Week |
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Wednesday |
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1 |
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1-20 Introduction |
1-22 Continue discussion of what
is "geometry"? Start on Euclid- Definitions, Postulates, and Prop 1. |
2 |
1-25 Euclid- Definitions, Postulates,
and Prop 1. cont'd |
1-27 Pythagorean plus. |
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3 |
2-1 Euclid early Props/ Pythagoras/ |
2-3 Euclid early Props/ Pythagoras/ |
2-5 Equidecomposable polygons |
4 |
2-8 More on Details for triangulation of planar polygonal regions and adding parallelograms. |
2-10 Begin Constructions and the real number line M&I's Euclidean Geometry |
2-12 Constructions from M&I. |
5 |
2-15 Begin Constructions and the real number line. | 2-17 Similar Triangles.Intro to Proportions. Construction of rational numbers. Constructions and The real number line.- Continuity | 2-19
Coordinate based proofs.
Inversion and Orthogonal Circles . Isometries:Classification of Isometries |
6 |
2-22 Inversion and Orthogonal Circles | 2-24 Continuity applied.. Isometries:Groups and Classification of Isometries. |
2-26 Convexity of intersections. Start Proof of classification result for plane isometries |
7 | 2-29 Finish Classification of Line Isometries and coordinates | 3-2 Finish
Classification
of Plane
Isometries and coordinates |
3-4 Isometries
and
symmetries Begin Affine Geometry |
8 | 3-7 Proportion and Similarity | 3-9 Euclidean (Eudoxus) Proportion. . |
3-11 Euclidean (Eudoxus) Proportion.and similarity, |
9 |
3-14 No class Spring Break! | 3-16 No class Spring Break! | 3-18 No class Spring Break! |
10 |
3-21 Euclidean Proportion and ratios of lengths with numbers. Central Similarity as a transformation. |
3-23 The Affine Line and
Affine
Geometry
(planar
coordinates). Affine geometry- Homogeneous coordinates for aa affine line |
3-25 The Affine Line and
Affine
Geometry
(planar
coordinates). Affine geometry- Homogeneous coordinates and Visualizing the affine plane. |
11 |
3-28 Watch Non-Euclidean Universe (Open Univ. Video)- Lib. |
3-30 The 7 point Geomoetry. Axioms and Stuctures |
4-1 Watch Conics (Open Univ. Video)- Lib. |
12 |
4-4 Affine geometry- Homogeneous coordinates and Visualizing the affine plane. | 4-6 More on Affine Projective Geometry - Algebra! |
4-8 Desargues!
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13 |
4-11 Review and start of Synthetic Projective Geometry |
4-13 More Review and start of Synthetic Projective Geometry | |
14 |
4-18 Proofs - planes in Projective Geometry |
4-20 A start on transformations with homogeneous coord.s. The Planar Duality Principle |
4-22 Complete quadrangles. 3 dimensions, Sections and perspectivities. |
15 |
4-25 Projectivities and harmonics |
4- 27 | 4-29 |
16 |
5-2 |
5-4 | 5-6 |
Note that Euclid's treatment in its statement or its "proof"
never refers
the traditional equation, a2+b2=c2.
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 P0 and P1 for any real number x where a<x<b there is a point Px where Px is between Pa and Pb 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
semi-circle 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 line-circle:
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.
2-26
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 side-angle-side 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. |
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Graph of transformation. Example: A reflection. |
Coordinate function. x -> x' = f (x)
Examples: P x -> Px+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 Ta
:P x -> Px+a
would have Ta(x)
=
x + a and
Reflection about the origin can be denoted R0
, R0(x)
=
-x . What about a general reflection about the point with
coordinate c? Rc(x)
=?
Use
a translation by -c, then reflect about 0, and
translate back to c. So Rc(x)
=
Tc(R0(T-c(x)))
= Tc(R0((x
-c))=Tc(-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 non-colinear 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 |
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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 X-axis 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
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Matrix for RY
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Matrix for R
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[ | 1 | 0 |
] | [ | x | ] | = | [ | x | ] |
0 | -1 | y |
-y | |||||||
Matrix for RX |
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[ | -1 | 0 |
] | [ | x | ] | = | [ | -x | ] |
0 | 1 | y |
y | |||||||
Matrix for RY |
---|
[ | 0 |
1 |
] | [ | x | ] | = | [ | y |
] |
1 | 0 |
y |
x |
|||||||
Matrix for R |
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Hint: |
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3-4
See discussion above. |
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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) |
]=[ |
cos2(t)-sin2(t) | 2cos(t)sin(t) |
]=[ |
cos(2t) |
sin(2t) |
] |
sin(t) |
cos(t) |
0 |
-1 |
-sin(t) |
cos(t) |
2cos(t)sin(t) |
sin2(t)-cos2(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.
3-7
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=
an
given
equilateral
triangle. We looked at the six symmetries of this
figure and the table indicating the multiplication for these six
isometries.
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R120 |
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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.]
3-21 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(x-3) 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
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After
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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 A-parallel (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:
4-4
Projective Geometry and Homogeneous Coordinates for the Plane:
See linked notes. Informal projective geometry and algebra
4-6
Projective Geometry with a field with two elements:
Another model for the 7 point geometry. Instead of using real numbers for the prjective plane - use
Z2 = F2 = {0,1}.
+ | 0 | 1 | * | 0 | 1 | ||
0 | 0 | 1 | 0 | 0 | 0 | ||
1 | 1 | 0 | 1 | 0 | 1 |
A projective plane using F2 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 indicates the
point of the column is an element of the line of the row.
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Euclidean Line: `P_0` and `P_1` 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: `P_0`, `P_1`, and `P_{oo}`. Affine Plane: `P_{(0,0)}=<0,0,1>, P_{(1,1)}=<1,1,1>,P_{(oo,0)}=<1,0,0>`, `P_{(0,oo)}=<0,1,0>` The Horizon or Ideal line. Parallel lines meet at an infinite point on the horizon. |
Projective Line: `P_{oo` 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!
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Points:Ordinary coordinates`(x,y)` Lines: `Ax+By+C=0` |
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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]` |
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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 transformations, 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. |
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 |
|
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 nn 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 nn b` then there are three points `a nn b`, `a nn c`, and `b nn c`. |
The lines AA', BB' and CC' are incident to the point O. | The points `a nn a'`, `b nn b'`, and `c nn 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 nn a', b nn b'` and `c nn c'`.
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} . |
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 , `{AL,BL, CL}`.
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 nn 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.
For Reading: A side
trip to planar graphs and dual graphs. The term duality is used in many different mathematical contexts.In the study of planar graphs we can find an example of the use of "duality" that illustrates some of the aspects of mathematical duality. 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. The feature that duality exposes here in these statements is the replacement of the word "vertex" in the statement about G with the word "region" in the statement about D(G) and the word "region" in the statement about the graph G with the word "vertex" in the statement about D(G). |
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 . |
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(Px) = Px+1 can be realized as a product of
perspectivities in the affine plane.
First use the perspectivity that transforms the point Px
on the X axis to the point P' = Qx+1 on the Y axis.
Then use the perpectivity that transforms the point Qx+1
on the Y axis back to Px+1 on the X axis.
Then if T represents the product of these two perspectivities, T(Px)
=
Px+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].
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 quadrangle. [In the figure: The line XZ would determined two other points, `XZ nn AD=R` and `XZ nn 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. Notice the relation between the "double points" in the figure and the "single points". Proposition: H(RT,SU) is equivalent to H(SU,RT) I.e., if H(RT,SU) then H(SU,RT). Proof: (adapted from Meserve & Izzo).
|
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.