Projects in Geometry for High School Students Goal: Our goal in more detail will be expressed on the next page. Our journey will force us to understand plane and three-dimensional geometry. We will take the vector approach. Approaching geometry through the language of vectors proves to be much easier than classical geometry. When appropriate we will hint at other interesting questions which emerge from the mathematics we create trying to solve our problem. 1
Introduction: Our goal is to learn how to make maps the easy way. It is much easier to draw curves in the plane than ones on a sphere. Suppose I told you that there was a way knowing how pictures drawn in the plane will look on the sphere. That would be amazing, right? It is this relationship we will spend our times trying to construct (mathematically) in this talk. Structure of Talk: Throughout the talk I will also make side comments about other interesting mathematics which are the result of us trying to answer our question. Beginning: We first need the idea of a coordinate system. This will just be a model space that helps us visualize our objects i.e points, lines and curves. 2D-Coordinate System: Take a vertical line and a horizontal line and denote their point of int. as (0, 0). Then we will infer that the point (x, y) means move horizontally x-units and vertically y-units on the respective lines. Let us try a few exercises. Exercise 1: Draw a dot denoting the location of the points (1, 2), ( 5, 7) and (3, 4) in the coordinate system constructed above. 3D-Coordinate System: Using the idea behind the construction above, we take 3, pairwise perpendicular lines L x, L y, L z and denote their intersection as (0, 0, 0). Then we will infer that the point (x, y, z) means move x-units along L x, y-units along L y and z-units along L z. Let us try a few exercises. Exercise 2: Draw a dot which will denote the location of the following three points (1, 2, 3), ( 4, 3, 1) and ( 3, 2, 4). Comment: In our construction of model space we only said take some lines and intersect them, we never specified in which direction do these line travel! Hence, there are infinitely many ways to represents those points. 2
Middle 1-Discussion: We may have all drawn the canonical intersection of the lines x = 0 and y = 0. The immediate question should be, what about any other pair of orthogonal lines? Similarly, why the need for orthogonal lines, can we just use skew lines? Answer: We don t! And that s what s so cool about mathematics. If you want to use skew lines, go ahead, so long as it becomes a coordinate system, you are good to go. Comment: You may now say, Hey, wait a minute. I thought coordinate systems were just things you made by intersecting lines? Well, that s part of it, but there s slightly more. The extra part is that it is a bijection with R R. To understand what this means, we would have to improve our definition of a coordinate system. Basically, what the addition means is that if your coordinate system is represented by (u, v) then (p, q) is at location (u(p), v(q)). The original (p, q) is given by the canonical system i.e the one in which we intersect the lines x = 0, y = 0. We will choose to represent this coordinate system by the pair (x, y). Exercise 3: What is the relation ship between the coordinate system (2x, y) = (u, v) and canonical coordinate system? Exercise 4: Could you think of a coordinate system that shifts all points in the canonical coordinate system by the points (2, 3)? Comment: You can now start thinking about how coordinate systems relate to each other. Such a relationship can be given by a function i.e a rule. For example, in the above we have f(x, y) = (2x, y) = (u, v) i.e the rule say s, to get the coordinates for the system (u, v), we just multiply the canonical horizontal coordinate x by 2 and leave the vertical coordinate in the canonical coordinate system alone. 3
Middle-2 Discussion: The topic here is direction or displacement. We introduce the idea of a vector. A vector v is a quantity which denotes displacement by the following rule, the statement p is displaced from the point q by the vector v means, if you go in the direction v from p, then you arrive at q. In this way, vectors are independent of coordinate system. We want this because there is no time to write down a definition w.r.t each different coordinate system, simply because there are infinitely many! We now need a way of mathematically expressing a vector. Since we have the liberty to define things any which way we like, let us infer that in a coordinate system (u, v) then v = (u 2, v 2 ) (u 1, v 1 ) is the vector which goes from (u 1, v 1 ) to (u 2, v 2 ). We do this since the natural action (in which we define) of this vector gives v + (u 1, v 1 ) = (u 2, v 2 ). Again, the abstraction above is just give, because of the fact that it can be done. We will be working with the canonical system, always! Now since vectors aren t dependent on the initial point in which they emanate from, the vector w = w 1, w 2 moves any point (x, y) to (x + w 1, y + w 2 ). We now have all the ingredients to solve our problem. Exercise 5: We define the addition (subtraction) of vectors just as we do for points (for obvious reasons). Can you then think of where the tail, which results as the addition of two vectors, must lie? Exercise 6: Once you have that, can you thing how to represent the equation of a line using vectors. Comment: The last exercise is asking you to (in mathematical language) parametrize the equation of a line with a given direction. A parametrization of an object is a specific way to describe it which is via parameters and a function or rule as we say. You can now spend some time thinking about how you would wish to formalize this in a meaningful way. 4
Middle-3 Discussion: We now need to come up with a notion of distance in our coordinate system. We start with the distance function on the real line i.e d(x, y) = x y. The immediate question becomes, how do we describe the distance between two points p = (x 0, y 0 ) and q = (x 1, y 1 )? Answer: We use the Pythagorean Theorem! The points p, q are vertices of a right triangle and the distance between them is the hypotenuse. The legs have length x x 0 and y y 0 respectively which implies we should define, d(p, q) = x 1 x 0 2 + y 1 y 0 2 = (x 1 x 0 ) 2 + (y 1 y 0 ) 2 Exercise 7: Show that if p = (x 0, y 0, z 0 ) and q = (x 1, y 1, z 1 ) then they are the vertices of a right triangle and so by Pythagorean theorem, d(p, q) = (x 1 x 0 ) 2 + (y 1 y 0 ) 2 + (z 1 z 0 ) 2 Exercise 8: Use the above to define the length of the position vector associated to a point (x, y, z). From there, think about how to define a sphere of radius R in 3D. It may help to first think about how to define a circle in 2D. Hint: Use the metric! Comment: You now see that sometimes an object can be described through the metric, resulting in a relation on the coordinates in our system. We call such objects loci or if just one object, a locus. It actually turns out that all conic sections are loci! i.e parabolas, circles, hyperbolas, and ellipses. 5
Final Discussion: Now this is where we need to become mathematicians. We have a ton of new information and supposedly we know enough to solve our problem. Again, the goal in mind is to come up with a functional relationship which connects curves in the plane with those on a sphere. 6