1 §23.1 Hyperbolic Models 1 Intro to Poincare's Disk Model. A Point – Any interior point of circle C (the ordinary points of H or h-points) Line – Any diameter of C or any arc of a circle orthogonal to C in H are h-lines.

2 Poincare's Disk Model 2 Generally it is not easy to find a circle orthogonal to another circle, i.e. an h-line. So this model is somewhat difficult to use. However, the famous Canadian geometer H.S.M. Coxeter in conversations with M. C. Escher suggested its use in his art when Escher asked about demonstrating an infinite plane in a finite area.

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5 Hyperbolic Half-Plane Model 5 We can make a slight modification to the disc model that makes life a bit easier for us. We will transform the disc into a half-plane model.

6 Hyperbolic Half-Plane Model 6 The half-plane model can be thought of as taking the disk model at a point on the circle and separating the disk and stretching the boundary circle into a straight line. In essence it is all points P(x, y) for which y > 0 or the upper half-plane. This half-plane used to be the interior of the circle of the disk model. All such points are the h-points in the model. The x- axis is not part of the geometry.

7 Hyperbolic Half-Plane Model 7 A h-Point – any point P(x, y) where y > 0. h-Lines – vertical rays and semicircles with center on x-axis. (x - a) 2 + y 2 = r 2 Ideal points. Ultra-ideal points.

A Lambert quadrilateral. Notice that it denies the obtuse angle option.

99 Hyperbolic Half-Plane Model Finding h-lines. Find the h-line through the h-points A (2, 1) and B(2, 5). B (2, 5) A (2, 1) M (2, 0) The h-line AB* is a vertical ray, hence the equation is x = 2, y > 0.

10 Hyperbolic Half-Plane Model Finding h-lines. Find the h-line through the h-points A (2, 3) and B(9, 4). B (9, 4) M (11, 0) A (2, 3) N (1, 0) A semicircle with center on the x-axis has equation x 2 + y 2 + ax = b Plug the coordinates of A and B into the above and solve. a = - 12 and b = - 11 x 2 + y x = - 11 Check your work! a2 = b and a9 = b and solving

11 Hyperbolic Half-Plane Model Finding h-lines. Find the h-line through the h-points A (2, 3) and B(9, 4). B (9, 4) M (11, 0) A (2, 3) N (1, 0) A semicircle with center on the x-axis has equation x 2 + y 2 + ax = b You could also find the center of circle O by the perpendicular bisector of AB. The radius is then OA or OB and the equation follows from that info.

12 Hyperbolic Half-Plane Model Finding h-lines. Find the h-line through the h-points A (2, 3) and B(9, 4). B (9, 4) M (11, 0) A (2, 3) N (1, 0) The Midpoint of AB is (5 1/2, 3 1/2) and the slope is 1/7. We need the line thru (5 1/2, 3 1/2) with slope -7. y = - 7x If y = 0 then x = 6 so the center of the circle O is (0, 6). The radius is then OA or 5. And the circle with center O of radius 5 is x 2 – 12x = - 11

13 Hyperbolic Half-Plane Model 13 Distance Vertical ray Semicircle A B B A M MN

14 Hyperbolic Half-Plane Model 14 One of the requirements for distance is that if point C is between points A and B on segment AB then AC* + CB* = AB*. This is easy to establish. AC* + CB* = ln (AC,MN) + ln (CB,MN) N B M A Note: These are Euclidean distances.

15 Hyperbolic Half-Plane Model 15 Distance example Vertical ray B (2, 5) A (2, 8) M MN

16 Hyperbolic Half-Plane Model 16 Distance example Semicircle B (9, 4) M (11, 0) A (2, 3) N (1, 0)

17 Hyperbolic Half-Plane Model 17 Angle Measure Angle Measure – Let m  ABC * = m  A’BC’ where BA’ and BC’ are the Euclidean rays tangent to the sides of  ABC as in the disk model. A B C A’ C’

18 Hyperbolic Half-Plane Model 18 Angle Measure A B C A’ C’ Given two rays with slopes m 1 and m 2, the angle between those rays is Given lines AB: x 2 + y 2 = 9 and BC: x 2 + y 2 – 10x = - 9 Find the angle between them. We will solve them simultaneously to find the point of intersection B. x 2 + y 2 = 9 x 2 + y 2 – 10x = - 9 Yields x = 1.8 and y = 2.4 B (1.8, 2.4)

19 Hyperbolic Half-Plane Model 19 A B C A’ C’ The derivative of line AB: x 2 + y 2 = 9 is y’ = -x(9 – x 2 ) – ½ evaluated at x = 1.8 yields m 1 = The derivative of line BC: x 2 + y 2 – 10x = - 9 is y’ = (5 – x) (10x – x 2 – 9) evaluated at 1.8 yields m 2 = 1.333…

20 Verifying Axioms of Absolute Geometry in Half-Plane Model (a) Two points determine a line. If the two points are of the form (x 1, y 1 ) and (x 1, y 2 ) the line has the equation x = x 1. That is, it is a vertical ray. If the two points are of the form A(x 1, y 1 ) and B(x 2, y 2 ), consider the perpendicular bisector. It intersects the x- axis at the center of the semicircle which has radius equal to the distance from the center to either point. B r A C

21 Verifying Axioms of Absolute Geometry in Half-Plane Model (b) Each plane contains three noncollinear points and each line contains at least three points. Clear by construction.

22 Verifying Axioms of Absolute Geometry in Half-Plane Model (c) For any two h-points A and B, AB* > 0, with equality when A = B. By definition, distance is always positive or 0. Let A = B. Then (d) AB* = BA*

23 Verifying Axioms of Absolute Geometry in Half-Plane Model (e) Ruler Postulate. The points of each h-line l are assigned to the entire set of real numbers, called coordinates, in such a manner that (e) Ruler Postulate. The points of each h-line l are assigned to the entire set of real numbers, called coordinates, in such a manner that (1) Each point on l is assigned to a unique number (2) No two points have the same number (3) Any two points on l may be assigned the coordinates zero and a positive real number respectively (4) If points A and B on l have coordinates a and b, then AB =  a - b 

24 Verifying Axioms of Absolute Geometry in Half-Plane Model (e) Ruler Postulate. The points of each h-line l are assigned to the entire set of real numbers, called coordinates. Let P be any point on the h-line AB. Assign the real number x to P by the following x = ln (AP, MN) or if P is on a ray let x = ln (PM/AM) Note that we have omitted absolute value giving the full range of the reals. A P A P M MN x < 0 x > 0

25 Verifying Axioms of Absolute Geometry in Half-Plane Model (e) Ruler Postulate. (1) Each point on l is assigned to a unique number (2) No two points have the same number Q M P N Let P  Q then their coordinates are unique. If there coordinates were the same then (AP, MN) = (AQ, MN) which reduces to PN/QN = PM/QM which is impossible, because one of the ratios is greater than 1 and the other is less than 1.

26 Verifying Axioms of Absolute Geometry in Half-Plane Model (e) Ruler Postulate. (3) Any two points on l may be assigned the coordinates zero and a positive real number respectively Let P be 0 and Q be 1. One can always multiply the distance by a factor to arrange for this to happen. Q M P N

27 Verifying Axioms of Absolute Geometry in Half-Plane Model (e) Ruler Postulate. (4) If points P and Q on l have coordinates x and y, then PQ* =  x - y  Q [y] M P [x] N x = ln (AP, MN) and y = ln (AQ, MN) so

28 Parallel Postulate Hyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x 2 + y 2 = 25, and the point P (1, 7), find the two h-lines through P parallel to the h-line. 1. Find M and N from the equation of the h-line. Let y = 0 and then x =  5 and M (-5, 0), N (5, 0) 2. Find the equation of the line PM a = b a = b Hence a = and b = N M P [x] One parallel is x 2 + y 2 – x = 45.83

29 Parallel Postulate Hyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x 2 + y 2 = 25, and the point P (1, 7), find the two h-lines through P parallel to the h-line. 3. Find the equation of the line PN a = b a = b Hence a = 6.25 and b = N M P [x] The other parallel is x 2 + y x = 56.25

30 Assignment: §23.1