SIMO Retiree Blog

Glen & Sheldon here. At last, our schedules aligned enough for us to try this year's SMO Open together. Here's an outline of our thought process while solving the problems, featuring some screenshots from our Cocreate whiteboard. Problem 1 In the triangle $ABC$, $\angle B>90^\circ$, the incircle touches the sides $BC$ and $CA$ at $D$ and $E$, respectively. The lines $ED$ and $AB$ intersect at $P$. The incircle of the triangle $AEP$ touches the sides $PE$ and $AP$ at $D_1$ and $E_1$, respectively. The lines $E_1D_1$ and $AE$ intersect at $P_1$. Suppose $P,C,E,B$ are concyclic. Prove that $BE$ is parallel to $PP_1$. Weird-looking problem. The concyclic condition means $\angle ABC = \angle AED$. Maybe it could be less weird if we write it in terms of the angles of the triangle, $\angle ABC = 90^\circ + \frac{\angle ACB}2$. Actually, the equal angles imply $\triangle ABC \sim \triangle AEP$. The parallel condition is equivalent to $\frac{AB}{AE} = \frac{AP}{AP_1}$, and the s...

 (Glen here.) The blog has been somewhat inactive because everyone has been busy, so sorry if we kept you waiting! I got sent this year's SMO Senior Round 2 paper this morning, so I thought I'd give it a shot. I don't have paper with me, so that's an extra challenge, I guess. Problem 1 Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$ draw a line meetin [sic] the side $AB$ at $P$ sand [sic] the side $CD$ at $Q$. Find all the quadrilaterals so that there exists the segment $PQ$ that divides the triangles $ABM$ and $CDM$ into $4$ similar triangles. ??? What on earth is this??? Here's a picture:  Ok, we probably need $\angle APM = \angle MPB = 90^\circ$? Yeah, if not, then each triangle has two distinct angles that add up to $180^\circ$, which is bad. This also tells us that $AB||DC$, as $PQ$ is perpendicular to both. If $\angle APM = 90^\circ$, then the only two possibilities for $APM$ and $BPM$ to be similar are if $MA=MB$...

(Dylan here.) I realised the general content on this blog is pitched at a rather high level. Here's an attempt to balance it.

(David here.) Recently, a problem of mine came out as Q4 on SMO Open 2024. I'll go over how I proposed the problem and the theory behind it, which involves some machine learning ideas (!).

 Hi! It's Glen again. So I was supposed to write a post for next weekend, but I woke up this morning to pictures of this year's SMO Open Round 2 in my messages, and I figured I might as well give them a go. And then I figured that I should just release my solutions to beat everyone else to the chase. Hence, this blog post is a week early.

(This is Lucas.) Hi guys. It's been a long couple weeks (for those who are still active students, at least), let's wind down with some combi. In this post, I'll try to (at least partially) de-mystify colouring/invariants. As always, I'd encourage those of you who are still students to try the questions yourself to get a feel for them before reading the explanations that follow.  Question 1 Each cell of a 726 by 726 grid is initially filled with a "$+1$". In a move, we may flip the signs of a 3 by 3 grid of our choice, less two opposite corners (so we flip 7 cells in each move). Can we reach a state where every cell contains a "$-1$"? Effectively, we affect 7 cells in this formation (or its reflection). Initial Thoughts The setup is practically begging us to apply some colouring. However, the usual suspects don't yield any results, and thinking broadly it's hard to imagine a coloring applying nicely to a set of 7 cells.  Maybe we can pla...