§23.1 hyperbolic models

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.

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


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


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 ModelFinding 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.

1010



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 = bPlug the coordinates of A and B into the above and solve.

a = - 12 and b = - 11

x 2 + y 2 - 12x = - 11

Check your work!

2 2 + 3 2 + a2 = b and

9 2 + 4 2 + a9 = b and solving

1111



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.

1212



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 + 42.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


13

DistanceVertical ray

AMAB* ln

BM

Semicircle

AM BN

AB* lnAN BM

A B

B

A

M MN

• •

•

•

14


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• •

AM CN CM BNln ln

AN CM CN BM

AM CN CM BNln

AN CM CN BM

ln(AB,MN) AB*

AM BN

lnAN BM

Note: These are Euclidean distances.

15


15

Distance example Vertical ray

AMAB* ln

BM

8AB* ln 0.47

5

B (2, 5)

A (2, 8)

M MN

•

•

16


16

Distance example Semicircle

AM BN

AB* lnAN BM

B (9, 4)

M (11, 0)

A (2, 3)

N (1, 0)

••

1 12 2

1 12 2

90 80AB* ln

10 20

1AB* ln 36 1.792

2

12 2 2AM (11 2) (3 0) 90 12 2 2BN (9 1) (4 0) 80 12 2 2AN (2 1) (3 0) 10

12 2 2BM (11 9) (4 0) 20

17


17

Angle MeasureAngle 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


18

Angle Measure

A

BC

A’

•

C’

Given two rays with slopes m 1 and m 2 , the angle between those rays is 1 1 2

1 2

m mtan

1 m m

Given lines AB: x2 + y2 = 9 and BC: x2 + y2 – 10x = - 9 Find the angle between them.

We will solve them simultaneously to find the point of intersection B.

x2 + y2 = 9

x2 + y2 – 10x = - 9

Yields x = 1.8 and y = 2.4 B (1.8, 2.4)

19


19

A

BC

A’

•

C’

1 1 2

1 2

m mtan

1 m m

The derivative of line AB: x2 + y2 = 9 is y’ = -x(9 – x2) – ½

evaluated at x = 1.8 yields m1 = - 0.75

The derivative of line BC: x2 + y2 – 10x = - 9 is y’ = (5 – x) (10x – x2 – 9) evaluated at 1.8 yields m2 = 1.333…

1 0.75 1.33tan

1 ( 0,75)(1.333)

1 0.5833tan 90

0

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


(b) Each plane contains three noncollinear points and each line contains at least three points.

Clear by construction.

22


(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 AM AN

AB* ln ln 1 0AN AM

(d) AB* = BA*

AM BN

AB* lnAN BM

BM AN

lnBN AM

BA*

23


(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


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



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

x > 0

x > 0

25



(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



(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

• •

PM QNPQ* k ln 1 where

PN QM

1k

PM QNln

PN QM

27



(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

PQ* x y ln(AP,MN) ln(AQ,MN)

1(AP,MN)ln ln(PQ,MN)

(AQ,MN)

ln(PQ,MN) PQ*

28

Parallel PostulateHyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x2 + y2 = 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. 1 + 49 + a = b

25 + 0 - 5a = bHence a = - 4.1666 and b = 45.833

NM

P [x]•

One parallel is x2 + y2 – 4. 17 x = 45.83

29

Parallel PostulateHyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x2 + y2 = 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.

1 + 49 + a = b

25 + 0 + 5a = b

Hence a = 6.25 and b = 56.25

NM

P [x]•

The other parallel is x2 + y2 + 6.25 x = 56.25

30

Assignment: §23.1

§23.1 hyperbolic models

Documents