This year the International Mathematical Olympiad took place in Slovenia. I participated as one of the organizers (problem selection and coordination). It was probably one of the busiest and most exciting times of my life, particularly on the last day of coordination (grading) when just a couple of hours before the final meeting there were still about a dozen countries unfinished because of misunderstandings about the grading criteria. This is something that always happens, as it is quite hard to grade partially correct solutions written by *very* smart students in langauges only few people at the oympiad understand. Everything worked out well in the end, mostly because I had a wonderful team of graders, although we could have handeled a few details better. It seems though that ours was one of the best organized olympiads. After the olympiad I wanted to just sleep for two days. I was mentally drained.

This year’s selection of 6 problems was characteristic in two ways: there was one and a half problem from each of the four traditional areas (algebra, combinatorics, geometry, number theory) since problems 2 and 5 were a mix of two areas. The second characteristic was that this was one of the hardest math olympiads. The international jury, which selects the problems, selected the hardest available problem as problem 6, selected a very hard inequality as problem 3, and then it turned out that the “medium” problem 5 was also quite hard for the contestants. You should definitely try to solve some of the problems in your favorite language. Yes, every contestant competes in their mother tounge. Mind you, these problems can be solved with only advanced high-school math knowledge. In fact, they are designed so that fancy math theorems will not help you! Problem 6 caught the most attention and sparked the most ideas and conversations. It is this:

Assign to each side `b` of a convex polygon `P` the maximum area of a triangle that has `b` as a side and is contained in `P`. Show that the sum of the areas assigned to the sides of `P` is at least twice the area of `P`.

I tried to solve some of the problems with Mathematica and symbolic computation. It seems that problem 1 could be solved, but I did not have enough time to go through with it. All other problems stand no chance of being solved by present symbolic computation and theorem proving programs. Well, at least I would be amazed if someone showed my a computer solution to one of them.