# Homework 2

Lori Thomson
Math 5337

### Part A. Question 3 from the Geometer's Sketchpad homework

1. Given two points A and B, what is the set of points P such that AP + BP = a constant K? ("AP" is my shorthand for "the distance from point A to point P." Also, K must be greater than the distance AB, or there are no points P.) In this exercise, you use Sketchpad to look at this set and formulate a conjecture about its shape. (Do you remember what the shape will be from previous geometry courses?)

Turn in your final sketch. Briefly describe the mathematics of how you constructed it. Write your conjecture for the shape traced in part d. Answer the questions in part e.

Circles were constructed by "center and radius," where centers were at A and B, and radii were equal to the length of segments EC and DE, respectively. Points F and G were constructed at the intersections of the two circles. The ellipse is the result of tracing F and G while sliding E up and down DC.

1. Trace Locus of F and G as you vary E. What does this set look like?
It looks like an ellipse with foci A and B.

2. Experiment: move A and B around. Change the constant K (by moving D). Now trace the locus in part d again. How does the value of K effect the set of points you trace? What would happen if K were exactly distance AB?
As K gets longer, the resulting shape gets larger and looks more circular (less flat); as K gets shorter, the figure gets smaller and flatter, until eventually K equals distance AB exactly, in which case the figure would just be a line segment connecting A and B.

3. By now you should have a conjecture about the traced set. Can you prove your conjecture? Place the origin at the midpoint of the line segment between A and B. Write equations involving the coordinates for points in the traced set. Write an equation describing the shape that you conjecture. Are the these equivalent?
Let P be any point in the traced set. Let a = K/2 and c = AB/2. The sum of the distances is given by:
AP + BP = 2a
The coordinates P(x,y) on the traced set satisfy the equation
sqrt([x+c]2 + y2) + sqrt([x+c]2 + y2) = 2a
To simplify this expression, transpose the second radical to the right side of the equation, square and simplify twice to obtain
```  x2       y2
----- + ------ = 1
a2     a2-c2
```
Since the sum AP + BP = 2a of the two sides of the triangle ABP is greater than the third side AB = 2c the term "a2-c2" is positive and has real a positive square root, which we denote by b:
b = sqrt(a2-c2)
Then the equation for the traced region becomes:
``` x2    y2
--- + --- = 1
a2    b2
```
which is the equation of an ellipse with a and b as diameters in the x and y directions, respectively.

### Part B. Questions 3, 4 and 5 from the assignment about Monge's Theorem

1. From Part 3: State Monge's Theorem explicitly.
Given three circles of different radii, the intersections of each of the three pairs of external tangent lines to these circles are colinear.

Turn in your sketch of the three pairs of external tangents to three circles, illustrating Monge's Theorem, as in the spoiler.

My sketch demonstrating Monge's theorem is accompanied by a script that draws a tangent line to a circle through a given point not on the circle, and a second script that draws an external tangent line to two circles. (Warning: Loading the scripts causes error in Sketchpad.)

2. From Part 4: Given two points A and A', does there exist a dilation such that A' is the image of A under dilation? If not, what are conditions for such a dilation to exist? If so, is such a dilation unique? Explain.
There always exists a dilation that will map any point A to any other point A' and this dilation is not unique. Pick a point C on the line AA' as the center of dilation. The ratio of dilation should be the ratio A'C / AC. Any two segments whose lengths of this ratio may be used.

Given four points A, A', B, and B', does there exist a dilation such that A' is the image of A and B' is the image of B? If not, what are conditions for such a dilation to exist? Is such a dilation unique? Explain.

If such a dilation exists, the center C of the dilation must be at the intersection of the lines AA' and BB'. The ratio of dilation must be equal to both A'C / AC and B'C / BC. Therefore, such a dilation exists if only if A'C / AC = B'C / BC.

Turn in a sketch of dilating a polygon by marked ratio and a sketch of the composition of two dilations.

Marked dilation:

Double dilation:

3. Based on your observations, you should see that the composition of two dilations is a third dilation. Explain why the three centers of dilation are collinear.
[...]

When are two circles related by a dilation? Is such a dilation unique? Relate this to Part II.

A circle can be considered a dilation of another circle if (1) they have the same center and radius, or (2) their external tangents intersect. The second happens only if the circles are of different radii (otherwise the tangent lines would be parallel). The point where the tangent lines intersect is the center of dilation.

Combine these results to give a logical argument justifying Monge's Theorem.

In Monge's Theorem, all three circles have different radii. So, each can be constructed as a dilation of another. Suppose circle 2 is a dilation of circle 1, and circle 3 is a dilation of circle 2, making it a double dilation of circle 1. We know from question #4 that the composition of two dilations is a third dilation (Circle 3 is also a dilation of circle 1) and that the three centers of dilation are colinear. Since the centers of dilation are the intersections of the tangents, we can say that the intersections of the tangents are also colinear.

Done.