\documentstyle[12pt]{article}
\pagestyle{empty}
\textwidth=497pt
\textheight=690pt
\voffset=-80pt %-50pt for A4 paper
\oddsidemargin=-34pt
\evensidemargin=-34pt

\begin{document}
%\Large
%\abovedisplayskip=4pt plus 2pt minus 2pt
%\belowdisplayskip=4pt plus 2pt minus 3pt

\newcounter{zadanie}
\setcounter{zadanie}{0}

\arraycolsep=2pt

\begin{center}
{\bf 49-th Mathematical Olympiad in Poland}\\
Final Round, April 24--25, 1998
\end{center}

\begin{list}{\bf\arabic{zadanie}.}{\usecounter{zadanie}}

\centerline{\bf First Day}

\item 
Find all integers $(a,b,c,x,y,z)$ which satisfy the system of equations
$$\left\{
\begin{array}{rcl}
a+b+c & = & xyz\\
x+y+z & = & abc
\end{array} \right. $$
and the conditions $a \geq b\geq c \geq 1\,, x \geq y\geq z \geq 1\,.$ 

\item  
The Fibonacci sequence $(F_n)$ is given by:
$$ F_0=F_1=1\,,\quad F_{n+2}=F_{n+1}+F_n \quad\mbox{{\rm for}}\; n=0,1,2\ldots .$$
Determine all pairs $(k, m)$ of integers, with $m > k \geq 0$, for which
the sequence $(x_n)$ defined by
$$x_0=\frac{F_k}{F_m}\,,\quad 
x_{n+1}= \left\{
\begin{array}{ll}
{\displaystyle \frac{2x_n-1}{1-x_n}} &\;\mbox{{\rm for}}\; x_n\neq 1\,,\\[2ex]
1                    &\;\mbox{{\rm for}}\; x_n= 1\
\end{array} \right.
\quad (n=0,1,2\ldots )$$
contains the number $1$.

\item 
The convex pentagon $ABCDE$ is the base of the pyramid $ABCDES$. 
A plane intersects the edges $SA$, $SB$, $SC$, $SD$, $SE$ in points $A'$,
$B'$, $C'$, $D'$, $E'$ respectively which differ from the vertices of the
pyramid. Prove that the intersection points of the diagonals of the
quadrangles $ABB'A'$, $BCC'B'$, $CDD'C'$, $DEE'D'$, $EAA'E'$ are coplanar.

\vspace{20pt}

\centerline{\bf Second Day} 

\item
Prove that the sequence $(a_n)$ defined by
$$a_1=1\,,\quad a_n=a_{n-1}+a_{[n/2]}\quad\mbox{{\rm for}}\; n=2,3,4\ldots $$
contains infinitely many integers divisible by $7$.\\
{\it Note:} $[n/2]$ denotes the biggest integer not bigger
than $n/2$.

\item 
Points $D$, $E$ lie on the side $AB$ of the triangle $ABC$ and satisfy
$$\frac{AD}{DB}\cdot\frac{AE}{EB}=\left(\frac{AC}{CB}\right)^2\,.$$
Prove that $\angle ACD=\angle BCE$.

\item
Consider unit squares in the plane whose vertices have integer coordinates.
Let $S$ be the chessboard which contains all unit squares 
lying entirely inside the circle
$x^2+y^2\leq 1998^2$. In all the squares of the chessboard $S$ we write $+1$.
A move consists of reversing the signs of a row or of a column or of a diagonal
of the chessboard $S$. 
(Diagonals are formed by the squares of $S$ whose centers lie on the 
 lines intersecting rows and columns under the angle $45^{\circ}$). 
The goal is to reach $-1$ in exactly one unit square of $S$. 
Find out whether it is possible.

\end{list}

\end{document}
