SIMO Retiree Blog

(Shi Cheng here.) Hi, it's been a long time since I've been involved in the Maths Olympiad sphere, mostly because of uni, but recently I found out about this blog and I wanted to contribute. This is not a post about Math, but rather, I wanted to share some important mistakes and lessons I learnt from my time as a Math Olympiad student, so that hopefully you will not make them yourself and can maximise your potential as a student. Now, it is possible some of these lessons may already be obvious to you, and perhaps you are judging me for not having realised them earlier. Maybe I should have. But there could be others who may not have realised them, so hopefully this post will stop them from falling into the same traps I did, and for those who are already aware of them, this could serve as a reinforcement. Focus on the big idea(s) of a question. When I was still a Math Olympiad student, I remembered the way I would read solutions is by examining each line, making sure I understo...

 (Drew here.) Three weeks ago, we had a post about visualising combi, and last week, it was about the anatomy of an FE. Out of sheer coincidence, here I am now talking about visualising FEs! A nice way to avoid the notoriously grindy nature of FEs and favour a combi main like me :) Let's begin with a familiar one: Cauchy's functional equation. (Positive real Cauchy) Find all $f:\mathbb R^+\to\mathbb R^+$ satisfying $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R^+$. The solutions to this are $f(x)=cx$ for $c\in\mathbb R^+$, which are all linear functions. For any integer $n$, we can calculate $$\begin{align*}f(nx)&=f((n-1)x+x)\\&=f((n-1)x)+f(x)\\&=f((n-2)x)+f(x)+f(x)\\&=\dots\\&=\underbrace{f(x)+f(x)+\dots+f(x)}_n\\&=nf(x).\end{align*}$$ So given a rational number $r=\frac mn$ we have that \[f(rx)=f\left(\frac mnx\right)=mf\left(\frac xn\right)=\frac mnf\left(\frac xn\cdot n\right)=\frac mnf(x)=rf(x).\] Now back to the main topic, let's graph this functi...

Etienne here. Today I want to talk about Problem 6 of the IMO, and my thought process while solving it. The official IMO document provides 7 solutions to this problem, and to me all of them feel strange and unmotivatable at first read (especially the construction!). So I hope to demystify this problem and present it in the simplest and most intuitive way possible. The Problem P6. Let $\mathbb{Q}$ be the set of rational numbers. A function $f: \mathbb{Q} \to \mathbb{Q}$ is called aquaesulian if the following property holds: for every $x,y \in \mathbb{Q}$, $$ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). $$ Show that there exists an integer $c$ such that for any aquaesulian function $f$ there are at most $c$ different rational numbers of the form $f(r) + f(-r)$ for some rational number $r$, and find the smallest possible value of $c$. Solution This problem statement should jump out to anyone as unusual. First, for each pair $(x,y)$, $f$ can choose between two equat...

(This week: a special guest post from Andrew, a member of the IMO 2024 PSC!) This post is going to be a discussion of everything that could possibly be construed as related to Turbo , i.e. IMO 2024/5, from my perspective. It will be long. Apparently I will also be up against basically all of the Singaporean old people, so this should be fun. (Spoilers for IMO 2024/5. The opinions represented here are those of an individual and not necessarily the opinion shared by all or even a majority of PSC members.)  Turbo (IMO 2024/5) Turbo the snail plays a game on a board with $2024$ rows and $2023$ columns. There are hidden monsters in $2022$ of the cells. Initially, Turbo does not know where any of the monsters are, but he knows that there is exactly one monster in each row except the first row and the last row, and that each column contains at most one monster. Turbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in t...