SIMO Retiree Blog

(Drew here.) I've done a post on each of A, C and N, and decided it's finally time to finish off with G. My style of geometry is very unconventional, as people have pointed out throughout the years. If I had a nickel for every time I used inversion after I failed to perform an angle chase... I'd probably have over half a dollar by now. So here's a compilation of some of my most unique and/or elegant geometry solutions, and comments on my style. 2022 IMO Mock 2 P1 Let $APBCQ$ be a cyclic pentagon. A point $M$ inside triangle $ABC$ is such that $\angle MAB=\angle MCA$, $\angle MAC=\angle MBA$, and $\angle PMB=\angle QMC=90^\circ$. Prove that $AM$, $BP$, and $CQ$ concur. Being in my inversion addiction phase, after trivial angle chasing didn't work, I immediately attempted inversion. Point $M$ is where all the angle conditions are, so why not invert at $M$? I'll leave you to check that this just returns the same problem statement. At times even that could produce ...

(Guest author Andrew here!) A good exercise for Olympiad students is to prove the following: A real polynomial $f(x)\in \R[x]$ which is everywhere non-negative ( psd , short for positive semi-definite) must be a sum of squares ( sos ).  This is in some sense an algebraic witness to the analytic property of being non-negative, and is an example of what is known as a Positivstellensatz . One might reasonably ask whether this extends to the case of multiple variables, and it turns out the answer is no .  Consider the two variable polynomial $g(x,y)=x^4y^2+x^2y^4-3x^2y^2+1$ (the Motzkin polynomial ). Non-negativity follows from the AM-GM inequality. But how can we show that it isn't a sum of squares? We now introduce an object known as the Newton polytope. Given a polynomial $f$,  To every monomial with non-zero coefficient, assign a point whose $i^{th}$ coordinate is the degree of the $i^{th}$ variable in that monomial, e.g. assign to $x_1^n$ the point $(n, 0)$. The Newton p...

(It's Glen again.) Last week, I wrote about my experience test-solving EGMO 2025/3 . As it turned out, I had some free time at EGMO (the night before Day 1, i.e. after we made the mark scheme for Problem 3, and the morning of Day 1 itself) to try the other problems, so I figured I'd write about them as well. I will try to be a little more brief than last week, so the post doesn't turn into a saga. Problem 1 ( EGMO 2025/1 ) For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$ for all $1 \leqslant i \leqslant m-1$ Disclaimer: this was a pretty embarrassing case of wrong reasoning leading to the right answer. Try to spot where I went wrong! (Hint: there are many such places.) Let's try small cases: $N=3,4$ work. $N=5$ fails. $N=6$ works. If $N$ is odd, then the sequence starts with $N=1,2,\ldots$, so $3|N$. $N=3^k$ works: a...

(Glen here.) Last week, I had the privilege of coordinating for the European Girls' Mathematical Olympiad , which meant that I was involved in creating the marking scheme and grading scripts for one of the problems. But before doing any of that, I first had to test-solve the problem I was assigned. This post will be about my thought process during this test-solve. Diagrams are scanned in from my rough work, so I'm sorry if they look horrible. First, the problem: ( EGMO 2025/3 ) Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic. Initial thoughts Here's a d...