Download 3379 NonE hw

Homework:
Non-Euclidean Geometries, Math 3379
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1.
The Three Point Geometry:
Prove Theorem 2:
There are exactly 3 distinct lines in this geometry.
Part A, Suppose there are fewer than 3 lines…
Part B, Suppose there are more than 3 lines…
2.
Using the finite geometries as a “go by”, write a set of axioms that would do as
the ones that produced the following model. Add an axiom that details the
situation with respect to parallel lines.
This is an octahedron;“points” are interpreted as vertices and
“lines” are the edges.
D
E
C
A
B
F
3.
An Incidence Geometry with 7 points.
Fano’s Geometry is an incidence geometry with exactly seven points.
Find Fano’s Geometry on the internet: give the reference here:
Copy out the whole geometry:


Undefined terms
Axioms
Copy out the model with the points lettered. Write an explanation of how the
model matches each axiom exactly.
Write out a list of possible
 definitions – at least 3.
Find and learn the proofs for at least 2
 theorems. Include these in your homework. Legibly.
4.
Find 10 same/different pairs for TCG and EG. An example is:
the same:
different:
they have the same undefined terms and axioms up to SAS
SAS isn’t an axiom in TCG and it IS in EG.
Find 9 more pairs.
You may NOT list axioms singly. Each similarity and difference must be
substantially different from the other ones on the list.
5.
In Euclidean Geometry, circles can share no points (ie, they don’t intersect),
intersect in one point (tangent, internally tangent) or intersect in two points.
How do TCG circles interact?
Clearly there can be “no intersection”.
What about tangency – are there both internally tangent and externally tangent?
Look up internally tangent circles if you’ve never heard the term before.
Intersect in two points, clearly this exists.
Are there intersections of more than two points?
Write a 1 page illustrated essay about two TCG circles and how they interact.
6.
Write an essay about Saccheri Quadrilaterals and why they are important.
Include some details comparing and contrasting them across all 3 of the “big
geometries”: Euclidean, Spherical, and Hyperbolic.
Fill in the following chart
Geometry
Summit angles type
(acute, right, obtuse)
Summit to base
comparison (<, >, =)
Euclidean
Spherical
Hyperbolic
A Saccheri Quadrilateral in Hyperbolic Geometry is illustrated below:
Poincaré Disk Model
Disk Controls
mIGH = 51.1
mJHG = 51.0
GH = 2.59
G
H
JH = 0.93
GI = 0.93
mGIJ = 90.0
mHJI = 90.0
J
I
IJ = 1.97
This is the only kind of quadrilateral that you can have in all 3 of the big geometries:
a Saccheri Quadrilateral.
7.
Here is a list of statements that are equivalent to the Euclidean Parallel Postulate:








The area of a triangle can be made arbitrarily large.
The angle sum of all triangles is a constant.
The angle sum of any triangle is 180.
Rectangles exist.
A circle can be passed through any 3 noncollinear points.
Given an interior point of a angle, a line can be drawn through that point
intersecting both sides of the angle.
Two parallel lines are everywhere equidistant.
The perpendicular distance from one of two parallel lines to the other is always
bounded.
Consider each statement in turn and find a counterexample in Spherical or Hyperbolic
Geometry.
Illustration:

Two parallel lines are everywhere equidistant.
Poincaré Disk Model
This sketch depic ts the hyperbolic plane H2 us in g the Poincaré disk model. In this model, a line through
tw o poin ts is def ined as the Euc lidean arc pas sing through the points and perpendic ular to the c irc le .
Us e this document's custom tools to perform c onstructions on the hyperbolic plane, comparing y our findings
to equivale nt constructions on the Euc lidean plane.
Dis k Controls
Lines LK and IJ are parallel. I built perpendiculars
to LK and points at the intersection of the
perpendiculars on IJ. In measuring the distance -the "perpendicular" distance -- note that the lines
are NOT equidistant. Equidistance parallel lines
are a Euclidean phenominon.
I
D
J
L
F
LI = 0.76
G
K
E
KJ = 1.27
8.
Parallel Lines in Hyperbolic Geometry
F
Disk Controls
B
G
A
D
E
H
H AB is the line we’re interested in. Point D is our point NOT on H AB .
To which line is H AB “asymptotically parallel”?
Which lines are “divergently parallel” to H AB ?
9.
Vertical angles are congruent…in Euclidean Geometry for sure, but what about
the other 2 “big ones”?
Analyze the picture below for your consideration of Hyperbolic geometry:
F
Disk Controls
mGDI = 57.5
B
G
I
A
D
J
H
E
mEDJ = 57.5
Sketch in an illustration that is an example or is a counterexample in Spherical Geometry:
10.
Triangles and Exterior Angles in Hypebolic Geometry
F
Disk Controls
L
mBAH = 35.3
B
mAHB = 40.5
mABH = 35.6
A
M
mLBM = 144.4
m1+m2+m3 = 111.44
m1+m2 = 75.80
H
Yes, we still have triangles. No, the sum of the interior angles is not equal to 180; it is
LESS THAN 180 as promised. The difference between 180 and the sum of the interior
angles of a given Hyperbolic triangle is called the DEFECT of the triangle. In Spherical
geometry the difference between the sum of the interior angles of a spherical triangle and
180 is called the EXCESS of the triangle. Note that the defect of the triangle above is
58.6. (The sum of the angles is 121.4) We use the lower case Greek letter delta for defect
(  ).
Given that Exterior Angle Inequality Theorem in EG is true, is the exterior angle
(LBM) in the picture above demonstrating that the Theorem is true in Hyperbolic
Geometry?
Is it true in Spherical Geometry that an Exterior Angle is greater than either remote
interior angle? How do you know?
11. Right triangles – we’ve got them in all 3 “Big geometries”.
Disk Controls
mCDB = 90.0
C
mDCB = 17.6
mCBD = 18.1
D
B
DC = 1.81
DB = 1.79
A
CB = 2.96
Distance2+Distance2 = 6.48
Here’s a right triangle – the measures of the angles are shown. I built it using an H AB
and the Hperpendicular bisector of that line (again: Hyperbolic tools on the left menu).
 = 54.3
Does the Pythagorean Theorem hold in Hyperbolic Geometry?
Does it hold in Spherical Geometry?
12.
Distance
Poincaré Disk Model
Disk Controls
A
B
M
Measure the following distance in Euclidean Geometry:
AM
AN
BM
BN
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
is called the “cross product”.
AN  BM
The absolute value of the natural log of the “cross product” is a very clever way to
measure distances.
N
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
MABN
BMAN

= 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.
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 Euclidean 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
Euclidean Distance
AB = 0.09 in.
Hyperbolic distance
AB = 0.12
B
M
N
Euclidean distance .09
Hyperbolic distance .52
Poincaré Disk Model
Disk Controls
A
B
N
M
Euclidean distance
Hyperbolic distance
.09
.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.
We will use the Unit Circle, and the x-axis as our Hline.
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.
Putting these together in the cross product:
4 2

3 51
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?
How is this like or different from Spherical geometry and Euclidean geometry?
Problem 13
Find the Hdistances
d1 from (0, 0) to ( ½, 0)
d2 from (1/2, 0) to (15/16, 0)
d3 from ( ½, 0) to (½,½)
What does this tell you about Hyperbolic distances?
Sketch the Unit disc in Euclidean Geometry, put these points in (as “to scale” as you can)
and put both the Hyperbolic and Euclidean distances on your sketch in textboxes. By
hand is just fine.