In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.

In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

Click for solution This is in my Top 10 of all IMO problems. Not only because I was a competitor then, but also because I was present to the "problem solving session" organized by the hosts and I wached the Czech student Martin Chadek presenting his solution, which made the jury K.O. Here it is: Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$. The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$. We then have: $0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$ a contradiction. Therefore, the sequence cannot have $17$ terms. In order to show that $16$ is the answer, just take 16 real numbers satisfying $s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$. We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$. Thus we found all sequences with the given properties. As far as I know, Martin only solved this problem and got a special prize for it Other solutions are possible, including one using linear algebra. I'll post them if anyone is interested.

Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)

Let $a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). \] Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$

Let $\mathbb{N}$ be the set of positive integers. Let $f$ be a function defined on $\mathbb{N}$, which satisfies the inequality $f(n + 1) > f(f(n))$ for all $n \in \mathbb{N}$. Prove that for any $n$ we have $f(n) = n.$