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.
Theorem Pappus (parallel case):
Hexagon? If two pairs of opposite sides are parallel, then the third are also parallel.
Draw equilateral triangles on the sides of any triangle, then the centers of those triangles create an equilateral triangle.
If you have any triangle, draw squares on two of the sides and draw lines from the midpoints of these squares to the midpoint of the side you didn’t draw a square off of. These two lines are equal length.
Wedge product (2D):
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 is: u1v2 - u2v1
u ^ u = 0
u ^ v + v ^ u = 0
ABC collinear <==> A^B + B^C + C^A = 0.
REVISIT TRIPLE PRODUCT OF 3 3D VECTORS.
ABC \in R^2 (and z=1): A B C = 0 <==> ABC collinear.
Where ABC \in R^2: 1/2(A B C) = area of triangle. Put z=1.
A B C = Volume of parallelepiped.
1/6 (A B C) = volume of pyramid with triangle base.
w1 w2 w3
det(u1 u2 u3 ) = can do the diagonal lines + - thing.
v1 v2 v3
Law of Cosines:
Given an angle, theta, between b and a, c^2 = a^2 + b^2 - 2abcos(theta).
Median: line from vertex that intersects midpoint of opposite side. Intersection of medians is centroid.
Orthocenter: draw lines from vertex that make 90 degree with opposite side (altitudes), their intersection is the orthocenter.
Showing perpendicularity: just need to show dot product is zero. Really just use algebra.
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.
A, B, C three non-collinear points, any vector P may be expressed as P = xA + yB + zC, where x + y + z = 1.
Black hole problem: two lines and a point, all intersecting at a point, describe that point of intersection by using Desargues. 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.
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.
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.
Polar triangle of triangle x,y,z is defined as x’y’z’ where x’ = pole of great circle yz, on the same side as x. y’, z’ defined similarly.
T_xS^2 is tangent space to S^2 at vector x. v is unit tangent vector to segment xy if ||v|| = 1, and y=alpha*x + beta*v.
Length of Spherical Segment xy is arccos(x \dot y). Length = angle.
Side lengths of triangle abc are pi - angles of a’b’c’.
Law of Cosines: cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma).
Dual law of Cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cos(c). Note that the angle, gamma, depends on the opposite side, not just the other angles!
Law of sines: sin(a)/sin(alpha) = sin(b)/sin(beta) = sin(c)=sin(gamma)
Theta-lune: two intersecting great circles at angle theta, that area (for both sides) = 4*theta. Could think about 2pi case.
(alpha + beta + gamma) = pi + area
Sum of angles go between pi< and <3pi
H^2 is all vectors u such that Phi(u,u)=-1.
Length of curve d(x,y) = arccosh(-\Phi(x,y))
phi: t—> xcosh(t) + usinh(t) where Phi(u,u)=1, phi stays in H^2 for all t. Also, d(x,s(t))=t.
Law of cosines: cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(gamma).
Dual law of cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cosh(c). Same thing as in S^2: gamma depends on opposite side, not just the other two angles.
Pi(d) = arcsin(1/ch(d)), where d is the vertical distance between a line and its limiting parallel. Pi defines the angle opposite the line.
The area of 3x ideal triangles is finite, and its area is pi.
(alpha + beta + gamma) = pi - area.