SIMO Retiree Blog

It's Glen again. Last week , I mentioned how the relative difficulty levels of the ISL 2018 C4, C5 and C7 felt off to me. Out of curiosity, I looked up the C6 to see where it'd fit, and subsequently ended up solving it in the shower. (Yes, recurring theme, I know.) Anyway, this post will be about my thought process in solving this problem (sans paper, because shower). ( ISL 2018 C6 ) Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board. If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$. If no such pair exists, we write two times the number $0$. Prove that, no matter how we make the choices in (i), operation (ii) will be performed only finitely many times. Preliminary thoughts Here's a way of visualising this problem: we can think of representing our numbers as chips on a number line. Then, operation (i) corresponds to...

(This is Glen.) There hasn't been a post written for this week, so I figured I'd scroll through AoPS and try a problem that looked interesting. This ended up being: ( ISL 2018 C7 ) Consider $2018$ pairwise crossing circles no three of which are concurrent. These circles subdivide the plane into regions bounded by circular edges that meet at vertices . Notice that there are an even number of vertices on each circle. Given the circle, alternately colour the vertices on that circle red and blue. In doing so for each circle, every vertex is coloured twice - once for each of the two circle that cross at that point. If the two colours agree at a vertex, then it is assigned that colour; otherwise, it becomes yellow. Show that, if some circle contains at least $2061$ yellow points, then the vertices of some region are all yellow. In theory, the 2018 shortlist was the one that I had early access to (since I was an Observer in 2019), but I don't remember trying this problem. I was p...

(This is Yan Sheng.) Today's post started with this question: IMC 2015 Q6 : Prove that $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n(n+1)}<2$. The interested reader can pause here to try it out, while I helpfully fill the next few sentences with flavour text to delay spoilers. One of my favourite hobbies is to cosplay as Euler. No, not literally (though I'm sure some of you freaks who read this blog would love to see it), but by trying to manipulate infinite series in all sorts of fun ways. I've already written here before about hand computation and zeta(2) , and this post will be more of the same mucking around. Theodorus's constant For today's problem, seeing that the terms are of order $n^{-3/2}$, the first instinct should be to compare with a telescoping series whose tail sums are of order $n^{-1/2}$. Indeed, it's completely routine to check that$$2\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)-\frac1{\sqrt n(n+1)}=\frac{(\sqrt{n+1}-\sqrt n)^2}...