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\begin{center}
{\bf 50-th Mathematical Olympiad in Poland}\\
Final Round, April 14--15, 1999
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\begin{list}{\bf\arabic{zadanie}.}{\usecounter{zadanie}}

\centerline{\bf First Day}

\item 
Point $D$ lies on the side $BC$ of the triangle $ABC$ and is chosen so that
$AD > BC\,.$ Point $E$ lies on the side $AC$ and 
satisfies
$$\frac{AE}{EC}=\frac{BD}{AD-BC}\,.$$
Prove that $AD > BE\,.$

\item  
Given positive integers $a_1<a_2<a_3<\ldots <a_{101}$ less than $5050$.
Prove that there exist four different numbers $a_k\,,a_l\,,a_m\,,a_n$
such that the number $a_k+a_l-a_m-a_n$ is divisible by $5050$.

\item 
Prove that there exist positive integers $n_1<n_2<\ldots <n_{50}$
such that 
$$n_1+S(n_1)=n_2+S(n_2)=n_3+S(n_3)=\ldots=
n_{50}+S(n_{50})\,,$$
where $S(n)$ denotes the sum of the digits of $n$.

\vspace{20pt}

\centerline{\bf Second Day} 

\item
Find out for which integers $n\geq 2$ the system of equations
$$\left\{
\matrix{
    x_1^2+x_2^2+50=16x_1+12x_2 \cr
    x_2^2+x_3^2+50=16x_2+12x_3 \cr
    x_3^2+x_4^2+50=16x_3+12x_4 \cr
    \dotfill\cr
    x_{n-1}^2+x_n^2+50=16x_{n-1}+12x_n \cr
    x_n^2+x_1^2+50=16x_n+12x_1 \cr
}\right.
$$
has integer solutions $\,x_1,x_2,
x_3,\ldots,x_n$.

\item 
Let $a_1,a_2,...,a_n,b_1,b_2,...,b_n$ be given integers.
Prove that
$$\sum\limits_{1\le i<j\le n}\!\!
(|a_i-a_j|+|b_i-b_j|)\le\!\!
\sum\limits_{1\le i,j\le n}\!\!
|a_i-b_j|\ .$$

\item
In a convex hexagon $ABCDEF$ the following equalities hold:
$$\angle A +\angle C +\angle E=360^\circ\ , \qquad
%\mbox{oraz}\quad\ 
{AB\over BC}\cdot{CD\over DE}\cdot{EF\over FA}=1\ .$$
Prove that 
$\;\displaystyle{{AB\over BF}\cdot{FD\over DE}\cdot{EC\over CA}=1\;.}$

\end{list}

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