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\begin{document}

\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip

\centerline{\large \bf Christchurch, 1998}
\bigskip
\medskip

\centerline{\Large \bf Problem Set 4}

\vspace{1cm}
\centerline{\Large \bf Problems}
\vspace{1cm}

{\bf 1.} The sequence $\row an,\dots $  is defined as follows:
$a_1=2$, and $a_n$ is the greatest prime divisor of $a_1a_2\ldots
a_{n-1}+1$. Prove
that this sequence does not contain $5$.\par
\medskip

{\bf 2.} A finite sequence $\gamma = b_1b_2\ldots b_n$ of zeros and ones
satisfies
the properties:\par
(i) For any $i\ne j$ the subsequence $b_ib_{i+1}b_{i+2}b_{i+3}b_{i+4}$
is
different
from the 
subsequence $b_jb_{j+1}b_{j+2}b_{j+3}b_{j+4}$;\par
(ii) If we add to $\gamma$ a digit zero or one on the right then the
extended
sequences $b_1b_2\ldots b_n0$ and $b_1b_2\ldots b_n1$ will not have
property (i) any
more.\par

Prove that the first four digits in $\gamma $ are the same as the last
four
digits of this sequence.
\medskip

{\bf 3.} Let $A_1$ be a point on the side $BC$ of a triangle $ABC$ such
that
incircles of the triangles $ABA_1$ and $ACA_1$ touch $AA_1$ at the same
point. In a
similar way, points $B_1$ and $C_1$ are chosen on the sides $AC$ and
$AB$,
respectively. Prove that the lines $AA_1$, $BB_1$, $CC_1$ are
concurrent.
\medskip

{\bf 4.}  Let $P(x,y)$ and $Q(x,y)$ be two polynomials satisfying\par
\indent (i) for every fixed $y\ge 0$ the functions $P(x,y)$ and $Q(x,y)$
are
increasing functions of $x$ for $x\ge 0$.\par
\indent (ii) for every fixed $x\ge 0$ the function $P(x,y)$ is
increasing and
the function $Q(x,y)$ is decreasing function of $y$ for $y\ge 0$.\par
\indent (iii) $P(x,0)=Q(x,0)$ for all $x\ge 0$ and $P(0,0)=0$.\newline
Determine for which $a\ge 0$ and $b\ge 0$ the system
\begin{eqnarray*}
P(x,y)&=&a\\
Q(x,y)&=&b
\end{eqnarray*}
has a unique solution $(x,y)$ with $x\ge 0$, $y\ge 0$.

\newpage
%============================

\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip

\centerline{\large \bf Christchurch, 1998}
\bigskip
\medskip

\centerline{\Large \bf Problem Set 4}
\vspace{1cm}
\centerline{\Large \bf Solutions}
\vspace{1cm}

{\bf 1.} Since $a_1=2$ and $a_2=3$, all other numbers will be relatively
prime to
$6$. Also the number $a_1a_2\ldots a_{n-1}$ is divisible by $2$ but not
$4$. Thus if
$a_n=5$ then the number $a_1a_2\ldots a_{n-1}+1$ is a power of $5$. But
since every
number $5^m-1$ is divisible by $4$ we get a contradiction.
 
 
{\bf 2.} Take the last four digits $abcd$. Since addition of either $0$
or
$1$ on
the right destroys the property, both subsequences $abcd0$ and $abcd1$
occur in
$\gamma$. Thus the combination of four digits $abcd$ occur at least
three times.
Suppose that none of these is positioned in the beginning. Then each of
them has
either $0$ or $1$ in front of it. Hence one of the subsequences $0abcd$
or
$1abcd$ will occur twice, which is impossible.

{\bf 3.} Let us denote points as shown in the figure:\begin{figure}[h]
\centerline{\picture 3.8in by 2.2in (fig4 scaled 800)}
\caption{}
\end{figure}

Now, $BA_1=BJ+JA_1=BK+LA_1=(AB-AK)+LA_1$, and similarly
$A_1C=LA_1+(AC-AM)$. As
$AK=AL=AM$, we have $BA_1-A_1C=AB-AC$. On the other hand,
$BA_1+A_1C=BC$,
so we get
$$
BA_1=\frac{AB+BC-AC}{2},\qquad A_1C=\frac{AC+BC-AB}{2}.
$$
Similarly,
$$
AB_1=\frac{AB+AC-BC}{2},\qquad B_1C=\frac{AC+BC-AB}{2}.
$$
and 
$$
BC_1=\frac{AB+BC-AC}{2},\qquad C_1A=\frac{AB+AC-BC}{2}.
$$
We see that $AB_1=AC_1$, $BA_1=BC_1$ and $CA_1=CB_1$ and 
$$
\frac{BA_1}{A_1C}\cdot \frac{CB_1}{B_1A}\cdot \frac{AC_1}{C_1B} =1;
$$
hence by Ceva's Theorem the lines $AA_1$, $BB_1$, $CC_1$ are
concurrent.\par
\medskip 

{\bf Remark:} the point of intersection is the Gergonne point of the
triangle $ABC$.
\bigskip

{\bf 4.} As increasing polynomials, $P(x,0)$ and $Q(x,0)$ are strictly
increasing
and tend to $\infty $ as $x\rightarrow \infty $. So, for every $a\ge 0$
and
$b\ge 0$,
there exist unique $x_1$, $x_2$ such that $P(x_1,0)=a$ and $Q(x_2,0)=b$.
Also there
is a unique $y_1$ such that $P(0,y_1)=a$. The graph of $P(x,y)=a$ is a
strictly
decreasing curve joining $(0,y_1)$ and $(x_1,0)$. On the other hand, 
the
graph of 
$Q(x,y)=b$ is a strictly increasing curve starting at $(x_2,0)$. As
polynomials are
continuous functions, from this it is clear that the system has a
solution,
and a
unique one, if and only if $x_2\le x_1$. Since $P(x,0)$ is increasing,
and
$P(x_2,0)=Q(x_2,0)=b$, this is equivalent to $a\le b$.
\end{document}