Careful readers of my blog might have heard about plans to have a second edition of Napkin out by the end of February. As it turns out I was overly ambitious, and (seeing that I am spending the next week in Romania) I am not going to make my self-imposed goal. Nonetheless, since I did finish a decent chunk of what I hoped to do, I decided the perfect is the enemy of the good and that I should at least put up what I have so far.

So since this is someplace between version 1 and the (hopefully eventually) version 2, it seems appropriate to call it version 1.5. The biggest changes include a complete rewrite of the algebraic geometry chapters, new parts on real analysis and measure theory, and a reorganization of many of the earlier chapters like group theory and topology, with more examples and problems …

There’s a recent working paper by economists Ruchir Agarwal and Patrick Gaule which I think would be of much interest to this readership: a systematic study of IMO performance versus success as a mathematician later on.

Here is a link to the working paper.

Despite the click-baity title and dreamy introduction about the Millennium Prizes, the rest of the paper is fascinating, and the figures section is a gold mine. Here are two that stood out to me:

There’s also one really nice idea they had, which was to investigate the effect of getting one point less than a gold medal, versus getting exactly a gold medal. This is a pretty clever way to account for the effect of the prestige of the IMO, since “IMO gold” sounds so much better on a CV than “IMO silver” even …

In the previous post we defined $p$-adic numbers. This post will state (mostly without proof) some more surprising results about continuous functions $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$. Then we give the famous proof of the Skolem-Mahler-Lech theorem using $p$-adic analysis.

1. Digression on $\mathbb C_p$

Before I go on, I want to mention that $\mathbb Q_p$ is not algebraically closed. So, we can take its algebraic closure $\overline{\mathbb Q_p}$ — but this field is now no longer complete (in the topological sense). However, we can then take the completion of this space to obtain $\mathbb C_p$. In general, completing an algebraically closed field remains algebraically closed, and so there is a larger space $\mathbb …$

I think this post is more than two years late in coming, but anywhow…

This post introduces the $p$-adic integers $\mathbb Z_p$, and the $p$-adic numbers $\mathbb Q_p$. The one-sentence description is that these are “integers/rationals carrying full mod $p^e$ information” (and only that information).

The first four sections will cover the founding definitions culminating in a short solution to a USA TST problem.

In this whole post, $p$ is always a prime. Much of this is based off of Chapter 3A from Straight from the Book.

1. Motivation

Before really telling you what $\mathbb Z_p$ and $\mathbb Q_p$ are, let me tell you what you might expect them to do.

In elementary/olympiad number theory, we’re already well-familiar …

I’ve added a new Euclidean geometry handout, Constructing Diagrams, to my webpage.

Some of the stuff covered in this handout:

• Advice for constructing the triangle centers (hint: circumcenter goes first)
• An example of how to rearrange the conditions of a problem and draw a diagram out-of-order
• Some mechanical suggestions such as dealing with phantom points
• Some examples of computer-generated figures

Enjoy.

1. Synopsis

One of the major headaches of using complex numbers in olympiad geometry problems is dealing with square roots. In particular, it is nontrivial to express the incenter of a triangle inscribed in the unit circle in terms of its vertices.

The following lemma is the standard way to set up the arc midpoints of a triangle. It appears for example as part (a) of Lemma 6.23.

Theorem 1 (Arc midpoint setup for a triangle)

Let $ABC$ be a triangle with circumcircle $\Gamma$ and let $M_A$, $M_B$, $M_C$ denote the arc midpoints of $\widehat{BC}$ opposite $A$, $\widehat{CA}$ opposite $B$, $\widehat{AB}$ opposite $C$.

Suppose we view $\Gamma …$

Median Putnam contestants, willing to devote one of the last Saturdays before final exams to a math test, are likely to receive an advanced degree in the sciences. It is counterproductive on many levels to leave them feeling like total idiots.

— Bruce Reznick, “Some Thoughts on Writing for the Putnam”

Last February I made a big public apology for having caused one of the biggest scoring errors in HMMT history, causing a lot of changes to the list of top individual students. Pleasantly, I got some nice emails from coaches who reminded me that most students and teams do not place highly in the tournament, and at the end of the day the most important thing is that the contestants enjoyed the tournament.

So now I decided I have to apologize for 2016, too.

The story this time is that I inadvertently sent over 100 students home having solved two …

I recently had a combinatorics paper appear in the EJC. In this post I want to brag a bit by telling the “story” of this paper: what motivated it, how I found the conjecture that I originally did, and the process that eventually led me to the proof, and so on.

This work was part of the Duluth REU 2017, and I thank Joe Gallian for suggesting the problem.

1. Background

Let me begin by formulating the problem as it was given to me. First, here is the definition and notation for a “block-ascending” permutation.

Definition 1. For nonnegative integers $a_1$, …, $a_n$ an $(a_1, \dots, a_n)$-ascending permutation is a permutation on $\{1, 2, \dots, a_1 + \dots + a_n\}$ whose descent set is …

I wanted to quickly write this proof up, complete with pictures, so that I won’t forget it again. In this post I’ll give a combinatorial proof (due to Joyal) of the following:

Theorem 1 (Cayley’s Formula)

The number of trees on $n$ labeled vertices is $n^{n-2}$.

Proof: We are going to construct a bijection between

• Functions $\{1, 2, \dots, n\} \rightarrow \{1, 2, \dots, n\}$ (of which there are $n^n$) and
• Trees on $\{1, 2, \dots, n\}$ with two distinguished nodes $A$ and $B$ (possibly $A=B$).

This will imply the answer.

Let’s look at the first piece of data. We can visualize it as $\mathrm{n\; \dots }$

I’m reading through Primes of the Form $x^2+ny^2$, by David Cox (it’s good!). Here are the high-level notes I took on the first chapter, which is about the theory of quadratic forms.

(Meta point re blog: I’m probably going to start posting more and more of these more high-level notes/sketches on this blog on topics that I’ve been just learning. Up til now I’ve been mostly only posting things that I understand well and for which I have a very polished exposition. But the perfect is the enemy of the good here; given that I’m taking these notes for my own sake, I may as well share them to help others.)

1. Overview

Definition 1. For us a quadratic form is a polynomial $Q=Q(x,y)=a{x}^{\mathrm{2\; \dots }}$