclic, then the opposite angles of the quadrilateral are supplementary, i.e., the sum of opposite angles is 180. [ This will provide half of the proof of Theorem 2.4.4. ] Exercise 2.5.3. Give a synthetic proof of the Inscribed Angle Theorem 2.4.1 using the properties of isosceles triangles in Theorem 1.4.6. Hint: there are three cases to consider: here y is the angle subtended by the arc and q is the angle subtended at the center of the circle. The problem is to relate y to q. 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.  Case 3: The center of the circle lies in the exterior of the inscribed angle.  End of Exercise 2.5.3. For Exercises 2.5.4, 2.5.5, and 2.5.6, 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 2.5.4. Use the Inscribed Angle Theorem to devise a Sketchpad construction that will construct the tangents to a given circle from a given point P outside the circle. Carry out your construction. (Hint: Remember Corollary 2.4.2). Exercise 2.5.5. In the following figure  the line segments EMBED Equation.2  and  EMBED Equation.2  are the tangents to a circle centered at O from a point P outside the circle. Prove that  EMBED Equation.2  and  EMBED Equation.2  are congruent. Exercise 2.5.6. Let l and m be lines intersecting at some point P and let Q be a point on l. Use the result of Exercise 2.5.5 to devise a Sketchpad construction that constructs a circle tangential to l and m that passes through Q. Carry out your construction. For Exercises 2.5.7 and 2.5.8, we consider regular polygons again, that is, polygons with all sides congruent and all interior angles congruent. If a regular polygon has n sides we shall say it is a regular n-gon. For instance, the following figure  is a regular octagon above, i.e., a regular 8-gon. By Theorem 2.2.2 the interior angle of a regular n-gon is  EMBED Equation.2 . The measure of any central angle is  EMBED Equation.3 . In the figure  EMBED Equation.3  is an interior angle and  EMBED Equation.3 is a central angle. Exercise 2.5.7. Prove that the vertices of a regular polygon always lie on a circumscribing circle. (Be careful! Dont 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 2.5.8. Now suppose that the edge length of a regular n-gon is l and let R be the radius of the circumscribing circle for the n-gon. The Apothem of the n-gon is the perpendicular distance from the center of the circumscribing circle to a side of the n-gon.  EMBED Word.Picture.8  The Apothem (a) With this notation and terminology and using some trigonometry complete the following R = l _____________ , l = R ______________, Apothem = R__________ . Use this to deduce (b) area of n-gon =  EMBED Equation.2 nR2 sin EMBED Equation.2 , (c) perimeter of n-gon = 2nR sin EMBED Equation.2 . (d) Then use the well-known fact from calculus that  EMBED Equation.3  to derive the formulas for the area of a circle of radius R as well as the circumference of such a circle. Exercise 2.5.9. Use Exercise 2.5.8 together with the usual version of Pythagoras theorem to give an algebraic proof of the generalized Pythagorean Theorem (Theorem 2.3.6). Exercise 2.5.10 Prove the converse to the Pythagorean Theorem stated in Theorem 2.3.5. 2.6 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 1.8 while exploring Pedal Triangles. 2.6.1 Theorem (Simson Line). If P lies on the circumcircle of  EMBED Equation.3 , 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 figure below. Why do P, D, A, and E all lie on the same circle? Why do P, A, C, and B all lie on another circle? Why do P, D, B and F all lie on a third circle? Verify all three of these statements using Sketchpad.  In circle PDAE,  EMBED Equation.3 . Why? In circle PACB,  EMBED Equation.3 . Why? In circle PDBF,  EMBED Equation.3 . Why? Since  EMBED Equation.3 , points D, E, and F must be collinear. QED In Exercise 1.9.4, the Miquel Points of a triangle were constructed. 2.6.2 Theorem (Miquel Point) If three points are chosen, one on each side of a triangle, 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 figure below.  Let D, E and F be arbitrary points on the sides of  EMBED Equation.3 . Construct the three circumcircles. Suppose the circumcircles for  EMBED Equation.3  and  EMBED Equation.3  intersect at point G. We need to show the third circumcircle also passes through G. Now, G may lie inside  EMBED Equation.3 , on  EMBED Equation.3 , or outside  EMBED Equation.3 . We prove the theorem here in the case that G lies inside  EMBED Equation.3 , and leave the other two cases for you (see Exercise 2.8.1).  EMBED Equation.3  and  EMBED Equation.3  are supplementary. Why?  EMBED Equation.3  and  EMBED Equation.3  are supplementary. Why? Notice  EMBED Equation.3 . Combining these facts we see the following.  EMBED Equation.3 . So  EMBED Equation.3  or  EMBED Equation.3  and  EMBED Equation.3  are supplementary. Thus C, E, G, and F all lie on a circle and the third circumcircle must pass through G. QED The proof of Theorem 2.6.3 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 1.5.4). The second results states that the point which is 2/3 the distance from a vertex (along a median) to the midpoint of the opposite side is the centroid of the triangle (see Theorem 1.5.6). 2.6.3 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 a triangle is called the Euler Line. Proof. In the diagram below,  EMBED Equation.3  is the midpoint of the side opposite to A and O, G, and H are the circumcenter, centroid, and orthocenter, respectively. Since A, G, and  EMBED Equation.3 are collinear, we can show that O, G, and H are also collinear, by showing that  EMBED Equation.3 . To do this, it suffices to show that  EMBED Equation.3 . If we also show that the ratio of similiarity is 2:1, then we will also prove that G trisects  EMBED Equation.3 .  The proof that  EMBED Equation.3  with ratio 2:1 proceeds as follows: Let I be the point where the ray  EMBED Equation.3  intersects the circumcircle of  EMBED Equation.3 . Then  EMBED Equation.3  (why?). It follows that  EMBED Equation.3  with ratio 2:1 (why?) It is also true that AIBH is a parallelogram (why?) and  hence  EMBED Equation.3 . Since G is the median, we know that EMBED Equation.3 . Thus we have two corresponding sides proportional. The included angles are congruent because they are alternate interior angles formed by the parallel lines  EMBED Equation.3  and  EMBED Equation.3  and the transversal  EMBED Equation.3 . (Why are  EMBED Equation.3  and  EMBED Equation.3  parallel?) Thus,  EMBED Equation.3  with ratio 2:1 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, i.e. triangles in which we cant form the triangles  EMBED Equation.3  and  EMBED Equation.3 . What sorts of triangles arise? You should find two special cases. Finish the proof of Theorem 2.6.3 by proving the result for each of these cases (see Exercise 2.8.2). 2.7 THE NINE POINT CIRCLE. Another surprising triangle property is the so-called Nine-Point Circle, sometimes credited to K.W. Feuerbach (1822). Sketchpad is particularly well adapted to its study. The following Demonstration will lead you to its discovery. 2.7.0 Demonstration: Investigate the nine points on the Nine Point Circle. The First set of Three points: In a new sketch construct  EMBED Equation.3 . Construct the midpoints of each of its sides; label these midpoints D, E, and F. Construct the circle that passes through D, E, and F. (You know how to do this!) This circle is called the Nine-Point Circle. Complete the statement: The nine-point circle passes through _________________________________. The Second set of Three points: In general the nine-point circle will intersect  EMBED Equation.3  in three more points. If yours does not, drag one of the vertices around until the circle does intersect  EMBED Equation.3  in three other points. Label these points J, K, and L. Construct the line segment joining J 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 point M and complete the following statement: The nine-point circle also passes through ____________________________. . The Third set of Three points: The red segments intersect the circle at their respective endpoints (J, K, or L). For each segment there exists a second point where the segment intersects the circle. Label them N, O and P. To describe these points measure the distance between M and each of A, B, and C. Measure also the distance between M and each of N, O, and P. What do you observe? Confirm your observation by dragging the vertices of  EMBED Equation.3 . Complete the following statement: The nine-point circle also passes through ___________ You should create a Nine Point Circle tool from this sketch and save it for future use. End of Demonstration 2.7.0. To understand the proof of Theorem 2.7.1 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. 2.7.1 Theorem (The Nine-point Circle) The midpoints of the sides of a triangle, 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 nine-point circle. Your final figure should be similar to  Proof: (See Figure 1) In  EMBED Equation.3  label the midpoints of  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3 , by A', B' and C' respectively. There is a circle containing A', B' and C'. In addition, we know A'C'AB' is a parallelogram, and so A'C'=AB'. (See Figure 2) Construct the altitude from A intersecting  EMBED Equation.3  at D. As  EMBED Equation.3 is parallel to  EMBED Equation.3  and  EMBED Equation.3  is perpendicular to  EMBED Equation.3 , then  EMBED Equation.3  must be perpendicular to EMBED Equation.3 . Denote the intersection of  EMBED Equation.3  and  EMBED Equation.3  by P. Then EMBED Equation.3 ,  EMBED Equation.3  and  EMBED Equation.3 . (See Figure3) Consequently,  EMBED Equation.3  by SAS. So AB'=B'D. By transitivity with A'C'=AB' we have B'D=A'C'. Thus A'C'B'D is an isosceles trapezoid. Hence, by the remarks preceding this theorem, A', C', B', and D 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.  Figure 1  Figure 2  Figure 3  Figure 4   (See Figure 5) Let J denote the midpoint of the segment joining vertex A and the orthocenter H. Then, again by the connection of midpoints of the sides of a triangle,  EMBED Equation.3  is parallel to  EMBED Equation.3 . (See Figure 6) Now  EMBED Equation  but  EMBED Equation . Hence  EMBED Equation . (See Figure 7) Therefore C' lies on a circle with diameter  EMBED Equation.3 . A similar argument shows that B' lies on the circle with diameter  EMBED Equation.3 , and hence J lies on the circle determined by A', B', and C'. Likewise, the other two midpoints of the segments joining the vertices with the orthocenter lie on the same circle. QEDFigure 5  Figure 6  Figure 7   2.8 Exercises. In this exercise set, Exercise 2.8.3 2.8.7 are related to the nine point circle. Exercise 2.8.1. Using Sketchpad, illustrate a case where the Miquel Point lies outside the triangle. Prove Theorem 2.6.2 in this case. Exercise 2.8.2. Prove Theorem 2.6.3 for the two special cases: (a) The triangle is isosceles. (b) The triangle is a right triangle. Exercise 2.8.3. For special triangles some points of the nine-point circle coincide. Open a new sketch and draw an arbitrary EMBED Equation.3 . Explore the various possibilities by dragging the vertices of  EMBED Equation.3 . Describe the type of triangle (if it exists) for which the nine points of the nine-point circle reduce to: 4 points: _____________________ 5 points: ________________________ 6 points: _____________________ 7 points: ________________________ 8 points: _____________________ Exercise 2.8.4. Open a new sketch and draw an arbitrary triangle  EMBED Equation.3 . 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? Measure the length of  EMBED Equation.3 ,  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3 . What results for a general triangle do your calculations suggest? Measure the radius of the nine-point circle of  EMBED Equation.3 . Measure the radius of the circumcircle of  EMBED Equation.3 . What results for a general triangle do your calculations suggest? Drag the vertices of the triangle around. Do your conjectures still remain valid? Exercise 2.8.5. Open a new sketch and draw an arbitrary  EMBED Equation.3 . Let H be the orthocenter and O be the circumcenter of  EMBED Equation.3 . Construct the nine-point circles for  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3 . Use sketchpad to show that these nine-point 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 A, B, and C, then the locations of the midpoints P, Q, and R of the sides of  EMBED Equation.3  are uniquely determined. Similarly, the locations of the feet of the altitudes D, E, and F will be determined once A, B, 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, i.e., we construct the vertices A, B, and C knowing the midpoints or the feet of the altitudes. Use the notation from the following figure.  Exercise 2.8.6. (a) Prove the line segment  EMBED Equation.2  is parallel to side  EMBED Equation.2 . (b) Given points P, Q, and R, show how to construct points A, B, and C so that P, Q, and R are the midpoints of the sides of  EMBED Equation.3 . (c) Formulate a conjecture concerning the relation between the centroid G of  EMBED Equation.3  and the centroid of  EMBED Equation.3 . Exercise 2.8.7. (a) Assume  EMBED Equation.3  is acute (to ensure the feet of the altitudes lie on the sides of the triangle). Prove that PC = PB = PE = PF and that P lies on the perpendicular bisector of the line segment  EMBED Equation.2 . (b) Given points D, E, and F, show how to construct points A, B, and C so that D, E, and F are the feet of the altitudes from the vertices of  EMBED Equation.3  to the opposite sides. (Hint: remember the nine-point circle). 2.9 THE POWER OF A POINT AND SYNTHESIZING APOLLONIUS. Another application of similarity will be to a set of ideas involving what is often 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 2.9.0. Discover the formula for the power of point P with respect to a given circle. 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 l through P. Compute the lengths PA, PB of  EMBED Equation.2 , and EMBED Equation.2  respectively; then compute the product PAPB of PA and PB. Drag l while keeping P fixed. What do you observe? Investigate further by considering the case when l is tangential to the given circle. Use this to explain your previous observation. What happens to the product PAPB when P is taken as a point on the circle? Now let P be a point inside the circle, l a line through P and A, B its points of intersection with the circle. Again compute the product PAPB of PA and PB. Now vary l. Investigate further by considering the case when l passes through the center of the given circle. Can you reconcile the three values of the product PAPB for P outside, on and inside the given circle? Hint: consider the value of OP 2 r 2 where O is the center of the given circle and r is its radius. End of Demonstration 2.9.0. The value of PAPB in Demonstration 2.9.0 is often called the power of P with respect to the given circle. Now complete the following statement. 2.9.1 Theorem. Let P be a given point,  EMBED Equation.3 a given circle, and l a line through P intersecting  EMBED Equation.3 at A and B. Then the product PAPB of the distances from P to A and B is ________________ whenever P is outside, whenever it is inside or when it is on  EMBED Equation.3 ; the value of the product PAPB is equal to _________________ where O is the center of  EMBED Equation.3  and r is the radius of  EMBED Equation.3 . The proof of part 2 of Theorem 2.9.1 in the case when P 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  EMBED Equation.2  is a tangent to the circle. By the Pythagorean Theorem  EMBED Equation.3  so it suffices to show that PAPB = PC 2. 2.9.2 Theorem. Given a circle  EMBED Equation.3  and a point P outside  EMBED Equation.3 , let l be a ray through P intersecting  EMBED Equation.3  at points A and B. If C is a point on the circle such that  EMBED Equation.2  is a tangent to  EMBED Equation.3  at C then PAPB = PC2. Proof. The equation PAPB = PC2suggests use of similar triangles, but which ones?  Let  EMBED Equation.2  be a diameter of the circle. By the Inscribed Angle Theorem  EMBED Equation.3  and  EMBED Equation.3  is a right angle. Thus  EMBED Equation.3  and as EMBED Equation.2  is tangent to the circle  EMBED Equation.3 . Therefore,  EMBED Equation.3 . By AA similarity  EMBED Equation.3  is similar to  EMBED Equation.3  proving PA/PC = PC/PB or PAPB = PC2. QED Theorem 2.9.3. Given a circle  EMBED Equation.3  and a point P inside  EMBED Equation.3 , let l be a line through P intersecting  EMBED Equation.3  at points A and B. Let  EMBED Equation.3  be the chord perpendicular to the segment  EMBED Equation.3 . Then the value of the product PAPB is equal to  EMBED Equation.3  where O is the center of  EMBED Equation.3  and r is the radius of  EMBED Equation.3 . Proof.  By AA similarity  EMBED Equation.3  is similar to  EMBED Equation.3  so that PA/PC = PD/PB. Thus PA PB=PCPD. By HL,  EMBED Equation.3  is congruent to  EMBED Equation.3  so that PC=PD. By the Pythagorean Theorem  EMBED Equation.3 . Re-arranging and substituting, we obtain EMBED Equation.3  Therefore,  EMBED Equation.3  as desired. QED There is a converse to theorem 2.9.2 that also will be useful later. You will be asked to provide the proof in Exercise 2.11.1 below. 2.9.4 Theorem. Given a circle  EMBED Equation.3  and a point P outside  EMBED Equation.3 , let l be a ray through P intersecting  EMBED Equation.3  at points A and B. If C is a point on  EMBED Equation.3  such that PAPB = PC2, then  EMBED Equation.2  is a tangent to  EMBED Equation.3  at C. In Chapter 1 we used Sketchpad to discover that when a point P moves so that the distance from P to two fixed points A, B satisfies the condition PA = 2PB then the path traced out by P is a circle. In fact, the locus of a point P such that  EMBED Equation.3  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, this is a misnomer since Aristotle who had used it to give a mathematical justification of the semicircular form of the rainbow had already known the result. That this locus is a circle was confirmed 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 of Apollonius with respect to fixed points A, B through an arbitrary point P we shall go through the proof now. The proof requires several lemmas, which we consider below. 2.9.5 Lemma Given  EMBED Equation.3 , let D be on EMBED Equation.2 , and E on  EMBED Equation.2  such that EMBED Equation.2  is parallel to EMBED Equation.2 . Then  EMBED Equation.3  and  EMBED Equation.3 .Proof. Let F be the intersection of EMBED Equation.2  with the line parallel to EMBED Equation.2  passing through E. Then  EMBED Equation.3 by AA similarity and  EMBED Equation . The quadrilateral EFBD is a parallelogram, therefore EF=DB and  EMBED Equation . A similar argument shows EMBED Equation . QED 2.9.5a Lemma (Converse of Lemma 2.9.5). Given  EMBED Equation.3 , let D be on EMBED Equation.2 , and E on  EMBED Equation.2  such that  EMBED Equation.3  or  EMBED Equation.3  (see figure below), then EMBED Equation.2  is parallel to EMBED Equation.2 .  Proof. Assume  EMBED Equation.3 . The line through D parallel to  EMBED Equation.2  intersects  EMBED Equation.2  at point F with  EMBED Equation.2  parallel to EMBED Equation.2 . By Lemma 2.9.5,  EMBED Equation.3 . But  EMBED Equation.3  also, so  EMBED Equation.3  which implies that F = E. Thus  EMBED Equation.2  =  EMBED Equation.2  is parallel to EMBED Equation.2 . If  EMBED Equation.3 , the proof is similar. QED 2.9.6 Theorem The bisector of the internal angle  EMBED Equation.2  of  EMBED Equation.3  divides the opposite side  EMBED Equation.2  in the ratio of the adjacent sides  EMBED Equation.2  and  EMBED Equation.2 . In other words,  EMBED Equation . Proof. Suppose  EMBED Equation.3  bisects  EMBED Equation  in  EMBED Equation.3 . At C construct a line parallel to  EMBED Equation , intersecting  EMBED Equation  at E, producing the figure below. But then  EMBED Equation  and  EMBED Equation  since they are corresponding angles of parallel lines. In addition,  EMBED Equation  since they are alternate interior angles of parallel lines. Hence  EMBED Equation.3  is isosceles and BE = BC. By the previous lemma  EMBED Equation . But BE=BC, so  EMBED Equation . This completes the proof. QED 2.9.7 Exercise. The converse to Theorem 2.9.6 states that if  EMBED Equation , then  EMBED Equation.3  bisects  EMBED Equation  in the figure above. Prove this converse. You may use the converse to Lemma 2.9.5, proven in Lemma 2.9.5a. 2.9.8. Theorem The bisector of an external angle of  EMBED Equation.3  cuts the extended opposite side at a point determined by the ratio of the adjacent sides. That is to say, if  EMBED Equation  is extended and intersects the line containing the bisector of the exterior angle of C at E, then  EMBED Equation . Proof: There are two cases to consider. Either m EMBED Equation  < m EMBED Equation  or m EMBED Equation  > m EMBED Equation . (If m EMBED Equation  =m EMBED Equation , then the bisector of the exterior angle at C is parallel to  EMBED Equation .)  Assume that m EMBED Equation  < m EMBED Equation . Then (as shown in the figure) the bisector of  EMBED Equation  will intersect the extension of  EMBED Equation  at E, and AE > AB. At B, construct a line parallel to  EMBED Equation , intersecting  EMBED Equation  at F. Then  EMBED Equation  since they are corresponding angles of parallel lines; And  EMBED Equation  since  EMBED Equation  bisects  EMBED Equation ; and  EMBED Equation  since they are alternate interior angles of parallel lines. Hence  EMBED Equation.3 BFC is isosceles and FC = BC. Now by a previous lemma,  EMBED Equation . But FC=BC; so  EMBED Equation . This proves the assertion for the case when m EMBED Equation  < m EMBED Equation . If m EMBED Equation  > m EMBED Equation , then the line containing the bisector of  EMBED Equation  intersects the extension of  EMBED Equation  at point E on the other side of A, with A between E and B. A similar argument proves the assertion for this case as well and the theorem is proved. QED 2.9.9 Exercise. The converse to Theorem 2.9.8 states that if  EMBED Equation  in the figure above, then  EMBED Equation  bisects the external angle of  EMBED Equation.3  at C. Prove this conjecture. We are now able to complete the proof of the main theorem. 2.9.10 Theorem (Circle of Apollonius). The set of all points P such that the ratio of the distances to two fixed points A and B (that is  EMBED Equation.3 ) is constant (but not equal to 1) is a circle.  Proof: Assume the notation above and that  EMBED Equation.3 where  EMBED Equation.3  is a constant. There are two points on  EMBED Equation.3  indicated by C and D in the figure with the desired ratio. By the converse to Theorem 2.9.6 and the converse to Theorem 2.9.8,  EMBED Equation.3 and  EMBED Equation.3  are the internal and external angle bisectors of the angle at P. Thus they are perpendicular (why?), so  EMBED Equation.3 is a right angle. This means that P lies on a circle with diameter  EMBED Equation.3 . QED In the previous proof what happens in the case where  EMBED Equation.3 ? Also, see Exercise 2.11.2 . 2.10 TILINGS OF THE EUCLIDEAN PLANE. The appeal of many of the most interesting decorations or constructions we see around us, whether manufactured or in nature, is due to underlying symmetries. Two good illustrations of this are the so-called Devils and Angels designs by the Dutch graphic artist M. C. Escher. Underlying both is the idea of tilings of the plane, in the first example the Euclidean plane, in the second example the hyperbolic plane.   But examples can be found everywhere from floor coverings, to wallpaper, to the mosaics of Roman villas and to decorations of structures as varied as Highway 183 in Austin and Islamic mosques. An understanding of the geometry underlying these designs and their symmetries increases our understanding and appreciation of the artistic design as well as geometry itself. The classification of these symmetries is actually a fascinating problem linking both algebra and geometry, as we shall see later. Some of the simplest, yet most striking designs come from tilings by regular polygons or by congruent polygons. Examples can be found everywhere in Islamic art because of the ban imposed by the Koran on the use of living forms in decoration and art. This style of ornamentation is especially adapted to surface decoration since it is strongly rooted in Euclidean plane geometry. Sketchpad will enable us to reproduce these complicated and colorful designs. Once the underlying geometry has been understood, however, we can make our own designs and so learn a lot of Euclidean plane geometry in the process. Four examples illustrate some of the basic ideas.  Example  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 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 edge-to-edge. To describe all these possibilities at once what we want is a general definition of coverings of the plane by polygons without overlaps. Specializations of this definition can then be made when the polygons have special features such as the ones in the first four examples. 2.10.2. Definition. A tiling or tessellation of the Euclidean plane is a collection T1, T2, ... , Tn, of polygons and their interiors such that no two of the tiles have any interior points in common, 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 different 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 first is not regular since neither the six-pointed polygon nor the rhombus is regular. To distinguish the second example from the others we shall make a crucial distinction. 2.10.3. Definition. 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 edge-to-edge, whereas example two is not edge-to-edge. The point of this edge-to-edge 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 a tiling will always mean an edge-to-edge 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 well-known from floor coverings and was given already in example 3 above. 2.10.4. Demonstration. Investigate which regular polygons could be used to create an edge-to-edge regular tiling of order one. Use the 3/Triangle (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 6/Hexagon (By Edge) script - what in nature does your picture remind you of? Now use the 5/Pentagon (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 patterns you can make. One example is given below; can you find others?  End of Demonstration 2.10.4. Can you tile the plane with a regular pentagon? To see why the answer is no we prove the following result. 2.10.5. 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 p-sided polygon tiles the plane with q tiles meeting at each vertex. Since the interior angle of a regular p-sided polygon has measure  EMBED Equation.2 , it follows that  EMBED Equation.2 . But then  EMBED Equation.2 , i.e.,  EMBED Equation.2 . The only integer solutions of this last equation that make geometric sense are the pairs  EMBED Equation.3 = (3,6), (4,4), or (6,3). 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 - thats a design problem we shall return to later. What we shall do first is try to make the tiling more attractive by using more than one regular polygon or by using polygons that need not be regular. Lets look first at the case of an equilateral triangle and a square each having the same edge length. Demonstration 2.10.5a. Construct a regular tiling of order 2 where the order of the polygons is preserved at each vertex. 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. Heres one such example.  Notice that the use of colors can bring out a pattern to the ordering of the polygons at each vertex. As we move in counter-clockwise order around each vertex we go from S(green)  EMBED Equation.2 S(yellow)  EMBED Equation.2 T(white)  EMBED Equation.2 T(blue)  EMBED Equation.2 T(white) (and then back to S(green)) where S = square and T = equilateral triangle. This is one example of an edge-to-edge regular tiling of order two. Consider how many are there. End of Demonstration 2.10.5a. 2.10.6 Theorem. Up to similarity there are exactly eight edge-edge regular tilings of order at least 2, where the cyclic order of the polygons is preserved at each vertex. Keeping the order S EMBED Equation.2 S EMBED Equation.2 T EMBED Equation.2 T EMBED Equation.2 T of squares and triangles produced one such tiling. Convince yourself that S EMBED Equation.2 T EMBED Equation.2 T EMBED Equation.2 S EMBED Equation.2 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 edge-to-edge 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 p-gon, a q-gon, and an r-gon at each vertex, you get the equation  EMBED Equation.2  You can check that (4,8,8), (4,6,12), and (3,12,12) are solutions. (There are a few other solutions as well, but they will not make geometric sense.) Thus S EMBED Equation.2 O EMBED Equation.2 O, S EMBED Equation.2 H EMBED Equation.2 D, and T EMBED Equation.2 D EMBED Equation.2 D all produce tilings, where O stands for Octagon, H for hexagon, and D for Dodecagon. We are still missing three tilings, but you can have fun looking for them! (See Exercise 2.11.3.) 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. Lets consider two instances of this, the second being Pythagoras theorem. The first instance is a theorem known familiarly as Napoleons 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, Napoleons theorem says that the centers of these three equilateral triangles themselves form an equilateral triangle, as we saw in Exercise 1.8.5. The figure below makes this result clearer.  Here D, E, and F are the centers of the three equilateral triangles where by center is meant the common circumcenter, centroid and orthocenter of an equilateral triangle. Napoleons theorem says that  EMBED Equation.3  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 2.11.5. 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 which polygon appears. Now we will see how we can continue the figure above indefinitely and thus tile the plane. One should notice that the edge-to-edge condition imposes severe restrictions on the angles that can occur at a vertex. Label the angles in the original figure as follows.  Of course, the angles of the equilateral triangles are all the same but we have used different 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 of the right-angled triangle and one copy of each of the three different sizes of equilateral triangle will fit around a vertex with no gaps or overlaps. The figure can thus be constructed indefinitely by maintaining the same counter-clockwise order  EMBED Equation.2  at each vertex. Now draw the figure 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. 2.10.6a Demonstration. Open a new sketch and in the top left-hand corner of the screen draw a right-angled triangle as shown in the figure above. Make sure that your construction is dynamic in the sense that the triangle remains right-angled 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 right-angled triangle. Continue adding congruent copies of the right-angled triangle and the equilateral triangles to the sides of the triangles already in your figure. (One way to add congruent copies of the right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) Experiment a little to see what figures can be produced. Check that your construction is dynamic by dragging the vertices of the first right-angled triangle you drew. End of Demonstration 2.10.6a. Heres one that looks as if it might tile the plane if continued indefinitely.  The figure above of the Napoleon Tiling has an overlay of hexagons over it. To see where it came from, apply Napoleons 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 Napoleons 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 figure. The same design could have been obtained by putting a design on each regular hexagon and then tiling the plane with these patterned regular hexagons. This brings out a crucial connection between tilings and the sort of designs that are used for covering walls, floors, ceilings or any flat 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 2.10.1 is a wall-paper design because the portion of the design inside the initial regular hexagon will tile the plane as the figure below clearly shows.  2.10.7. Exercise. Find a square portion of Example 4 in Seection 2.10 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 a translation of two adjacent non-congruent squares. This tiling occurs often 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. 2.11 Exercises. Exercise 2.11.1. Prove Theorem 2.9.4. Given a circle  EMBED Equation.3  and a point P outside  EMBED Equation.3 , let l be a ray through P intersecting  EMBED Equation.3  at points A and B. If C is a point on  EMBED Equation.3  such that PAPB = PC2, then  EMBED Equation.2  is tangent to  EMBED Equation.3  at C. Exercise 2.11.2: Given points A, B and P use Sketchpad to construct the Circle of Apollonius passing through P. In other words, construct the set of points Q such that  EMBED Equation.3  where  EMBED Equation.3 . Exercise 2.11.3. Produce two different order-preserving edge-to-edge regular tilings of order 2, just using triangles and hexagons. Produce an order-preserving edge-to-edge regular tiling of order 3 using triangles, squares, and hexagons. (We now have the eight tilings mentioned in Theorem 2.10.5!) Exercise 2.11.4. Using Sketchpad construct the Napoleon Tiling. Choose a regular hexagon in your figure 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 2.11.5. While the tiling above makes a very convincing case for the truth of Napoleons theorem it doesnt 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 D, E, and F are the centers of the equilateral triangles constructed on the sides of the right-angled triangle  EMBED Equation.3 . Show that length EMBED Equation.2 . Determine also the lengths of  EMBED Equation.2  and  EMBED Equation.2 . (b) If  EMBED Equation.2  and  EMBED Equation.2 , write the values of  EMBED Equation.2 ,  EMBED Equation.2 ,  EMBED Equation.2 , and  EMBED Equation.2  in terms of a, b, and c. (c) Write down the addition formulas for sine and cosine.  EMBED Equation.2  = ,  EMBED Equation.2 = . (d) Let the lengths of  EMBED Equation.2 ,  EMBED Equation.2 , and  EMBED Equation.2  be x, y and z respectively. Use the Law of Cosines to show that  EMBED Equation.2 . Determine corresponding values for x and y. Deduce that x = y = z.  Use all the previous results to finish off a coordinate geometry proof of Napoleons theorem. Exercise 2.11.6. Instead of starting with a right-angled triangle, start with an arbitrary  EMBED Equation.3  and draw equilateral triangles on each of its sides and repeat the previous construction. Open a new sketch and draw a small triangle near the top corner of the screen; label the vertices A, B, and C. By using the Circle By Center+Radius tool you can construct congruent copies of this triangle. Draw one congruent copy of  EMBED Equation.3  in the center of the screen. Draw an equilateral triangle on each of its sides. Continue this construction preserving cyclic order at each vertex to obtain a tiling of the plane. The following figure is one such example. Construct the centers of all the equilateral triangles and draw hexagons as in the case of right-angled triangles. Do you think Napoleons theorem remains valid for any triangle, not just right-angled triangles?  Exercise 2.11.7. Can the plane be tiled by copies of the diagram for Yagloms Theorem (given below) as in the manner of the tiling corresponding to Napoleons Theorem? If so, produce the tiling using Sketchpad. Recall that Yagloms Theorem said if we place squares on the sides of a parallelogram, the centers of the squares also form a square.  2.12 One Final Exercise. Exercise 2.12.1. To the left in the figure 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 figure to the right.  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. Construct the circumcenters of the three outwardly pointing obtuse triangles on the sides of one of the right-angled 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 right-angled triangle? Use Sketchpad if necessary to check any conjecture you make. (Dont forget to drag!) Investigate what happens if you construct instead the three circumcenters of the right-angled 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 right-angled 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. (Dont forget to drag!)  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 FMicrosoft Equation 3.0DNQE Ole nCompObj(*ofObjInfo+qEquation Native r1Equation.3n.39q DABCT|,    .  & _1005402302o0Fǣ<=ȣ<Ole sPIC -/tTMETA vFTimes New Roman6w -!DE--$  & 'dxpr MTHUGrphbj " currentpoint ",Times .+DEt2 /MTsave save def 40 dict begin currentpoint 3 -1 rolPICT .2|CompObjRObjInfo14Ole10Nativel sub neg 3 1 roll sub 608 div 480 3 -1 roll exch div scale currentpoint translate 64 43 translate newpath 38 72 moveto 69 129 lineto 0 64 lineto 69 0 lineto 38 57 lineto 461 57 lineto 430 0 lineto 499 64 lineto 430 129 lineto 461 72 lineto closepath fill /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (DE) -7 405 sh end MTsave restore d"MATH /DEurFMicrosoft Equation 3.0DNQE Equation.3 /DE Equation /DEFMicrosoft Equation 3.0DNQE Equation.3 BCOle10FmtProgID 35 Equation Native 2_1008338312?8F=ȣ<=ȣ<Ole CompObj79RObjInfo:Equation Native 2_947429776 B=F=ȣ<=ȣ<Ole PIC <?TMETA PICT >@ TF@F    .  & Times New Roman6w -!P Times New Roman6w -!1Times New Roman6w -!P Times New Roman6w -!2- & ' dxpr  " currentpoint ",Times .+ P +1 ( P +2"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 704 div 512 3 -1 roll exch div scale currentpoint translate -554 -581 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 618 648 moveto 588 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (P) 624 965 sh (P) 900 965 sh 224 /Times-Roman f1 (1) 774 1061 sh (2) 1068 1061 sh end MTsave restore d;MATH/ P 1 P 2`/ P 1 P 2ObjInfoAEquation Native K_947429846DF=ȣ<=ȣ<Ole TaTa    .  & Times New Roman6w -!P Times New Roman6w -!2Times New RomaPIC CFTMETA PICT EG ObjInfoHn6w -!P Times New Roman6w -!3- & ' dxpr  " currentpoint ",Times .+ P +2 ( P +3"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 736 div 544 3 -1 roll exch div scale currentpoint translate 64 -1669 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 1736 moveto 611 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (P) 6 2053 sh (P) 318 2053 sh 224 /Times-Roman f1 (2) 174 2149 sh (3) 483 2149 sh end MTsave restore d;MATH/ P 2 P 3/ P 2 P 3Equation Native K_947429938;$KF=ȣ<=ȣ<Ole PIC JMTTFTF    .  & Times New Roman6w -!P Times New Roman6w -!nTimes New RomaMETA PICT LN ObjInfoOEquation Native Kn6w -!P Times New Roman6w -!1- & ' dxpr  " currentpoint ",Times .+ P +n ( P +1"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 704 div 544 3 -1 roll exch div scale currentpoint translate 64 -2980 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 3047 moveto 588 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (P) 6 3364 sh (P) 318 3364 sh 224 ns (n) 175 3460 sh 224 /Times-Roman f1 (1) 468 3460 sh end MTsave restore d;MATH/ P n P 1/ P n P 1_947447639RF=ȣ<=ȣ<Ole PIC QTTMETA T>>     .  & Times New Roman6w -!n Symbol`X``6w -! Times New Roma      !"#$%&'(*+,-./0123456789;>ACDEGHIJKLMNOPQRWYZ[]^_`abcdefghmopqstuvwxyz{|}~n6w -!3  & 'w dxpr  "  currentpoint ",Times .+ n, Symbol) ) 3/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 832 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy lengthPICT SUwObjInfoVEquation Native ._136644663YF=ȣ<=ȣ< 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (n) -8 261 sh 384 /Symbol f1 (\263) 273 261 sh 384 /Times-Roman f1 (3) 570 261 sh end MTsave restore dMATH  n3m n3FMicrosoft Equation 3.0DNQE Equation.3n.39q' (n-2)180Ole CompObjXZfObjInfo[Equation Native C_944831838y^Fң<ң<Ole PIC ]`TMETA BT4r4r   < .  & Times New Roman6w -!a Times New Roman6w -!2Symbol`X``6w -!+ Times New Roman6w -!b Times New Roman6w -!2Symbol`X``6w -!= 'Times New Roman6w -!c 1Times New Roman6w -!27 & '+<dxpr  <"< currentpoint ",Times .+ a (2, Symbol ++) b (2 +=) c (72/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1920 div 448 3 -1 roll exch div scale currentpoint translate 64 60 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 PICT _b)+CompObj:YObjInfoac<Equation Native =Z-1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (a) -5 324 sh (b) 727 324 sh (c) 1519 324 sh 224 /Times-Roman f1 (2) 203 152 sh (2) 935 152 sh (2) 1706 152 sh 384 /Symbol f1 (+) 434 324 sh (=) 1197 324 sh end MTsave restore dJMATH>C a 2 +b 2 =c 2۠FMicrosoft Equation Editor 2.0DNQE Equation.2> a 2 +b 2 =c 2_944834015kfFң<ң<Ole ?PIC eh@TMETA BT p    .  & Times New Roman6w -!c & ' dxpr  "  currentpoint ",Times .+cp/PICT giFObjInfojSEquation Native T(_944834195mFң<ң<MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 256 div 288 3 -1 roll exch div scale currentpoint translate 64 -37 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (c) -12 261 sh end MTsave restore dMATH   c`  cTOle UPIC loVTMETA XPICT np\ p    .  & Times New Roman6w -!a & ' dxpr  " currentpoint ",Times .+ao/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 288 div 288 3 -1 roll exch div scale currentpoint translate 64 -37 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (a) -5 261 sh end MTsave restore dMATH  a  aT>> p    .  & ObjInfoqiEquation Native j(_944834231dtFң<ң<Ole kPIC svlTMETA nPICT uwrObjInfoxTimes New Roman6w -!b  & ' dxpr  " currentpoint ",Times .+ bn/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 288 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (b) -9 261 sh end MTsave restore dMATH  bshEquation Native (_944832700{Fң<ң<Ole PIC z}T  bT4r4r   < .  & Times New Roman6w -!a Times New Roman6w META BPICT |CompObjYObjInfo~-!2Symbol`X``6w -!+ Times New Roman6w -!b Times New Roman6w -!2Symbol`X``6w -!= 'Times New Roman6w -!c 1Times New Roman6w -!27 & '< <dxpr  <"< currentpoint ",Times .<<+ a (2, Symbol ++) b (2 +=) c (72/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1920 div 448 3 -1 roll exch div scale currentpoint translate 64 60 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (a) -5 324 sh (b) 727 324 sh (c) 1519 324 sh 224 /Times-Roman f1 (2) 203 152 sh (2) 935 152 sh (2) 1706 152 sh 384 /Symbol f1 (+) 434 324 sh (=) 1197 324 sh end MTsave restore dJMATH>C a 2 +b 2 =c 2FMicrosoft Equation Editor 2.0DNQE Equation.2Equation Native Z_1000620405F(ԣ<(ԣ<Ole CompObjf> a 2 +b 2 =c 2FMicrosoft Equation 3.0DNQE Equation.3n.39q DABCObjInfoEquation Native 1_1008347889Fң<ң<Ole CompObjRObjInfoEquation Native 2_1008347890Fң<ң<FMicrosoft Equation 3.0DNQE Equation.3 CDFMicrosoft Equation 3.0DNQE Equation.3Ole CompObjRObjInfoEquation Native 2 ABFMicrosoft Equation 3.0DNQE Equation.3n.39q DCABFMicrosoft Equation 3.0DNQE _1000619911Fң<ң<Ole CompObjfObjInfoEquation Native 1_1000619941 Fң<ң<Ole CompObjfEquation.3n.39q DDACFMicrosoft Equation 3.0DNQE Equation.3n.39q DCABObjInfoEquation Native 1_1000619993Fң<ң<Ole CompObjfObjInfoEquation Native 1_1000620004Fң<ң<Ole CompObjfObjInfoEquation Native 1FMicrosoft Equation 3.0DNQE Equation.3n.39q DDCBTr|_1001317722Fң<ң<Ole PIC TMETA r    .  & Times New Roman6w -!CD - & 'dxpr MTHUGrphbj " currentpoint ",Times .+ CD"/MTsave save def 40 dict bePICT CompObjRObjInfoOle10Nativegin currentpoint 3 -1 roll sub neg 3 1 roll sub 608 div 448 3 -1 roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 508 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (CD) -12 325 sh end MTsave restore d"MATH CDurFMicrosoft Equation 3.0DNQE Equation.3 CDOle10FmtProgID  Equation Native 2_944835049Fң<ң<Ole  Equation BDT p    .  & Times New Roman6w -!xPIC TMETA PICT ObjInfo & ' dxpr  " currentpoint ",Times .+xp/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 288 div 288 3 -1 roll exch div scale currentpoint translate 64 -613 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (x) 11 837 sh end MTsave restore dMATH  x۠  xEquation Native (_1005403040Fң<ң<Ole PIC T     !"#$%&()*+,-./012349;<=>?@ACDEFGHIJKLMNOPQRSVY[\]^_`abcdfghijklmnopqrstuvwy~Tr|r    .  & Times New Roman6w -!CD - & 'META PICT CompObjRObjInfodxpr MTHUGrphbj " currentpoint ",Times .+ CD"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 608 div 448 3 -1 roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 508 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (CD) -12 325 sh end MTsave restore d"MATH CDurFMicrosoft Equation 3.0DNQE Equation.3 CD Equation ADTOle10NativeOle10FmtProgID  Equation Native 2_944835092Fң<ң<Ole PIC TMETA  PICT 'i     .  & Times New Roman6w -!cSymbol`X``6w -!- Times New Roman6w -!x & 'i dxpr  "  currentpoint ",Times .+c, Symbol)-) x/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 832 div 288 3 -1 roll exch div scale currentpoint translate 64 -1189 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (c) -12 1413 sh (x) 553 1413 sh 384 /Symbol f1 (-) 241 1413 sh end MTsave restore dMATH c-x۠ c-xObjInfo5Equation Native 6._944835272Fң<ң<Ole 7PIC 8TMETA :PICT BMObjInfoTTO0O     .  & Times New Roman6w -!x !a-  Symbol`X``6w -!= Times New Roman6w -!a !c   & 'M dxpr   " currentpoint ",Times .+ x(a" , Symbol( =( a+c" I/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1024 div 896 3 -1 roll exch div scale currentpoint translate 64 -2141 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 2586 moveto 247 0 rlineto stroke 673 2586 moveto 241 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (x) 43 2444 sh (a) 30 2978 sh (a) 700 2444 sh (c) 707 2978 sh 384 /Symbol f1 (=) 355 2685 sh end MTsave restore d6MATH* xa=ac* xa=acEquation Native UF_1000620206Fң<ң<Ole WPIC XTT0 J  1 .  & Times New Roman6w -!c Symbol`X``6w -!- Times New Roman6w -!x !b META ZPICT eCompObjxRObjInfoz-  Symbol`X``6w -!=Times New Roman6w -!b (!c( ' . & '1dxpr  1"1 currentpoint ",Times .+ c, Symbol)-) x( b" (=( (b*c" 'v/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1568 div 896 3 -1 roll exch div scale currentpoint translate -725 -1110 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 789 1555 moveto 789 0 rlineto stroke 2004 1555 moveto 237 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (c) 809 1413 sh (x) 1374 1413 sh (b) 1088 1947 sh (b) 2027 1413 sh (c) 2036 1947 sh 384 /Symbol f1 (-) 1062 1413 sh (=) 1686 1654 sh end MTsave restore dC a 2 +b 2 =c 2FMicrosoft Equation Editor 2.0DNQE Equation.2> a 2 +b 2 =c 2T>#># p    .  & _944835663Fң<ң<Ole PIC TMETA Symbol`X``6w -!  & ' dxpr  " currentpoint ", Symbol .+ Сk/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 384 div 352 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def PICT  ObjInfoEquation Native (_947576056@ Fң<ң</ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh end MTsave restore dMATH i yd  T>>     .  & Times New Roman6w -!n Symbol`X``6w Ole PIC TMETA PICT w-! 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ABFMicrosoft Equation 3.0DNQE Equation.3 ABObjInfoEquation Native 2_1009202386}cFGnɣ<Gnɣ<Ole CompObjRObjInfoEquation Native 2_1009202436!FGnɣ<Gnɣ<Ole CompObj "RObjInfo#Equation Native FFMicrosoft Equation 3.0DNQE Equation.3* APB=90 oTXh_947449533P+&FGnɣ<Gnɣ<Ole PIC %(TMETA !X     .  & Times New Roman6w -!PA - & ' dxpr  "  currentpoint ",Times .+ PAPICT ')&ObjInfo*5Equation Native 62_947449558-FGnɣ<Gnɣ<" /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 576 div 416 3 -1 roll exch div scale currentpoint translate 64 -1708 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 1775 moveto 456 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (PA) 6 2092 sh end MTsave restore d"MATH PA PAOle 7PIC ,/8TMETA :PICT .0?TXhX     .  & Times New Roman6w -!PB - & ' dxpr  "  currentpoint ",Times .+ PB" /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 576 div 416 3 -1 roll exch div scale currentpoint translate 64 -2348 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 2415 moveto 455 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (PB) 6 2732 sh end MTsave restore d"MATH PB M 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/Times-Roman f1 (2) 768 293 sh (180) 1152 542 sh 224 /MT-Extra f1 (o) 1739 370 sh end MTsave restore dLMATH@ n-2n()180 oFMicrosoft Equation Editor 2.0DNQE Equation.2@ n-2n()180 oFMicrosoft Equation 3.0DNQE Equation.3n.39q_1001854468JFGnɣ<Gnɣ<Ole CompObjIKfObjInfoL! 360nFMicrosoft Equation 3.0DNQE Equation.3n.39q DEFEquation Native =_1001854479OFGnɣ<Gnɣ<Ole CompObjNPfObjInfoQEquation Native 1_1001854491MTFGnɣ<Gnɣ<Ole CompObjSUfObjInfoVEquation Native 1_1062931949Y FGnɣ<ʣ<FMicrosoft Equation 3.0DNQE Equation.3n.39q ABC  FMicrosoft Word Picture MSWordDocWord.Picture.89q1Table=CompObjX[hObjInfoWordDocumentZ\K8@8 Normal_HmH sH tH DAD Default Paragraph FontVi@V  Table Normal :V 44 la (k@(No List  4 4 @V4)+,./12500M90f0M90f0M90yM90y443l*+,2$ $ dU$AXO@)*n(  JB @ # S"'JB  # S"&JB  # S"%JB  # S"$JB  # S"#JB @ # S""JB  @ # S"!JB   # S" JB   # S"JB   # o"JB   # S"JB  # S"JB  # S"JB  # S"JB  # S"2  CENGl;HYJ9;QS `T<u`T<u9;<u9;"  6" JB  # S"  6" JB  # S"JB  # tE"JB  # tE"JB  # tE"JB  # tE"JB  # S"  6" JB  # S" JB @ # B^" JB  # tE" JB   # B^" JB ! # tE" P2 " 3 S"P2 # 3 S"P2 $ 3 S"P2 % 3 S"P2 & 3 S"P2 ' 3 S"P2 ( 3 S"P2 ) 3 S"P2 * 3 S"B S  ?  !"#$%&'4*t): a t( &Gt': a t&Ovt% Gt$Ovt#  t": &a t!t _Mt  tF  Gt tTtt  tne  t&KRt&tt,/to  t+<tcCtt^>_?t_`I t_t I J t I t ^/t /0t  I tI J t^I t^_t/ t  tI t)5),-5:(5)5|h)5@~4@UnknownG:Times New Roman5Symbol3& :ArialA"GenevaArial"hYFYF##!>4((3H(?|hDepartment of Mathematics Frank Shirley5@ 4bjbj22 XX) ....  @$$$$$===$8Rv====..$$e=.l$$=,$ V'EM0@i....======= A l R ()*,-/04h|hB*CJ OJQJhphh|hB*CJOJQJhphh|hjh|hUmHnHu)+,./1234 v:5/GV9/ v: )3N N!{""{"#"$"%SummaryInformation(]UDocumentSummaryInformation8]_947451120l`Fʣ<ʣ<Ole ZOh+'0p  , 8 DPX`hssDepartment of Mathematics.epa Normal.dotoFrank ShirleyMa2anMicrosoft Word 10.0@F#@ME@\E#Z՜.+,0 hp  "The University of Texas at Austina(A  TitlePIC _bTMETA "PICT acObjInfodT#0#     .  & Times New Roman6w -!1 !2-   & ' dxpr   " currentpoint ",Times .+ 1*2" /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 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BbJ:0 0T6?6)\-+lUJ:0 (U&?ΡB벃P.u ͥC+WWz;|7yo*UvP<y|GW PIC CEvTPICT xkCompObjDHRObjInfo k %dxpr MTHU %Grphbj %" % currentpoint ", Symbol .+ ,Times) ABC/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1184 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh 384 /Times-Roman f1 (ABC) 285 261 sh end MTsave restore d!MATH ABCcuFMicrosoft Equation 3.0DNQE Equation.3 ABC Equation ABCFMicrosoft Equation 3.0DNQE Equation.3Ole10NativeGIOle10FmtProgID J Equation Native 1_1008360120BPMF2Σ<2Σ<Ole CompObjLNRObjInfoOEquation Native 1 DABCTX dxpr MTHU Grphbj "  currentpoint ",Times .+ BD"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 640 div 416 3 -1_1008360121TF2Σ<Σ<Ole PIC QSTPICT  roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 513 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (BD) -7 325 sh end MTsave restore d"MATH BDurFMicrosoft Equation 3.0DNQE Equation.3 BD EquationCompObjRVRObjInfoOle10NativeUWOle10FmtProgID X Equation Native 2_1008360122K]FΣ<Σ<Ole PIC Z\T BDTX| dxpr MTHU Grphbj "  currentpoint ",Times .+ AB"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 608 div 416 3 -1PICT CompObj[_RObjInfoOle10Native^` roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 500 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AB) -5 325 sh end MTsave restore d"MATH ABurFMicrosoft Equation 3.0DNQE Equation.3 AB EquationOle10FmtProgID a Equation Native 2_1008360123fFΣ<Σ<Ole  ABT> Tdxpr MTHU TGrphbj T" T currentpoint ", Symbol .+ ,Times) ABD)@) ) CBD/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3PIC ceTPICT CompObjdhRObjInfo 1 roll sub 2688 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh (@) 1190 261 sh (\320) 1499 261 sh 384 /Times-Roman f1 (ABD) 285 261 sh (CBD) 1793 261 sh end MTsave restore d0MATH$ ABD@CBD-1FMicrosoft Equation 3.0DNQE Equation.3$ ABD@CBDOle10Nativegi(Ole10FmtProgID j Equation Native @_1008360124btoFΣ<Σ< Equation$ ABD@CBDTz>h Rdxpr MTHU RGrphbj R" R currentpoint ", Symbol .+ ,Times) BEC)@) ) ABDOle PIC lnTPICT CompObjmqR/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 2624 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh (@) 1127 261 sh (\320) 1436 261 sh 384 /Times-Roman f1 (BEC) 285 261 sh (ABD) 1730 261 sh end MTsave restore d0MATH$ BEC@ABD-1FMicrosoft Equation 3.0DNQE Equation.3ObjInfoOle10Nativepr(Ole10FmtProgID s Equation Native @$ BEC@ABD Equation$ BEC@ABDTz>h Rdxpr MTHU RGrphbj R" R currentpoint ", Symb_1008360125xFΣ<Σ<Ole PIC uwTPICT                    ! " # $ % ' , / 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A C H K M N O P Q R S T U V W X Y Z [ \ ] _ d g l n o p q r s t u v w x y z | ol .+ ,Times) BCE)@) ) CBD/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 2624 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh (@) 1134 261 sh (\320) 1443 261 sh 384 /Times-Roman f1 (BCE) 285 261 sh (CBD) 1737 261 sh end MTsave restore d0MATH$ BCE@CBD-1FMicrosoft Equation 3.0DNQE Equation.3CompObjvz RObjInfo Ole10Nativey{ (Ole10FmtProgID | $ BCE@CBD Equation$ BCE@CBDFMicrosoft Equation 3.0DNQE Equation.3n.39qEquation Native  @_1009110761;FΣ<Σ<Ole  CompObj~ fObjInfo Equation Native  1_1008360126kFΣ<Σ<Ole   DCBETz$0k5dxpr MTHU 5Grphbj 5"5 currentpoint ",Times .+ AB+BE" , Symbol(=( !AD+DC" D/MTsave save def 40 dict begin currentpPIC  TPICT  kCompObj& RObjInfo( oint 3 -1 roll sub neg 3 1 roll sub 1696 div 896 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 564 0 rlineto stroke 990 403 moveto 600 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AB) 27 261 sh (BE) 36 795 sh (AD) 1017 261 sh (DC) 1026 795 sh 384 /Symbol f1 (=) 672 502 sh end MTsave restore dBMATH6 ABBE=ADDC1 FMicrosoft Equation 3.0DNQE Equation.36 ABBE=ADDC Equation6 ABBE=ADDCTz$0Ole10Native) :Ole10FmtProgID * Equation Native + R_1008360127FΣ<Σ<Ole - PIC . TPICT 0 jCompObjB Rj5dxpr MTHU 5Grphbj 5"5 currentpoint ",Times .+ AB*BC" , Symbol(=( !AD+DC" D/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1696 div 896 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 564 0 rlineto stroke 990 403 moveto 600 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AB) 27 261 sh (BC) 29 795 sh (AD) 1017 261 sh (DC) 1026 795 sh 384 /Symbol f1 (=) 672 502 sh end MTsave restore dBMATH6 ABBC=ADDC1 FMicrosoft Equation 3.0DNQE Equation.36 ABBC=ADDC Equation6 ABBC=ObjInfoD Ole10NativeE :Ole10FmtProgID F Equation Native G RADDCTz$0j5dxpr MTHU 5Grphbj 5"5 currentpoint ",Times .+ AB*BC" , Symbol(=( !AD+DC" D/MTsave save def 40 dict begin currentpo_1008358605)FΣ<Σ<Ole I PIC J TPICT L jint 3 -1 roll sub neg 3 1 roll sub 1696 div 896 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 564 0 rlineto stroke 990 403 moveto 600 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AB) 27 261 sh (BC) 29 795 sh (AD) 1017 261 sh (DC) 1026 795 sh 384 /Symbol f1 (=) 672 502 sh end MTsave restore dBMATH6 ABBC=ADDC1 FMicrosoft Equation 3.0DNQE Equation.3CompObj^ RObjInfo` Ole10Nativea :Ole10FmtProgID b 6 ABBC=ADDC Equation6 ABBC=ADDCFMicrosoft Equation 3.0DNQE Equation.3Equation Native c R_1008358526,FΣ<Σ<Ole e CompObjf RObjInfoh Equation Native i 2_1008358527FΣ<Σ<Ole j  BDT>k %dxpr MTHU %Grphbj %" % currentpoint ", Symbol .+ ,Times) ABC/MTsave save def 40PIC k TPICT m kCompObj{ RObjInfo}  dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1184 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh 384 /Times-Roman f1 (ABC) 285 261 sh end MTsave restore d!MATH ABCcuFMicrosoft Equation 3.0DNQE Equation.3 ABC Equation!Ole10Native~ Ole10FmtProgID  Equation Native  1_1008360129Fϣ<ϣ<  ABCTX| dxpr MTHU Grphbj "  currentpoint ",Times .+ AB"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 608 div 416 3 -1Ole  PIC  TPICT  CompObj R roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 500 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AB) -5 325 sh end MTsave restore d"MATH ABurFMicrosoft Equation 3.0DNQE Equation.3 AB EquationObjInfo Ole10Native Ole10FmtProgID  Equation Native  2 ABTz$0k5dxpr MTHU 5Grphbj 5"5 currentpoint ",Times .+ AC+BC" , Symbol(=( "AE*BE" !E/MTsave save def 40 dict begin currentpo_1008360130FΣ<Σ<Ole  PIC  TPICT  kint 3 -1 roll sub neg 3 1 roll sub 1696 div 896 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 580 0 rlineto stroke 1006 403 moveto 565 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AC) 27 261 sh (BC) 37 795 sh (AE) 1033 261 sh (BE) 1042 795 sh 384 /Symbol f1 (=) 688 502 sh end MTsave restore dBMATH6 ACBC=AEBE1 FMicrosoft Equation 3.0DNQE Equation.3CompObj RObjInfo Ole10Native :Ole10FmtProgID  6 ACBC=AEBE Equation6 ACBC=AEBET_>TEquation Native  R_1063424436Fϣ<ϣ<Ole  PIC  TCompObj fObjInfo Ole10Native (Ole10FmtProgID   FMicrosoft Equation 3.0 DS Equation Equation.39q$ BFC@ECG EquationtII  "BAC T_>TEquation Native  8_1063424453Fϣ<ϣ<Ole  PIC  T FMicrosoft Equation 3.0 DS Equation Equation.39qIXI  "ABCIT_>TCompObj fObjInfo Equation Native  8_1063424480Fϣ<ϣ<Ole  PIC  TCompObj fObjInfo  FMicrosoft Equation 3.0 DS Equation Equation.39qtII  "BAC T_>T FMicrosoft Equation 3.0 DS EqEquation Native  8_1063424479Fϣ<ϣ<Ole  PIC  TCompObj fObjInfo Equation Native  8_1063424932Fϣ<ϣ<uation Equation.39qIXI  "ABC T_>T FMicrosoft Equation 3.0 DS Equation Equation.39qOle  PIC  TCompObj fObjInfo $ BFC@ECG EquationtII  "BCGTrhdxpr MTHUGrphbj Ole10Native (Ole10FmtProgID  Equation Native  8_1008360131 F(dϣ<(dϣ<Ole  PIC  TPICT  CompObj R" currentpoint ",Times .+ CE" /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 576 div 448 3 -1 roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 473 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (CE) -12 325 sh end MTsave restore d"MATH CEurFMicrosoft Equation 3.0DNQE Equation.3ObjInfo Ole10Native Ole10FmtProgID  Equation Native  2 CE Equation CETrdxpr MTHUGrphbj _1008360132F(dϣ<(dϣ<Ole  PIC  TPICT              " $ + - . / 0 1 2 3 4 5 6 7 8 9 : ; = D I K L M N O P Q R S T U V W X Y Z [ ] b e g h i j k l m n o p q r s t u v w y ~ " currentpoint ",Times .+ AC"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 640 div 448 3 -1 roll exch div scale currentpoint translate 64 59 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 8 moveto 516 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AC) -5 325 sh end MTsave restore d"MATH ACurFMicrosoft Equation 3.0DNQE Equation.3CompObj RObjInfo Ole10Native Ole10FmtProgID   AC Equation ACT_>T FMicrosoft Equation 3.0 DS EqEquation Native  2_1071747430F(dϣ<(dϣ<Ole  PIC  TCompObj fObjInfo Ole10Native (Ole10FmtProgID  uation Equation.39q$ BFC@ECG Equation-xxs  "BFCE" "ECGT_>TEquation Native  I_1071747371F(dϣ<(dϣ<Ole  PIC  T FMicrosoft Equation 3.0 DS Equation Equation.39q$ BFC@ECG Equation-hT FMicrosoft Equation 3.0 DS Equation Equation.39q$ BFC@ECG EquationCompObj# fObjInfo% Ole10Native& (Ole10FmtProgID ' Equation Native ( 5_1008360134  F(dϣ<ϣ<Ole ) PIC   * Tt  "BCGTE>@ Pdxpr MTHU PGrphbj P" P currentpoint ", Symbol .+ ,Times) BCE)@) ) CBF/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3PICT , CompObj < RObjInfo> Ole10Native? ( 1 roll sub 2560 div 384 3 -1 roll exch div scale currentpoint translate 64 59 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 261 sh (@) 1134 261 sh (\320) 1443 261 sh 384 /Times-Roman f1 (BCE) 285 261 sh (CBF) 1737 261 sh end MTsave restore d0MATH$ BCE@CBF-1FMicrosoft Equation 3.0DNQE Equation.3$ BCE@CBFOle10FmtProgID @ Equation Native A @_980526997UFϣ<ϣ<Ole B  Equation$ BCE@CBF FMicrosoft Equation 3.0 DS Equation Equation.39qa Ԍ CompObjC fObjInfoE Equation Native F )_1008360135Fϣ<ϣ<Ole G PIC H TPICT J kCompObj\ RTz$0k5dxpr MTHU 5Grphbj 5"5 currentpoint ",Times .+ AC+FC" , Symbol(=( "AE*BE" !E/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1696 div 896 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 580 0 rlineto stroke 1006 403 moveto 565 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (AC) 27 261 sh (FC) 60 795 sh (AE) 1033 261 sh (BE) 1042 795 sh 384 /Symbol f1 (=) 688 502 sh end MTsave restore dBMATH6 ACFC=AEBE1 FMicrosoft Equation 3.0DNQE Equation.36 ACFC=AEBEObjInfo^ Ole10Native_ :Ole10FmtProgID ` Equation Native a R Equation6 ACFC=AEBETz$0k5dxpr MTHU 5Grphbj 5"5 currentpoint ",Time0xx]=0 ЊJ q -gL0;) Xr9y pw b!)" ze"~^bSa2]qxnkonҞ:K{M@:]4)^/3ǨR*HxDdB  S A? 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" # $ % & ' ( ) * + , - . / 0 3 6 8 9 : ; < = > ? @ A B C D E G H I J K L M N O P Q R S T U V W Z ] b d e f h i j k l m n o p q r s x z { | ~  Times New Roman6w -!q Symbol`X``6w -!p Times New Roman6w -!( !1 Symbol`X``6w -!- Times New Roman6w -!2 #!/ +Times New Roman6w -!p 1Times New Roman6w -!) 8Symbol`X``6w -!= >Times New Roman6w -!2 HSymbol`X``6w -!p N & 'FMicrosoft Equation 3.0DNQE Equation.3n.39q6 q180(1-2/p)=360T`X`   4 .ObjInfou Equation Native  R_979679210xFUУ<UУ<Ole  PIC wz TMETA  NPICT y{ ObjInfo|1   & Times New Roman6w -!1 Times New Roman6w -!p-  Symbol`X``6w -!+ Times New Roman6w -!1 Times New Roman6w -!q  Symbol`X``6w -!=!Times New Roman6w -!1 ,!2, + 2 & '4dxpr  4"4 currentpoint ",Times .+ 1*p" , Symbol( +( 1*q" (!=( ,1*2" +/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1664 div 960 3 -1 roll exch div scale currentpoint translate 64 42 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 403 moveto 270 0 rlineto stroke 665 403 moveto 242 0 rlineto stroke 1333 403 moveto 235 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (1) 39 261 sh (1) 690 261 sh (1) 1354 261 sh (2) 1354 795 sh 384 /Times-Italic f1 (p) 60 795 sh (q) 690 795 sh 384 /Symbol f1 (+) 363 502 sh (=) 1015 502 sh end MTsave restore dHMATH< 1p+1q=12at< 1p+1q=12Trr   W .Equation Native 2 X_979679279vFUУ<UУ<Ole 4 PIC ~5 TMETA 7 PICT F gObjInfoX Equation Native Y I  & Times New Roman6w -!( Times New Roman6w -!p Symbol`X``6w -!- Times New Roman6w -!2 !)( Times New Roman6w -!q &Symbol`X``6w -!- /Times New Roman6w -!2 8!) >Symbol`X``6w -!= ETimes New Roman6w -!4 O & 'gWdxpr  W"W currentpoint ",Times .+ ()p, Symbol) -) 2))()q) -) 2)))=) 4V/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 2784 div 448 3 -1 roll exch div scale currentpoint translate 64 -1047 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (\() -18 1367 sh (\)\() 917 1367 sh (\)) 1951 1367 sh 384 /Times-Italic f1 (p) 156 1367 sh (q) 1183 1367 sh 384 /Symbol f1 (-) 426 1367 sh (-) 1460 1367 sh (=) 2168 1367 sh 384 /Times-Roman f1 (2) 716 1367 sh (2) 1750 1367 sh (4) 2482 1367 sh end MTsave restore d9MATH- 3 (p-2)(q-2)=4de- (p-2)(q-2)=4 FMicrosoft Equation 3.0 DS Equation Equation.39q8v (p,q)_980074079&wFUУ<UУ<Ole [ CompObj\ fObjInfo^ Equation Native _ 9_979886237FUУ<UУ<Ole ` PIC a TTr r  p    .  & Symbol`X``6w -! & ' dxpr  " META c PICT g ObjInfot Equation Native u ( currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Tr _979886250FUУ<UУ<Ole v PIC w TMETA y r  p    .  & Symbol`X``6w -! & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 PICT } ObjInfo Equation Native  (_979886273FUУ<UУ< roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Tr r  p    .Ole  PIC  TMETA  PICT    & Symbol`X``6w -! & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G ObjInfo Equation Native  (_979886289FUУ<UУ<Ole    Tr r  p    .  & Symbol`X``6w -!PIC  TMETA  PICT  ObjInfo  & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Equation Native  (_979886539F-ѣ<-ѣ<Ole  PIC  TTr r  p    .  & Symbol`X``6w -! & ' dxpr  " META  PICT  ObjInfo Equation Native  ( currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Tr _979886556F-ѣ<-ѣ<Ole  PIC  TMETA  r  p    .  & Symbol`X``6w -! & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 PICT  ObjInfo Equation Native  (_979886563F-ѣ<-ѣ<roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Tr r  p    .Ole  PIC  TMETA  PICT    & Symbol`X``6w -! & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G ObjInfo Equation Native  (_979886571F-ѣ<-ѣ<Ole    Tr r  p    .  & Symbol`X``6w -!PIC  TMETA  PICT  ObjInfo                    " # & + - . / 0 1 2 3 4 5 6 7 8 9 : ; < = ? @ A B C D E F G H I J K L M N O R U Z _ d i n s u v w x y z { | } ~  & ' dxpr  "  currentpoint ", Symbol .+l/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 448 div 320 3 -1 roll exch div scale currentpoint translate 64 -5 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\256) -18 261 sh end MTsave restore dMATH G   Equation Native  (_136653956\F-ѣ<-ѣ<Ole  PIC  TTr8r U  6 .  & Symbol`X``6w -!p Times New Roman6w -!( !1 META  CompObj fObjInfo Equation Native ! 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PTimes New Roman6w -!b ^Symbol`X``6w -! gTimes New Roman6w -!d v & '_~dxpr  ~"~ currentpoint ",Times .+ a, Symbol) )e))c))f))b) )d8/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 4032 div 448 3 -1 roll exch div scale currentpoint translate 64 -1222 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (a) -5 1542 sh (e) 738 1542 sh (c) 1458 1542 sh (f) 2249 1542 sh (b) 2973 1542 sh (d) 3724 1542 sh 384 /Symbol f1 (\256) 271 1542 sh (\256) 990 1542 sh (\256) 1713 1542 sh (\256) 2502 1542 sh (\256) 3249 1542 sh end MTsave restore d6MATH*  aecfbd* aecfbdFMicrosoft Equation 3.0DNQE Equation.3n.39q  S_1001494131F-ѣ<-ѣ<Ole S CompObjT fObjInfoV Equation Native W (_1001494151F-ѣ<-ѣ<Ole X CompObjY fFMicrosoft Equation 3.0DNQE Equation.3n.39q  SFMicrosoft Equation 3.0DNQE Equation.3n.39qObjInfo[ Equation Native \ (_1002954934F-ѣ<-ѣ<Ole ] CompObj^ fObjInfo` Equation Native a (_1001408674F-ѣ<-ѣ<  SFMicrosoft Equation 3.0DNQE Equation.3n.39q QA=mQBOle b CompObjc fObjInfoe Equation Native f 7_1001408656F-ѣ<-ѣ<Ole g CompObjh fObjInfoj FMicrosoft Equation 3.0DNQE Equation.3n.39q$ )PAPB=mFMicrosoft Equation 3.0DNQE Equation.3n.39qEquation Native k @_1001507713\F-ѣ<-ѣ<Ole l CompObjm fObjInfoo Equation Native p 1_951078194F-ѣ<-ѣ<Ole q  DABCT`@ M  8 .  & Times New Roman6w -!BF -PIC r TMETA t PICT  ObjInfo Symbol`X``6w -!= Times New Roman6w -!c Times New Roman6w -!3 0 ( )- * ,- ,//5& # & '8dxpr  8"8 currentpoint ",Times .+ BF" , Symbol)=) c)3" (#" ,#"& /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1792 div 512 3 -1 roll exch div scale currentpoint translate 64 -2172 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 2239 moveto 477 0 rlineto stroke /sqr { 3 index div /thick exch def gsave translate dup dup neg scale dup 4 -1 roll exch div 3 1 roll div 0 setlinewidth newpath 0 0 moveto dup .34 mul 0 exch lineto .375 .214 rlineto dup thick add dup .375 exch lineto 2 index exch lineto dup thick 2 div sub dup 3 index exch lineto .6 exch lineto .375 0 lineto clip thick setlinewidth newpath dup .34 mul 0 exch moveto .15 .085 rlineto .375 0 lineto thick 2 div sub dup .6 exch lineto lineto stroke grestore } def 441 378 384 1239 2593 16 sqr /stb { newpath moveto 0 setlinewidth 2 copy rlineto } def /enb { rlineto neg exch neg exch rlineto closepath fill } def /hb { stb 0 thick enb } def /vb { stb thick 0 enb } def -136 424 1193 2215 vb /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (BF) 10 2556 sh (c) 891 2556 sh 384 /Symbol f1 (=) 585 2556 sh 384 /Times-Roman f1 (3) 1483 2556 sh end MTsave restore d=MATH1 BF=)c 3 ؠ1 BF=)c 3 TX|Equation Native  M_951078247}F-ѣ<-ѣ<Ole  PIC  TX     .  & Times New Roman6w -!AD - & ' dxpr  "  currentpoint ",Times .+ ADMETA  PICT  ObjInfo Equation Native  2"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 608 div 416 3 -1 roll exch div scale currentpoint translate 64 -2812 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 2879 moveto 508 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (AD) 17 3196 sh end MTsave restore d"MATH ADDe AD_951078295F-ѣ<ѣ<Ole  PIC  TMETA  TXhX     .  & Times New Roman6w -!BE - & ' dxpr  " PICT  ObjInfo Equation Native  2_951076993 Fѣ<ѣ< currentpoint ",Times .+ BE" /MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 576 div 416 3 -1 roll exch div scale currentpoint translate 64 -3452 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 3519 moveto 477 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (BE) 10 3836 sh end MTsave restore d"MATH BE M BEOle  PIC   TMETA  PICT    T>8>    6 .  & Symbol`X``6w -! Times New Roman6w -!ABC Symbol`X``6w -!= $Symbol`X``6w -!q - & ' 6dxpr  6" 6 currentpoint ", Symbol .+ ,Times) ABC)=) q/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1728 div 384 3 -1 roll exch div scale currentpoint translate 64 56 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 264 sh (=) 1111 264 sh 384 /Times-Italic f1 (ABC) 285 264 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (q) 1391 264 sh end MTsave restore d'MATH ABC=qnd ABC=qTzr$ObjInfo  Equation Native  7_951077049Fѣ<ѣ<Ole  PIC  TMETA  PICT  ObjInfo zr   5 .  & Symbol`X``6w -! Times New Roman6w -!CAB Symbol`X``6w -!= $Symbol`X``6w -!f - & '                        % ' ( ) * , - . / 0 1 2 3 4 5 6 7 8 9 > @ A B C E F G H I J K L M N O P Q R W Y Z [ \ ^ _ ` a b c d e f g h i j k p r s t u v w x y z | } ~  5dxpr  5"5 currentpoint ", Symbol .+ ,Times) CAB)=) f/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1696 div 448 3 -1 roll exch div scale currentpoint translate 64 -520 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Symbol f1 (\320) -9 840 sh (=) 1098 840 sh 384 /Times-Italic f1 (CAB) 285 840 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (f) 1389 840 sh end MTsave restore d'MATH CAB=f CAB=fT{>{>     .  & Times New Roman6w -!siEquation Native  7_951077119Fѣ<ѣ<Ole  PIC  TMETA  @PICT  ObjInfo! Equation Native " 1n Symbol`X``6w -!q  & ' dxpr  "  currentpoint ",Times .+ sin, Symbol)q/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 768 div 384 3 -1 roll exch div scale currentpoint translate 64 -1102 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (sin) -19 1422 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (q) 450 1422 sh end MTsave restore d!MATH sinq sinqT>>     .  & _951077155*Fѣ<ѣ<Ole # PIC  $ TMETA & @Times New Roman6w -!cos Symbol`X``6w -!q  & ' dxpr  "  currentpoint ",Times .+ cos, Symbol)q/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 832 div 384 3 -1 roll exch div scale currentpoint translate 64 -1681 translate /cat { dup length PICT !+ ObjInfo": Equation Native ; 1_951077179%Fѣ<ѣ<2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (cos) -9 2001 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (q) 512 2001 sh end MTsave restore d!MATH  cosq  cosqT{r{r    .Ole < PIC $'= TMETA ? @PICT &(D   & Times New Roman6w -!sin Symbol`X``6w -!f  & 'dxpr  " currentpoint ",Times .+ sin, Symbol)f/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 768 div 448 3 -1 roll exch div scale currentpoint translate 64 -2257 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (sin) -19 2577 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (f) 461 2577 sh end MTsave restore d!MATH sinf  sinfTrObjInfo)S Equation Native T 1_951077204#,Fѣ<ѣ<Ole U PIC +.V TMETA X @PICT -/] ObjInfo0l r    .  & Times New Roman6w -!cos Symbol`X``6w -!f  & 'dxpr  " currentpoint ",Times .+ cos, Symbol)f/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 832 div 448 3 -1 roll exch div scale currentpoint translate 64 -2836 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (cos) -9 3156 sh /f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def 384 /Symbol f2 (f) 523 3156 sh end MTsave restore d!MATH cosf  cosfTr8Equation Native m 1_102378513593Fѣ<ѣ<Ole n PIC 25o Tr   6 .  & Times New Roman6w -!cos( Times New Roman6w -!A Symbol`X``6w -!+ Times New Roman6w -!META q BPICT 47{ CompObj RObjInfo68 B *Times New Roman6w -!) 1 & '6dxpr  6"6 currentpoint ",Times .+ cos()A, Symbol) +) B))/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1728 div 448 3 -1 roll exch div scale currentpoint translate 64 -3988 trans late /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (cos\() -9 4308 sh (\)) 1520 4308 sh 384 /Times-Italic f1 (A) 665 4308 sh (B) 1286 4308 sh 384 /Symbol f1 (+) 974 4308 sh end MTsave restore d-MATH! cos(A+B)FMicrosoft Equation 3.0DNQE Equation.3! cos(u+v)T`rEquation Native  =_1023785174;Fѣ<ѣ<Ole  PIC := T4dxpr  4"4 currentpoint ",Times .+ sin()A, Symbol) +) B))/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 1664 div 448 3 -1 roll exch div scale currentpoint translate 64 -4564 translate /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDiPICT  CompObj<> RObjInfo? Equation Native  =rectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Roman f1 (sin\() -19 4884 sh (\)) 1446 4884 sh 384 /Times-Italic f1 (A) 591 4884 sh (B) 1212 4884 sh 384 /Symbol f1 (+) 900 4884 sh end MTsave restore d-MATH! sin(A+B)aFMicrosoft Equation 3.0DNQE Equation.3! sin(u+v)TXh_951078449NBFѣ<ѣ<Ole  PIC AD TMETA  X     .  & Times New Roman6w -!FE - & ' dxpr  "  currentpoint ",Times .+ FE" /MTsave save def 40 dict begin currentpoint 3 -1 rolPICT CE ObjInfoF Equation Native  2_951078484IFѣ<ѣ<l sub neg 3 1 roll sub 576 div 416 3 -1 roll exch div scale currentpoint translate 64 -4092 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 4159 moveto 473 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (FE) 6 4476 sh end MTsave restore d"MATH FE M FETX              _          ! " # $ % & ' ( ) * + , - . / 0 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A B C D E F G H I J K L M N O P Q R S T U z W X Y Z [ \ ] ^ h a b c d e f g j k l m n o p r s t u v w x { | } ~  Ole  PIC HK TMETA  PICT JL X     .  & Times New Roman6w -!DF - & ' dxpr  "  currentpoint ",Times .+ DF"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 640 div 416 3 -1 roll exch div scale currentpoint translate 64 -4732 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 4799 moveto 520 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (DF) 10 5116 sh end MTsave restore d"MATH DF DFObjInfoM Equation Native  2_951078519GxPFѣ<ѣ<Ole  PIC OR TMETA  PICT QS ObjInfoT TXX     .  & Times New Roman6w -!DE - & ' dxpr  "  currentpoint ",Times .+ DE"/MTsave save def 40 dict begin currentpoint 3 -1 roll sub neg 3 1 roll sub 640 div 416 3 -1 roll exch div scale currentpoint translate 64 -5372 translate /thick 0 def /th { dup setlinewidth /thick exch def } def 16 th 0 5439 moveto 520 0 rlineto stroke /cat { dup length 2 index length add string dup dup 5 -1 roll exch copy length 4 -1 roll putinterval } def /ff { dup FontDirectory exch known not { dup dup length string cvs (|______) exch cat dup FontDirectory exch known {exch} if pop } if findfont } def /fs 0 def /cf 0 def /sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def /f1 {ff dup /cf exch def sf} def /ns {cf sf} def /sh {moveto show} def 384 /Times-Italic f1 (DE) 10 5756 sh end MTsave restore d"MATH DE M DEEquation Native  2_980074489WFѣ<ѣ<Ole  PIC VY TTx 0x    .  & Times New Roman6w -!zTimes New Roman6w -!2 Symbol`X``META  CompObjXZ fObjInfo[ Equation Native    b $).369<?BEHKPUX[`cfkprstuvwxy{~6w -!=Times New Roman6w -!1 !3-  Times New Roman6w -!a%Times New Roman6w -!2 +Symbol`X``6w -!+3Times New Roman6w -!b<Times New Roman6w -!2 BSymbol`X``6w -!+ITimes New Roman6w -!2STimes New Roman6w -!abYTimes New Roman6w -!cosf!30w!Symbol`X``6w -!(!!) & ' FMicrosoft Equation 3.0 DS Equation Equation.39qb z 2 =13a 2 +b 2 +2abcos30Times New Roman()FMicrosoft Equation 3.0DNQE Equation.3n.39q DABCFMicrosoft Equation 3.0DNQE _1001507771a^Fѣ<ѣ<Ole CompObj]_fObjInfo`Equation Native 1_1001507795cFѣ<ѣ<Ole CompObjbdfEquation.3n.39q DABC  FMicrosoft Word Picture MSWordDocWord.Picture.89q FMicrosoft Word DocumentNB6W  FMicrosoft Word Picture MSWorObjInfoeEquation Native 1_1071748966h Fѣ<_ң<1TableV +@-0T%ԏՋjf o 1 p&+@" !x}SKo1$JZRA<\=;xd^4Q9 nؤɅzw|3kK?VF`-g.D,͌^0Ь%|*o8y.w L=:2&t*cuq1o)>8dPS -9đcu\>0t{7k1enϫw澼(}.BBk<;3-;*xEAKBA1B[ lՎ",^>‹W/"CҥlnYfog3 $$7%Y%̧V3.K`yӍ' + {jXviO$^x/L$LqE,!e>kPژcZZ;{ֱG"p[NOSZJDc"=l3)מQZ;RkUR띊n`ydzd?h0|5>*WxڅJ@M+kY(H+'} "衚DB8E!p4D Tak7G=[*%o#g9{ϬWtlb PA9&SO3cw,GLG2-|]zӤTܾ|bDd,$R ] c .AL pict@B) (j0T%z񋏈53) (j?dmo<[9xW}lޖRmP*c(H+X_2! ؜ED=A6LJnXq*(;6}GX1OeV|O٤}NQMЯ?U%¡NӖKO Î@R/{gvYTJ6e=Ӝlɮ3j67eu69JYo>]^Rr5g/p?-Qɻv,Xl64<.q?-7|h9Ug&guH}ω,Ylv&M=w99y(_`ݐ]v2w@_}6DWpzs': lL4'>_FcꅧQYe= {T_ww?Dp`Ow2Xv 3PXqXj%,NΘhy~ . 82?"ؓЧ~NCQcvEvEn27HEcwS)VaT?3; 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This means that you must prove that there is a p""oint equidistant from all the vertices of the regular polygon." "Don'""Custom Tools"       _ ,>>Iz1 T@$jPWr{7X @cIf* hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo(hh^h`. D^`D5o(.0^`05o(..88^8`5o(... 88^8`5o( .... `^``5o( ..... `^``5o( ...... ^`5o(....... ^`5o(........@h^`.@hh^h`.@hh^h`.^`5o(. 0^`05o(..0^`05o(... 88^8`5o( .... 88^8`5o( ..... `^``5o( ...... `^``5o(....... ^`5o(........\^`\5o(\^`\5o(.0^`05o(..0^`05o(... 88^8`5o( .... 88^8`5o( ..... `^``5o( ...... `^``5o(....... ^`5o(........ hh^h`OJQJo(@h^`.8 T@ T@,9_ {7X>I>I9@cjPW8 ̙@h h^h`OJQJo(<9`b@hh^h`.9 @hh^h`. i:@: NormaldhCJ_HmH sH tH B@B Heading 1$$<@&a$ 5CJKH@@@ Heading 4$$dh<a$56<A@< Default Paragraph Font8>@8 Title$<a$ 5CJ$KH8Y@8  Document Map-D OJQJGTimes New Roman5Symbol3 Arial3Times5 Tahomah"from the Custom Tools menu, choose Create New Tool" "and "`"in the dialogue box, check Show Script View. "j"a dialogue box will appear. Check the box labeled " ". "P"Repeat the process for points B and C."*"name your tool and "."In the Script View, d""tool, "" Notice that the givens which are automatically matched are now listed as """objects""cally m""o objects in" "Use " "tool""are related.""in Chapter 3 " "tool""In order"" to investigate, we will need tools to construct other regular pentagons given one edge." "con"T" Sketchpad contains a folder entitled """If you haven'":" Polygons.gsp into the Tool"" Folder and restart Sketchpad or simply open the document to make its tools available."("regular pentagons." " " "Tool"N", the generalized Pythagorean Theorem"4"were proven in Chapter 1""result""result""is half""Corollary ""1.5.4""results""is the""Theorem 1.5.6"")."" Con"0"In the diagram below, "V" is the midpoint of the side opposite to " "and ""O, G, " "and "" are the circumcenter, centroid, and orthocenter, respectively. To prove the theorem, it "2"suffices to prove that ""To show that "V" are collinear, it suffices to show that ""SInce ""A, G, " "and "6"are collinear, in order t"" We can ""we can""also ""by showing"N"To do this, it suffices to show that "h" with ratio 2:1. Note that this also proves that "" trisects ""will " "THe ":" proceeds as follows: Let "<" be the point where the ray "D" intersects the circumcircle of "" Then "B" (why?). It then follows that "^" with ratio 2:1 (why?) It also follows that " "AIBH"@" is a parallelogram (why?) and""hence " ". ""Since ">" is the median, we also have "". THus """parallel lines "" and "" and the "". (Why are "" parallel?)"" THsu "" by SAS."F"for which our proof breaks down, "" and " "the "". What sort of triangles arise? You should find two special cases. Prove "P"Theorem 2.6.3 for each of these cases."". ","(see Exercise 2.8.2)"pL       _ ,>>Iz1 T@$jPWr{7X @cIf* ^`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo( hh^h`OJQJo(hh^h`. 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S@````x(XXZ``````````,PXXZ S@````l*YZ_QddQdQdQddQdQdx!(YZQddQddQdQdQddQdQdl*YZeggggggggx(YZPdQdggggggggx ( hrfR9x(XrfR9,PXrf S@z, aFR9C`x(}Fx(}F~~ ~~~~ ~~~~x(nF  x(nF  x(}F  x(}F6 6 6x(}Ftttttttx(}Fx(}F  x(}F/ / /x(}F9N9N999N99x(}F9N9N999N99x(}F  x(}F  3 @jbjb^^ xh<h<@l>>T-f*(/////zL0,x0fffffffg 7if0//00f"@//"@"@"@0//f"@0f"@"@"@"@,|Eϼl7"@"@%-f-f"@7i"@7i"@"@ SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4 SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4 EMBED Word.Picture.8  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4. The mathematical notion of similarity describes the idea of change of scale that is found in such forms as map making, perspective drawings, photographic enlargements and indirect measurements of distance. Recall from high school that geometric figures are similar when they have the same shape, but not necessarily the same size. More precisely, triangles  EMBED Equation.3  and  EMBED Equation.3  are said to be similar, written  EMBED Equation.3 , when all three pairs of corresponding angles are congruent and the lengths of all three pairs of corresponding sides are proportional. To establish similarity of triangles, however, it is not necessary to establish congruence of all pairs of angles and proportionality of all pairs of sides. The following results are part of high school geometry. 2.5.1 Theorem. (AA) If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. 2.5.2 Theorem. (SSS) If three sides of one triangle are proportional respectively to three sides of another triangle, then the triangles are similar. 2.5.3 Theorem. (SAS) If two sides of one triangle are proportional respectively to two sides of another triangle and the angles included by these sides are congruent, then the triangles are similar.PAGE 13  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4isPAGE 4 are aPAGE 4 PAGE 4 sPAGE 4 . Exercise 2.4.1 studies further the formula for the sum of the angles of polygons having holes in them. Exercises 2.4.2 2.4.6 are related to the Inscribed Angle Theorem. Exercises 2.4.7 and 2.4.8 are for fun. PAGE 9 PAGE 9 . PAGE 9 Use Sketchpad as in Section 2 to complete the following conjecture: ifPAGE 9 PAGE 9 ., thenPAGE 9 is (n ) 180.PAGE 9 Note that your answer must depend on both n and h . PAGE 9 h .PAGE 9 sisting of a polygon containing PAGE 9 PAGE 9 itsPAGE 9 PAGE 12 PAGE 12 This means'PAGE 12 that PAGE 12 provePAGE 12 PAGE 12 PAGE 12 scriptsPAGE 15  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4hing --------------------------PAGE 15 This demonstration presents an opportunity to explain another feature of Custom Tools called Auto-Matching. We will be using this feature when we use Sketchpad to explore the Poincar 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. --------------------------- Script Composition using Auto-Matching PAGE 15 PAGE 15 Script Composition using Auto-Matching PAGE 15 sketchThen open a new script and press the record button. PAGE 15 Open a new sketch and construct  EMBED Equation.3  with the vertices labeled. Open a new sketch and construct  EMBED Equation.3  with the vertices labeled. Open a new sketch and construct  EMBED Equation.3  with the vertices labeled. Next construct the line (not a segment)  EMBED Equation.3 . (You should have pressed the record button before you do this.) Your sketch should look like the figure above. Select the vertices B-A-C in order and choose "Mark Angle B-A-C" 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 line EMBED Equation.3 . Choose Rotate from the Transform Menu and then rotate by Angle B-A-C. (You should have pressed the record button before you do this.) PAGE 15 Your sketch should look like the figure above. PAGE 15 PAGE 15 in the Script WindowPAGE 16 Custom ToolsCreate New ToolPAGE 16 PAGE 16 Show Script ViewdPAGE 16 PAGE 16 "PAGE 16 "Automatically Match Sketch ObjectPAGE 16 PAGE 16 ge the label from A to Auto-A. chanPAGE 16 Also change the label on Point B from B to Auto-B and the label on Point C from C to Auto-C.PAGE 16 BCPAGE 16 DPAGE 16 mments to your script, and save it.Now you can stop recording, add co and save it.mments to your script, PAGE 16 --------------------- End of Script Composition using Auto-Matching ------------------------PAGE 16 PAGE 16 script PAGE 16 PAGE 16 "PAGE 16 givensPAGE 16 cally matched are now listed as AssumedPAGE 16 "PAGE 16 "PAGE 16 "PAGE 16 "PAGE 16 o objects in the sketch labeled PAGE 16 lablesPAGE 16 "PAGE 16 "PAGE 16 After setting the Script Directory appropriately, use PAGE 16 scriptPAGE 16 combine together.PAGE 16 PAGE 15 7PAGE 17  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 8PAGE 17  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 42.5 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 first is perhaps the best known of all results in Euclidean plane geometr                          ! " # $ % & ' ( ) * + , - . / 0 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z [ \ ] ^ _ ` a b c d e f g h i j k l m n o p q s t u v w x y z { | } ~  y, namely Pythagoras theorem. This is frequently stated in purely algebraic terms as  EMBED Equation.2 , 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 algebraic/synthetic 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. 2.5.4 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  EMBED Equation.2  and the lengths of the other two sides are EMBED Equation.2  and EMBED Equation.2 , then  EMBED Equation.2 . Proof: Let  EMBED Equation.3  be a right-angled triangle with right angle at C, and let  EMBED Equation.3  be the perpendicular from C to the hypotenuse  EMBED Equation.3  as shown in the diagram below.  Show EMBED Equation.3  is similar to  EMBED Equation.3 . Show EMBED Equation.3  is similar to  EMBED Equation.3 . Now let  EMBED Equation  have length  EMBED Equation.2 , so that  EMBED Equation  has length  EMBED Equation.2 . By similar triangles,  EMBED Equation.2  and  EMBED Equation.2  Now eliminate  EMBED Equation.2  from the two equations to show  EMBED Equation.2 . There is an important converse to the Pythagorean theorem that is often used. 2.5.5 Theorem. (Pythagorean Converse) Let  EMBED Equation.3  be a triangle such that  EMBED Equation.2 . Then  EMBED Equation.3  is right-angled with  EMBED Equation.2 ACB a right angle. 2.5.5a Demonstration (Pythagorean Theorem with Areas) You may be familiar with the geometric interpretation of Pythagoras theorem. If we build squares on each side of  EMBED Equation.3  then Pythagoras theorem relates the area of the squares. Open a new sketch and draw a right-angled triangle  EMBED Equation.3 . Using the 4/Square (By Edge) script 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. 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 right-angled triangle you have drawn. End of Demonstration 2.5.5a. 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 figure 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? 2.5.5b Demonstration (Generalization of Pythagorean Theorem) Draw a new right-angled triangle  EMBED Equation.3  and use the 5/Pentagon (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 1.5.5(b).) What do you notice about the areas in this case? Now complete the statement of Theorem 2.5.6 below for regular n-gons. End of Demonstration 2.5.5b. 2.5.6 Theorem. (Generalization of Pythagoras theorem) When similar copies of a regular n-gon,  EMBED Equation.2 , are constructed on the sides of a right-angled triangle, each n-gon having the same edge length as the side of the triangle on which it sits, then ____________________ ______________________________________________________. The figure below illustrates the case of regular pentagons.  2.5.6a Exercise. Use Exercise 2.4.8 together with the usual version of Pythagoras theorem to give an algebraic proof of Theorem 2.5.6. 2.5.7 Demonstration. Reformulate and prove the result corresponding to Theorem 2.5.6 when the regular n-gons constructed on each side of a right-angled triangle are replaced by similar triangles. This demonstration presents an opportunity to explain another feature of Custom Tools called Auto-Matching. We will be using this feature in Chapter 3 when we use Sketchpad to explore the Poincar 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. Script Composition using Auto-Matching Open a new sketch and construct  EMBED Equation.3  with the vertices labeled. Next construct the line (not a segment)  EMBED Equation.3 . Select the vertices B-A-C in order and choose "Mark Angle B-A-C" 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 line EMBED Equation.3 . Choose Rotate from the Transform Menu and then rotate by Angle B-A-C. Select the vertices A-B-C in order and choose Mark Angle A-B-C 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  EMBED Equation.3 . Choose Rotate from the Transform Menu and rotate by Angle A-B-C. Construct the point of intersection between the two rotated lines and label it F.  EMBED Equation.3  is similar to  EMBED Equation.3 . 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 labeled A, B, and C already constructed in your sketch where you want to construct the similar triangle. Then you only 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 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 which match those in the Assuming section, then Sketchpad will require you to match those objects manually, as if they were Given Objects. Now open a new sketch and construct a triangle with vertices labeled A, B, and C. In the same sketch, construct a right triangle. Use the similar triangle tool to build triangles similar to  EMBED Equation.3  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 2.5.7. PAGE 17 PAGE 7 4PAGE 7 5PAGE 7 2.5PAGE 7 2.5PAGE 8 2.5PAGE 8 . Using the 4/PAGE 8 Square (PAGE 8 By Edge)PAGE 8 scriptPAGE 8 End of Demonstration 2.5PAGE 8 This PAGE 8 2.5PAGE 9 PAGE 9 PAGE 9 PAGE 9 construct other regular pentaPAGE 9 PAGE 9 "PAGE 9 Sketchpad contains a folder entitled Custom'PAGE 9 PAGE 9 PAGE 9 5PAGE 9 5PAGE 9 5PAGE 9 5PAGE 9  PAGE 10 s the case of regular pentagons.PAGE 10 The figure below illustrates the case of regular  PAGE 10 PAGE 10 PAGE 10 regulaPAGE 10 r PAGE 10 PAGE 10 The figure below illustrates the case of regular pentagons.  The figure below illustrates the case of regular pentagons.  The figure below illustrates the case of regular pentagons.  PAGE 10 Now complete the statement of Theorem 2.3.6 below for regular n-gons. Repeat these constructions for an octagon instead of a pentagon. (Note: You can create an Octagon By Edge script from your construction for Exercise 1.3.5(b).) What do you notice about the areas in this case? Now complete the statement of Theorem 2.3.6 below for regular n-gons. PAGE 9 PAGE 9 5PAGE 10 5PAGE 10 2.3.6a Exercise. Use Exercise 2.4.8 together with the usual version of Pythagoras theorem to give an algebraic proof of Theorem 2.3.6. End of Exercise 2.4.8. Exercise6aPAGE 16 4PAGE 16 5PAGE 10 and prove PAGE 10 5PAGE 10 ScriptPAGE 10 5PAGE 11 3PAGE 11 3PAGE 12 3PAGE 12 3PAGE 13 3PAGE 13 3PAGE 13 3PAGE 13 4PAGE 13 4PAGE 13 3PAGE 13 4PAGE 13 4PAGE 13 4PAGE 13 3PAGE 13 4PAGE 13 3PAGE 13 tPAGE 13 4PAGE 14 For Exercises 2.4PAGE 14 .4, 2.4PAGE 14 .5, and 2.4PAGE 14 4PAGE 14 3PAGE 14 4PAGE 15 4PAGE 15 4PAGE 15 4PAGE 15 4PAGE 15 4PAGE 16 4PAGE 16 3PAGE 16 PAGE 16 Theorem 2.3.6, PAGE 16 ,PAGE 16 8PAGE 17 9PAGE 17 are proven in most geometry textsPAGE 18 theoremPAGE 18 theorem requiredPAGE 18 is PAGE 18 Exercise 2.8.6PAGE 18 PAGE 18 theorem requiredPAGE 18 is the centroid of the triangle.PAGE 18 PAGE 18 PAGE 18 Proof. PAGE 19 PAGE 19 ConConPAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 theorem, it su To prove the heorem, it sutffices to prove that PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 TPAGE 19 IPAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19 in order toPAGE 19 PAGE 19 it suffices to showPAGE 19 We canPAGE 19  EMBED Equation.3 PAGE 19 PAGE 19 PAGE 19 PAGE 19 also proves PAGE 19 If the triangle is equilateral, the three points coincide and so they are trivially collinear. We use the figure above. Here, H is the orthocenter and O is the circumcenter. Let G be the point where the line containing  EMBED Equation.3  intersects the line containing the median EMBED Equation.3  from vertex A, and so points O, G and H are collinear. This assumes that these two lines are distinct, but if they are not distinct, then the line containing  EMBED Equation.3  must (unless the triangle is equilateral) be distinct from the lines containing one of the other two medians. In that case, we would proceed as below using that median instead of  EMBED Equation.3 . We prove the theorem with the assumption that point G lies between O and H and in the interior of  EMBED Equation.3 . Although it is tedious to prove, this in fact always occurs unless O and H coincide, and in that case O, H and the centroid all coincide and  EMBED Equation.3  is equilateral. PAGE 19  EMBED Equation.3 PAGE 19 PAGE 19 HPAGE 19 PAGE 19 PAGE 19 PAGE 19  EMBED Equation.3 PAGE 19  EMBED Equation.3  EMBED Equation.3 PAGE 19 PAGE 19 PAGE 19 PAGE 19  PAGE 19 PAGE 19 PAGE 19 PAGE 19 PAGE 19  EMBED Equation.3  EMBED Equation.3 PAGE 19 PAGE 19 PAGE 19 PAGE 20 PAGE 20 PAGE 20 PAGE 20 HPAGE 20 congreconger=PAGE 20 PAGE 20 PAGE 20 PAGE 20 PAGE 20 PAGE 20 PAGE 20 Hhsu PAGE 20 that PAGE 20 PAGE 20 First notice that quadrilateral AHBI is a parallelogram. (Why?) Now,  EMBED Equation.3  (by AAA) . Since OA / AH = OA / IB = , the ratio of similarity is . Therefore GA / GA = . Thus, AG / AA = 2/3 . This implies that G is the centroid of  EMBED Equation.3 , by the remarks above. By similarity of triangles, OG / HG = , so the centroid G trisects the segment  EMBED Equation.3 , the segment joining the orthocenter and the circumcenter. Note: the assumption in the figure is that m EMBED Equation.3  < 90 . A similar argument can be used to prove the theorem for the case that m EMBED Equation.3  > 90 (see Exercise 2.6.3a) and the argument can easily be modified for the case that m EMBED Equation.3  = 90. PAGE 20 PAGE 20 PAGE 20 2.6.3a Exercise. Sketch the figure used in Theorem 2.6.3 for the case that m EMBED Equation.3  > 90 and then prove the theorem for that case. Using Sketchpad can help here.PAGE 20 PAGE 20 QEDPAGE 20 PAGE 20 for which our proof breaks down.PAGE 20 'PAGE 20  EMBED Equation.3 PAGE 20  EMBED Equation.3 tthePAGE 20 PAGE 20 PAGE 20 . What sort PAGE 20 PAGE 20 QEDPAGE 20 PAGE 20 m 2.6.3 for each of these cases. PAGE 20 PAGE 20 Exercise 2.8.2 Prove the converse to the Pythagorean Theorem stated in Theorem 2.5.5.PAGE 24 Prove Theorem 2.6.3 whenPAGE 24 and when the triPAGE 24 PAGE 24 the triangle is isosceles the triangle is a right triangle the triangle is isosceles the triangle is a right triangle the triangle is isosceles the triangle is a right triangle PAGE 24 (a) the triangle is isosceles PAGE 24 (a) the triangle is isosceles (b) the triangle is a right triangle tPAGE 24 sceles PAGE 24 tPAGE 24 PAGE 24 (a) PAGE 24 (b) PAGE 24 PAGE 16 Exercise 2.5.10 Exercise 2.5.10 PAGE 16 ProvePAGE 16 the conPAGE 16 erse to the Pythagorean Theorem.PAGE 16 notice that Sketchpad comes equipped with a Nine Point Circle script in the Sample Scripts Folder.PAGE 21 OPAGE 21 PAGE 21  EMBED Equation  EMBED Equation . End of Exercise 2.8.5. PAGE 25 PAGE 26  EMBED Word.Picture.8 PAGE 29 HPAGE 29  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4whichPAGE 11 thatPAGE 11 PAGE 11 in the sketch with labels whichPAGE 11 so we assume that G lies inside the triangle.PAGE 18 To complete the proof, the argument can be modified to prove the theorem for the cases where the Miquel point is outside of the triangle or on one of the sides. (See Exercise 2.8.1) PAGE 18 PAGE 18 PAGE 19 ra with PAGE 19 PAGE 19 Note that this will 2:1. PAGE 19 PAGE 19 e 2:1, then we havPAGE 19 nPAGE 19 (why?). It thenPAGE 19 atio 2:1 (why?) It also followsPAGE 19 is the median, we also have PAGE 19 PAGE 28 PAGE 19 PAGE 19 PAGE 20 PAGE 22 PAGE 22 PAGE 23 PAGE 24  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4PAGE 51 PAGE 2 PAGE 3 PAGE 4 PAGE 5 PAGE 6 PAGE 7 PAGE 8 PAGE 9 PAGE 10 PAGE 11 PAGE 12 PAGE 13 PAGE 14 PAGE 15 PAGE 16 PAGE 17 PAGE 18 QEDPAGE 20 PAGE 20 ProvePAGE 20 for eachPAGE 20 PAGE 30  EMBED Equation.3  PAGE 32 thaPAGE 31 t PAGE 31 PAGE 33 will PAGE 37 PAGE 44 PAGE 35 PAGE 35 PAGE 37 PAGE 37 PAGE 38 PAGE 40 hedrPAGE 41  PAGE 43 right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) PAGE 49 right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) right triangle is to use your Auto-Matching similar triangle script. Just label your original right triangle appropriately.) PAGE 42 except for the last one PAGE 44 because translations in two different directions of a regular hexagonal portion of the design will tile the plane precisely because of Napoleons theorem. PAGE 44 are wallpaper designs PAGE 44 yourPAGE 44 sPAGE 44 Construct several such squares. PAGE 45 PAGE 45  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4PAGE 2 PAGE 2 PAGE 4 PAGE 5 PAGE 6 PAGE 6 PAGE 8 PAGE 8 PAGE 9 PAGE 10 PAGE 11 PAGE 12 PAGE 14 PAGE 14 PAGE 15 PAGE 16 PAGE 18 PAGE 18 PAGE 19 PAGE 20 PAGE 22 PAGE 22 PAGE 23 PAGE 24 PAGE 25  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4PAGE 49 PAGE 2 PAGE 2 PAGE 4 PAGE 3 PAGE 4 PAGE 5 PAGE 6 PAGE 7 PAGE 8 PAGE 9 PAGE 10 PAGE 11 PAGE 11 PAGE 12 PAGE 14 PAGE 13 PAGE 16 PAGE 17 PAGE 18 PAGE 19 PAGE 20 PAGE 21 PAGE 22 PAGE 23 PAGE 24 PAGE 25 PAGE 26 PAGE 26 PAGE 28 PAGE 29 PAGE 30 PAGE 31 PAGE 32 PAGE 33 PAGE 34 PAGE 35 PAGE 36 PAGE 37 PAGE 38 PAGE 38 PAGE 40 PAGE 41 PAGE 41 PAGE 40 PAGE 40 PAGE 41 PAGE 43 PAGE 44 PAGE 45 PAGE 46 PAGE 47 PAGE 48 fPAGE 48  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4Open a new sketch and draw a right-angled triangle  EMBED Equation.3 . 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. 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 right-angled triangle you have drawn. qPAGE 17 gPAGE 17  EMBED Equation.3  and  EMBED Equation.3  are supplementary. Why?  EMBED Equation.3  and  EMBED Equation.3  are supplementary. Why?  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4 SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4PAGE 49  SEQ Example_1 \* ARABIC 1 The above example shows a typical Arabic design. This was drawn starting from a regular hexagon inscribed in a circle. Demonstration 2.10.0. Construct the design in Example 1 using Sketchpad. 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 figure like the one below.  To construct a second 12-sided regular polygon having one side adjacent to the first regular hexagon, reflect your figure in one of the sides of the first regular hexagon. Now complete the construction of the previous Arabic design. End of Demonstration 2.10.0. 2.10.1 Exercise. If the radius of the circumscribing circle of the initial regular hexagon is R, determine algebraically the area of the six-pointed star inside one of the circles. Continuing this example indefinitely will produce a covering of the plane by congruent copies of three polygons - a square, a rhombus and a six-pointed star. Notice that all these congruent copies have the same edge length and adjacent polygons meet only at their edges, i.e., the polygons do not overlap. The second example  Example  SEQ Example_1 \* ARABIC 2 if continued indefinitely 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  SEQ Example_1 \* ARABIC 3 is one very familiar one from floor coverings or ceiling tiles; when continued indefinitely 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  SEQ Example_1 \* ARABIC 4 EMBED Equation  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation  EMBED Equation  EMBED Equation  EMBED Equation  EMBED Equation.3  EMBED Equation.3  [\+-pq6 & F$a$$$a$h`hklmn+,56OPQSlmno VW$%>?@A}~j? 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