# Solutions

Lesson 1
For the first proof, a/c = AD/a, b/c = BD/b. Solving for a2 and b2, and noting that AD + BD = c, we get the desired equality.

For the first set of questions in the second proof, the angles of the enclosed quadrilateral are all right angles because the sum of the two acute angles in a right triangle is a right angle, and a line has 180 degrees. This would be true for any positive choice of a and b. The area of the enclosed quadrilateral is the length of the hypotenuse, squared; this is also the area not covered by triangles.

For the second set of questions in the second proof, the area is not covered by triangles is a2 + b2. This is because the blue quadrilateral has area a2, and the red one has area b2 in the following diagram. So c2 = a2 + b2.

Lesson 2
1. 8, 99
2. 4
3. |x1 - x2|
4. 17
5. |y1 - y2|, if the points are annotated as (x1, y1) and (x1, y2).
6. 3, 4, 5
7. sqrt( (x1 - x2)2 + (y1 - y2)2)
8. 6, 6, 9
9. absolute value of the difference in the position where coordinates differ
10. sqrt( (x1 - x2)2 + (y1 - y2)2)
11. sqrt( (x1 - x2)2 + (y1 - y2)2)
12. sqrt( (x1 - x2)2 + (y1 - y2)2 + (z1 - z2)2)

Lesson 3
1. Intersect the plane containing both points in consideration and the origin with the surface of the sphere. This will give you two arcs; take the shorter one. If the points are antipodal, there are infinitely many such planes, and given any one, the two arcs you get are the same length.
2. sqrt(2r - 2x1x2 -2y1y2 - 2z1z2)
3. 2arcsin( (1/2r)sqrt(2r - 2x1x2 -2y1y2 - 2z1z2))
In the diagram below, the blue line represents the path on the surface of the sphere, and the triangle and angle depicted are as in problem 3. The vertex between the two sides of length r is the origin.

4. [2arcsin( (1/2r)sqrt(2r - 2x1x2 -2y1y2 - 2z1z2)) / 360]*2*pi*r
5. Rotation does not change distance. The distance traveled is r*pi
6. cos(b')*r
z = sin(b')*r
7. x = cos(a)cos(b')*r
y = sin(a)cos(b')*r

Lesson 4
1. 18, 3, 17
2. |x1 - x2| + |y1 - y2|
(|x1 - x2| + |y1 - y2|) C |x1 - x2|
3. 4, (-1, -4), 9
4. min{|x1 - x2|, |y1 - y2|}
If |x1 - x2| = |y1 - y2|, and neither is 0, there are two ways, otherwise, there is one way.
5. (0, 0), (0, y), (x, 0), 8
6. 2max{x, y}
7. |z1 - z2|
8. For a completely discrete method of measurement, you could divide the area you were concerned with into square foot units, and measure the elevation at the center of each square. On your map, you would divide the space into a grid, and write the elevation number on each square. A more continuous method would be to pick an increment of elevation change, say 50 feet, and draw a series of curves, where two points being on the same curve means that they are at the same elevation. The most continuous method would be to shade your map according to estimated elevation change, where two points being the same color would represent them being at the same elevation.

The first two methods have the advantage of being perfectly accurate for the information they claim to represent. Also, they are easier to work with computationally because they contain only a finite amount of data. Their main disadvantage is information loss. Although very little information would be lost in the first method, in the second method there could be a forty foot cliff that you were unaware of because the so-called "isolines" only represented elevations at the given increments. Although the information this method portrays is very accurate, there may be gaps in the information it presents. Another disadvantage for the first method is that such a map takes a very long time to construct.

The last method has the disadvantage that any map that takes such a form is necessarily approximating much of the data it claims to represent. As such, no such map is perfectly accurate in the information it claims to represent. The map also has the disadvantage that direct computations can be tricky to complete; however, if the function that depicts the shading is easy to work with, computations may be less time-consuming than in the discrete case. A major advantage is that visually such a map is very easy to interpret.

Lesson 5
1. The first property is satisfied because |a - b| = |b - a|. The second and third properties are satisfied because the absolute value of a number is only zero if the number is zero, and is otherwise positive. The fourth property is satisfied because if A = (x1, y1), B = (x2, y2), C = (x3, y3), then d(A, B) + d(B, C) = 2(max{|x1 - x2|, |y1 - y2|} + max{|x2 - x3|, |y2 - y3|}) and d(A, C) = 2max{|x1 - x3|, |y1 - y3|}. So since 2|x1 - x2| + 2|x2 - x3| is at least as big as 2|x1 - x3|, then 2(max{|x1 - x2|, |y1 - y2|} + max{|x2 - x3|, |y2 - y3|}) is also. We can make a similar statement about |y1 - y3|.
2.

This corresponds to the town with one street situation.
3. This is not a distance function. The second property fails for the example A = (1, 2), B = (1, 3).
4.

This corresponds to the situation of the photographer.
5. The first property is satisfied because |a - b| = |b - a|. The second and third properties are satisfied because the absolute value of a number is only zero if the number is zero, and is otherwise positive. So the function is only zero if x1 = x2 and y1 = y2, that is, if A = B, and is otherwise strictly positive. The fourth property holds because |x1 - x2| + |x2 - x3| is at least as big as |x1 - x3|, and the same holds if the the x's are replaced with y's.
6.
This models the cab driver situation.
7. yes, yes
8. yes
9. yes
10. no

Lesson 6
1. The set of points of distance less than one from the origin is open. Say that a point is distance r from the origin, where r is less than 1. Then all the points within distance (1-r)/2 of the given point are less than 1 unit from the origin. The set of points of distance exactly 1 from the origin is not open. Take any such point. Then no matter how small you choose r, the set of points within distance r of the given point will have some points of distance more than 1 from the origin. The set of points of distance more than 1 from the origin is open. Given any such point, you can draw a small circle centered at it that does not intersect the circle of radius 1 centered at the origin.
2. The answer to both questions is yes. This is because given any point inside of a diamond, we can draw a small circle around it and remain inside the diamond AND given any point inside a circle, we can draw a small diamond around it and stay within the circle.
3. The intersection of two open sets is open. Given a point in the intersection, there are r1 and r2 so that all the points within distance r1 of the given point are in the first open set, and all the points within distance r2 are in the second one. So, consider the smaller of r1 and r2. Intuitively, this says that all the points that are close to both A and B are close to each other. The farthest apart S and T can be is 2r.
4. Yes. For a given point in the union, it is in one or the other of the open sets. So, there is a radius r so that all the points within this distance of the given point are in a given one of the two open sets. Using this radius, they are also is also in the union.
5. They will also be close together.
6. Yes to both questions. The function f sends each point back to itself, and we already showed that the same subsets of the plane are open whether you use the cab driver or the Euclidean distance function.
7. Yes! Given any open set in the plane, the points that a constant function sends there are either the entire plane, or the empty set, both of which are open.
8.1 Closed because the set points of distance less than 1 from the origin and the set of points of distance greater than 1 from the origin are both open, and the union of two open sets is open.
8.2 Closed because given any point in the complement we can draw a small circle around it that does not contain the origin or any of the points of distance 1 from the origin. Can you write this up more formally, giving precise distances?
8.3 Neither because no circle around the origin is contained in the set, and any circle around a point with 0 for its y-coordinate and a nonzero x-coordinate will contain some points with positive y-coordinate. So the set is not open, and its complement is not open.
8.4 Open because if the x-coordinate of a given point is strictly positive, then just divide it by 2 and consider the circle of that radius around the point.
9. The intersection of two closed sets is closed. Consider the complement of each set separately. These are both open. Their union is the complement of the intersection of the two closed sets and is open. The union of two closed sets is closed. Consider the complement of each separately. They are both open. Their intersection is the complement of the union of the two closed sets and is open. Intuitively this says that if two points are close together and a third point is far away from one of them, then it is also far away from the other one.