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Revision as of 09:08, 7 September 2024

$\huge\mathcal{JOHN}$

Favorites

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.

My Contributions

Redirects, homepage, talk pages, reverts, and reverted edits are excluded. Most edits are very minor.

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