Week 

Wednesday 

1 

120 Introduction 
122 Continue discussion of what
is "geometry"? Start on Euclid Definitions, Postulates, and Prop 1. 
2 
125 Euclid Definitions, Postulates,
and Prop 1. cont'd 
127 Pythagorean plus. 

3 
21 Euclid early Props/ Pythagoras/ 
23 Euclid early Props/ Pythagoras/ 
25 Equidecomposable polygons 
4 
28 More on Details for triangulation of planar polygonal regions and adding parallelograms. 
210 Begin Constructions and the real number line M&I's Euclidean Geometry 
212 Constructions from M&I. 
5 
215 Begin Constructions and the real number line.  217 Similar Triangles.Intro to Proportions. Construction of rational numbers. Constructions and The real number line. Continuity  219
Coordinate based proofs.
Inversion and Orthogonal Circles . Isometries:Classification of Isometries 
6 
222 Inversion and Orthogonal Circles  224 Continuity applied.. Isometries:Groups and Classification of Isometries. 
226 Convexity of intersections. Start Proof of classification result for plane isometries 
7  229 Finish Classification of Line Isometries and coordinates  32 Finish
Classification
of Plane
Isometries and coordinates 
34 Isometries
and
symmetries Begin Affine Geometry 
8  37 Proportion and Similarity  39 Euclidean (Eudoxus) Proportion. . 
311 Euclidean (Eudoxus) Proportion.and similarity, 
9 
314 No class Spring Break!  316 No class Spring Break!  318 No class Spring Break! 
10 
321 Euclidean Proportion and ratios of lengths with numbers. Central Similarity as a transformation. 
323 The Affine Line and
Affine
Geometry
(planar
coordinates). Affine geometry Homogeneous coordinates for aa affine line 
325 The Affine Line and
Affine
Geometry
(planar
coordinates). Affine geometry Homogeneous coordinates and Visualizing the affine plane. 
11 
328 Watch NonEuclidean Universe (Open Univ. Video) Lib. 
330 The 7 point Geomoetry. Axioms and Stuctures 
41 Watch Conics (Open Univ. Video) Lib. 
12 
44 Affine geometry Homogeneous coordinates and Visualizing the affine plane.  46 More on Affine Projective Geometry  Algebra! 
48 Desargues!

13 
411 Review and start of Synthetic Projective Geometry 
413 More Review and start of Synthetic Projective Geometry  
14 
418 Proofs  planes in Projective Geometry 
420 A start on transformations with homogeneous coord.s. The Planar Duality Principle 
422 Complete quadrangles. 3 dimensions, Sections and perspectivities. 
15 
425 Projectivities and harmonics 
4 27 Constructing Harmonics 
429 
16 
52 
54 Con ics  56 Pascal and Brianchon 
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 .
See also A New Approach to Hilbert's Third Problem  University of ...by D Benko.
First, consider some of the background results which were known to Euclid: (1) parallelograms results and (2) triangle results. The justifications for these results can be reviewed briefly.
(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.
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. 
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.
We reviewed the connection between the Euclidean constructions and the three transformations (translation,
rotation, and reflection) and the fact that if T is either a
translation, determined by a vector ( an oriented line segment), a
rotation determined by a center and an oriented angle, of a reflection,
determined by a line, the image if a triangle `T( Delta ABC) = Delta
A'B'C'` is a triangle that is congruent to `Delta ABC`.
Other properties of these transformations, that preserve measure of line
segments and congruence of angles, as well as similarities will be
discussed further.
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 `P_{n/(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.
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.]
The relation of the
inversion
transformation with respect to a circle and orthogonal circles.
Proposition: If
C2 is orthogonal to C1 (with center O) and A is a point on
C2 then the ray OA will intersect C2 at the point A' where A and A' are
inverses with respect to the circle C1. Click
here for the proof.
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.
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.
Definition: An isometry on a line l /plane π /space S is a function (transformation), T, with the property that for any points P and Q, d(T(P),T(Q))=d(P,Q) or m(PQ)=m(P'Q').
Comment: Other transformation properties can be the focus of attention replacing isometries as the key transformation in a geometrical structure. The connection between figures and the selected transformations is made by the fact that the transformations have a "group" structure under the operation of composition. When the transformations form a "group" there is a resulting equivalence relation structure on the figures of the geometry. This is illustrated by the relation between isometries and "congruence" in euclidean geometric structures in which distance and/or measure has a role.
226
Convexity: Another brief side trip into
the
world
of convex
figures.
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.
Some General Features of Isometries:
What information determines an isometry?
Proposition: For a planar isometry, T, where T(P) = P', when we know T(A), T(B), and T(C) for A,B, and C three noncolinear point , then T(P) is completely determined by the positions of A', B', and C'.
Proof: In fact we saw that T(B)=B' must be on the circle with center A' and radius= m(AB), and T(C)= C' must be on the intersection of the circles one with center at A' and radius = m(AC) and the other with center at B' and radius = m(BC). Once these points are determined, then for any point P, P' must be on the intersection of 3 circles, centered at A', B', and C' with radii = to m(AP), m(BP), and m(CP) respectively. These three circles do in fact share a single common point because the associated circles with centers at A,B, and C all intersect at P.
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.
Now consider Euclid's treatment of the sideangleside congruence [Proposition 4] and how it relates to transformations of the plane that preserve lengths and angles.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.
Such a transformation T: plane > plane, has T(P)=P', T(Q)=Q' and T(R)=R' with d(P,Q) = d(P',Q') [distance between points are preserved] or m(PQ)=m(P'Q') [measures of line segments are invariant].
Review briefly the outline of Euclid's argument for Proposition 4.
Notes:
Correspondence
figures on a single line: Example: A reflection. 

Graph of transformation. Example: A reflection. 
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.
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.]
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!
Orientation Preserving 
Orientation Reversing 

Fixed points  Rotations  Reflections 
No Fixed points  Translations  Glide reflections 
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 R_{X}

Matrix for R_{Y}

Matrix for R

[  1  0 
]  [  x  ]  =  [  x  ] 
0  1  y 
y  
Matrix for R_{X} 

[  1  0 
]  [  x  ]  =  [  x  ] 
0  1  y 
y  
Matrix for R_{Y} 

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

Matrix for R 



Hint: 


34
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:
As with line 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.
37
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).
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,
a
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.]
321 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.
We can
use coordinate geometry to describe a central similarity S of factor 5
with center at (3,2) as follows: (x,y) > (x
 3,y  2) > (5(x
3) ,5(y  2)) > (5(x
3) +3 ,5(y  2) +2).
Thus S(x , y) = (5x
12 ,5y  8).
Here is the
visualization of S(x,y) as a map in Winplot:
Before

After
