• Course Description.
• What is synthetic geometry?
• What is analytic geometry?

Other aspects of geometry briefly
• transformations
• geometry as an empirical science
• geometry as a formal system of information
• geometry focused on special objects like triangles, or special qualities like convexity.
• A figure C is called convex if given any two points in the figure, A and B, the line segment AB is a subset of the figure.
• Examples. An elliptical region in the plane is convex. A lunar region in the plane is not convex. [Here is an interesting piece of geometry related to the area of lunes.]
• Geometry on the web and using GSP and Wingeom  2D Introduction.
• The Pythagorean Theorem using GSP.

.

• Geometry has traditionally been interested in both results- like the Pythagorean Theorem- and foundations -  using  axioms to justify the result in some rigorous organization. We will be concerned with both results and foundations.
• In the distinction between synthetic and analytic geometry the key connecting concept is the use of measurements. Initially we will try to avoid the use of measurement based concepts when possible.
• To explore some of these issues, let's looked at the proofs of the Pythagorean Theorem.
• First consider Euclid's statement and proof of Proposition 47.
In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

Note that Euclid's treatment in its statement or its "proof" never refers the traditional equation, a²+b²=c².
In one alternative proof for this theorem illustrated in the java sketch below, we consider 4 congruent right triangles and 2 squares and then the same 4 triangles and the square on the side of the hypotenuse arranged inside of a square with side "a+b" . Can you explain how this sketch justifies the theorem?

Another proof using "shearing" illustrated in the Java sketch below taken from a Geometers' Sketchpad example can be connected to Euclid's proof..
Sorry, this page requires a Java-compatible web browser. 
(Based on Euclid's Proof)  D. Bennett 10.9.9
1. Shear the squares on the legs by dragging point P, then point Q, to the line. Shearing does not affect a polygon's area.
2. Shear the square on the hypotenuse by dragging point R to fill the right angle.
3. The resulting shapes are congruent.
4. Therefore, the sum of the squares on the sides equals the square on the hypotenuse.

In considering the Pythagorean theorem, what kind of assumptions were needed in the first proof with the triangles and squares?
Here are some considerations related to those assumptions:
• How could we justify identifying "equal" objects (congruent figures)?
• How do the objects fit together?
• How do movements effect the shapes of objects.

• Consider Euclid's Proposition 1 and  Proposition 2.
• These propositions demonstrate that Euclid did not treat moving a line segment as an essential property worthy of being at the foundations as an axiom. However, this is a fundamental tool  for all of geometry.
• Note that in the proofs of these propositions certain points of intersection of circles are presumed to exist without reference to any of the postulates. These presumptions were left implicit for hundreds of years, but were cleared up in the 19th century when careful attention was given again to the axioms as a whole system.

[An example of a geometry where circles do not intersect is given by using the rational coordinate plane. Points correspond to ordered pairs of rational numbers. then the circle with center (0,0) and radius 1 and the circle with center (1,0) also radius 1 meet in the ordinary plane at the points with coordinates (1/2, sqrt(3)/2) and (1/2, -sqrt(3)/2) . Since sqrt(3)/2 is not a rational number, this ordered pair does not correspond to a point in the rational coordinate plane, so the two circles do not have a point of intersection.
Another example of a point not in the rational coordinate plane is the point (sqrt(2),0). This point can be constructed in the ordinary plane with straight edge and compass using the circle with center (0,0) and radius determined by the points (0,0) and (1,1). This circle will meet the X-coordinate axis at the point (sqrt(2),0) ]

1. Construct a sketch with technology of Euclid's Proposition 1 in Book I.
2. Construct a sketch with technology of Euclid's Proposition 2 in Book I.
3. Construct a sketch with technology of one "proof" of the Pythagorean Theorem.

In the Next Section: We'll look at the possibilities of dissections .

• Use tangram pieces to make a square.
• Note that in putting the pieces together to form any other shape, the area of that shape would be the same as the area of the square unless there is some overlap of the pieces in the shape.

Comment: In a sense a positive (yes) response to this question means that one could create a set of smaller shapes with which one could make either of the two regions using precisely these smaller shapes. The answer to this question is yes (in fact this is a 20th century result), which is the basis for the remainder of this section
[The analogous problem in three dimensional geometry: volume equality of polyhedra is a necessary but not sufficient condition for a similar result. This was the third of the famous 23 "Hilbert problems" for the 20th century. This was first demonstrated by Dehn in the 1930's by using another invariant of polyhedra related to the lengths of the edges and the dihedral angles between the faces of the polyhedra. References to be supplied]

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 was discussed, but no proof was given. This is a question that will be considered in the next meeting on Tuesday.]
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.)

2/6 (i) Justification of the result (2)a on midpoints of the sides of a triangle;
This proposition had not been proven by anyone in class... so it was added to the problems to be done for the next class. [Either create a proof or find one... possibly in Euclid.]

(ii) The triangulation of any polygonal region in the plane was used in the proof of the equidecomposable polygon theorem.
The proof of this proposition examines a more careful characterization of the polygonal regions being considered. The key idea of the proof goes by induction on the number n = the number vertices = the number of sides in the polygon, as follows:
When n = 3 the result is trivial.
Suppose n> 3 and that for any polygon with k vertices/ sides, where k<n, the polygon can be triangulated.
Now proceed to consider the vertices, v1,  v2, ..., vn ordered so that vi is adjacent to vi+1 and vn is adjacent to v1.
Take a ray from v1 and rotate it from v1v2 so that it intersects the inside of the polygon. continue to rotate until it meets another vertex.
Case 1. This vertex is v3. Then consider the polygonal region Q1 = v1v3...vn which has n-1 vertices. By induction Q1 can be triangulated, so the original polygon is triangulated using the triangulation of Q1 and the triangle v1v2v3.
Case 2. The vertex is vn. Then consider the polygonal region Q2 = v2v3...vn which has n-1 vertices. By induction Q2 can be triangulated, so the original polygon is triangulated using the triangulation of Q2 and the triangle v1v2vn.
Case 3. The vertex is vk with k different from 3 or n. Then consider the polygonal regions Q3 = v1v2...vk which has k vertices (k<n) and Q4 = v1vkvk+1...vn which has n-(k-2)<n vertices. By induction Q3 and Q4 can be triangulated, so the original polygon is triangulated using the triangulations of Q3 and Q4.
For more discussion of proofs of this proposition see Triangulations and arrangements, Two lectures by Godfried Toussaint, transcribed by Laura Anderson and Peter Yamamoto.

The lecture also included a discussion briefly of how to "add parallelograms", which is used for the same proof of the equidecomposable polygon theorem.

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).

Laboratory Exercises: Due by Friday, Feb. 9.
1. Show how to "add" two arbitrary triangles to create a single parallelogram. <---This is now optional.
2. Do Construction 3, 4, 6, 7, and 8 from Meserve and Izzo Section 1.2.

Sorry, this page requires a Java-compatible web browser. Sorry, this page requires a Java-compatible web browser. 
After reviewing materials defining, rays, segments, angles, triangles, and planes,
review eight of the basic Euclidean constructions described in M&I section 1.2. Note that several of these constructions rely on some foundations that assert the existence of points of intersection of circles.

2/8: The midpoint proposition.
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, and well as the result that corresponding parts of congruent triangles are congruent (CPCTC).]

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.)
Sorry, this page requires a Java-compatible web browser. 
 
 

At this stage we should discuss the continuity axiom for a euclidean line:

• Any nonempty family of nested segments will have at least one point in the intersection of the family.

I.e., given P₀ and P₁   for any real number x there is a point P_x where a<x<b if and only if  P_x is between P_a  and P_b and the point Pk . corresponds to the number k  for any rational number k.
 

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.

Proof outline for the circle-circle: 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 circle-circle result fills in a hole in the proof of Proposition 1 in Book I of Euclid.
 

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>x² is convex using the tangent lines to the parabola y=x² and the focus of the parabola to determine a family of half planes whose intersection would be the described region.
 

Isometries:
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. 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:

• The Side-side-side (SSS) congruence of triangles (If Corresponding sides of two triangles are congruent, then the triangles are congruent), allows one to conclude that any isometry also transforms an angle to a congruent angle.
• The key connection between the congruence of figures in the plane and isometries: Figures F and G are congruent if and only if there is an isometry of the plane T so that T(F) = {P' in the plane where P'= T(P) for some P in F} = G.

You can read more about isometries by checking out this web site: Introduction to Isometries.

Line Isometries: Consider briefly isometries of  a  line.
1) translations and 2) reflections.
How can we visualize them?
Transformations figures (before and after lines); Correspondence figure on a single line; Graph of transformation; Coordinate function?
Can we classify them? Is every line isometry either a translation or a reflection?
 

There are at least four types of isometries of the plane: translation, rotation, reflection and glide reflection.
Lab Exercises Due  Friday Feb. 16.  1. Construct a scalene triangle using Wingeometry. Illustrate how to do i) a translation by a given "vector", ii) a rotation by a given angle measure, and iii) reflections across a given line..
2. Create a sketch that shows that the product of two reflections is either a translation or a rotation.

2/15 The classification of isometries.
Recall the informal definition of an isometry: A transformation (function) that preserves "distance."

Notice that 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.

Watch the video Isometries (Video # 2576 in Library) up to the place where it is demonstrated that 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.
 

Watch the remainder of the video on isometries, which classifies the product of 3 reflections as either a reflection or a glide reflection.
 

Note: The plane isometries form a group: identity, associativity, inverses... not commutative in general.
Symmetry of a figure: S is asymmetry 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).
Two interesting sites for looking at symmetries are the Symmetry Web page  and the Symmetry, Crystals and Polyhedra page.

The connection between Euclid's definition of proportionality (equal ratios) and real number equality of quotients.
We can show for line segments the following two propositions for segments A,B,C,and D with m(A)=a,m(B)=b, m(C)=c, and m(D)=d:

1. If A:B::C:D then a/b=c/d .
2. If a/b=c/d then A:B::C:D

2/27More on central similarities: In the plane a similarity T with factor m and center at (0,0) will transform  (a,b) to (ma,mb). This x' = mx, and y' = my are the equations for this transformation.
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'): T(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.

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.]

In the lab we discussed how to review constructions and
watched a video (in Library #209 cass.2) on similarity "On Size and Shape"  from the For All Practical Purposes Series.
 

We can use this fact to construct a circle C2 through  a given point B on a circle C1 and a point A inside the circle so that C2 is orthogonal to C1.
[ First construct the inverse 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.] Likewise we can construct an orthogonal circle C2 through two points A and B inside a circle C1, as shown in the sketch below.

• Line coordinates: The ordinary coordinates (x) corresponds to homogeneous coordinates <x,1>=<a,b> so x=a/b with b not 0. Homogeneous coordinates for a point on a line are related to lines through the origin. <a,b>=<d,e> if and only if  there is some k, not 0, where d=ka and  e=kb. These coordinates correspond to a line in the ordinary plane passing through (0,0). Except for the line Y=0, each of these  lines has exactly one point on it of the form (x,1). A point with homogeneous coordinate <a,0> corresponds to the ideal (infinite) point on the affine line. A similar treatment works for planar coordinates.
• Plane coordinates: The ordinary coordinates (x,y) corresponds to homogeneous coordinates <x,y,1>=<a,b,c> so x=a/c and y=b/c with c not 0. A point with homogeneous coordinate <a,b,0> corresponds to an ideal point in the affine plane.

• An example of vector methods in geometry: Show that the perpendicular bisectors of any triangle meet in a single point [whiich is equidistant from the vertices].
• An introductory discussion of the role of axioms and postulates in mathematics and the sciences.
• More on the affine plane:
• An affine line in an affine plane.
• Visualizing the horizon line on an image plane.
• Homogeneous coordiates for the affine plane.

 

•  Lab Exercise: Due 3-16.1. Draw a sketch of the affine plane showing the horizon line and label lthe lines X=1,2,-1, Y= 1,2,-1 and points (1,2) and (2,-1).

2. Draw a sketch of the seven point geometry model.
• Preface: Some examples of "structures."
• Musical structures. An atonal chord structure.
• Committee structures. A committee structure.
• Scheduling structures. A design for scheduling classes?
• Color Wheel/Triangle. (show with GSP?)
• A Geometric structure
• A graphical geometry.
• Visualizing this geometry. (GSP?)
• A Geometric structure
• A triangular geometry.
• Visualizing this geometry. (GSP?)
• Some projective geometric ideas based on a focus.
• A line through the focus appears as a point.
• A plane through the focus appears as a line.
• Two points determine a plane including the focus.
• Any two planes through the focus determine a line through the focus.
• Definitions:

A line through the focus is called a projective point (a P-point) and
A plane through the focus is called a projective line (a P-line).

A projective plane is the geometric object made up of the collection of P-points and P-lines.

• In a projective plane,

any two P-points determine a unique P-line and any two P-lines determine a unique P-point.
• Some algebraic descriptions of lines and planes in 3 dimensions.
• A plane through the origin has an equation of the form Ax + By + Cz  = 0, where [A,B,C] is not [0,0,0]. The triple [uA,uB,uC] will determine the same plane as long as u is not 0.

For example, [1,0,1] determines the plane with equation X + Z = 0.
• A line through the origin has the equation of the form (X,Y,Z) = (a,b,c) t where (a,b,c) is not (0,0,0).The triple (ua,ub,uc)will determine the same line as long as u is not 0.

For example, (1,0,-1) deteremines the line with equation (X,Y,Z)=(1,0,-1)t.
• A P-line has an equation of the form Ax + By + Cz  = 0, where [A,B,C] is not [0,0,0].

The triple [uA,uB,uC] will determine the same plane as long as u is not 0.

We'll call [A,B,C] homogeneous coordinates of the P-line.

For example, [1,0,1] are homogeneous coordinates for the P-line determined by the plane with equation X + Z = 0.

• A P-point has the equation (X,Y,Z) = (a,b,c) t where (a,b,c) is not (0,0,0).The triple (ua,ub,uc) will determine the same line as long as u is not 0.

We'll call <a,b,c> homogeneous coordinates of the P- point.

For example, <1,0,-1> are homogeneous coordinates for the P-point determined by the line with equation (X,Y,Z) = (1,0,-1) t.

• A P-point lies on a P line or a P -line passes through the the P-point if and only if  Aa+Bb+Cc= 0 where [A,B,C] are homogeneous coordinates for the P-line and <a,b,c> are homogeneous coordinates for the P-point.

For example, the P-point <1,0,-1> lies on the P-line [1,0,1].
• NOTE: All of the discussion works as long as the symbols A,B,C, a,b, and c represent elements of a field, that is, a set with two operations that work like the real or rational numbers in terms of addition and multiplication.
• A field with two elements:   F₂ = {0,1}.

 

+	0	1			x	0	1
0	0	1			0	0	0
1	1	0			1	0	1

A projective plane using F₂ has  exactly 7 points:

<0,0,1>, <0,1,0>, <0,1,1>,<1,0,0>,<1,0,1>,<1,1,0>,<1,1,1>.

A projective plane using F₂ has exactly 7 lines:

[0,0,1], [0,1,0], [0,1,1], [1,0,0], [1,0,1], [1,1,0], [1,1,1].

This projective plane satisfies the geometric structure properties.

Lines\Points	<0,0,1>	<0,1,0>	<0,1,1>	<1,0,0>	<1,0,1>	<1,1,0>	<1,1,1>
[0,0,1]		X		X		X
[0,1,0]	X			X	X
[0,1,1]			X	X			X
[1,0,0]	X	X	X
[1,0,1]		X			X		X
[1,1,0]	X					X	X
[1,1,1]			X		X	X

• We can visualize the projective plane  using F₂ using 7 points of the unit cube in ordinary 3 dimensional coordinate geometry.

The discussion reviewed the structure established 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.

Lab Exercise: Due 4-6.
1. Draw a sketch for Desargue's theorem in the plane.
[Cancel problem 2. 2. Using the 3d Wingeometry, draw a sketch of a cone with an ellipse, parabola and an hyperbola as planar sections.]
 

Consider a circle in a plane and a single point in space not on that plane (the vertex or apex). This will determine a cone, made from the lines passing through vertex and points on the circle.

i....related to 2 foci (ellipse and hyperbola) or 1 focus (parabola),
ii...related to a focus and directrix which was also related to eccentricity.

• Using two spheres tangent at two focii (F1 and F2) to the sectioning plane that also touch the sides of the cone, the ellipse is characterized by the distance from a point on the ellipse P having the sum of the distances PF1 + PF2 a constant,

while the hyperbola has the magnitude (absolute value) of the difference of these lengths |PF1-PF2| a constant. The parabola which in a sense lies between the ellipses and hyperbolae does not fit into these descriptions using two focii. The film explains and visualizes these relations.
• Again using the spheres and choosing one of these, consider the plane through the sphere at the circle where the sphere touched the cone. This plane meets the sectioning plane on a line, called the directrix, d. The film shows that for any conic section a point P on the conic is characterized by considering the ratio of PF1 to the distance from that point to the directrix, Pd . This ratio (called the eccentricity of the conic, e) is a constant for each conic section, in fact the ratio of the sines of the angle between the sectioning plane and the circle's plane and the angle between the cone and the circle's plane.

The following table summarizes the results related to this ratio. It is fully explained in the video.
 

ratio = e	Conic
e = 0	circle
0 < e < 1	ellipse
e = 1	parabola
e > 1	hyperbola

• Notice that from the ideas of projection from the vertex of the cone, any of the conic sections could be the shadow cast by the circle on a sectioning plane.

• Desargues' Theorem: (in projective 3 space). If two non co-planar triangles ABC and A'B'C' are perspectively related by the center O, then the points of intersection P=AB*A'B'; Q=AC*A'C' ; and R=BC*B'C' all lie on the same line.
• The proof of this result in space was covered in the previous section. Recall that this relied on the intersection of the planes determined by the two triangles. In a "projective spatial geometry" any pair of distinct planes would intersect in a line.
• Notice that if we use a central projection of the spatial configuration of lines and vertices from the spatial Desargues' Theorem we have a planar configuration of lines and vertices, satisfying the same hypotheses and illustrating the same conclusion. This basic idea of projecting a result from 3-space onto the plane is used in the proof of the planar version of Desargues' Theorem as provided in the book Geometry and The Imagination by Hilbert and Cohn-Vossen.
• Desargues' Theorem: (in the (projective) plane). If two coplanar triangles ABC and A'B'C' are perspectively related by the center O, then the points of intersection P=AB*A'B'; Q=AC*A'C' ; and R=BC*B'C' all lie on the same line.
• Proof: omitted here at this time. See the handout from H&C-V.
• Comments: This theorem is a result of projective geometry in its use of the fact that any pairs of line will have a point of intersection. This can be transferred to ordinary Euclidean Geometry by using the connecting geometry of the affine plane where parallel lines meet on the ideal line. The result allows that one pair of the lines determining P,Q, or R may be parallel, but if two of the pairs are parallel, then the line deteremined by those two pairs is the ideal line, so the third point of interesection is also an ideal point. The consequence in Euclidean geometry is these special cases of  Desargues' Theorem:
• Desargues' Theorem: (in the Euclidean plane). If two coplanar triangles ABC and A'B'C' are perspectively related by the center O,  then either

(i) there are three points of intersection P=AB*A'B'; Q=AC*A'C' ; and R=BC*B'C' which all lie on the same line;
(ii) one pair of sides, say AB and A'B' are parallel to the line determined by the points of intersection of the other pairs of sides Q=AC*A'C' ; and R=BC*B'C';
or (iii) if two pairs of sides , say AB and A'B' are parallel and  AC and A'C' are parallel, then the third pair of sides BC  and B'C' are also parallel.
[Hopefully I will have a java sketch here in the future to illustrate this from the projective, affine and euclidean views in the plane.]

• More examples of proofs :
• AB-C = AC-B

Proof: This is a statement of set equality. By the symmetry of the equation, it will be enough to show that if X is a point in AB-C, then X is a point in AC-B. So suppose X is a point in AB-C. Then by the definition of AB-C, there is a point P on the line AB so that X is on the line PC. What is required is to find a point Q on AC so that X is on QB. Now consider the triangle PAC. B is on PA, X is on PC, so by Axiom 6, there is a point of intersection, Q where BX meets AC.  End of Proof. [EOP]
• FG is a subset of AB-C

 

 
 






























Proof: Suppose X is on the FG with F and G in AB-C. Since F is in AB-C, there is a point P on AB where F is on PC. Also by work done previously (see 3-29) we have shown that FG intersects AB at a point Q. To show that X is in AB-C, it is enough to show that XC meets AB (at the point R). So consider the triangle PQF. X is on FQ, C is on PF, so by Axiom 6, XC meets PQ. But PQ = AB. EOP.
• If F,G, and Q (not on FG) are in AB-C then AB-C=FG-Q

This is done in two parts. (i) Show that FG-Q is a subset of AB-C.
(ii) Show that C is in FG-Q. [Then by previous work, A and B are also in FG-Q and by part (i) AB-C is a subset of  FG-Q.
Proof: (i) [ to be completed....]
• Any two distinct co-planar lines intersect in a unique point. [The proof of this proposition is an exercise.]

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).

Duality in Plane Projective geometry:

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:

One of the most important logical features of planar projective geometry is connected to the duality relation. Each of the dual statements for the postulates for planar projective geometry is a theorem of this geometry.
For example:
Postulate: Given two distinct points there is a line with those points on it.
Dual Statement and Theorem: Given two distinct lines there is a point with those lines on it.

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.
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.
Since Desargue's Theorem uses the hypothesis of a perspective relation between two triangles, we first look briefly at the dual concept.
 

[this will include a discussion of the dual concept for planar perspective, the statement of D's Theorem as S, the dual statement , S',  and the recognition that S' is the logical converse  of S. Thus we can say, "The converse of Desargues' theorem is true by the duality principle."]

So the isometry can be considered from the point of view of algebra with homogeneous coordinates to be the result of treating the coordinates as a column vector, multiplying by a square matrix, and then using the components of the resulting column vector as the homogenous coordinates for the transformed point.

The video Orthogonal Projection was shown in the lab hour.

More on  projectivities. The set of projectivities can be considered a group (as did isometries) , i.e. a set together with an operation which satisfies certain nice algebraic properties: closure, associativity, an identity and inverses.
 

4/24 More on line transformations with homogeneous coordinates .
These transformations can be identified with matrix multiplications since A(cv)=cAv for A a matrix, c a scalar, and v a column vector so that pairs of homogeneous coordinates for one point are transformed to homogeneous coordinates for a single point.
Translation: T(x)=x+3 became
 

.. Reflection R(x)= -x  uses the matrix

 Dilation by a factor of 5  M(x)=5x  uses the matrix

 and inversion I(x)=1/x uses

Consider how composition of these transformations corresponds to matrix multiplication and how these transformations interact with the ideal point at infinity using homogeneous coordinates.

The idea of a general projective transformation of a projective line using homogeneous coordinates and invertible 2x2 square matrices was also introduced. We have shown now that all the transformations of the line we had previously discussed were examples of this general type of projective transformation.
 

These two topic (harmonics and projective transformations) will be a major topic for the next few weeks.

The relation between the "double points" in the figure and the "single points". We prove using the following figure (adapted from Meserve & Izzo) that H(RT,SU) is equivalent to H(SU,RT). I.e., we show that if H(RT,SU) then H(SU,RT):
The new quadrangle used to show this harmonic relation is determined on the figure by the four points P3, P4, W and V.
[To complete the proof we need only show that WV meets SU at the point T.]
Notice in the figure that Triangle WP1P2 is perspectively related by the line SU to triangle VP3P4. Thus by converse of Desargues' Theorem we have that the triangles are perspectively related by a point. But this point must be T, so the line WV passes through the point T, completing the demonstration that H(SU,RT).

The key issue then becomes:
Was the point constructed from the points A, B and C uniquely determined by the fact that it was 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.)

With the existence and uniqueness of the point D established,
Examples of establishing a coordinate system for a projective line by choosing three distinct points to be P0, P1, and P¥.
We can construct P2, P-1, 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).
 

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=2x-1 with y(0)=3 has a solution curve determined by the tangent lines determined by the derivative: y=x^2-x+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.

We then discussed using {0,1,2} for homogeneous coordinates connected to the arithmetic given by the tables

The homogeneous coordinates for this set identify ordered triples, for example: <1,0,1>=<2,0,2> and <1,0,2>=<2,0,1>.  There are 27 possible ordered triples, and thus 26 when we exclude (0,0,0), and these are each paired by the factor 2 with another triple, so there will be exactly 13 points ( and by the comparable work with lines) and 13 lines in this geometry.
.... some details still to be reported.
 
 

[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.
 

1. Now that we've established how to connect homogeneous coordinates to a projective line using harmonics we continued by looking at some more algebraic projective lines- with coefficients of Z₂, Z₃, and Z₅, we saw that these projective lines would have 3,4, and 6 points respectively, corresponding to the 2,3,and 5 ordinary points and one ideal point. This lead to
2. A discussion of using the complex numbers, C, and the geometry of CP(1). The ordinary points on this "line" can be visualized as a plane or all the points of a sphere except one- which corresponds to the north pole or the point at infinity. [This was compared to the visualization of RP(1) as a circle.]
3. The discussion turned to algebraic projective transformations of RP(1). When these leave the ideal point fixed they are called affine transformations and form a group under composition. We showed that an affine transformation has a matrix of the form

a	b
0	1

and therefore T(Px)=Pax+b. Thus an affine transformation of the projective line is a dilation/reflection followed by a translation.

4. We looked at the idea of establishing the idea of a distance between ordinary points using absolute value-

d( <a,b>,<c,d>) = |a/b - c/d|.    We showed this is well defined and then discussed what the isometries for this distance would be. After some analysis we saw that this would mean that in the matrix of an isometry |a|=1, so a=1 or a=-1. I.e,  the matrix is

+/-1	b
0	1

or that T(Px)=P+/-x+b. Thus an isometry of RP(1) is a reflection followed by a translation.

5. We also looked further at Pascal's theorem and its planar dual Brianchon's theorem. Next class we'll prove Brianchon's theorem using an elliptic hyperbaloid.