User:Johnxyz1

$\huge\mathcal{JOHN}$

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Favorite topic: $\text{Counting \& Probability}$ for which I am reading AOPS intermediate book on

Favorite color: $\text{\textcolor{green}{Green}}$

Favorite software: $\mathit{Microsoft}\ \text{Excel}$

Favorite Typesetting Software: $\text{\LaTeX}$

Favorite Operating System: Linux (although I am rarely on one)

$\Large\text{\bfseries\LaTeX}$ typesetting

Below are some stuff I am doing to practice $\text{\LaTeX}$ . That does not mean I know all of it (actually the only ones I do not know yet is the cubic one and the $e^{i\pi}$ one)

$\text{If }ax^2+bx+c=0\text{, then }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $e^{i\pi}+1=0$ $\sum_{x=1}^{\infty} \frac{1}{x}=2$ $\begin{align*} x &= \sqrt[3]{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right) + \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} - \frac{b^2}{9a^2}\right)^3}} \\ & + \sqrt[3]{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right) - \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} - \frac{b^2}{9a^2}\right)^3}} - \frac{b}{3a} \\ &\text{(I copied it from another website but I typeset it myself;}\\ &\text{I am pretty sure those are not copyrightable. I still need \textit{years} to even understand this.)}\\ &\text{This is the cubic formula, although it is \textit{rarely} actually used and memorized a lot. The equation is}\\ &ax^3+bx^2+cx+d=0 \end{align*}$

Source code for equations:

https://1drv.ms/t/c/c49430eefdbfaa19/EQw12iwklslElg9_nCMh0f0BVthxSSl-BOJAwsXtGbbhPg?e=1LfZJm

Math Stuff (for my own)

Complementary casework example: https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_25

Representing Actions as Permutations

The idea is that if you must do a fixed number of operations of multiple types, you can make those operations letters, and permutate them. For example, if you have a grid of $4\times 6$ and you want to walk from one corner to the opposite one, WLOG you need to go up $4$ times and right $6$ times. You can do that in any order, so basically you are arranging

   UUUURRRRRR

which simplifies the problem.

Example: 2024 AMC 8 Problems/Problem 13. In this problem you can treat going up as $U$ and going down as $D$. Since you have to end up on the ground in $6$ steps you have $3$ U's and $3$ D's; same as above. There are some special cases --- begin with U end with D and invalid stuff.

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