Here I will post some interesting math things. This is mostly to serve as my own notebook and thought archive, so there's a good chance a lot of the content is wrong. Don't expect much from here for the first few years, unfortunately. Moving here from AoPS: https://artofproblemsolving.com/community/c1216112_blogologolog, will be a while before I set up latex and mathjax properly.
Counting the general number of tangents to a degree-n curve: through cursed moving points
This is a continuation of the first blog post and is slightly deranged.
Recall that a moving point in the projective plane can be given by a regular map from P1 into P2. A geometric condition (i.e. collinearity) can be given by a function in the function field of P2, which pulls back through some moving point, to a function on P1 with certain degree (defined as its number of zeroes).
This is a technique common in olympiad geometry, where degree+1 "cases" can be checked to prove the function on P1 is always zero.
For a degree-n smooth algebraic curve \(\mathcal{K}\), we can do the same thing.
However the annoying thing is due to Bezout's theorem, a degree-1 condition in P2 will give a degree-n condition on \(\mathcal{K}\) (since the zero set of the condition will intersect the curve n times). Now the pullback through a "moving point" (a regular map from K to P2) will give a rational function \(\varphi\) on the function field of K with degree \( nd \).
As an example, let P be a point in the plane and let's count the number of points A with the tangent at A passing through P. This tangent line can be given by the (n-1)th polar of A in K, which as a map from P2 into dual of P2 is a degree n-1 map. A line in P2 passing through a point in P2 is a degree 1 function. Therefore the function on \(\mathcal{K}\) corresponding to the condition has degree \(n(n-1)(1) \), so we are done.
This is all rigorizable with effective and principal divisors but whatever. oh and btw a linear system of a degree-2 divisor somehow characterizes all pairs of "involutions" on a curve but there's still a lot to figure out. (i don't actually know any intersection theory so I assume for smooth enough curves linear and rational equivalence are basically the same thing and rational equivalence of divisors is basically just two evaluations of the same moving point.)
A triangle is a very ugly cubic.
Many triangle geometry theorems are really just theorems about cubics.
Define the "second polar" as the first polar of the first polar, defined in the previous post. Then we have the following:
The first polar of a triangle wrt. a point P is the circumconic with perspector at P.
The second polar of triangle wrt. a point P is its trilinear polar.
These aren't hard to show geometrically, e.g. by cross-ratio properties. But Clairaut's theorem lets us swap the partial derivatives, getting the following property:
Any point Q on the circumconic with perspector at P has trilinear polar of Q passing through P.
This immediately proves how the trilinear polar of any point on the circumcircle passes through the symmedian point.
Counting the general number of tangents to a degree-n projective curve
Suppose you have a degree-n smooth algebraic curve \(\mathcal{K}\) and a sufficiently general point P in the complex projective plane. We will prove the following theorem in a very overkill way.
Theorem: There are generally (n)*(n-1) tangents to the curve which can be drawn through P.
The standard proof .
Let \(\frac{\partial K}{\partial x}, \frac{\partial K}{\partial y}, \frac{\partial K}{\partial z}\) be the three partial derivatives of \(\mathcal{K}\). Consider the first polar curve of P wrt. this algebraic curve, defined as \[P_x\frac{\partial K}{\partial x}+P_y\frac{\partial K}{\partial y}+P_z\frac{\partial K}{\partial z}=0\].
Since this curve is of degree (n-1) and intersections with it and the original curve correspond to tangent lines through P, we have n*(n-1) tangents by Bezout's theorem.
A proof with Riemann-Hurwitz
Define a rational map from the curve to the Riemann sphere via projection through P onto some unspecified line. Then, we will have each tangent (up to multiplicity) is actually a degree-2 ramification point of this map, so we can count it with Riemann-Hurwitz!
We have by Riemann-Hurwitz, where chi(K) is the Euler characteristic of our curve and e_P is the number of preimages of a point P: \[\chi(\mathcal{K})=N \cdot 2 - \sum_{P \in K} (e_P-1).\].
By the genus-degree formula and Euler characteristic formula, we will have
\[g(\mathcal{K}) = \frac{(N-1)(N-2)}{2} \implies \chi(\mathcal{K}) = 2-(N-1)(N-2) =N \cdot 2 - \sum_{P \in \mathcal{K}} e_P-1.\]
\[\sum_{P \in \mathcal{K}} e_P-1 = 2-(N-1)(N-2) -N \cdot 2 = N(N-1) .\]