[Math] Unsolved/Least Solved IMO Questions

Question

I recently read this article http://blog.mathfights.com/once-upon-a-time-on-imo/ where the author discusses an IMO problem from 2006 that only about 20 participants out of 600 were able to solve. So this got me wondering, have there ever been any IMO problems where no contestant was able to solve it.If not does any one know of any other problems with only a few number of correct solutions?

Answer 1

Migrated from a comment.

Note the IMO website under results allows you to click on the year where you will find statistics as to how well participants did on the questions in that year.

Also see problem 6 in 2009 - anyone who has seen an answer to this problem will find it hard to see what the difficulty was.

It is normal that the sixth problem in the IMO is the most difficult of the six, and that the third is very difficult too. However the judgment of the examiners is not perfect e.g. see 1996 when problem 5 turned out to be hardest, and problem $3$ in 2007 was also very difficult.

The Art of Problem Solving website has a section devoted to Olympiad problems.

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For $2009/6$ (spoiler)

! If we lose the mathematical fuzz, we can imagine a frog which has to find a safe way to get from $0$ to $N$, with $N$ being the sum of $n$ distinct integers $a_1\lt \dots \lt a_n$. An opponent can mine the interval from $0$ to $N$ in $n-1$ places, which are known to the frog in advance.

!We note that if $n$ places are mined they could be $a_1 \dots a_n$, in which case no safe first leap would be available. In the case of one leap and no mines, the frog is free. We attempt an induction on $n$, by deleting the rightmost mine (imagine the number line from zero), and reducing our leap set by the longest leap, so initially attempting to arrive at $N-a_n$ without using the leap $a_n$ and missing the rightmost mine.

! Now we know we can clear $n-2$ mines with $n-1$ distinct leaps, unless the last of them is at $N-a_n$. Assume first that this is not the case. We have beaten the first $n-1$ mines. If we put the last mine back, it will be the rightmost mine, and if there is a problem at all it will have been hit by a leap $a_r$ with $r\lt n$ so $a_r\lt a_n$. We replace $a_r$ with $a_n$ and have cleared all the mines and can use the remaining leaps in any order.

! In fact the only problem we have is if there is a mine at $N-a_n$ and we can't replace the last leap with $a_n$ because that also lands on a mine. We now know that there are two mines near the right-hand end, and our induction will work if we can dodge them both with two leaps. We know that two leaps $a_i, a_n$ from the right-hand end will clear the mine at $N-a_n$ and since there are $n-1$ choices for $a_i$ and only $n-2$ remaining mines one of these at least is available to make the induction work.