Transcript Hyperbolic Half

§23.1 Hyperbolic Models
Intro to Poincare's Disk Model.
Point – Any interior point
of circle C (the ordinary
points of H or h-points)
• A
Line – Any diameter of
C or any arc of a circle
orthogonal to C in H
are h-lines.
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Poincare's Disk Model
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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Hyperbolic Half-Plane Model
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.
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Hyperbolic Half-Plane Model
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 xaxis is not part of the geometry.
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Hyperbolic Half-Plane Model
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
A
Ideal points.
Ultra-ideal points.
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A Lambert quadrilateral.
Notice that it denies the obtuse angle option.
Hyperbolic Half-Plane Model
Finding h-lines.
Find the h-line through the h-points A (2, 1) and B(2, 5).
The h-line AB* is a vertical ray, hence the equation is
x = 2, y > 0.
B (2, 5)
•
A (2, 1)
•
M (2, 0)
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Hyperbolic Half-Plane Model
Finding h-lines.
Find the h-line through the h-points A (2, 3) and B(9, 4).
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.
2 2 + 3 2 + a2 = b and
9 2 + 4 2 + a9 = b and solving
a = - 12 and b = - 11
x 2 + y 2 - 12x = - 11
Check your work!
A (2, 3)
•
N (1, 0)
• B (9, 4)
M (11, 0)
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Hyperbolic Half-Plane Model
Finding h-lines.
Find the h-line through the h-points A (2, 3) and B(9, 4).
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.
A (2, 3)
•
N (1, 0)
• B (9, 4)
M (11, 0)
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Hyperbolic Half-Plane Model
Finding h-lines.
Find the h-line through the h-points A (2, 3) and B(9, 4).
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
A (2, 3)
•
N (1, 0)
• B (9, 4)
M (11, 0)
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Hyperbolic Half-Plane Model
Distance
Vertical ray
AM
AB*  ln
BM
Semicircle
AB * 
A
AM   BN 

ln
 AN   BM 
•
N
A
•
B
•
•B
M
M
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Hyperbolic Half-Plane Model
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)
Note: These
  AM   CN  
  CM   BN  
 ln 
  ln 
 are Euclidean
  AN   CM  
  CN   BM   distances.
   AM   CN  
 ln  
  AN   CM  


  AM   BN  
  CM   BN   


   ln 
  AN   BM  
  CN   BM   
 ln(AB, MN)  AB *
A
•
N
•B
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M
Hyperbolic Half-Plane Model
Distance example
Vertical ray
AM
AB*  ln
BM
8
AB*  ln
 0.47
5
N
A (2, 8)
•
B (2, 5)
•
M
M
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Hyperbolic Half-Plane Model
Distance example
Semicircle
AB * 
AM   BN 

ln
 AN   BM 
AB *  ln
90
10
1
1
2
2
80
20
1
1
2
1
AM 
(11  2)  (3  0)  90
BN 
(9  1)  (4  0)  80
1
AN 
(2  1)  (3  0)  10
1
BM 
(11  9)  (4  0)  20
2
2
2
2
2
2
2
2
2
2
2
1
2
2
1
AB*  ln 36  1.792
2
A (2, 3)
•
N (1, 0)
• B (9, 4)
M (11, 0)
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Hyperbolic Half-Plane Model
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’
C’
A
C
•
B
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Hyperbolic Half-Plane Model
Angle Measure
Given two rays with slopes m 1 and m 2 , the angle
between those rays is   tan 1 m1  m2
1  m1m2
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 + y 2 = 9
x2 + y2 – 10x = - 9
Yields x = 1.8 and y = 2.4 B (1.8, 2.4)
A’
C’
A
• B
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C
Hyperbolic Half-Plane Model
  tan 1
m1  m2
1  m1m2
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…
  tan
1
  tan
 0.75  1.33
1  (  0, 75)(1.333)
1
0.5833
 90
0
A’
C’
A
• B
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C
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 xaxis at the center of the
semicircle which has radius
equal to the distance from the
center to either point.
A
•B
•
r
C
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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.
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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
AB * 
AM   AN 

ln
 AN   AM 
 ln 1  0
(d) AB* = BA*
AB * 
AM   BN 

ln
 AN   BM 

BM   AN 

 ln
 BN   AM 
 BA *
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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
(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
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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.
P•
x<0
A•
A• x > 0
x<0
•Px > 0
N
M
M
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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
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.
P
N
•
• Q
M
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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.
PQ *  k
PM   QN 

ln
 PN   QM 
1
k
PM   QN 

ln
 PN   QM 
 1 where
P
N
•
• Q
M
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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
x = ln (AP, MN) and y = ln (AQ, MN) so
PQ *  x  y  ln(AP, MN)  ln(AQ, MN)
(AP, MN)
 ln
 ln(PQ, MN) 1
(AQ, MN)
P [x]
•
• Q [y]
  ln(PQ, MN)  PQ *
N
M27
Parallel Postulate
Hyperbolic 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
• P [x]
25 + 0 - 5a = b
Hence a = - 4.1666
and b = 45.833
M
One parallel is x2 + y2 – 4. 17 x = 45.83
N
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Parallel Postulate
Hyperbolic 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
• P [x]
M
N
The other parallel is x2 + y2 + 6.25 x = 56.25
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Assignment: §23.1
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