SIMO Retiree Blog

Sometimes, we forget how far we’ve come. This blog is now a little more than a year old! We’ve done well to keep up our weekly publishing streak, with 52 posts this year (and 17 more in the tail of 2023). Let’s celebrate the turn of the year with a consolidation of our writings.  Algebra Choo Ray (8) broke down Vieta Jumping, showing us that beyond identifying the key idea, it was also important to double down on the analysis, and draw connections between features of the problem and of the solution. Wee Kean (14) gave a concise introduction to linear algebra, and explained a geometric (Combi-Nullstellensatz-flavoured) application via dimension counting of a space of polynomials.  Dylan (39) showcased ISL 2023 A7, exposing the geometry hidden behind a discrete averaging of the square root function. Gabriel (46) explained his approach towards a piecewise inequality problem: investigating small cases motivated an induction proof, with terms grouped by size. Drew (35) demonstrat...

(David here!) Putnam 2024 was released recently, so I decided to try a bunch of the problems while on the plane ride from NYC to Chicago. (A3) Let $T$ be a (uniformly) random bijection $$ T: \{1,\,2,\,3\}\times\{1,\,2,\,\ldots,\,2024\}\to\{1,\,2,\,\ldots,\,6072\} $$ conditioned on $T$ being increasing in both arguments. Does there exist indices $(a,b)$ and $(c,d)$ such that $$1/3 \le P(T(a,b)\lt T(c,d)) \le 2/3?$$ I took some liberties with reformulating it probabilistically here. My initial reaction was that this was probably true. My first idea was that if $(a,b) \lneq (c,d)$, then the probability is 0, and flipped around it is 1. So maybe we can take some intermediate sequence of indices that slowly switches them? Didn't get very far with this. Well, thinking slightly meta, the problem can't possibly be that hard - there is no way we're figuring out exactly how many $T$ satisfy $T(a,b) \lt T(c,d)$, so maybe we should be looking for bijections (or partial bijecti...

 (This is Glen.) At some point, Ker Yang wrote a post on Reim's theorem, which he used to solve two past year IMO problems. I remember commenting to him that I had solved neither of them with Reim (and no, I did not bash them). Later on, I tried reconstructing my solution to IMO 2017/4, and I noticed something interesting that made me find another (slightly weird) solution that (I think) isn't on AoPS. So that's what I'll be writing about today. First, the problem: ( IMO 2017/4 ) Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$...

(This is Yan Sheng.) In my mind, J. Kenji López-Alt is the greatest nerdsniper chef: he has made claims about cooking methods and techniques, based on experimental evidence, which have become the inspiration for later theoretical research. For instance, on the subject of grilling meat , he writes that "...flipping steak repeatedly during cooking can result in a cooking time about 30% faster than flipping only once"; this was confirmed by a model of Thiffeault (2022) (here are some slides from a talk ). Today's post, however, is about onions: Question : What is the optimal way to cut an onion such that the size variation among the diced pieces is minimised? After removing the ends and halving the onion, we can view it as approximately a half-cylinder with concentric layers. This reduces the dimension of the problem, so now we're cutting concentric semicircles: (In the following diagrams, I'm viewing the arcs themselves as the layers, and not the spaces bet...