Martin Flashman's Courses - Math 371 Spring, '16

Geometry Notes
  Geometric Structures for the Visual

[Work in Progress DRAFT VERSION Based on notes from 09 and 11]

Notes on Java and JavaSketchpad Malfunction:
The use of Java has become a browser and machine dependent issue.
In particular, the Java used in David Joyce's version of Euclid does not work uniformly in Firefox on HSU computers, but does seem to work on Chrome. Other Java works in Firefox, but not in Chrome. I will try to indicate these dependencies when possible.
Readers who might have difficulty running the Java applets are advised to use a book marklet that converts JavaSketchpad sketches on this page (or anywhere else on the internet) to work completely independently of Java.
Go to this site to install the small tool in your web browser OR
click on the following link to "fix" the java. Convert JavaSketch
Please give feedback, whether bug reports or other, at the following email address:
Green sections indicate tentative plans for those dates.

1-20  Introduction 1-22 Continue discussion of what is "geometry"? 
Start on Euclid- Definitions, Postulates, and Prop 1.
 1-25 Euclid- Definitions, Postulates, and Prop 1. cont'd
1-27 Pythagorean plus.
1-29 Euclid Postulates/ Pythagoras
  2-1 Euclid early Props/ Pythagoras/
2-3 Euclid early Props/ Pythagoras/
 2-5 Equidecomposable polygons
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.
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
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,
3-14 No class Spring Break! 3-16 No class Spring Break! 3-18 No class Spring Break!
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.
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.
4-4  Affine geometry- Homogeneous coordinates and Visualizing the affine plane. 4-6 More on Affine Projective Geometry - Algebra!
 4-8 Desargues!
4-11 Review and start of Synthetic Projective Geometry
4-13 More Review and start of Synthetic Projective Geometry
4-15 RP(2)
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.
4-25 Projectivities and harmonics
4- 27 Constructing Harmonics
5-4 Con ics
5-6 Pascal and Brianchon

1-25     Still to be covered is the course project on the assignments webpage.

  • In one alternative proof for this theorem illustrated in the GeoGebra sketch below, follow the movie to consider 4 congruent right triangles and the square on the side of the hypotenuse arranged inside of a square with side "b+c" and then moving through the movie the same 4 triangles and 2 squares arranged inside of a square with side "b+c"  . Can you explain how this sketch justifies the theorem?
  • 1-27 Review of previous "proof of the PT.

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

    2-3 Some Comments on Problem Set #1:Intersections for families of convex figures:
           Using the notation of M  & I:
                                `cap \{ [P_0P_r : r>0\} = \{P_0\} ; cap \{ [P_{-1-r}P_{1+r} : r>0\} = \{P_{-1}P_1\}`
           To show that `\D = `{`P` in plane where `d(P,P^*) le 1`} is convex where `P^*` is a point in the plane,
            recognize that `D = cap` { half planes determined by tangent lines to the boundary of `D` that contain `P^*`}. These half planes are all convex.

    [Side Trip?] Moving line segments: Can we move a line segment without changing its length.

    We can look further at the foundations of the proofs of the Pythagorean Theorem in two ways:

    See also A New Approach to Hilbert's Third Problem - University of 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.
    Another component of the proof of the equidecomposable polygon theorem is the ability to "add two parallelograms to form a single parallelogram which is scissors congruent to the two separate parallelograms". Here's how:
    Intersect two pairs of parallel lines, l and l' with m and m'- one from each of the given parallelograms. Draw a diagonal HI in the resulting parallelogram.
    Cut and translate one parallelogram so that it is scissors congruent to a parallelogram HIJK within the same parallel lines l and l' with one side being the diagonal.
    Cut and translate the other parallelogram so that it is scissors congruent to a parallelogram HINO within the same parallel lines m and m' with one side being the diagonal and on the other side of the diagonal HI from the transformed first parallelogram. Now draw the parallel NO to the diagonal in the second transformed parallelogram HINO so that it intersects the parallels  l and l' from the first parallelogram at the points P and Q. This makes one larger parallelogram JKPQ which is scissors congruent to the original two parallelograms.
    Compare this with  Euclid Proposition  45.

    This is a Java Applet created using GeoGebra from - it looks like you don't have Java installed, please go to

            Follow this link for a proof of the equidecomposable polygon theorem. or here is a slightly different approach.
    Footnote on Equidecomposable Polygon Theorem:
    The relation of being "scissors congruent" (sc) is an equivalence relation: 
    1.  A   sc  A
    2. If A sc B then B sc A.
    3. If A sc B and B sc C then A sc C.
    The theorem can be interpreted by saying for any A and B, polygonal regions in the plane, A sc B if and only if Area(A) = Area(B).

    General discussion of invariants:
    Geometry (and Math in general) studies objects and how they are transformed.
    Some features of the objects are preserved by the transformations. These features are often described as invariants of the transformation.
    In studying  equidecomposable polygons, the objects are polygonal regions in the plane, and the transformations are scissors congruences.
    The area of the initial polygon is the same as the area of the transformed polygon, so area is an invariant of scissor congruence.
    The general question: Given some objects and transformations, what are the invariants of these transformations?  Is there a collection of invariants for the transformations so that if two objects have the same invariants then they are equivalent in some sense based on the transformations being studied.
    The main impact of the Equidecomposable Polygon Theorem: For planar polygons and scissors congruence, the area of the polygon is an invariant that is sufficient to determine when two polygons are or are not scissors congruent
    Another geometric relation connected to transformations is congruence for triangles. 
    We will study further the transformations that are key to the congruence relation: translation. rotation, and reflection.
    There are many invariants for these transformation: they include the length of segments and the measurement of angles as well as area of the triangles.
    It should be clear that the area of two triangles is not enough to determine whether they are congruent. 
    There are some key invariants of a triangle which can determine congruence:

              Read  the definitions in M&I section 1.1
    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

    [Review the construction of a line though a given point parallel to a given line. See Euclid  I.31.]
    Brief Introduction to Similar Triangles: [More details later in the course.]
    Basic Result: Given `Delta ABC` and `Delta A'B'C'` with `<Acong<A'` and  `<B cong <B'` then `<Ccong<C'` and `{m(AB)}/{m(A'B')} = {m(BC)}/{m(B'C')} = {m(AC)}/{m(A'C')}` and we say that `Delta ABC` is similar to  `Delta A'B'C'` and write `Delta ABC ~ Delta A'B'C'`.
    This fact is used repeatedly in M&I to justify the construction of `P_{k/n}`, a rational point on the number line for the rational number `k/n` from an integer lattice on the euclidean plane. 
    Using the rational number points and the continuity axiom for a euclidean line we can construct a unique point `P_x` on the euclidean line for every real number `x` by using the (infinite) decimal representation of the number. With this correspondence the order relation of the real numbers corresponds to the between-ness relation of points on a line segment and there is a geometric construction for the arithmetic operations with real numbers that matches all the arithmetic properties of the real numbers.
    For example we can "add" `P_x` to `P_y` to obtain a new point `P_s` and `s = x+y`. This can be done with circles or with parallel lines.
    We will look at multiplication and division further in the next few sessions. Historically this "isomorphism" between the real number arithmetic and geometry constructions was identified by Descartes in his work The Geometry.(1637)

    Examples of using coordinates  (and Vectors) for proofs in euclidean geometry:

    Proposition: An angle inscribed in a semicircle is a right angle.
    Restated: If `AC` is the diameter of a circle and `B` is a distinct point on the circle, then `<ABC` is a right angle.

    Coordinate Proof: Using coordinates, assume the center of the circle has coordinates `(0,0)` and that the point `A` has coordinates `(1,0)` and `C` has coordinates `(-1,0)` while `B` has coordinates
    `(x,y)` where `x^2 + y^2 = 1`.
    To show `<ABC` is a right triangle by application of the converse to the Pythagorean theorem it suffices to show that `[(x - 1)^2 + y^2 ]+ [(x + 1)^2 + y^2] = 2^2 = 4`.
    This can be verified by expanding the squared terms involving `x` giving :  ` [x ^2 + 1  + y^2 ]+ [x ^2  + 1 + y^2] = 2+2 = 4`. EOP.

    Vector Proof: Let `v` be the vector from `A` to `C` and `w` be the vector from `B` to `C`. It suffices to show the dot product of these vectors is `0`. 
    Now `v cdot w =  ( (x - 1)(x+1) + y^2 ] = x^2 - 1 + y^2  = 0`. EOP.
      Introducing Orthogonal Circles and The inverse of a point with respect to a circle. Convexity of a geometric figure.

    Note on Three Historical Problems of Constructions with Straight Edge and Compass (Euclidean Constructions)
    (1) Trisection of an angle: Since it possible to bisect and trisect any line segment and bisect any angle, the issue is, is it possible to trisect any angle?
    Note: If you use folding and Origami construction it is possible to trisect an angle. [How to Trisect an Angle with Origami - Numberphile ...]

    (2) Duplication of a cube [Wikipedia]: Since it is possible to construct a square with twice the area of a given square, the issue is, is it possible to construct a cube with twice the volume of a given cube?

    (3) Squaring a circle: Since it is possible to construct a square the same area as any given polygonal region, the issue is, is it possible to construct a square with the same area as a given circle.
    [Squaring the Circle]

    We can use this proposition in the following
    Constructions: 1. 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.

    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.

    Sorry, this page requires a Java-compatible web browser. 
    Further notes on the continuity axiom.

    The continuity axiom leads to the important result due to G. Cantor that  any list (possibly infinite) of points in a given segment of a euclidean line will not have every point in that segment on the list. [Note  that we can make a list of points corresponding to the rational numbers once a unit length had been established.  `1/1,1/2,2/1,1/3,2/2,3/1,1/4,2/3,3/2,4/1, ....`]

    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 the hole in the proof of Proposition 1 in Book I of Euclid.


    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.

    Facts: (i) If  T and S are isometries then ST is also an isometry where ST(P) = S(T(P))= S(P')  [T(P)=P'].

    (ii) If T is an isometry, then the transformation S defined by S(P') = P when T(P) = P' is also an isometry.

    Proof : (i) Consider d(ST(P),ST(Q)) = d( S(T(P)),S(T(Q)) ) = d(S(P'),S(Q')) = d(P',Q') = d(P,Q) .

    (ii) Left as an exercise for the reader.

    Comment: If G ={ T: T is an isometry of a structure}, then since Id(P)=P is an isometry, Facts (i) and (ii) together the fact that composition (product) of transformations is an associative operation make G together with the operation of composition a group structure.

    Convexity: Another brief side trip into the world of convex figures.

          Recall the Definition: A figure F is convex if whenever A and B are points in F, the line segment AB is a subset of F.

          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.

    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.

    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.


    • The Side-side-side (SSS) congruence of triangles

    • (If Corresponding sides of two triangles are congruent, then the triangles are congruent)
    • This 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:

    • Proposition: 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.

    We begin a more detailed study of isometries with a look at
    Line Isometries:
    Consider briefly isometries of  a  line.
    1) translations and 2) reflections.
    How can we visualize them?
    • Mapping Diagrams (before and after lines) T: P -> P';

      Correspondence figures on a single line:
      Example: A reflection.
      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.
    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.

    Line isometries and coordinates:

      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) = -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?

      • Mapping Diagrams (before and after planes); [GeoGebra] [Add figures here.]
      • Correspondence figures on a single plane; [GeoGebra]
      • Graph? [Visualizing 4 dimensions! ]

      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) f
      our 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.]

    • 3-2

    • Proposition: Any plane isometry  is either a reflection or  the product of two or three reflections.
      • 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.

        We can do this geometrically or using analytic geometry and the matrices!

      • Note: The four types of isometries can be characterized completely by the properties of orientation preservation/reversal and the existence of fixed points. This is represented in the following table:

        Fixed points Rotations Reflections
        No Fixed points Translations Glide reflections

        Note: The plane isometries form a group: A set together with an operation (the product or composition) that is
        (0) Closed under the operation - the product of two isometries is an isometry;
        (1) The operation is associative - R(ST) = (RS)T  or for any point P,  (R(ST))(P) = R( ST(P))= R(S(T(P))) = (RS)(T(P)) = ((RS)T)(P).
        (2) The identity transformation (I) is an isometry -  I(P)=P.
        (3) For any isometry T there is an (inverse) isometry S so TS = ST = I.
        The operation of composition is not necessarily commutative in the sense that ST is not always the same transformation as TS:
        For example If R1 and R2 are reflections in intersecting lines l1 and l2 then the isometry R1R2 is a rotation about the point of intersection in the opposite direction to R2R1.

    • What about coordinates and plane isometries?

    • Coordinates: (a la M&I I.3) Use any two non-parallel lines in the plane with coincident 0. Then you can determine "coordinates" for any point by using parallelograms. [As indicated previously, all rational coordinates can be constructed from establishing a unit. This is outlined more thoroughly in the reading in I.3.]

      Isometry examples with coordinates in the plane: (See M&I I.5 and I.6)

      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

      (x,y) (

      1 0
      0 -1
      ) = (x,-y)
        Matrix for RX 
      (x,y) (

      -1 0
      0 1
      ) = (-x,y)
        Matrix for RY
      (x,y) (

      ) = (y,x)
      Matrix for R 

      Or writing the vectors as column vectors

      [ 1 0
      ] [ x ] = [ x ]
      0 -1 y
        Matrix for RX 
      [ -1 0
      ] [ x ] = [ -x ]
      0 1 y
      Matrix for RY
      [ 0
      ] [ x ] = [ y
      1 0
      Matrix for R 


    If R90 is rotation about (0,0) by 90 degrees, then R90(x,y) = (-y,x).
    Question: What is rotation about (0,0) by t degrees: R(t)?
    Hint: What does the rotation do to the points (1,0) and (0,1)? Is this rotation a "linear transformation?"
    More general Question: What about reflection R(A,B) about the line AX+BY = 0?

    [ 0
    ] [ x ] = [ -y
    1 0
    Matrix for R90


    [ a
    ] [ x ] = [ ax+by
    R(t):Matrix for rotation by t degrees? 

    Consider that P(1,0) will be transformed to 
    P'(cos(t), sin(t))= (a,c
    and Q(0,1) will be transformed to 
    Q'(-sin(t), cos(t))=(b,d)

    [ a
    ] [ x ] = [ ax+by
    Matrix for R(A,B)?

    More general Question: What about reflection R(A,B) about the line AX+BY = 0?

    [ cos(t)
    ] [ x ] = [ xcos(t)-ysin(t)
    R(t):Matrix for rotation by t degrees.

    See discussion above.

    [ a
    ] [ x ] = [ ax+by
    Matrix for R(A,B)?
     Hint: Rotate, reflect, and rotate back!
    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)
    ] [ 1
    ] [ cos(t) sin(t)
    `cos^2(t)-sin^2(t)` 2cos(t)sin(t)

    Here is the visualization of R(A,B) as a map in Winplot using: (x,y)==>(cos(2t)x+sin(2t)y,sin(2t)x-cos(2t)y)

    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.

        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.

        The Group Table of Symmetries of an Equilateral Triangle.
        Two interesting sites for looking at symmetries are the Symmetry Web page  and the Symmetry, Crystals and Polyhedra page.
      • Discussion: What about isometries in three dimensions?

      • These are generated by spatial reflections in a plane.
        A spatial isometry is determined by the transformation of 4 points not all in the same plane (which determine the simplest 3 dimensional figure- a tetrahdron!)
        Any isometry of space can be expressed as the product of at most 4 reflections.
        These results are proven in the same fashion as the comparable results were proven in the plane.

    • Proportions and similarity:
    An example of the use of similar triangles and proportions to constructing "square roots":
    Mean Proportions in right triangles and Inverses:
    Consider a right triangle ABC with hypotenuse AB. Notice that if the altitude CD is constructed with the hypotenuse AB as the base, the figure that results has 3 similar right triangles. ABC, ACD, and DCB. Using similarity of these triangles we see that there is a proportion of the segments of the hypotenuse AD, DB and the altitude CD given by AD:CD::CD:BD. If we consider the lengths of these segments respectively as a,h, and b then the numerical proportion may be expressed as a/h=h/b or using common algebra $ab=h^2$
    . Notice this says that $ h= \sqrt{ab}$.

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    Application of this construction: Choosing a = 1, this proportion becomes $h = \sqrtb$

        A look at problems caused by the diagonal of a square and the issue of finding a unit that would measure both the side and the diagonal.
        First look at the Euclidean algorithm for finding a common segment with which to measure two segments.
          Euclid's original treatment of the "division algorithm" :
          If  n and d>0 are integers
          then there are integers q and r with r=0 or 0<r<d where n = q*d +r.

          Using q*OD to represent a segment that is made of q segments all congruent to OD,
          Euclid's division algorithm is stated in geometry that of ON is a segment and OD is a segment that is contained as a subsegment of ON, then ON is congruent to q*OD  with  possibly a remaining segment RN which is congruent to a subsegment of OD.

          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.

        In the application of the euclidean algorithm to the diagonal and side of a square, the procedure appears to stop.
        However, we can show that because of the fundamental theorem of arithmetic, it would be impossible to find a segment with which to measure both the side and the diagonal of a square.

        The impact of this on geometry was that one could not presume that all of geometry could be handled by using simple ratios of whole numbers for measurements.

        [A geometry based on ratios alone would not permit one to accomplish proposition 1 of Book I of Euclid! since this would mean that the geometry would have to be able to have ratio involving the sqr(3) - which like the sqr(2) is also an irrational number.]

      A major part of geometry before Descartes was Euclid's (Eudoxus') resolution of the issue in Book V def'ns 1-5.(Joyce)
      Definition 1
      A magnitude is a part of a magnitude, the less of the greater, when it measures the greater.
      Definition 2
      The greater is a multiple of the less when it is measured by the less.
      Definition 3
      A ratio is a sort of relation in respect of size between two magnitudes of the same kind.
      Definition 4
      Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another.
      Definition 5 (Byne's)
      Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when,
      if any equimultiples whatever are taken of the first and third,
      and any equimultiples whatever of the second and fourth,
      the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order.

        Look at these definitions and note some key items:
        * Ratios exist only between magnitudes of the same type. (This is usually described as "Homogeneity".)
        * For ratios to be equal the magnitudes must be capable of co-measuring.
        * Euclid's axioms do not deny the existence of infinitesimals- but will not discuss equality of ratios that use them.

        Briefly: An infinitesimal segment is a segment AB that is so short that it would appear to be coincident points and for any k, k*AB cannot contain any ordinary segment determined by points that we can see as distinct!
        n the history of mathematics, infinitesimal segments we used in the early developments of the calculus. For example: Consider the ratio of {the change in the area of a square, A(s), when the length of a side, s, is changed by an infinitesimal, ds} to {the change in the length of the side, ds}. That is: What is [A(s + ds) - A(s)]/ds ?  Answer: [2s*ds + ds*ds]/ds = 2s +ds. Thus the ratio would be indistinguishable as a number from 2s. ]

        * 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.)

        The connection [between Euclid's definition of proportionality (equal ratios) and real number equality of quotients].
        We can show the following
        Proposition: 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

      • The concept of similarity as a transformation.

      • Euclidean Constuction for a Central Similarity:

        Proposition 1 (Book VI) of Euclid provides the tool for the central similarity transformation in Euclidean geometry.  To perform the transformation of the plane with center O and using two given line segments to determine the magnification, first draw a ray from O and locate P and P' on the ray so the given segments are congruent to OP and OP'. To transform a point Q
        draw the ray OQ, then the segment PQ. Through P' construct a line l that is parallel
        to PQ. Then l will meet OQ at a point Q' and by proposition 1, Q' is the appropriate point to which Q is transformed by the central similarity as required.

        It is not difficult to show from this construction that a central similarity will transform a line  l to a line l' that is parallel to the original line l. Thus one can prove the related results for euclidean geometry:
        Corollary: If T is a central similarity then
        (i) T preserves the measure of angles and
        (ii) T transforms a pair of parallel lines to parallel lines.

        Since isometries also preserve the measure of angles and transform a pair of parallel lines to parallel lines (since a reflection does) we see that any composition of central similarities and isometries will likewise preserve the measure of angles and transform a pair of parallel lines to a parallel lines. 
        Exercise: Show such compositions will transform a triangle to a similar triangle and ABC and A'B'C' is any pair of similar triangles, the there is an isometry T and a central similarity S so that TS transforms ABC to A'B'C'.

    Interlude (not covered in Lecture): Using Vectors in a geometric structure:
    The segment connecting midpoints of the sides of a triangle proposition: A Vector Proof

        Similarity Transformations in Coordinate Geometry. [Not covered in class.]
        Consider a similarity on a line with a center of similarity and a given positive magnification factor. This leads to a consideration of the effect of a similarity on the coordinates of a point on the line.
        If we use the point P0 for the center, then we see that the similarity T with magnification factor of 2 would transform Px to P2x, or T(Px)=P2x. Removing the P from the notation we have T(x)=2x. Using T(Px)=Px', we find that T is described by the correspondence where x'=2x.

        [More generally: If we use the point P0 for the center, then we see that the similarity T with magnification factor of m would transform Px to Pmx, or T(Px)=Pmx. Removing the P from the notation we have T(x)=mx. Using T(Px)=Px', we find that T is described by the correspondence where x'=mx. ]

        With the center at another point, say P3, the transformation T* is controlled by the fact that T*(Px) - P3 = 2(Px-P3).
        So that  for T* we have x' = 3 + 2(x-3).

        [Again generally: With the center at point Pa, the transformation T* is controlled by the fact that T*(Px) - Pa = m(Px-Pa). So that  for T* we have x' = a + m(x-a).]

        We can also observe that if we let S(Px)=P(x-3) and S-1(Px)=P(x+3) then

        S-1(T(S(Px))) = S-1(T(P(x-3)))
        = S-1(P2(x-3))
        = P(3+2(x-3))
        = T*(Px),
        so S-1TS=T*.

        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
        ] [ x ] = [ mx ]
        0 m
        Matrix for Tm
        Other central similarities T* with factor m and center at (a,b) can be recognized as related to Tm by using the translation S(x,y) = (x - a,y - b) and seeing that  S-1TmS=T*.
        [Try this yourself!]

        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:

    S(x , y) = (5x -12 ,5y - 8).



        View video (in Library #4376 ) on "Central similarities" from the Geometry Film Series. (10 minutes)
        View video (in Library #209 cass.2) on similarity (How big is too big? "scale and form")  "On Size and Shape"  from the For