Chapter 6
Chapter 6 homework
6.1
04
05
12 points
10 points
6.2
01
10 points
6.5
04
10 points
6.7
01
20 points
6.8
01
10 points
1
6.1
Basic Theorems of Hyperbolic Geometry
We still have our 6 axioms of Neutral Geometry and now for our seventh, we will adopt the
Hyperbolic Parallel Postulate:
For every line l and for every point P that does not lie on l, there are at least two lines m and n
such that P lies on both m and n and both lines are parallel to l.
This is a negation of the Euclidean Parallel Postulate.
Theorem 6.1.1
For every triangle ABC ,  (ABC )  180 .
Corollary 6.1.2
For every triangle ABC , 0°<  (ABC )  180 .
Theorem 6.1.3
For every convex quadrilateral,
Corollary 6.1.4
The summit angles in a Saccheri quadrilateral are acute.
Corollary 6.1.5
The fourth angle in a Lambert quadrilateral is acute.
Theorem 6.1.6
There does not exist a rectangle.
ABCD ,  ( ABCD)  360 .
Since the Saccheri quadrilateral summit angles are acute, it is not a rectangle!
Theorem 6.1.7
In a Lambert quadrilateral, the length of a side between two right angles is
strictly less than the length of the opposite side.
Corollary 6.1.9
In a Saccheri quadrilateral, the length of the altitude is less than the length
of a side.
The altitude of a Saccheri Quadrilateral is the segment joining the midpoint of the summit to the
midpoint of the base. It is, by an earlier proof, perpendicular to each. The length of the altitude
is called the height.
Corollary 6.1.8
In a Saccheri quadrilateral, the length of the summit is greater than the
length of the base.
Theorem 6.1.11
AAA
If ABC is similar to DEF , then ABC is congruent to DEF .
2
Strategy:
We will show one pair of sides is congruent and have the triangles
congruent by ASA. We’ve got the angles all congruent by hypothesis,
we will assume no side pairs are congruent and show a contradiction.
Proof:
Let ABC and DEF be 2 similar triangles. This gives us all 3
cooresponding angle measures are congruent.
Suppose AB  DE , BC  EF , and AC  DF . Now there are 3 side pairs
so 2 pairs have one side longer or 2 pairs have one side shorter than
in ABC . We will assume that AB  DE and AC  DF . If this is not the
case, a similar proof will work with the 2 pairs that are longer.
Chose points on ABC so that a smaller internal triangle congruent to
DEF is created… AB ' C ' .
C
F
C'
A
B'
B
D
E
Now AB ' C '  DEF by SAS and quadrilateral C’B’BC is convex (Thm. 4.6.7).
Further AB ' C '  ABC and AC ' B '  ACB . Now, note that the angles at vertices C’ and
B’ are linear. Thus the convex quadrilateral is a rectangle with the angle
sum = 360°.
This contradicts that we are in Hyperbolic geometry! (see Thm 6.1.6). So at least one of the
inequalities on the sides. So our original triangles are congruent. QED
Thus there is no such concept of similarity in Hyperbolic geometry. All triangles are congruent.
Similarly, all Saccheri Quadrilaterals are congruent.
3
Theorem 6.1.12
If ABCD and A ' B ' C ' D ' are two Saccheri Quadrilaterals such that
 ( ABCD)   ( A ' B ' C ' D ') , then ABCD  A ' B ' C ' D ' .
Read and understand the proof, please.
6.1.A
The Poincaré Disc is a 2D model of Hyperbolic Geometry.
The points are those for which x 2  y 2  1, known as the Unit Disc (as opposed to the Unit
Circle).
Lines are arcs of orthogonal circles. These are circles that intersect the Unit Circle with tangents
at the point of intersection that are at 90° to the circle.
Poincaré Disk Model
This sketch depicts the hyperbolic plane H2 using the Poincaré disk model. In this model, a line through
two points is defined as the Euclidean arc passing through the points and perpendicular to the circle.
Use this document's custom tools to perform constructions on the hyperbolic plane, comparing your findings
to equivalent constructions on the Euclidean plane.
Disk Controls
B
A
F
E
C
Radius
D
4
6.2
Common Perpendiculars
We assume that parallel lines are equidistant from one another…this is not the case in
Hyperbolic Geometry
Disk Controls
A N
E
M
B
H
G
I
HI  0.75
F
mBHI  90.0
NM  0.94
C
mHNM  90.0
D
Theorem 6.2.1
If l is a line, P is an external point, and m is a line such that P lies on m,
then there exists at most one point Q such that Q  P , Q lies on m and
d(Q, l) = d(P, l).
Let’s look at what this is really saying!
This says there are at most 2 distinct points from one parallel line to another at a given distance
between the parallel lines. Note, too, unstated – one of the angles of intersection must be 90° for
each point pair.
Proof:
Let l and m be two lines. Suppose there are 3 points P, Q, and R on m such that
d ( P, l )  d (Q, l )  d ( R, l ) . Let P’, Q’, and R’ denote the feet of the perpendiculars on l,
respectively. As these distances are all non-zero, none of P, Q, or R lies on l. Thus at least 2 of
the 3 points must lie on the same side of l, suppose these are P and Q. We then see that
quadrilateral PP ' Q ' Q is a Saccheri Quadrilateral. Thus l and m are parallel (Thm 4.8.10, part
4) and further all 3 points lie on the same side of l since they are collinear.
5
Assume that P*Q*R (if not, then rename the points). Now both PP ' Q ' Q and QQ ' R ' R are
Saccheri Quadrilaterals. From this we see that PP ' Q and RQQ ' are acute. This
contradicts the fact that they are supplements! So our supposition that all three distances are the
same is not true.
P
Q
P'
R
Q'
R'
Definition:
Lines l and m admit a common perpendicular if there exists a line n such that n is perpendicular
to both l and m.
Theorem 6.2.3
If l and m are parallel lines and there exist two points on m that are
equidistant from l, then l and m admit a common perpendicular.
6
Theorem 6.2.4
If lines l and m admit a common perpendicular, then that common
perpendicular is unique.
Theorem 6.2.5
Let l and m be parallel lines cut by a transversal t. Alternate interior
angles formed by l and m with transversal t are congruent if and only if l
and m admit a common perpendicular and t passes through the midpoint of
the common perpendicular.
Let’s sketch this!
Disk Controls
A
E
H
G
I
B
HI  0.75
F
mBHI  90.0
mHIF  89.9
HG  0.38
GI  0.37
CLASSWORK #1 here
7
Some familiar items from Chapters 3 and 4:
Disk Controls
C
D
mADC = 90.0°
mCDB = 90.0°
AD = 1.29
DB = 1.29
A
B
Radius
AC = 2.37
CB = 2.37
Isosceles triangle with a bisector forms two
right subtriangles
8
Vertical angles are still congruent:
Disk Controls
C
mCEB = 104.2°
A
E
C1
B
D
mAED = 104.2°
Remember: your eyes are trained to see Euclidean!
9
The Exterior Angle Theorem in Hyperbolic Geometry:
Disk Controls
C
mCBG = 101.9°
C1
G
m1 + m2 + m3 = 157.98°
E
B
mCEB = 46.8°
mECB = 33.1°
mEBC = 78.1°
EA is strictly GREATER THAN in Hg
Deficit!
Exterior Angle Thm in the Big Three review:
CourseWork #2
10
6.3
The Angle of Parallelism
Construction: Let l be a line and let P be an external point. Drop a perpendicular from P to l and
call the foot of the perpendicular A. Let B  A be a point on l. For each real number r with
0  r  90 there exists a point Dr on the same side of PA as B such that m (APDr )  r  .
Define a set K:
K  {r PDr  AB  }
K is called the INTERSECTING SET for P and AB .
P
r°
A
Dr
B
mPAB = 90.00°
Note that K is a subset of [0, 90).
Theorem 6.3.1
Let K be the intersecting set for P and AB . If r  K , then
1.
s  K for every 0 < s < r.
2.
there exists a t  K such that t > r.
K is an open interval of the form [0, r0 ) where r0 is called the critical number for P and AB .
Any ray from P that makes an angle less than r0 intersects l. And any ray that makes an angle
more than r0 doesn’t intersect l on the AB side. This is why r0 is critical.
11
Disk Controls
F
M
Radius
A
Definition: Angle of parallelism
FM is the HLine just past the one that would define r0 .
Suppose P, A, and B are as given in the definition of intersecting set. Suppose r0 is the critical
number for P and AB . Let D be a point on the same side of PA as B such that m (APD)  r0 .
The angle APD is called the angle of parallelism for P and AB .
P
Dt
Dr
A
B
Ds
R
S
T
As you would expect there is another set of rays on the other side of PA that are symmetric to the
ones shown. Think of r0 as a kind of limit as the angle measure goes to 90°.
12
Disk Controls
C
A
B D
E
Radius
Theorem 6.3.5
The critical number depends only on d(P, l).
13
As the distance from P to l increases the critical number decreases.
Thus we view it as a function on the real numbers.
Disk Controls
J
F
Radius
A
Theorem 6.3.7
 : (0, )  (0,90] is a nonincreasing function; that is a < b implies
 (a)   (b) .
Theorem 6.3.8
Every angle of parallelism is acute and
every critical number is less than 90.
This is every bit as much a distinguishing feature between Euclidean Geometry and Hyperbolic
Geometry as angle defect. In Euclidean geometry the angle of parallelism is exactly 90°. In
Hyperbolic geometry it is strictly less than 90° and varies with the location of P with respect to l.
14
mYXP  17.7
X
In fact, the angle of parallelism is a
measure of how much deviation from
the Euclidean ideal a parallel line has.
Just as defect measures how far from a
Euclidean triangle a Hyperbolic triangle
is.
mQVP  46.9
U
V
W
P
mVPW  90.1
Y O
CourseWork #3
15
6.4
Limiting Parallel Rays
Definition:
Two rays PD and AB are called limiting parallel rays, denoted PD AB
If B and D are on the same side of PA and PD  AB   and every ray
between PD and PA intersects AB .
The angle of parallelism provides the basic example of limiting parallel rays.
Theorem 6.4.2
If PD AB , then PD AB .
If two rays are limiting parallel rays, then they are parallel.
(asymptotically)
Theorem 6.4.3
Symmetry of Limiting Parallelism
If PD AB , then AB PD .
Theorem 6.4.4
Endpoint Independence
If AB is a ray, and P, Q, and D are points such that Q*P*D, then
PD AB if and only if QD AB .
The points need to be collinear.
Theorem 6.4.5
Existence and Uniqueness of Limiting Parallel Rays.
If AB is a ray and P is a point that does not lie on AB , then there exists a
unique ray PD such that PD AB .
Theorem 6.4.7
Transitivity of Limiting Parallelism
If AB , CD , EF are three rays such that AB CD and CD EF , then
either AB EF or AB and EF are equivalent rays.
16
Lemma 6.4.8
If AB , CD , EF are three rays such that AB CD and CD EF , then
either AB and EF are equivalent rays or AB EF .
Lemma 6.4.9
If AB , CD , EF are three rays such that AB CD and CD EF , then
there exists a line l such that l intersects each of AB , CD , and EF .
CourseWork #4
17
6.5
Asymptotic Triangles
Asymptotic triangles are also called Omega triangles.
Disk Controls
P
D
P. Disk
B
A
Radius
It’s not really a triangle because at least one vertex is not really in our space.
Notation: check the book: an open triangle followed by 4 letters for 2 vertices and 2 points on
the limiting rays.
The Exterior Angle Theorem for Asymptotic Triangles
Theorem 6.5.2
If DPAB is an asymptotic triangle and C*A*B, then
m (CAP)  m (APD) .
18
Disk Controls
P
D
P. Disk
B
A
C
Radius
Corollary 6.5.3
Angle Sum Theorem
If PAB is an asymptotic triangle, then m (APD)  m (PAB)  180 .
We cannot measure an “angle” with a “vertex” that is not a point in our space!
19
Theorem 6.5.4
Side-Angle Congruence Condition
Let EPAB and FQCD be two asymptotic triangles. If APE  CQF
and AP  CQ then PAB  QCD .
Disk Controls
mAPE = 62.8°
mPAB = 24.0°
P
AP = 2.04
E
CQ = 2.04
A
B
P. Disk
F D
Q
mCQF = 62.5°
mQCD = 24.4°
Radius
C
20
6.6
The Classification of Parallels
Theorem 6.6.2
Classification of Parallels
Let l and m be parallel lines.
Part 1 If l and m are asymptotically parallel, then l and m do not admit a
common perpendicular.
Part 2 Either l and m admit a common perpendicular or they are
asymptotically parallel.
Theorem 6.6.3
Suppose l m . Let P, Q, and R be points on m such that P*Q*R and let
A, B, and C be the feet of the perpendiculars from P, Q, and R to l.
1.
If PA  m , then PA  QB  RC
2.
If PQ AB , then PA  QB  RC
1.
Disk Controls
mPAB = 90.0°
mAPQ = 90.0°
mQBA = 90.0°
PA = 0.36
mRCB = 90.0°
R
QB = 0.43
RC = 1.21
m
l
P Q
A B
C
Radius
21
2.
Controls
mPAB = 90.0°
mQBA = 90.0°
mRCB = 90.0°
PA = 0.99
QB = 0.13
RC = 0.17
l
Pm
A
Q
RR
CB
Radius
Disk Controls
P
K
I
M'
mKIJ = 89.8°
mIJA = 89.8°
M
A
J
Radius
22
Theorem 6.6.4
If l and m are parallel lines that admit a common perpendicular, then for every positive number
d 0 there exists a point P on m such that d ( P, l )  d0 . Furthermore P may be chosen to lie on
either side of the common perpendicular.
Disk Controls
mPEF = 90.1°
C
P
E
F
A
B
Radius
What does this say about distances being bounded in the Poincare Disc?
Summarizing:
For every line l and point P not on that line, there are two lines through P asymptotically parallel
to l. There is no common perpendicular between l and these two lines. Every line between these
two particular parallels is also parallel to l, but will allow a common perpendicular. These are
the divergently parallel lines. See illustration below.
CourseWork #5
23
Disk Controls
P
K
I
M'
mKIJ = 89.8°
mIJA = 89.8°
M
A
J
Radius
24
6.7
Properties of the Critical Function
Construction: Let l be a line and let P be an external point. Drop a perpendicular from P to l and
call the foot of the perpendicular A. Let B  A be a point on l. For each real number r with
0  r  90 there exists a point Dr on the same side of PA as B such that m (APDr )  r  .
Define a set K:
K  {r PDr  AB  }
K is called the INTERSECTING SET for P and AB .
Theorem 6.7.1
 : (0, )  (0,90] is a strictly decreasing function; that is a < b implies
 (a)   (b) .
This is a strictly Hyperbolic result.
K is nonincreasing in Neutral Geometry.
Theorem 6.7.2
If l and m are asymptotically parallel lines, then there exists a point T on m
such that d (T , l )   .
Theorem 6.7.3
lim  ( x)  0 .
Theorem 6.7.4
Theorem 6.7.5
x 
lim  ( x)  90 .
x  0
K is onto; that is, for every number y  (0,90) there exists an x  (0, )
such that K(x) = y.
25
6.8
The defect of a triangle
Theorem 6.8.1
For every   0 there exists an isosceles triangle ABC such that
 (ABC )    and  (ABC )  (180   ) .
Proof
Let   0 be fixed and given. Select a large enough so that

K(a )  . Let PAB be an isosceles right triangle with A being the right
4
angle and both legs having measure a. Now PB intersects AB thus the

measure of APB is less than the critical number ( K(a )  ). The
4
measure of ABP is also less than the critical number because the
triangle is isosceles. Place a copy of this triangle adjacent to the one we
constructed, sharing side AP . Now we have another isosceles triangle


with 2 angles measuring  and one angle measuring  so the angle
4
2
sum is less than   0 .
Theorem 6.8.2
For every   0 there is a right triangle APB such that
 (APB)  (180   ) and  (APB)    .
This means we may have triangles with a small defect.
Theorem 6.8.3
For every positive number  there is a positive number  such that if
ABC is a triangle in which every side has length less than d, then
 (ABC )   .
This means that when we have a triangle with a very short side, the defect will be small. It turns
out that defect is proportional to the area of the triangle.
Theorem 6.8.4
For every pair of points A and B and for every positive number  there is
a number d > 0 such that if C is any point not on AB with AC < d, then
 (ABC )   .
Theorem 6.8.5
Continuity of Defect
The function f :[0, c]  R defined in the text is a continuous function
26
Enrichment:
Circles and Arcs:
Hyperbolic Circles are Euclidean circles (with a different center) that are entirely within the
space. Let’s look at one of two of these
Horocycles are “circles” that are tangent to the boundary and thus are OPEN in the space.
What other geometry has horocycles?
Hypercycles are arcs of Euclidean circles that are NOT orthogonal…more than one point lies
outside the space. Secant lines are hypercycles with their centers at
infinity
CourseWork #6
27
6.9
Is the Real World Hyperbolic?
Read this in preparation for your term paper.
28