Here is an interesting geometrical problem given in the recently held Bulgarian National Math Olympiad (NMO) 2024. It is actully a converse version of a well-known fact, which makes it curious. Here it is. Problem (Bulgarian NOM 2024, p2). A triangle is given. The points and are respectively on and such that and Let and Find (Nikolay Nikolov, Aleksander Ivanov) Solution. Answer: Denote . We interpret the configuration as follows see fig. 1. We have an isosceles triangle with and Initially Then, the three points start moving with equal velocities, that is, It ends when reaches ..read more

I want to comment here on a problem I gave to Bulgarian TST for BMO 2024. Here is the story of this problem. About two years ago I came across a problem, (I don’t remember it’s statement – it was about a bipartite graph) and my idea was to delete its edges one by one, so that when one deletes an edge incident to a vertex, one has to delete another edge incident to the same vertex within a certain number of steps, say . It was interesting to estimate the minimum (as a function of the number of vertices ). Of course it depends on the given graph. For example, if you have a complete graph of ve ..read more

Two nice number theory (NT) flavored problems were given at the recent RMM (Romanian Master of Mathematics) competition. The only needed knowledge of NT that one has to have in order to solve them consists of the following two facts. For any prime 1) If then 2) For any integers with we have or Yes, a fifth grader knows it. Still the problems are not so easy. Some people want strict boundaries between different subjects in math competitions. They want for example some “pure” NT problems. Of course, mathematics has different subjects. But they share common methods and technique called Mat ..read more

Here we consider problem 6 of the recent Romanian Master of Mathematics competition – a very difficult problem. Nobody solved it during the competition. Let me point out a very similar problem – RMM Shortlist 2020, A1 – see [3], proposed by the same author. I think it’s of similar difficulty. In my opinion both are not suitable for a high school Olympiad level, although I like them because it’s real math. Maybe they fit for Miklos Schweitzer competition, but on the other hand both problems are known results – the current one by Filaseta and Konyagin – see [2] – so the papers can easily be foun ..read more

This is a continuation of the previous post. I’ll show a more elegant way that will allow us to find all solutions to the functional equation we considered, but it requires a tool called the Krein – Milman theorem. The outline of this post is to first recall the problem, then Krein – Milman theorem and the intuitive motivation behind it. We will finally apply it. Problem (China MO). Let be a function satisfying Prove that In the previous post – see [1] – we proved it using finite differences and making some estimates. Krein – Milman Theorem. Let be a convex set inside some real vector spac ..read more

A year or two ago I saw an interesting problem. The main point was to prove an estimate for a function that has a kind of mean value property – it’s defined on the 2-dimensional lattice grid and its value at each lattice point equals the average of its values on some neighboring knots. It had no solution, just a vague hint. I think it’s still unsolved in AoPS forum, but unfortunately, I lost the link and I lack patience to search that site. It was given at some Chinese Olympiad or some selection test, not sure. It is a tough problem, although the statement is just one short sentence, and it se ..read more

I didn’t expect that Combinatorial Nullstellensatz theorem could be applied to many Math Olympiad problems. In two recent post we discussed some of its applications – see [1], [2]. Yesterday I saw (see [3] ) a solution to a nice problem that used this method. It was given at the Kürschák Math competition, 2019. Here is the statement Problem (Kürschák MO, 2019, p2). Find all families of subsets of such that for any nonempty subset , exactly half of the elements satisfy that is even. I give here two solutions to this problem, which I borrowed from [3]. The first one uses double counting, the ..read more

The problem we are about to comment on was published 15 years ago in Crux Mathematicorum magazine and not long after appeared as Problem 2 in Romania TST 3, 2009. Equilateral triangles have the property of appearing where we least expect them … remember Morley’s trisector theorem. We will deal with exactly this situation. A significant problem, and it’s useful and pleasant to revisit it. Problem. Prove that the circumcircle of a triangle contains exactly 3 points whose Simson lines are tangent to the triangle’s Euler circle and these points are the vertices of an equilateral triangle. Let be ..read more

Here is a problem that I saw in the AoPS site [1]. It was given at the second round of the Math Olympiad held in Kyiv, 2024. I thought that the second round was not the final one, but maybe I was wrong, because this problem is nice, and I don’t think it’s that easy. The first thing that comes to mind upon seeing the problem is invariant/monovariant technique – see [2], [3], [4]. In this case that means finding some quantity that strictly decreases as the process goes on. If you manage to find it, it will mean that the process stops at some point. For example, this quantity could be the number ..read more

It’s time for geometry again … with Nevena Sabeva! Knowing the properties of the harmonic quadrilateral has been a required part of geometry preparation for math competitions in recent years, even at the junior level! I think that problems like the one that follows deserve to be considered in order to improve the geometric skills of eighth-graders. It was given at the 2024 Japan Junior MO Final. Problem (2024 Japan Junior MO Final, p5). Let be a point on arc of circumcircle of acute triangle (not containing ), such that . Let be a reflection point of with respect to line , be a reflectio ..read more