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]


Blue sections indicate tentative plans for those dates.
Week
Monday
Wednesday
Friday
1

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... Dissections ?
1-29 Dissections- equidecomposeable polygons
3
  2-1 Begin Constructions and the real number line. 2-3 M&I's Euclidean Geometry
 2-5 More on Equidecomposeable polygons
4
2-8
 2-10
2-12
5
2-15
 2-17
2-19
6
2-22
2-24
2-26
7
2-29
 3-2
3-4
8
3-7
3-9
3-11
9
3-14 No class Spring Break!
 3-16 No class Spring Break!   3-18 No class Spring Break!
10
3-21   3-23
3-25
11
3-28
3-30
4-1
12
4-4  4-6 
 4-8
13
4-11
4-13
4-15
14
4-18 4-20
4-22
15
4-25
4- 27 4-29
16
5-2
5-4














[Side Trip] Moving line segments:


 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 ...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.


        Follow this link for a proof of the equidecomposable polygon theorem. or here is a slightly different approach.
        

            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
 




            Introducing Orthogonal Circles and The inverse of a point with respect to a circle. Convexity of a geometric figure.


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.

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