Class discussed in groups the possible kinds of intersection of a (complete) cone with a plane. Six types were found (seven, if circles are taken as a separate type). Conic material will mostly come from BEG, Ogilvy and GTC. LAB 1 introduced several construction methods for conics, include the Director Circle method.
Five definitions of conics are introduced or reviewed. Then connections were made for 2--4 in the case of the parabola and ellipse.
More about equations of conics from definition to completing the square. Optical properties of conics.
Dandelin spheres were used to connect conic definition #1 with definition #3. The visualization of cones and spheres as surfaces or rotation was used to discuss the construction of Dandelin spheres, as related to the construction of circles tangent to 3 lines (incircles and excircles from Math 444). A simple construction of Dandelin spheres for plane sections of cones was also presented. LAB 2 introduced rotated Cartesian (Euclidean) coordinates
Worksheet introducing coordinates relative to rotated coordinate system. This leads to the matrix formula for rotation.
Class worked through a worksheet that investigated visualization of affine transformations f(x) = Ax+b. In this case x can be viewed as affine coordinates with origin b and unit points on axis equal A1+b and A2+b. Lab 3 investigated affine coordinates and also introduced an important figure in a triangle that will lead to barycentric cooridinates and Ceva's theorem.
Before introducing parallel projection, some basic properties of 3-dimensional space were introduced.
Class then moved to an explanation of the properties of parallel projection (using some helpful solar rays). We learned that lines map to lines because the parallel rays through one line form a plane, which intersects the image plane in a line. A similar argument showed that ratios on a line are preserved, i.e., if A, B, C are collinear with images A', B', C', then AC/AB = A'C'/A'B'. Finally it was shown that parallel lines project to parallel lines. This shows that parallel projection is an affine transformation. Not all affine transformations are parallel projections, since a parallel projection preserves length on at least one line (the line of intersection of the domain plane and the image plane -- see the text).
In class it was shown how a mass m at A and a mass n at B have center of mass at C= (1/(m+n))(mA + nB). If we think of n = t, a parameter, then we can also write C(t) = (1-t)A + tB.
Then using the triangle from the homework, it was shown than any point P in a triangle ABC can be located by three, numbers which add up to 1. These are called the barycentric coordinates of P with respect to ABC. If we intersect line AP with line BC at point A', the ratio BA'/A'C = a ratio of two of the barycentric coorinates. If we write this equation for the intersection of BP and CP with the opposite sides of the triangle, we are led to the theorem of Ceva, that the product of these 3 ratios = 1.
More information about barycentric coordinates is on the web at this Berkeley Math Circle site and also a site at UC Davis.
In a triangle ABC, let point P be on BC with BP/BC = 1/2 and let Q be on CA with CQ/CA = 1/4. What are the barycentric coordinates of X = intersection of line AP with line BQ? In other words, what masses at A, B, C will make the center of mass be X? We start with any mass we like at B, say m_B = 1. The we use Archimedes to figure out the mass at C from the location of the center of mass for B and C at P. Then we take m_C and use this to find the mass at A by the center of mass Q. The class in small groups did the example of a triangle ABC each of whose sides are divided into 3 equal parts, BC by A1, A2, CA by B1, B2, AB, by C1, CC2. We set P = intersection of AA2 and BB1 and let Q = intersection of AA1 and BB2 and found barycentric coordinates of P = (1/4,1/4,1/2) and barycentric coordinates of Q = (2/5,2/5,1/5).
The end of the class was devoted to introducing the construction of the Bezier quadratic curve by means of the De Casteljau algorithm. There was a handout from a book by Farin. References on the web can be found at a site at UC Davis and a site at WPI as well as others you can find in a web search. In the afternoon, Lab 4 explored Bezier curves and area of parallelograms via shear transformations.
The class was mostly devoted to Quiz 1. Answers were given after the quiz.
Here are some Review problems on Euclidean transfomations for the upcoming Midterm.
At the beginning of the class, there was more discussion and review of barycentric coordinates, from several points of view. Given a reference triangle ABC, a point P can be located by 3 numbers x, y, z satisfying the relation x + y + z = 1. These numbers appear in many ways in figures involving P and ABC.
- The three ratios of areas (BCP)/(ABC) = x, (CAP)/(ABC) = y, (ABP)/(ABC) = z.
- If the line p1 is the line through P parallel to BC, then if p1 intersects line AB at P1 then BP1/BA = x, etc. The three lines through P parallel to the 3 sides of ABC divide the triangle ABC 4 parallelograms and into 3 triangles similar to ABC with ratios of similitude x, y, z. Each side is divided into 3 segments. For example BC has length a and is divided into segments of length za, xa, ya.
- The "cevian lines" AP, BP, CP, intersect the sides in A', B', C' with ratios such as BA'/A'C = z/y. etc. The product of these 3 side ratios = 1 (see Ceva's Theorem in the text).
In each of these cases above. since there is a relationship among the numbers, if we are given two, we can find the third. For example, if we are given x and z, we can find y = 1 - x - z. If we are given two of the 3 cevian side ratios, then the third ratio is the reciprocal of the product of the other two.
You may want to refer to these Notes on Ratios from Parallel Lines.