Hello! You might be wondering (if you don't know me already) why there's a section all on math. Here's the catch: I'm sorry to upset you, but I'm a geo main.

Basically, I'm going to give you a remote gallery walk (not really) of some of the most intriguing questions I've done. This is quite a lot, and most of them are geo, so be prepared, and if you're anti-geo you might want to exit this page fast. Like, really fast.

Right now, I'm kinda low on the AMC 10 sheets, but I'm inching towards 2024 at a speed of a snail chopped in half. Still, it's wiser to think more than to speed-run, so I normally just try to think more deeply on the questions that catch me in the fish net. Here are some examples of questions I did, and they'll probably include diagrams.

A bug travels from A to B along the segments in the hexagonal lattice pictured above. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?

To solve this, we can split the diagram into 5 sectors, knowing that each arrow not filled in is "restricted" in a way. I drew vertical partitions along the arrow verticals that were only filled, and found the pathWhat an AFK Joker Does All Day possibilities for each section in turn. After connecting each sector based on the answers I recieved, the problem resulted in an answer of 2400. This one was strangely similar to a programming question.

Let $T_1$ be a triangle with sides $2011, 2012,$ and $2013$. For $n \ge 1$, if $T_n = \triangle ABC$ and $D, E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB, BC$ and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $( T_n )$?

Basically, you have to find a pattern with the sides by calculating the next triangle sides in the sequence using the incircle and its tangency. You'll eventually discover that the sequence continues with a very convenient pattern. In order to have a nonexisting triangle, it has to not fit the triangle inequality. If you continue working with the pattern of the sequence, you come up with an answer of $\frac{1509}{128}$.