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Chapter 3 Other Geometries Prequel A Axioms for Neutral Geometry Prequel B Revisiting the Four Point Geometry 3.1 The Concept of Parallelism 3.2 Points, Lines, and Curves in Poincare’s Model of Hyperbolic Space 3.3 Polygons in Hyperbolic Space 3.4 Congruence in Hyperbolic Space Homework Chapter 3 on the website separately 1 The Axioms for Neutral Geometry Undefined terms: point, line, plane, space Axioms: A1 To each pair of points (A, B) is associated a unique real number, denoted AB, with least upper bound, What do the SMSG Axioms for Euclidean Geometry start with? Which SMSG A matches this…how is it alike and how is it different? SMSGA1. Given any two distinct points there is exactly one line that contains them. SMSGA2. The Distance Postulate: To every pair of distinct points there corresponds a unique positive number. This number is called the distance between the two points. 2 A2 For all points A and B, AB 0, with equality only when A = B. Where is this in SMSG A? Why do we have this and A3, below A3 For all points AB = BA. A4 Given any four distinct collinear points A, B, C, and D such that A – B – C, then either D – A – B, A – D – B, B – D – C, or B – C – D. SMSG A? 3 A5 Each two points A and B lie on a line, and if AB < , that line is unique. How is this different from SMSG A1? SMSGA1. Given any two distinct points there is exactly one line that contains them. A6 Each three noncollinear points determine a plane. SMSGA7. Any three points lie in at least one plane, and any three noncollinear points lie in exactly one plane. 4 A8 If two planes meet, their intersection is a line. SMSGA8. If two planes intersect, then that intersection is a line. A9 Space consists of at least four noncoplanar points, and contains three noncollinear points. Each plane is a set of points of which at least three are noncollinear, and each line is a set of at least two distinct points. SMSGA5. A. B. Every plane contains at least three non-collinear points. Space contains at least four non-coplanar points. 5 A10 (Ruler Postulate): Given line L and two points P and Q on L, the points of L can be placed into one – to – one correspondence with the real numbers x such that x (called coordinates) in such a manner that 1. points P and Q have coordinates 0 and k > 0, respectively 2. if A and B on the line have coordinates a and b, then AB is a b if 2 a b a b if a b SMSGA3. The Ruler Postulate: The points of a line can be placed in a correspondence with the real numbers such that A. To every point of the line there corresponds exactly one real number. B. To every real number there corresponds exactly one point of the line, and C. The distance between two distinct points is the absolute value of the difference of the corresponding real numbers. SMSGA4. The Ruler Placement Postulate: Given two points P and Q of a line, the coordinate system can be chosen in such a way that the coordinate of P is zero and the coordinate of Q is positive. 6 COORDINATES: First in Euclidean Geometry Coordinates A close reading of SMSG Axiom 3 shows that in pure geometry we have one number that specifies a point on a line, not two as in Cartesian Plane. We call this number the geometric coordinate to distinguish it from the Cartesian coordinates of a given point. We will spend some time exploring this notion. We start with a horizontal number line in one dimension. Identify points on the line with the real numbers A = 5, B = 1, C = 0, D = 3, and E = 7. Note that each of these locations is specifically identified with exactly one point that has one real number associated with it. Now in order to calculate the distance from E to B, take the absolute value of the difference of the real numbers in either order: 7 1 8 . Note that EB = BE because we are using the absolute value of the difference. This is exactly what the postulate stipulates. This works perfectly well with vertical number lines, too. Try it with points Q = 12 and R = 5. QR = 5 12 = 12 5 . There’s no real difference between the 7 number lines except orientation. The horizontal line has slope = 0 and the vertical line has an undefined slope. When we combine the vertical number line and the horizontal number line, intersecting them at the point 0 of each line, we have the structure to move into the Cartesian plane and begin discussing coordinate geometry. NOTE THAT the postulate applies to the lines we are looking at as axes AND to every other line in the Cartesian Plane. In fact, reading it closely shows that we must be able to find a real number for each point on any line. (especially those with slopes that are any real number, m > 0). Further, the axiom specifically states that each point on any line is associated with a single real number that can be combined (absolute value of the difference) with the single real number that is the coordinate of any other point on the line to get the distance. Happily there is a nice formula to work with that does for us this perfectly. To find the geometric coordinate, take the x coordinate from the Cartesian coordinates and multiply it times the square root of ( 1 + the slope squared). Suppose we have a point P on the line y = mx + b and it’s coordinates are (s, q): 8 We take the x coordinate, s, and multiply it by 1 m 2 to get the real number we will assign to P. If we do this with every point on the line, we will have individual real numbers assigned to each point. P = (s, q) P = s 1 m2 The Cartesian coordinates for point P are (s, q) and the geometric coordinate is s 1 m 2 . If we work out the geometric coordinates of two points and if we take the absolute value of the difference of these numbers we will have the distance between two points on the line. Example 1: Suppose we have A = ( 1, 2) and B = ( 1, 6) and AB = 2 5 when we calculate the distance using the Euclidean Distance Formula: 9 d ( x2 x1 )2 ( y2 y1 )2 The slope of the line containing A and B is found using the slope formula m= y 2 y1 so for this example m = 2. We need the factor x 2 y1 1 m2 which is 5 . The geometric coordinate for A is 1( 5) 5 and the geometric coordinate for B is 5 (the x coordinate for B is 1). The absolute value of the difference is 5 5 which is 2 5 as promised. Example 2: Suppose we take the line y = 3x + 1. If we pick two points on the line we can use the distance formula to find the distance between them. Let’s use ( 1, 4) and ( 3, 10). Using the traditional distance formula we find that the distance between them is 40 2 10. Now our axiom asserts that we can find a single real number for each point that can be used and will result in this same distance. Note, that 1 m 2 = 10 for this line. 10 Using the formula above the first point will be assigned the real number 1( 10 ) and the second point will be assigned the real number 3 ( 10 ). The postulate says the absolute value of the difference between these numbers is the distance between the points. Which is exactly 2 10 . So this method is an easy way to find distances when you have the point coordinates or the equation of the line. It does point out that a standard geometric approach is slightly different that that of coordinate geometry or of algebra. 11 Example 3: It is possible to get nice whole number distances if you use lines with irrational slopes. For example: The points (1, 0) and (0, 3 ) are on the line y 3 x 3 . Here’s a sketch of the points and the line done in Math GV: 12 What are the geometric coordinates for the points and the distance between them? 1 m2 2 which is a very nice number So the geometric coordinate for (1, 0) is 2 and the geometric coordinate for (0, 3 ) is 0 and the distance between the two points is 2 units. Coordinate Exercise: Find the geometric coordinates and the distance between the points (1, 2) and (3, 10), Check your work using the Euclidean distance formula. 13 Application Exercise: An application of this idea is to explore the following situation: Two segments are said to be congruent segments if they have the same length. Are the two segments defined by the following endpoints congruent? Segment A has endpoints ( 5, 2) and (7, 2) Segment B has endpoints (0, 5) and (12, 9) Find the coordinates of each endpoint. Get the distances: 14 What do these segments look like when graphed on the coordinate plane? In the other 2 of the Big Three: Hyperbolic and Spherical We will learn about the formulas for the coordinates in these two geometries as we work with them in the upcoming weeks! One last look with an eye toward all 3 geometries. Let’s look at this “alpha” A10 (Ruler Postulate): Given line L and two points P and Q on L, the points of L can be placed into one – to – one correspondence with the real numbers x such that x (called coordinates) in such a manner that 1. points P and Q have coordinates 0 and k > 0, respectively 2. if A and B on the line have coordinates a and B, then AB is a b if 2 a b a b if a b 15 Appendix to Coordinates Why Does the Geometric Coordinate Formula Work? First take two points: P1 ( x1 , y1 ) P2 ( x2 , y2 ) The formula for the line containing them is y = mx + b so the coordinates are really: P1 ( x1 , mx1 b) P2 ( x2 , mx2 b) Now calculate the distance between them using the Euclidean distance formula: ( x2 x1 )2 (mx1 b (mx2 b))2 Distribute the minus sign in the second summand and the b’s cancel out: ( x2 x1 )2 (mx2 mx1 )2 16 Now, the “m” can be factored out, and then the difference of the x’s squared can be factored out: ( x2 x1 ) 2 m 2 ( x2 x1 ) 2 ( x2 x1 ) 2 (1 m 2 ) Take the square root of the first factor to get – note the POSITIVE square root, guaranteed by absolute value signs: x2 x1 ( 1 m2 ) This is, exactly, the formula we’re using in to meet the terms of A3. 17 A11 (Plane Separation Postulate): Let L be any line lying in any plane P. The set of all points in P not on L consists of the union of two subsets H1 and H2 of P such that 1. H1 and H2 are convex sets 2. H1 and H2 have no points in common 3. if A lies in H1 and B lies in H2 such that AB < , line L intersects segment AB . SMSGA9. The Plane Separation Postulate: Given a line and a plane containing it, the points of the plane that do not lie on the line form two sets such that A. each of the sets is convex, and B. if P is in one set and Q is in the other, then segment PQ intersects the line. SMSGA10. The Space Separation Postulate: The points of space that do not line in a given plane form two sets such that A. each of the sets is convex, and B. if P is in one set and Q is in the other, then the segment PQ intersects the plane. 18 A little bit on Convexity: A set is called convex, if for every pair of points in the set, the points on the segment joining them are also in the set. Examples: Circle: Interior of a circle: Planes: 19 A12 Each angle ABC is associated with a unique real number between 0 and 180 denoted m ABC. No angle can have measure 0 or 180 A13 (Angle Addition Postulate): If D lies in the interior of ABC, then m ABD + m DBC = m ABC. A14 (Protractor Postulate): The set of rays AX lying in a plane and on one side of a given line AB , including ray AB , may be placed into one – to – one correspondence with the real numbers x such that 0 x 180 (called coordinates) in such a manner that 1. ray AB has coordinate 0 2. if rays AC and AD have coordinates c and d, then m CAD = c d . SMSGA11. The Angle Measurement Postulate: To every angle there corresponds a real number between 0 and 180. SMSGA12. The Angle Construction Postulate: Let ⃗⃗⃗⃗⃗ 𝐴𝐵 be a ray on the edge of the half-plane 𝐻. For every 𝑟 between 0 and 180 there is exactly one ray ⃗⃗⃗⃗⃗ 𝐴𝑃 with 𝑃 in 𝐻 such that 𝑚∠𝑃𝐴𝐵 = 𝑟. SMSGA13. The Angle Addition Postulate: If 𝐷 is a point in the interior of ∠𝐵𝐴𝐶, then 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 + 𝑚∠𝐷𝐴𝐶. A15 (Linear Pair Axiom): A linear pair of angles is a supplementary pair. SMSGA14. The Supplement Postulate: If two angles form a linear pair, then they are supplementary 20 A16 (SAS Postulate): If two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of another, the triangles are congruent. SMSGA15. The SAS Postulate: Given an one-to-one correspondence between two triangles (or between a triangle and itself). If two sides and the included angle of the first triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence. Full stop. Now we have to make a choice about parallelism! 21 The SMSG Axioms for Euclidean Geometry – A categorical system A1. Given any two distinct points there is exactly one line that contains them. A2. The Distance Postulate: To every pair of distinct points there corresponds a unique positive number. This number is called the distance between the two points. A3. The Ruler Postulate: The points of a line can be placed in a correspondence with the real numbers such that A. To every point of the line there corresponds exactly one real number. B. To every real number there corresponds exactly one point of the line, and C. The distance between two distinct points is the absolute value of the difference of the corresponding real numbers. A4. The Ruler Placement Postulate: Given two points P and Q of a line, the coordinate system can be chosen in such a way that the coordinate of P is zero and the coordinate of Q is positive. A5. A. B. A6. If two points line in a plane, then the line containing these points lies in the same plane. A7. Any three points lie in at least one plane, and any three non-collinear points lie in exactly one plane. A8. If two planes intersect, then that intersection is a line. A9. The Plane Separation Postulate: Given a line and a plane containing it, the points of the plane that do not lie on the line form two sets such that A. each of the sets is convex, and B. if P is in one set and Q is in the other, then segment PQ intersects the Every plane contains at least three non-collinear points. Space contains at least four non-coplanar points. line. 22 A10. The Space Separation Postulate: The points of space that do not line in a given plane form two sets such that A. each of the sets is convex, and B. if P is in one set and Q is in the other, then the segment PQ intersects the plane. A11. The Angle Measurement Postulate: To every angle there corresponds a real number between 0 and 180. A12. The Angle Construction Postulate: Let ⃗⃗⃗⃗⃗ 𝐴𝐵 be a ray on the edge of the halfplane 𝐻. For every 𝑟 between 0 and 180 there is exactly one ray ⃗⃗⃗⃗⃗ 𝐴𝑃 with 𝑃 in 𝐻 such that 𝑚∠𝑃𝐴𝐵 = 𝑟. A13. The Angle Addition Postulate: If 𝐷 is a point in the interior of ∠𝐵𝐴𝐶, then 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 + 𝑚∠𝐷𝐴𝐶. A14. The Supplement Postulate: If two angles form a linear pair, then they are supplementary A15. The SAS Postulate: Given an one-to-one correspondence between two triangles (or between a triangle and itself). If two sides and the included angle of the first triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence. Neutral axioms stop here! A16 The Parallel Postulate: Through a given external point there is at most one line parallel to a given line. A17. To every polygonal region there corresponds a unique positive number called its area. A18. If two triangles are congruent, then the triangular regions have the same area. 23 A19. Suppose that the region R is the union of two regions R1 and R2. If R1 and R2 intersect at most in a finite number of segments and points, then the area of R is the sum of the areas of R1 and R2. A20. The area of a rectangle is the product of the length of its base and the length of its altitude. A21. The volume of a rectangular parallelpiped is equal to the product of the length of its altitude and the area of its base. A22. Cavalieri’s Principal: Given two solids and a plane. If for every plane that intersects the solids and is parallel to the given plane, the two intersections determine regions that have the same area, then the two solids have the same volume. Revisiting The Four Point Geometry Undefined terms Axioms point, line, on A1 There are exactly four points. A2 Any two distinct points have exactly one line on both of them. A3 Each line is on exactly two points. Models – 24 Possible Definitions: Parallel lines – what about parallel lines here: Collinear points Theorems 1. The four point geometry has exactly 6 lines. 2. Each point of the geometry has exactly 3 lines on it. 3.1 The Concept of Parallelism We’ve seen many examples of parallelism from earlier in the semester: no parallel lines in the 3 Point Geometry and Spherical Geometry to Playfair’s setup: …”exactly one line through the point parallel” to “…several lines through the point parallel” in the 6 Point Geometry and in the Klein Disc. In the Big Three: Euclidean is the “exactly one” Hyperbolic is the “many” Spherical is the “none” Now let’s discuss triangles. All three have them! And the Tricotomy Law is hard at work on 25 the sum of the interior angles of a triangle. Euclidean geometry: = 180 degrees Spherical geometry: > 180 degrees Hyperbolic geometry: < 180 degrees Let’s look at Spherical geometry: We measure angles with a Euclidean protractor. Select the vertex and put the protractor “center” there. Project the legs up into the plane tangent to the sphere at the vertex and measure. Let’s do that. Now, did everybody get the SAME sum? 26 Did anybody get 270 degress? This is very non-Euclidean, no? Let’s look at Hyperbolic geometry with Sketchpad… 27 Saccheri – Legendre Theorem for interior angles states that the sum of the interior angles of a triangle is less than or equal to 180 degrees. It is a theorem about HG and EG, but not SG and was an historic breakthrough when it was proved. Let’s look at a very important construct with a history: Saccheri Quadrilateral This is a quadrilateral with right angles at each base vertex and congruent upright legs. And it is a geometric object in EACH of the Big Three. It exists in each of the Big Three and has some properties that are the same in each and some properties that are different in each. It was first constructed in the early 1700’s by a Jesuit monk named Giovanni Girolamo Saccheri in an effort to prove that one could derive the parallel postulate from Euclid’s other four postulates. Of course, his attempts didn’t work all the way. He did eliminate the possibility of Spherical Geometry, though using Euclid’s axioms, not Neutral Geometry axioms (those came MUCH later on). Here’s a picture of a Saccheri Quadrilateral B Summit leg A C leg Base BAD and CDA are right angles. BA AD and CD AD D Thus we define a Saccheri Quadrilateral as a quadrilateral with 2 opposite sides congruent and two adjacent right angles at the legs. One all encompassing phrase is “biperpendicular quadrilateral”. 28 Note that we simply do not discuss the length of the summit or the measures of the summit angles – that’s because those parts have different properties in each of the “Big Three” geometries. Here’s a summary of the properties of a Saccheri Quadrilateral in the 3 geometries: Euclidean Geometry Hyperbolic Geometry Spherical Geometry summit length = base summit longer than base summit shorter than base summit angles = 90 summit angles are acute summit angles are obtuse In Euclidean geometry, a Saccheri Quadrilateral is called a rectangle. Note the Trichotomy Law showing up in the summit base length and the measure of the summit angles. Lemma: A Saccheri Quadrilateral is a convex polygon. This is true in all three geometries. 29 Theorem : The summit angles of a Saccheri Quadrilateral are congruent. Proof : Let ABCD be a Saccheri Quadrilateral. Construct diagonals BD and AC B C k A D Now consider BAD and CDA) (this is called decomposing the SQ. They are congruent by SAS. This means that CAD is congruent to BDA by CPCF*. From properties of a quadrilateral and the Angle Addition Postulate, we have the following equations and we will subtract them: mBAC + mCAD = 90 mBDA + mBDC = 90 Thus mBAC mBDC = 0 . So these are congruent. We have also that BD CA by CPCF. Thus BDC is congruent to CAB by SAS. This makes the summit angles congruent by CPCF. *Congruent parts of congruent figures are congruent! I find that this proof gives a good feel for how to work with Saccheri Quadrilaterals. 30 Saccheri was an Italian Jesuit priest whose grasp of logic was impeccable. His work was complimented by the Englishman Lambert who also has a quadrilateral named for him. The Lambert quadrilateral has 3 right angles and the measure of the 4th angle varies with the geometry it’s in. Saccheri’s plan was to eliminate two of the possible cases for the summit angles of a quadrilateral (obtuse and acute), thereby proving that Euclidean geometry was the only true geometry. He failed at that, but he did eliminate the Hypothesis of the Obtuse angle. This is because Spherical Geometry is quite different from Hyperbolic and Euclidean. Unfortunately, and unknown to Saccheri, Hyperbolic Geometry is a model for our axioms and it is the geometry that fulfills the Hypothesis of the Acute Angle. The Hypothesis of the Obtuse Angle is not valid in Absolute Geometry (Hyperbolic and Euclidean). Absolute geometry groups HG and EG; Neutral groups all three. There was a LOT of tidying that went on in the early 20th century on geometry. We will look at the Triangle Associated with a Saccheri Quadrilateral to eliminate the Hypothesis of the Obtuse Angle. The construction is to take an arbitrary triangle and orient it base horizontally. Pick the midpoints of the two non-horizontal legs and connect them with a segment. Extend the segment to the left and right in such a way that the new endpoint is beyond the vertex of the base on each side and then connect the base and the extended segment with a perpendicular to the extended segment (ie “RAISE” and not drop a perpendicular). Here’s the illustration and a summary of what is given or true by construction: ABC is arbitrary AM = MB CN = NB B’B B’C’ C’C B’C’ 31 raise a perpendicular from A to B'C' and call the point of intersection D. A T3 T2 M B' N C' D T1 T4 B C T1 T2 and T3 T4 because of AAS (right angle, vertical angle, half the side). This makes BB’ CC’. Thus BB’C’C is a Saccheri Quadrilateral by definition – it’s just upside down from the usual picture. I’ve used greek symbols for the congruent angles in each; these are congruent by CPCF. (and 32 Now the sum of the interior angles of the triangle is less than or equal to 180 by the Saccheri-Legendre Theorem..…true in Hyperbolic and Euclidean but NOT Spherical. ABC x 180 Note that A = So mA + mB + mC 180, which with substitution and a little reorganizing becomes mB + m + mC + m = x 180. Now mB + m S1 is the measure of a summit angle and mC + m S2 is the measure of the other summit angle. So we have S1 + S2 180. Since we know that the summit angles of a Saccheri Quadrilateral are congruent S1 = S2 = 90 and we have 2S = S1 + S2 180. So S 90. Thus the summit angles are not obtuse. Historical context matters here… 33 Another theorem: Show that the summit length of a Saccheri Quadrilateral is greater than or equal to the base length. Proof Let ABCD be a Saccheri Quadrilateral D C B A This means that mA = m1 + m2 = 90 by definition. By properties of quadrilaterals, point C is an element of intDAB. By Saccheri-Legendre, ABC 180 . Which is to say that m2 + m3 + 90 180 so that m2 + m3 90. Since m1 + m2 = 90, we know that m1 + m2 m2 + m3, thus m1 m3. In considering DAC and BCA, note that we have 1 and 3 enclosed by congruent sides (they share the diagonal and the outer sides are legs of the Saccheri Quadrilateral) and the measures of the angles are not equal. By the Hinge Theorem, then, we have DC AB. 34 Note that a refinement of the Saccheri Quadrilateral is the Lambert Quadrilateral. A Lambert Quadrilateral has three right angles. You will find Lambert Quadrilaterals near discussions of Saccheri Quadrilaterals. Let’s discuss those summit angles again. In EG, we have a rectangle: What is the sum of the interior angles of a rectangle? How do you know? Let’s look at a Saccheri Quadrilateral on a sphere: What is the sum of the interior angles of a SQ in SG? How do you know? 35 Here’s a sketch of one in Hyperbolic geometry from Sketchpad: What is the sum of the interior angles of a SQ in HG? How do you know? 36 3.2 Points, Lines, and Curves in Poincare’s Model of Hyperbolic Space Points: Euclidean: of the plane or of 3-space, the usual Spherical: {( x, y, z ) x 2 y 2 z 2 1} surface points only Embedded in 3-space, using the center (0, 0, 0) as a reference! Using Euclidean planes tangent to the sphere to measure angles! Hyperbolic: our model will be 2-dimensional, the Poincaré Disc {( x, y ) x 2 y 2 1} . This is a slice of the 3-D interior of the unit circle model. Lines: Euclidean: the usual Spherical: Great Circles, as opposed to circles on the surface The intersection of the surface with a plane that contains the center of the sphere. e.g. the equator or any Great circle through the North pole and South pole…though any orientation of a GC is allowed Hyperbolic: Orthogonal Circles to the (disallowed) Unit Circle. A circle that is orthogonal to the given circle intersects it in two points and tangent lines to each circle at the point of intersection are perpendicular. Note that diameters of the disc are also lines in this space even though they don’t look like arcs; each diameter is said to be an arc of a circle with a center at infinity. Let’s look at how to find orthogonal circles to our unit circle. We’ll take the center and radius form of the equation: ( x h)2 ( y k )2 r 2 And foil it out. Getting all the terms with variables on the left and all the constants on the right, we have 37 x 2 y 2 2hx 2ky r 2 h2 k 2 We have another general form for a circle: x 2 y 2 ax by c By coefficient matching we find a h b k c r 2 h2 k 2 Now let’s look at our unit circle and an orthogonal circle. Note that the radii are perpendicular to one another at each intersection of the circles. Thus we have a right triangle when we connect the circle centers. x^2+y^2+ax+by=c D (h, k) r x^2+y^2=1 O 1 P 38 OD 2 h2 k 2 By the Pythagorean Theorem, then, h2 k 2 r 2 1 Note that OD 2 OP 2 PD 2 r 2 1 OD Now from above: c r 2 h2 k 2 h2 k 2 r 2 c So c 1 Now we see that we can use x 2 y 2 ax by 1 to find orthogonal circles to the unit circle all day long! With one point in the unit disc we’ll get one equation and with a second point in the unit disc, we’ll get a second equation. Solve for a and b.* *I am indebted to David C. Kay’s discussion in his excellent College Geometry, second edition, pages 448 and 449 for this equation. 39 To show that two points determine a line: You pick two points, A and B, from the interior of the Unit Circle and substitute their Cartesian coordinates into x 2 y2 ax by 1 . You will get two equations with two unknowns, a and b. If you solve the system, you will have an equation for a circle that is orthogonal to the unit circle and includes points A and B. The arc of this orthogonal circle determined by A and B is a line in the Hyperbolic geometry of the Unit Disc. For example, suppose you have the points (½, 0) and (½, ½). Let’s get the equation of the orthogonal circle to the Unit Circle that contains these two points. If you substitute the coordinates of the first point in the equation you get 1 1 a 1 4 2 a 5 2 Now substituting the coordinates of the second point and the value for a, you get 1 1 5 1 1 ( ) b( ) 1 4 4 2 2 2 b 1 2 40 So our equation is 5 1 x 2 y 2 x y 1 2 2 5 1 5 (x ) 2 (y ) 2 4 4 8 Here’s a graph of this circle from Sketchpad. Note the perpendicular tangent lines (you can draw these in – at both intersections). fx = 1-x2 1.4 gx = - 1-x2 5 hx = 8 qx = - - x- 5 8 - x- 5 4 5 4 2 + 2 + 1 1.2 4 1 1 4 0.8 0.6 0.4 0.2 -2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5 -0.2 -0.4 -0.6 -0.8 -1 -1.2 -1.4 41 Angles of Parallelism page 104 Let’s look at a line and a point not on that line. First we’ll run a perpendicular through the point to the line. Then we’ll run lines through the point that intersect on the left…all the way over to the ONE asymptotically parallel line which intersects our line at an omega point (a circle point). This is the left sensed parallel to the line. The angle from the omega point to the point not on the line to the base of the perpendicular is called an angle of parallelism. The left and right angles of parallelism are congruent and acute. The proofs of this are on pages 105 and 106. In paper 2, select one of the proofs and explicate it as if you are teaching it to a bright high school student. In the end you must make sure I know you know how the prove works. Especially concentrate on answer the “why?” question. 42 Circles: A circle is a locus of points equidistant from a given point, the center. True in each of the Big Three. You should be familiar with Euclidean circles. Let’s look at Spherical circles first. When a plane intersects the surface of the sphere and does NOT include the center of the sphere, it makes a circle not a line. It becomes a matter of convenience to eliminate one of the TWO centers by choosing the one that has the shorter radius. TWO? For this we’ll need to discuss distance. In Spherical geometry: d r where theta in in rads and is the measure of the central angle subtended by the two points. Now, for us, except in the homework, the radius is one (it is not ALWAYS one, though). So what is the distance from (0, 1, 0) to (0, 0, 1)? Where are these on the sphere? Work it out. What is the 3D distance formula? Where is the circle centered at the first point with a radius of this distance? Where is the second center? 43 And from ( 2 2 3 1 , , 0) to ( , , 0) 2 2 2 2 Where are these on the sphere? Where is the circle centered at the first point with a radius of this distance? Where is the second center? Now these are called antipodal points. Let’s talk a minute about them. 180 degrees apart. Every point has an antipodal Circle centers. Between A and B nonantipodal. Between A and B antipodal. 44 Now let’s look at Hyperbolic circles. What do you see vs what do you know? 45 An artifact of the Hyperbolic distance formula. Let’s look at that for a few minutes. Here’s line AB. How far is it from A to B. We’ll need the omega points at the ends of the lines to make the calculation. Poincaré Disk Model Disk Controls A M B N Measure the following distance in Euclidean Geometry: AM AN BM BN 46 You use these Euclidean measurements to calculate the Hyperbolic distance with a formula: AM BN the Hdistance from A to B is ln AN BM AM BN AN BM is called the “cross product”. The absolute value of the natural log of the “cross product” is a very clever way to measure distances. Let’s look at some consequences of this formula. Poincaré Disk Model Disk Controls Euclidean distances MA = 0.86 in. BM = 2.07 in. AN = 1.91 in. BN = 0.69 in. Hyperbolic distances MA = 35.48 HDAB MB = 37.37 ln MABN BMAN = 1.89 NA = 36.38 NB = 33.50 EG AB = 1.25 in. A AB = 1.89 M B N Here’s a picture with Hyperbolic distances on the left and Euclidean distances on the right. The hyperbolic distances were measured using the hyperbolic distance tool on the left and the Euclidean distances were measured using the Euclidean distance tool on the top menu. On the right, the calculation for Hyperbolic distance is shown. Remember that the calculation uses Euclidean distances. The Hyperbolic tools do the calculation automatically for you. 47 In Euclidean Geometry the distance between points on a segment is fixed and independent of location in the plane. In Hyperbolic Geometry, however, you have an interesting stretching of calculated distances that depends on whether the points are close to the center of the disc or close to the edge of the disc. Points can be the SAME Euclidean distance apart and have different Hyperbolic distances depending on their location in the disc. This is a function of the distance formula. Here’s an illustration with two points that are .09 apart in Eucildean geometry and located in two different spots in the disc. Note that the Hyperbolic distances are different and the points are further apart out near the edge of the disc. Poincaré Disk Model Disk Controls A Euclidean Distance AB = 0.09 in. Hyperbolic distance AB = 0.53 B M Euclidean distance .09 Hyperbolic distance .52 N 48 Poincaré Disk Model Disk Controls A B N Euclidean Distance AB = 0.09 in. Hyperbolic distance AB = 0.12 M Euclidean distance .09 Hyperbolic distance .12 Notice that the Hyperbolic distance depends on WHERE you are in the disc. Points that are the same Euclidean distance apart in different locations on the disc are different Hyperbolic distances apart. 49 Let’s do a problem together to look at this situation: We will use the Unit Circle, and the x-axis as our Hline. It’s a diameter: Great circle with it’s center at infinity! Poincaré Disk Model Disk Controls C1 (-1,0) A (1,0) P. Disk Center D E B P. Disk We will find the distance from D to E using the formula. I’ll do one as an example. If D has the Cartesian coordinates (1/3, 0) and E has the Cartesian coordinates (3/5, 0), then they are 4/15 .267 apart in Euclidean Geometry. DA EB DA EB The Hdistance from D to E is ln = ln DB EA DB EA DA is 4/3. EB is 2/5. DB is 2/3. EA is 8/5. 50 Putting these together in the cross product: 4 2 3 51 2 8 2 3 5 And the absolute value of ln (.5) is approximately .69, the Hdistance. Now, I’m going to give you 5 points. Fill in the following chart with both the Euclidean Distances and the Hyperbolic Distances and draw a conclusion about Hyperbolic Distance and location in the disc. Points Distances P0 (0, 0) NA P1 (1/3, 0) P0 to P1 P2 (3/5, 0) P1 to P2 P3 (7/9, 0) P2 to P3 P4 (15/17, 0) P3 to P4 P5 (31/33, 0) P4 to P5 Euclidean d Hyperbolic d What do you observe about the distances as you move further from the center of the disc? 51 3.3 Polygons in Hyperbolic Space We’ve looked at triangles and Saccheri quadrilaterals. Each of the Big Three has polygons, of course. The EXCESS of a Spherical triangle is Sum of the Int. Angles – 180 degrees. The DEFECT of a Hyperbolic triangle is 180 degrees – Sum of the Int. Angles. There is a Pythagorean Theorem in each of Spherical and Hyperbolic. In each geometry it involves angle measures. In Spherical, given the three angles measuring A, B, and C of a right triangle with C being the right angle and R is the measure of the sphere’s radius: cos(C / R) cos( A / R) cos( B / R) In Hyperbolic, “cosh” is hyperbolic cosine. cosh C cosh A cosh B Note that if your hyperbolic angle measures are large, you will get approximately: C A B log 2 Area of triangles: In Spherical: Area ( A B C ) R 2 In Hyperbolic: R 2 ( A B C ) Note that all the formulas are based on angle measures! This means that if two triangles have the same angle sum, then they have the same area. In Hyperbolic geometry, there is a largest triangle. It has angle sum zero and an area of 180 degrees. Can you guess which triangle this is? It has 3 omega points as vertices! 52 LUNES! A two-sided polygon exclusive to Spherical geometry. Where do these show up? The sides are congruent. The summit angles are congruent. Area? A fraction of the surface area of the sphere. ( 4 r 2 ) Polygons in general: Can we have 5 sided figures in each geometry? Let’s restrict ourselves to convex figures here. The sum of the interior angles of a convex polygon in EG is (n 2)(180) . What is it for Hyperbolic? What is it for Spherical? 53 3.4 Congruence in Hyperbolic Space If 3 angles of one triangle are congruent respectively to 3 angles of a second triangle the triangles are congruent…in SG and HG How is this different than in EG? What does it tell you about similarity? Saccheri Quadrilaterals with congruent summits and summit angles are congruent. Where is this true? 54 Let’s discuss some similarities between the Big Three: 55 And now some differences: 56