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.
A triangle is a very ugly cubic.
Idea from Li4
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 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.
Thm. 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
\[g(\mathcal{K}) = \frac{(N-1)(N-2)}{2} \implies \chi(\mathcal{K}) = 2-(N-1)(N-2) =N \cdot 2 - \sum_{P \in \mathcal{K}} e_P-1.\]
\[\sum_{P \in \mathcal{K}} e_P-1 = 2-(N-1)(N-2) -N \cdot 2 = N(N-1) .\]