For the first proof,

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

1. 8, 99

2. 4

3. |

4. 17

5. |

6. 3, 4, 5

7. sqrt( (

8. 6, 6, 9

9. absolute value of the difference in the position where coordinates differ

10. sqrt( (

11. sqrt( (

12. sqrt( (

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(2

3. 2arcsin( (1/2

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

4. [2arcsin( (1/2

5. Rotation does not change distance. The distance traveled is

6. cos(

7.

1. 18, 3, 17

2. |

(|

3. 4, (-1, -4), 9

4. min{|

If |

5. (0, 0), (0,

6. 2max{

7. |

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.

1. The first property is satisfied because |

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 |

6.

This models the cab driver situation.

7. yes, yes

8. yes

9. yes

10. no

1. The set of points of distance less than one from the origin is open. Say that a point is distance

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

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

5. They will also be close together.

6. Yes to both questions. The 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

8.4 Open because if the

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.