If C is a point on the line AB, then
C = (1-t)A + tB.
If A, B, C are collinear, then real numbers x, y, z not all zero such that (bidirectional)
x+y+z=0 and xA + yB + zC = 0
Best said: if a line cuts a triangle (you have to extend one of its sides), then the product of the ratio of each of the cuts = -1.
Best said: In a triangle, if lines are drawn from each of the edges to the opposite side and the lines are concurrent, then the product of the ratios of each of the cuts is 1. The centroid is an instance of Ceva, where each of the ratios in 1 because it’s (1/2 / 1/2 ). Converse is also true. If product is 1, lines are concurrent.
Desargues (parallel case):
If the sides of two triangles are parallel, then the joins of the corresponding vertices are either concurrent or parallel. This point of concurrence is the center of perspective.
Two triangles, distinct vertices, if they’re perspective with respect to a point, they’re perspective with respect to a line.
Two triangles, point inside of one. If the three lines connecting that point to each vertex are parallel to the sides of the other triangle, then there exists a point in the other triangle that has the same property.
Dot (Inner) Product:
A . B is equal to |A| |B| cos(angle between). Which is essentially the product of the projection of one vector onto the other.
Exterior Product (wedge product):
Length: |u| * |v| * sin (angle)
Geometrically: it’s the area of the quadrilateral created by the two vectors.
Line is perp to uv plane.
Orientation Right hand rule.
Algebraic, u ^ v :
i j k
det(u1 u2 u3 ) = can do the diagonal lines + - thing.
v1 v2 v3
REVISIT TRIPLE PRODUCT OF 3 3D VECTORS.
Also written wedge (^). Is equal to twice the area of the triangle formed by the two vectors.
A ^ B ^ C = 1/2 the area of the triangle ABC in the Z=1 plane.
u ^ u = 0
u ^ v + v ^ u = 0
if u,v,w in R^2, then u ^ v ^ w = 0.
A,B,C collinear (3 points in the plane) <==> A ^ B + B^C + C^A = 0.
So essentially what you do is when you have three vectors
Need to get the whole collinearity and determinant down.
Law of Cosines:
Given an angle, theta, between b and a, c^2 = a^2 + b^2 - 2abcos(theta).
If you know there’s a certain angle between two vectors, then you can write one of the vectors as R(other). You can also write one vector in terms on another two, then do a rotation on that vector, and then split up the rotations to be rotations of the the two that made it. So, say you had R(u + v / 2), well that’s = (R(u) + R(v))/2. And say you knew something about R(u) and R(v), then you’ve just made a little jump.
Theorem Pappus (parallel case):
Hexagon? If two pairs of opposite sides are parallel, then the third are also parallel. Consider the exterior product for a proof.
A, B, C three non-collinear points, any vector P may be expressed as P = xA + yB + zC, where x + y + z = 1.
Tool: divide by sum of coefficients.
Generally, for proceeding, it’s good to write down what you know, and perhaps a diagram and look for what you think will be true. Then, you know you want to get certain terms into an equation at the end, so you try to shove those terms into one equation, and play around with the algebra.
Black hole problem: two lines and a point, all intersecting at a point, describe that point of intersection by using Disargues. Short version: create two triangles with these lines, show that they’re perspective with respect to a line and therefore they are perspective with respect to a point. Essentially what you do is create one full triangle with your lines and then another side of a triangle. You show that these two have an axis of perspective, MAY WANT TO REVISIT.
Altitudes of triangles are concurrent proof: Draw the triangle, draw two of the altitudes and the point that they intersect at, and then you’ll say “let’s show that the line that goes through this intersection from the remaining vertex is actually an altitude”. So you just need to show that some dot product is zero. Should be algebra.
REVISIT THEOREMS FROM CLASS I MISSED.
feuerbach’s circle theorem: Just take a brief look.
Centroid proof: So like, for the centroid of a triangle being the average of the three sides, you connect the edges to the midpoint of their opposite side, and you notice that the intersection seems to divide these lines in a ration of 2:1, so you write the midpoint as two of the sides and then you’re tasked with incorporating the last side, so you do so by considering this point on the line with a ratio of 2:1, and then you see that the algebra simplifies to sum of all the sides over three.
Menelaus Proof: Use Theorem 2.
Medians of a triangle are concurrent proof: Really, you just observe the median for one of the sides, and you consider the point along the anchor that is 2:1 farther from the vertex it drops from. Then you’ll see that because it bisects the opposite side, you can write it as sum of halves of the other two sides. It’s algebra from there. This is the triangle’s centroid.