
Wednesday 

1/17 Introduction 
Start on Euclid Definitions, Postulates, and Prop 1. 

1/22 Begin The Pythagorean Theorem plus... 
1/24 Pythagorean plus... Dissections
? 
1/26 Guest Lecture Here’s looking at Euclid 
1/29 Dissections equidecomposeable polygons Begin Constructions and the real number line. 
1/31 M&I's Euclidean Geometry Constructions and The real number line. Continuuity 

2/5 Construction of rational numbers.  2/7 Inversion and Orthogonal Circles . 
2/9. Odds and ends, More on Inversion. Isometries 
2/12 Classification of Isometries & More :) 
2/14 Isometries. Coordinates and Classification 

2/19 More Isometries 
2/21 Recognizing Isometries 
2/23
Symmetry. 
2/26 Similarity & Proportion (Euclid V) More on Proportion and Measurement  2/28 More on Similarity and transformations. 
Inversion and Beginning to See The Infinite. The Affine Line and Homogeneous Coordinates. 
3/5 The Affine Plane and Homogeneous Coordinates More on seeing the infinite. The Affine Line and Homogeneous Coordinates. 
3/10 More on The Affine Line and Homogeneous Coordinates and the Affine Plane 
3/9 Homogeous coordinates and the Affine Plane

3/12 No class Spring Break! 
3/17 No class Spring Break! 

3/19 Introduction to projective geometry with homogenous coordinates. Lab time: Video "A nonEuclidean Universe." Introduction to Desargues' Theorem a result of projective geometry. 
3/21Another vector geometry Proof. Axioms Connecting Axioms to Models. 

3/26 Z2 and
Finite Projective Geometry. Models for Affine and Projective Geometries. 
3/28
Axioms for Synthetic Projective Geometry (see M&I) Examples of proofs in synthetic projective geometry. 

4/2 Quiz 2 More examples of proofs in synthetic projective geometry. . 
4/4
Desargues' Theorem Proof of Desargues' Theorem in the Plane A look at duality and some applications. 
Sections and Perspectively related Figures Some key configurations. space duality, perspective reconsidered. 
4/9
Conics. Introduction to Pascal's Theorem Video: Orthogonal Projection 
4/11 Perspectivies as transformations. 
Sections in Space The complete Quadrilateral. 
4/16 Start Isometries
with Homog. Coord. Matrices 

Harmonics: uniqueness and coordinates for Projective Geometry. Planar transformations and Matrices 
4/23
An Inversion Excursion? Projectivities. Projective relations Projective Line transformations: Synthetic Projectivities; Inversion Video? Projective Conics Video? 
4/25 

4/30 Quiz #3?  5/2 

Note that Euclid's treatment in its statement or its "proof" never refers
the traditional equation, a^{2}+b^{2}=c^{2}.
A look at the possibilities of
dissections .
(1) a. Parallelograms between a pair of parallel lines and on the same
line segment are equal (in the sense of being able to decompose one to
reconstruct the other). Proposition
35.
b. Parallelograms between a pair of parallel lines and on congruent
segments are equal (in the sense of being able to decompose one to reconstruct
the other). Proposition
36.
(2) a. The line segment connecting the midpoints of two sides
of a triangle is parallel to the third side and is congruent to one half
of the third side.
[The justification of this result is left
as an exercise in traditional Euclidean Geometry.]
b. By rotating the small triangle created by connecting the midpoints
of two sides of a triangle 180 degrees about one of the midpoints, we obtain
a parallelogram. (This shows that the triangle's area is the area
of this parallelogram which can be computed by using the length of the
base of the triangle and 1/2 of its altitude which is the altitude of
the parallelogram.)
Compare this with Euclid Prop. 42 and Prop. 44.
Another component of the proof of the equidecomposable polygon
theorem is the ability to "add two parallelograms to form a single
parallelogram which
is scissors congruent to the two separate parallelograms". Here's how:
Intersect two pairs of parallel lines, l and l' with m and m' one
from each of the given parallelograms. Draw a diagonal HI in the resulting
parallelogram.
Cut and translate one parallelogram so that it is scissors congruent
to a parallelogram HIJK within the same parallel lines l and l' with one
side being the diagonal.
Cut and translate the other parallelogram so that it is scissors congruent
to a parallelogram HINO within the same parallel lines m and m' with one
side being the diagonal and on the other side of the diagonal HI from the
transformed first parallelogram. Now draw the parallel NO to the diagonal
in the second transformed parallelogram HINO so that it intersects the
parallels l and l' from the first parallelogram at the points P and
Q. This makes one larger parallelogram JKPQ which is scissors congruent
to the original two parallelograms.
Compare this with Euclid Proposition 45.
Follow this link for a proof of the equidecomposable polygon theorem. or here is a slightly different approach.
The film Equidecomposable Polygons also proves the result:
If two polygonal regions in the plane have the same area, then there
is a decomposition of each into polygons so that these smaller polygons
can be moved individually between the two polygons by translations or half
turns (rotations by 180 degrees).
Review materials defining rays, segments, angles, triangles, and planes in M&I.
Angle Bisection  Euclid Prop 9 
Line Segment Bisection 
Euclid Prop 10 
Construct Perpendicular to line at point on the line 
Euclid Prop 11 
Construct Perpendicular to line at point not on the line  Euclid Prop 12 
Move an angle  Euclid Prop 23 
Construct Parallel to given line through a point 
Euclid Prop 31 
Some comments about Constructions: It is important
to notice that constructions also require a justification (proof) that
the construction has in fact been achieved.
In proving the constructions
we use some basic euclidean results, such as the congruence of all corresponding
sides in two triangles is sufficient to imply the triangles are congruent
(SSS). [Other basic Euclidean results are SAS and ASA congruence conditions,
as well as the result that corresponding parts of congruent triangles are
congruent (CPCTC).]
For these particular constructions to be
justified by the same arguments given by Euclid in a geometric
structure, the structure will need both an equivalence relation called
congruence for line segments, angles, triangles and propositions that
connect congruence of triangles to sufficient conditions
like SSS, SAS, and ASA.
The constructions play two important but different roles in (euclidean) geometry:
(i) Construction allow us to "move", deconstruct and reconstruct
figures, while maintaining the magnitudes of the pieces and angles .
Thus constructions provide the tools for transformations such as
rotations and translations.
(ii) Constructions allow us to develop comparative measurements based on a "unit" segment and the "straight" angle.
It should also be noted that the three transformations
(translation, rotation, and reflection) commonly used in geometry are connected
to constructions as well. For example, to translate a figure by a vector
it would be useful to know how to construct parallelograms.
Note on midpoints: With the construction of midpoints in Euclidean Geometry, we can show that a Euclidean line segment has an infinite (not finite) number of distinct points. Furthermore, if we think of approximating real number distance with points on a segment after establishing a unit length, then we can construct the position of a euclidean point as close as we want to the position where a real number might correspond to a point in that position.
I.e., given P_{0 }and P_{1} for any real number x where a<x<b there is a point P_{x} where P_{x} is between P_{a} and P_{b} if and only if the point Pn/2^k corresponds to the number n/2^k for any integer n and natural number k .
For example: We can bisect or trisect a line segment, giving us the
ability to find points representing rational numbers with denominators
involving powers or 2 and 3, such as 5/6, 7/18, etc.
The figure below gives two ways to achieve these constructions. One
can see how to generalize these to allow one to construct points to represent
any rational number on the line so that the arithmetic of numbers is consistent
with the arithmetic of geometry. [Adding segments and adding numbers, etc.]
Consider M&I's constructions of the same correspondence of
integer and rational points. These also rely on the ability to construct
parallel lines.
The continuity axiom can also be used to prove: If a line, l, (or circle, O'A') has at least one point inside a given circle OA and one point outside the same given circle then there is of a point on the line (circle) that is also on the given circle.
Proof outline for the linecircle: Use bisection between the points on the line l outside and inside the circle OA to determine a sequence on nested segments with decreasing length approaching 0. The point common to all these segments can be shown to lie on the circle OA.
Proof outline for the circlecircle: Draw the chord between the inside and outside points on the circle O'A'. Use bisection on this chord to determine rays that by the previous result will meet the circle O'A'. The bisections can continue to determine a sequence of nested segments with decreasing length approaching 0 and with endpoints determining one outside point and one inside on O'A' . The point common to all the endpoints on the chord will determine a point on O'A' that can be shown to also lie on OA.
Note: The circlecircle result fills in a hole
in the proof of Proposition 1 in Book I of Euclid.
Tangents to circles.
In considering constructions of tangents to circles we use the characterization
of a tangent line as making a right angle with a radius drawn at the point
it has in common with the circle. ( Book
III Prop. 16.) In our construction, not Euclid's (Book
III Prop. 17), we also use the result that any angle inscribed in a
semicircle is a right angle. ( Book
III Prop. 31.)
Doing arithmetic with constructions in geometry.
Note
that the construction above allows one to construct a point Px' from a
point Px as long as x is not 0 so that x' x = 1.[ Use the circle of radius
1 with center at P0 to construct the inverse point for Px.]
INSERT Figure and discussion here for construction of "inverse" with respect to a circle.
The half plane example: Consider the half plane determined by a line l and a point P not on the line. This can be defined as the set of points Q in the plane where the line segment PQ does not meet the line l. Discuss informally why the half plane is convex.
Review the problems on convex figures in Problem Set 1.
Other convex examples: Apply the intersection property [The intersection
of convex sets is convex.] to show that the interior of a triangle is convex.
Show that the region in the plane where (x,y) has y>x^{2}
is convex using the tangent lines to the parabola y=x^{2} and the
focus of the parabola to determine a family of half planes whose intersection
would be the described region.
Solution: First construct the inverse A' of A with respect to C1 and then the tangent to C1 at B and the perpendicular bisector of AA' will meet at the center of the desired circle.
2. Construct a circle C2 through two points A and B inside
a circle C1 so that C2 is orthogonal to C1.
Solution: This solution is demonstrated in the sketch below.
Proposition: If T is an isometry, then T is 1:1 and onto as a function.
Proof: 1:1. Suppose that T(P)=T(Q). Then d( T(P),T(Q))=0=d(P,Q) so P=Q.
onto. Suppose R is in the plane. Consider A,B, and C in the plane where C is not on the line AB. Then the points T(A),T(B), and T(C) form a triangle and using the distances d(T(A),R), d(T(B),R), and d(T(C),R), we can determine a unique point X in the plane where d(A,X)=d(T(A),R), d(B,X) = d(T(B),R), and d(C,X)=d(T(C),R), so T(X) = R.
2 14
We begin a more detailed study of isometries with a look at
Line Isometries: Consider briefly isometries of a
line.
1) translations and 2) reflections.
How can we visualize them?
Coordinate function. x > x' = f (x)
Examples: P_{ x} > P_{x+5} a translation;
P_{ x} > P_{x} a reflection.
Can we classify them? Is every line isometry either a translation or
a reflection? Why?
Prop.: The only isometries of the line are reflections and translations.
Proof:
Given A and A', there are only two choices
for B '. One forces the isometry to be a translation, the other forces
the isometry to be a reflection.
Discuss further in class.
Use T to denote both the geometric
transformation and the corresponding function transforming the
coordinates of the points. So ... T(x) = x + 5 for the translation example and T(x) = x for the reflection example.
More generally, a translation T_{a} :P_{ x} > P_{x+a} would have T_{a}(x) = x + a and Reflection about the origin can be denoted R_{0}_{ }, R_{0}(x) = x . What about a general reflection about the point with coordinate c? R_{c}(x) =? Use a translation by c, then reflect about 0, and translate back to c. So R_{c}(x) = T_{c}(R_{0}(T_{c}(x))) = T_{c}(R_{0}((x c))=T_{c}(x+c)= x + 2c.
216
Remarks on line isometries:
(i) Any isometry of a line can be expressed as the product of at most 2 reflections.
(ii) The product of two line reflections is a translation.
Now we look at Plane Isometries:
Consider isometries of a plane.
1) translations 2) rotations and 3) reflections.
How can we visualize them?
Coordinate functions?
Remark: we have previously shown that an isometry of the plane is completely determined by
the correspondence of three noncolinear points.
The classification
of isometries.
There are at least four types of isometries of the plane: translation,
rotation, reflection and glide reflection. [In fact , we will show that any planar isometry is one of these four types.]
Translation: T: P_{ (x,y)} > P_{(x+5, y+2)}
is a translation of the plane by the vector <5,2>. If we use the coordinates
for the point and T(x,y) = (x',y') then x' = x+5 and y' = y+2. We
can express this with vectors <x',y'> = <x,y> + <5,2>. So translation
corresponds algebraically to the addition of a constant vector.
Reflections: Across Xaxis RX(x,y) = (x,y); Across Y axis RY(x,y)
= (x,y); Across Y=X , R(x,y)=(y,x). Notice that these can be accomplished
using a matrix operation. Writing the vectors as row vectors
Matrix for RX

Matrix for RY

Matrix for R

[  1  0 
]  [  x  ]  =  [  x  ] 
0  1  y 
y  
Matrix for RX 

[  1  0 
]  [  x  ]  =  [  x  ] 
0  1  y 
y  
Matrix for RY 

[  0 
1 
]  [  x  ]  =  [  y 
] 
1  0 
y 
x 

Matrix for R 



Hint: 


Proposition: Any plane
isometry is either a reflection or the product of two or three
reflections.
Proof:
Click here!
219
Rotations and reflections [review +]:
More general Question: What about reflection R(A,B) about the line AX+BY
= 0?
See discussion above. 

Composition of Isometries corresponds to Matrix multiplication! 
If B = 0 then the line of reflection is the Y axis and we know the matrix for RY already. For all other B, let t be the angle which has tan(t) =  A/B. Thus the reflection R(A,B) has a matrix that must be the product [from left to right]of the matrices for R(t), RX, and R(t).
[  cos(t)  sin(t)  ]  [  1  0 
]  [  cos(t)  sin(t) 
]=[ 
cos^{2}(t)sin^{2}(t)  2cos(t)sin(t) 
]=[ 
cos(2t) 
sin(2t) 
] 
sin(t) 
cos(t) 
0 
1 
sin(t) 
cos(t) 
2cos(t)sin(t) 
sin^{2}(t)cos^{2}(t) 
sin(2t) 
cos(2t) 
Before Reflection 
After Reflection 
More General Planar Isometries:
A key idea is do the work at the origin and then transfer the work elsewhere using "conjugacy": Any transformation T at a general point or about a general line can be investigated by first translating the problem to the origin, S, performing the related transformation at the origin,T', and then translating the result back to the original position, S^{1}. That is using the "conjugacy" operation: T = S^{1} T' S where T' is the relevant transformation
at the origin. This works as well for rotation and reflections through the origin because
the composition of these transformations corresponds to matrix
multiplication.
Here is a link to an example of the matrix for the isometry of the coordinate plane that is rotation by 90 degrees counterclockwise about the point (1,2).
Proposition:
(i) The product of two
reflections that have the lines of reflection intersect at a point O is
a rotation with center O through an angle twice the size of the angle between
the two lines of reflection.
(ii) The product of two reflections that have parallel lines of
reflection is a translation in the direction perpendicular to the two lines
and by a length twice the distance between the two lines of reflection.
(iii) The product of three reflections is either a reflection or
a glide reflection.
Proof: We can do this geometrically or using analytic geometry and the matrices!
Watch the video Isometries (Video
# 2576 in Library) .
Orientation
Preserving  Orientation
Reversing  
Fixed points  Rotations  Reflections 
No Fixed points  Translations  Glide reflections 
2/23 An Applications of Reflection:
(i) Here is a problem encountered frequently in
first semester of a calculus course.
The
Carom Problem Can you see how the solution is related to a "carom"
(angle of incidence=angle of reflection)?
(ii) Here is a similar problem about triangles
that is also related to reflections.
Fano's
Problem Can you find the relationship?
Symmetry of a figure: S is a symmetry
of a plane figure F if
S is an isometry with S(F)=F.
Given
a figure F, the symmetries of F
form a subgroup of all the plane isometries, denoted Sym(F).
Example: Consider the figure F=
an given equilateral triangle. We looked at the six symmetries of this
figure and the table indicating the multiplication for these six isometries.





 

































 R120 







Repeated use of this algorithm suggests the
Euclidean algorithm for finding a common unit to measure both n and d,
or ON and OD.
If r1=0 or R1N is a point, then OD will be a common unit.
If not apply
the division algorithm to d or OD and r1 or R1N.
If this works to give
r2= 0 or R2N is a point, then R1N will be the common unit.
If not
apply the division algorithm to r1 or R1N and r2 or R2N.
If this works
to give r3= 0 or R3N is a point, then R2N will be the common unit.
If not
continue.
In common arithmetic since each remainder that is not zero is
smaller than the previous remainder, eventually the remainder must be 0
and the process will end finding a common divisor of the original d and
n.
* The Axiom of Archimedes that says that for any two segments one can be used to measure the other. [There are no infinitesimals for an Archimedean geometry.]
Example: Euclid uses the theory of proportion in Book VI, Proposition
1 and Proposition
2.
(see also Byrne's Prop.1 and Prop.2.)
We can also observe that if we let S(Px)=P(x3) and S^{1}(Px)=P(x+3) then
Central similarities in the plane:
In the plane, a similarity Tm with factor m and
center
at (0,0) will transform (x,y) to (mx,my). Thus x' = mx,
and y' = my are the equations for this transformation. This transformation
can be represented using a matrix as follows:
[  m 
0 
]  [  x  ]  =  [  mx  ] 
0  m 
y 
my  
Matrix for Tm 

Proposition: If l is a line in the plane with equation
AX+BY=C with A and B not both 0, then the set l' = {(x',y'):
Tm(x,y)=(x',y') for some (x,y) on l} is a line in the plane that
is parallel to l (unless l = l' ).
Proof (outline): Show that the equation of l' is
AX+BY = mC. Suppose (x,y) satisfies the equation AX+BY=C. Then Tm(x,y) = (mx,my). But Amx+Bmy=m(Ax+By)=mC, so the line l ' has equation AX+BY=mC, and therefore is parallel to the line l.
Before

After

In this structure we consider all points and lines in the usual plane with the exception of one special line designated as the "horizon" line. Points on this line is not considered as a part of the geometrical points but are used in defining the class of parallel lines in the reamining plane.
A line in this structure is any line in the original plane with the exception of the horizon line.
Two lines are called Aparallel (A for affine) if (i) they are both parallel in the usual sence to the horizon line or (ii) they have a point in common that lies on the horizon line.
At this stage we do not have a correspondence for points in this plane and real number coordinates.
We can consider a figure to illustrate this geometry as below:
Informal projective geometry and algebra
Projective Geometry with a field with two elements:
Another model for the 7 point geometry.
Z_{2 }= F_{2} = {0,1}.
+  0  1  *  0  1  
0  0  1  0  0  0  
1  1  0  1  0  1 
A projective plane using F_{2} has exactly 7 lines: AX + BY +CZ = 0 or [A,B,C] is a line using homogeneous coordinates.
This projective plane satisfies the geometric structure
properties described previously in the "7 point geometry".
This can be verified using the following table where an X indcates the point of the column is an element of the line of the row.










































Euclidean Line: P0 and P1
Euclidean Plane: P(0,0), P(1,0), and P(0,1). Axes at right angles. Parallel lines have no point in common. 
Affine Line: P0, P1, and Pinf.
Affine Plane: P(0,0)=<0,0,1>, P(1,1)=<1,1,1>,P(inf,0)=<1,0,0>, P(0,inf)=<0,1,0> The Horizon or Ideal line. Parallel lines meet at an infinite point on the horizon. 
Projective Line: Pinf is treated as an ordinary point. A circle.
Projective plane: All points in the affine plane are treated as ordinary points. Homogenous coordinates determine all points and lines. There are no parallel lines. All lines meet. 
The structure established so far for the affine and projective planes.
In an affine plane there is a special line, the horizon or ideal,
line containing all the ideal infinite points for lines in the ordinary
euclidean plane.
Any ordinary line has exactly one ideal infinite point
on it.
Two lines in the ordinary plane are parallel if they do not meet.
Two lines in the affine plane are parallel if they meet at the same ideal
infinite point on the ideal line.
In the projective plane there are no
parallel lines!




Points:Ordinary coordinates(x,y)
Lines: Ax+By+C=0 


Points: Ordinary (x,y) or generally homogeneous <x,y,z> x,y,z not
all 0.
Lines: Ax+By+C=0 or Ax+By+Cz = 0 or [A,B,C] 


Points: <x,y,z> x,y,z not all 0.
Lines: Ax+By+Cz = 0 or [A,B,C] A,B,C not all 0. 
Where we're going in the next few weeks. We
will be discussing the following topics: the definition of the algebraic
model for the affine and projective plane and spaces, the connection of
these geometries to visualizations, and the construction of points with
corresponding coordinates in the affine line and plane using P0, P1 and
Pinfinity.
The text materials in Meserve and Izzo provides support for these issues and topics. The relevant sections are listed in the course reading assignments. 
The axioms for projective geometry in a plane uses two basic objects: points and lines, and a relation between those: a point is on a line, or a line passes through a point. The technical term for this relation is "incident", so we say a point is incident to a line and a line is incident to a point.
The dual of a statement or description in the context of a projective
plane replaces the word "point" with the the word "line" and the
word "line" with the word "point".
Here are some examples of statements and the corresponding dual statement:
Two distinct points A and B are (incident) on a unique lineAB.  Two distinct lines a and b are (incident) on a unique point a#b. 
If the point C is not (incident to) on the line AB then there are three lines AB, AC, and BC.  If the line c is not (incident to) on the point a#b then there are three points a#b, a#c, and b#c. 
The lines AA', BB' and CC' are incident to the point O.  The points a#a', b#b', and c#c' are incident to the line o. 
As a consequence of this feature, plane projective has a special result which is about the theorems of geometry and their dual statments.
The Principle of Plane Projective Duality: Suppose S is a statement of plane projective geometry and S' is the planar dual statement for S. If S is a theorem of projective geometry, then S' is also a theorem of plane projective geometry.
The proof of this principle is a proof about proofs.
The idea is that a proof consists of a list of statements about lines
and points.
Each statement in a proof is either one of the postulates,
a previously proven theorem, or a logical consequence of previous statements.
So if we have a proof of a statement S, we have a sequences of statements
A1,A2,...,AN=S.
Each of these statements is either one of the postulates,
a previously proven theorem, or a logical consequence of proevious statements.
Now one can construct the sequence of dual statements A1', A2', ...,
AN' = S'.
With a little argument it can be seen that each of these dual
statements is also either a postulate, a theorem,or a logical consequence
of previous statements.
Here is an application of the principle of duality to Desargues' Theorem.
Note on Duality and the concept of perspective: We called two
point triangles ABC and A'B'C' perspectively related
with respect to a point O, if O is
on the lines AA', BB' and CC'.
We define the dual concept by saying that two line triangles
abc and a'b'c' are perspectively related with
respect to a line o, if o passes through
the points a#a', b#b' and c#c'.
Notice that the Dual of Desargues' Theorem is also the logical converse
of Desargues' Theorem. Thus we can say, "The
converse of Desargues' theorem is true by the duality principle."
The dual concept for section: A section of a planar figure, F, by
a point L, where L is not in the figure F, is a new figure
consisting of the lines p where p = L*M for any M,
a point in the figure F.
Examples: The section of the triangle ABC by the point L, is
the set of three lines on L , {A*L,B*L, C*Ll}.
The section by a point L (not lying on a line o) of a
pencil of points on the line o is the set of lines through L with
one line for each point in the pencil.
Line Perspective in the Plane: Figure F is perspectively related
to Figure F' by the line o if there is a correspondence of the line
of F with those of F' so that for any corresponding lines, m in
F and m' in F', the line o passes through the point m#m'
.
Notes:1. If the corresponding lines in two figures F and F' are
parallel, then in the affine plane, these corresponding lines will meet
on the horizon line. Thus these figures are perspectively related in the
projective plane.
2. If the figure F and the figure F' are congruent because of a
reflection, then F and F' are prespectively related by the line of reflection.
A look at perspective of planar figures with respect to a center O and its dual: perspective of planar figures with respect to an axis o.
A planar graph consists of a finite set of points called vertices, line (straight or curved) segments with these vertices as endpoints called edges, enclosing planar sets called regions. We can think of these regions as geographic states, the edges as boundaries between land sections, and the vertices and places where these boundaries meet.
So a planar graph G is a set of vertices, edges, and resulting regions in the plane.The dual graph of G is another graph, which for now we'll denote D(G). D(G) consists of a vertex for each region in G, a region for each vertex in G, and an edge for each edge. If R is a region in G, we choose a point in R, call it r, as a vertex of D(G). For each edge, E, of G, with regions R1 and R1 bordering on E, choose an edge, e, between r1 and r2 that crosses E. Finally, suppose V is a vertex of G. consider the edges that end at V and the regions that border these edges. Then these regions and edges correspond to vertices and edges of D(G) that surround a region which we'll denote v.
The graph D(G) consists of the vertices, r, edges, e, and regions v
just described.
One aspect of the dual graph is that information about it is revealed
by knowing information about the graph G. For example, if G has 5
regions, the D(G) has 5 vertices.
If G has 7 edges then D(G) also has 7 edges, and if G has 4 vertices,
then D(G) has 4 regions.
Take a look at this: Encyclopedia of Polyhedra
Here is a table showing the five platonic solids and the
duality relation of Vertices, Edges (lines), and Faces (Planes):

V 



8 



6 



12 



20 






Example. The figure F is a tetrahedron. P is a point that is
distinct from the tetrahedron's elements. The section consists of 6 planes and the 4 lines where those lines join the vertices to the point P . 
4 points {A,B,C,D} determine 6 lines {AB,AC,AD, BC, BD, CD} and three additional points {X,Y,Z}. 
The Complete Quadilateral: 4 lines {AB, BC, CD,AD}determine 6 points {A,B, C, D, X,Y} with three additional lines{AC, BD, XY} . 
Definition: The product (composition) of two
or more perpectivities is a called a projectivity.
Is we compose two perspectivities we can transform points on a line
to a second line and then back to the original line.
Thus the map from a euclidean, affine, or projective line to itself
can be a projective transformation.
Example: Translation of a euclidean coordinate line by adding
1 unit.
T(P_{x}) = P_{x+1} can be realized as a product of
perspectivities in the affine plane.
First use the perspectivity that transforms the point P_{x}
on the X axis to the point P' = Q_{x+1} on the Y axis.
Then use the perpectivity that transforms the point Q_{x+1}
on the Y axis back to P_{x+1} on the X axis.
Then if T represents the product of these two perspectivities, T(P_{x})
= P_{x+1}.
Other examples of projective line transformations include central reflection [Use P' =Q _{x } for reflection about 0]and central similarities [Use P' =Q _{ax } for central similarity with scale a and center 0].
But what is the algebra for this isometry when using arbitrary homogeneous coordinates for an ordinary point?
In summary, T ( <a,b>) = <a+3b,b>.
Notice this formula works as well for the ideal point on the affine
line.
This work can be done using some ideas from linear algebra. Recall
that in matrix multiplication:
[  1  3  ]  [  x  ]  =  [  x+3  ] 
0  1  1  1 
[  1  3  ]  [  a  ]  =  [  a+3b  ] 
0  1  b 

[  1  3  ]  [  a  ]  =  [  a+3b  ] 
0  1  b 

[  1  0  ] 
0  1 
[  1  0  ]  [  x  ]  =  [  x  ] 
0  1  1  1 
[  1  0  ]  [  a  ]  =  [  a  ] 
0  1  b 

[  5  0  ] 
0  1 
[  5  0  ]  [  x  ]  =  [  5x  ] 
0  1  1  1 
[  5  0  ]  [  a  ]  =  [  5a  ] 
0  1  b 

[  0  1  ] 
1  0 
[  0  1  ]  [  x  ]  =  [  1  ] 
1  0  1  x 
[  0  1  ]  [  a  ]  =  [  b  ] 
1  0  b 

If we let the matrix of these transformation be denoted by
T= [  a  b  ] 
c  d 
[  a  b  ]  [  x  ]  =  [  ax+b  ] 
c  d  1  cx+d 
[  a  b  ]  [  x  ]  =  [  ax+by  ] 
c  d  y 

[  a  b  ]  [  kx  ]  =  k[  ax+by  ]. 
c  d  ky 

k[  a  b  ]  [  x  ]  =  k[  ax+by  ]. 
c  d  y 

T= [  1  b  ]. 
c  d 
[  1  b  ]  [  1  ]  =  [  x  ] 
c  d  0  0 
T= [  1  b  ]. 
0  d 
[  1  b  ]  [  x  ]  =  [  x/d+b/d  ] 
0  d  1 

[  1  b  ]  [  x  ]  =  [  x+by  ] 
0  d  y 

lines/planes 
lines/planes 
lines[/Planes?] 
lines/planes 
Axioms
Euclid Hilbert 
No Axioms Yet
A figure indicating an ideal point or line 
Axioms
7 points/7lines A figure. 
M & I Axioms
A figure indicating all points and lines 
Parallel lines don't meet  Parallel lines meet
at an ideal point. 
All pairs of lines
have a common point 
All pairs of lines
have a common point 
Coordinates
Analytic/Algebraic 
Ordinary coord's
w/ infinite (ideal) points Homogeneous Coordinates 
Homogeneous Coordinates
with coefficients in {0,1}= Z_{ 2} 
Homogeneous Coordinates
with real number coefficients 
Transformations
Isometries Similarities 
Similarities  ?  perspective and projective transformations. 
Harmonic Relation of 4 points on a line. [This will be
the chief tool used to introduce (homogeneous) coordinates into projective
geometry.]
Four points on a line l are harmonically related if the line is determined by a pair of points from the intersection of lines in a complete quadrangle and the intersection of that line with the other two sides of the complete quandrangle. [In the figure: The line XZ would determined two other points, XZ#AD=R and XZ#BC=S, so that the points XRZS are harmonically related.] This would be denoted H(XZ,RS). Four points on a line that are harmonically related: Using the text notation we can show that if H(AB,CD) then H(BA,CD), and also conversely if H(BA,CD) then H(AB,CD). This is the meaning of saying "H(AB,CD) is equivalent to H(BA,CD)". Similarly we can show H(AB,CD) equiv. to H(AB,DC) and H(BA,DC). One way to see this is to envision relabeling the original quadrangle to reverse the order in which the points on the line are organized. 
One key issue then is:
Is the point constructed from the points A,
B and C uniquely determined by the fact that it is in the harmonic relation
with A, B, and C? That is, if A,B, and C are three points
on a line and D and D* are points where H(AB,CD) and H(AB,CD*), then must
D=D*?
This is the question of the uniqueness of the point D. We can prove
that in fact the point D is uniquely determined.
The
proof follows the argument of Meserve and Izzo.
It used Desargues' theorem several times.
Discussion of the dual concept of a harmonic relation between
four lines passing through a single point.
Theorem: If A,B,C, and D are on a line l with H(A,B,C,D)
and O is a point making a section with these four points, consisting of
the four lines a,b,c and d, then H(a,b,c,d).
Proof: see M&I Theorem 5.4.
Corollary: (By Duality) If a,b,c, and d
are on a point O with H(a,b,c,d) and l is a
line making a section with these four lines, consisting of the four points
A,B,C and D, then H(A,B,C,D).
Application:
We can think of a perspectivity between points ABCD on line l
and A'B'C'D' on the line l' with respect to the point O as
being a section of the points ABCD by the point O followed by a section
by the line l' of the lines a,b,c, and d on the point
O.
Applying the previous theorem and its corollary we see that: If
H(AB,CD) then H(ab,cd) and thus H(A'B',C'D').
Note: This application shows that if four harmonically related points
on a line are perspectively related to four points on a second line, then
the second set of four points is also harmonically related. Furthemore,
this result can be extended easily to points that are projectively related.
Thus the transformations of projectivity in projective geometry
preserves the harmonic relationship between four points.
[This last note is comparable to the fact that in Euclidean geometry,
isometries preserve length, and in affine geometry that similarities preserve
proportions.]
In many ways this gives a transformation of the point on the line to
other points that is similar in its nature to reflections and inversions.
Notice that on a projective (affine) line two point will cut the line into
two disjoint pieces, as does a single point for reflection and the points
P_{R} and P_{R} for inversion, where the transformation
maps points in one set into the other while leaving the "boundary points"
fixed.
With the existence and uniqueness of the point D established, we can
now consider some examples illustrating how to establish a coordinate
system for a projective line by choosing three distinct points to be P0,
P1, and P¥.
We can construct P2, P1, P1/2, (in two different ways).
Exercise: Construction P3 and P1/3.
Show that with the choice of three points on a projective line
we can construct points using harmonics to correspond to all real numbers
(as in our informal treatment of the affine line).
We will see how this transformation is completely determined by the correspondence of three distinct pairs of points. In some cases the transformation transforms an ordinary point to an ordinary point, the ideal point to an ordinary point and an ordinary point to the ideal point, and in some cases the transformation will transform the ideal point to the ideal point.
A transformation that transforms the ideal point to the ideal point
is called an affine transformation. The composition of two affine transformations
is an affine transformation. The inverse of an affine transformation is
an affine transformation and clearly the identity transformation is an
affine transformation, so the affine transformations are also a group under
the operation of composition (matrix multiplication).
Central Similarities at (0,0) by a factor of M.
Reconstruction of a point on a conic from Pascal's Theorem.
The following materials are from previous course notes and have not been incorporated into the current course notes.
Consider lines connecting corresponding points in a pencil
of points on a line related by a projectivity (not a perspectivity) and
noticed that the envelope of these lines seemed to be a conic, a line
conic. Notice briefly the dual figure which would form a more
traditional point conic. [Also notice how line figures might be related
to solving differential equations e.g. dy/dx=2x1 with y(0)=3 has a solution
curve determined by the tangent lines determined by the derivative: y=x^2x+3
which is a parabola.]
In discussing the issue of whether the 5th axiom could be proven from the other four axioms, we looked at an example of another axiom system, with an axiom N (any pair of lines having at one point in common) and an axiom P (given a line l and a point P not on that line there is a line m where P is on m and m and l have no common points) which is a version of the parallel postulate (Playfair's  not Euclid's). We gave examples showing that the four axioms and P were possible as well as the four axioms and N were possible. This showed that one can not prove axiom P or N from the other four axioms since P and N are contradictory.
By a similar analysis of the axioms for the seven point geometry we showed that it not possible prove the 5th postulate from the other four. The analysis examines the example of the seven point geometry and notices that by including an 8th and 9th point the resulting geometry would satisfy the other 4 axioms.
The model we have for affine geometry still satisfies the parallel
postulate, since the ideal (infinite) points of affine geometry are not
considered as ordinary points of the geometry. However, be removing this
distinction between ordinary and ideal points and considering the geometry
that results we obtain a geometry in which there are no parallel lines.
(A projective plane.) This will be a major focus of discussion for the
remainder of the term especially using the homogeneous coordinates to
consider points in this geometry from an analytic/algebraic approach.
An examination of the problems of transferring an spatial image of a plane to a second plane using the idea of lines of sight we arrived at an understanding of how points in the plane would correspond to lines through a point (the eye).
In discussing the 7 point geometry we visualized it using vertices of a cube (besides (0,0,0)) with their ordinary coordinates in standard 3 dimensional coordinate geometry and identified the 7 points . This allowed us to identify "lines" using the homogeneous coordinate concepts and their relation to planes in three dimension through (0,0,0). We identified all but one of the lines easily the last plane has ordinary equation X + Y + Z = 2... but in this arithmetic for {0,1} we have 1+1=0, so 2=0 and the vertices of that satisfy this equation in ordinary coordinates {(1,1,0), (1,0,1),(0,1,1)} form a line as well.
+  0  1  2  *  0  1  2  
0  0  1  2  0  0  0  0  
1  1  2  0  1  0  1  2  
2  2  0  1  2  0  2  1 
[Moved from earlier: The lab time was spent working on sketches showing ways to understand that result of the CAROMS film about the inscribed triangles of minimum perimeter.In Lab: Discuss some visual features such as trace and animation and start to look at the use of coordinates. We'll do more with coordinates, along with the use of traces and locus to see some aspects of coordinate geometry next week.]
We also watched the film on projective generation of conics which
introduced Pascal's theorem and its converse about hexagons inscribed in
a conic and showed how to use this result to construct a conic curve passing
through any 5 points. This work was also related to projectivities between
pencils of lines. We will be considering this further in the course. After
a short break we continued using metric ideas to construct an ellipse
as a locus on sketchpad and discussed how to do a parabola as well. By
next Thursday students should construct examples of the three conics on
sketchpad using metric ideas.
a  b 
0  1 
+/1  b 
0  1 
We spent the remainder of the lecture time going over the proof of Brianchon's Theorem using the proof of Hilbert and CohnVossen based on hexagons lying on the surface of an elliptic hyperbaloid (which is a ruled surface).