When I want to do a little bit of math, but not too much math, I answer questions on Math StackExchange (and occasionally ask them, too). Some highlights, judged entirely by how fun I think they are.
Did you know that you can 4-color a map of the US and use the color green only twice?
In general, coloring problems are often hard to solve. But this one turns out to be easy to answer by hand, and as a bonus, we get to draw a nice picture.
Several teams of varying size play team-on-team matches. When one team beats another, the leader of the losing team joins the winners, and the rest of the losing team splits into 1-person teams. Starting from n 1-person teams, how many games does it take on average to form one n-person team?
Here is a solution using Markov chains.
I am a fan of this style of Markov chain argument, where we get an answer without doing almost any calculations.
Is it possible to decompose K_{12,12} into four edge-disjoint copies of 3(K_{4,4}-I)?
Okay, I admit that the description sounds rather technical, and might not be exciting to someone who isn't a graph theorist. Personally, I like this question because we can solve it using simulated annealing, and this was my first attempt at using Mathematica to do that sort of thing.
There are two related questions found here (also answered by me) and here (not answered by me).
Start with n little heaps of sand, arranged in a row. Then, repeat the following: randomly pick two adjacent heaps, and bring them together into one bigger heap.
Eventually, we'll end up with all the sand in one large pile, but we can build a binary tree that represents the order we merged the heaps in. This concept was new to me, but apparently this is called a fusion binary tree.
What I answered here was the question: what is the distribution of the distance, in the binary tree, between two leaves representing heaps that were initially adjacent?
My answer was later cited in a physics paper, which is pretty much the coolest thing that can happen when you answer a question on StackExchange.
The Petersen graph (Wikipedia link) has a very famous drawing with a five-pointed star inside a pentagon, joined by a matching. But there are other interesting ways to draw it!
We can embed it in 3 dimensions with non-crossing unit line segments as the edges. Actually, there are lots of ways to do it. But I am happy with the very symmetric version I found.
We can draw it in a periodic lattice, relying on the idea that the Petersen graph can be drawn without crossings on the torus. There are lots of symmetric solutions here, mostly not found by me, though I did draw some pictures.
This is a question I asked myself.
Some Pythagorean triples a^{2}+b^{2}=c^{2} also satisfy φ(a^{2})+φ(b^{2})=φ(c^{2}). Is it just the ones we expect, or are there more?
This question has its origin in an embarrassing mistake of mine. Initially, I thought the question had a very nice solution, so I put it on Mathcamp's weekly Team Problem Solving competition. Turns out, my solution was wrong. I still wanted to know how to solve the problem, so I asked MSE. But apparently, MSE doesn't know the answer either. Do you?
This is a question I saw on MSE, but could not answer.
Put n dots in a row at the points (1,0), (2,0), (3,0), ..., (n,0), and n more dots in a row at the points (1,1), (2,1), (3,1), ..., (n,1). Then match the top row of dots to the bottom row however you like, so that each dot is connected by a straight line segment to a dot in the other row. How many times can you make the segments intersect?
OEIS sequence A332774 has the answer up to n=13. I wrote some code to compute the next few terms: the sequence definitely continues 79, 90, 103, 117 and the next two terms are at least 133 and 148.
(Update: as of September 2022, someone put up three of these terms, but for some reason not the fourth, up on the OEIS on my behalf.)
If those last two terms are correct, they obey the following pattern. The differences between consecutive terms in the sequence are 1, 3, 3, 5, 4, 6, 7, 9, 8, 10, 10, 12, 11, 13, 14, 16, 15. The differences between those differences are 2, 0, 2, -1, 2, 1, 2, -1, 2, 0, 2, -1, 2, 1, 2, -1. And here, it looks like the block "2, 0, 2, -1, 2, 1, 2, -1" is repeated twice; could it be repeated forever?
Maybe you'll figure it out :)