
\documentclass{article}
\pagestyle{empty}
\usepackage{amssymb}
\usepackage{latexsym}
\title{Pan African Mathematical Olympiad 2000}
\author{Paper 2 Solutions}

\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item[4.] Let $a, b$ and $c$ be real numbers such that
\[ a^2 + b^2 = c^2.\]
Solve the system
\[
\begin{array}{rcl}
x^2 + y^2 & =&z^2\\
(x+a)^2 + (y+b)^2 &= &(z + c)^2.
\end{array}
\]
\noindent {\bf Solution } {\em
The third equation expands to 
\[ (x^2 + y^2) + (a^2 + b^2) + 2ax + 2by = c^2 + z^2 + 2cz.\]
Subtracting the first two equations from this gives
\[ 2ax + 2by = 2cz \Leftrightarrow ax + by = cz
\Leftrightarrow ax + by + cz = 2cz.\]
Also $a^2 + b^2 + c^2 = 2c^2$ and $x^2 + y^2 + z^2 = 2z^2.$
Now apply the Cauchy-Schwarz inequality to the vectors
$(a,b,c)$ and $(x,y,z)$:
\[ \begin{array}{rcl}
2cz & = & ax + by + cz\\
& \leq & \sqrt{a^2 + b^2 + c^2} \cdot \sqrt{x^2 + y^2 + z^2} 
= \sqrt{2c^2} \cdot \sqrt{2z^2}\\
&=& 2cz\end{array}.\]

We have equality in the Cauchy-Schwarz inequality, which means
that one vector must be a multiple of the other. This means that if
$a = b = c = 0$, then $x,y,z$ could be any numbers such that
$x^2 + y^2 = z^2$ (or equivalently $z =
\pm \sqrt{x^2 + y^2}$), and such numbers clearly satisfy the conditions.

If $(a,b,c) \not = (0,0,0)$, then $x,y,z$ must be a multiple
of $(a,b,c)$. Thus $\exists \lambda \in \mathbb R$ for which $x =
\lambda a,$ $y = \lambda b$ and $z = \lambda c$. We can check whether 
this satisfies the conditions for all $\lambda$:
\[
\begin{array}{rcl}
a^2 + b^2 = c^2 & \Rightarrow & \lambda^2 (a^2 + b^2) = \lambda^2(c^2)
\Leftrightarrow x^2 + y^2 = z^2.\\
a^2 + b^2 = c^2 & \Rightarrow & (\lambda + 1)^2 (a^2 + b^2) = (\lambda + 1)^2c^2\\
& \Leftrightarrow & (x+a)^2 + (y+b)^2 = (z + c)^2
\end{array}
\]}
\item[5.] From a point $P$ outside a circle, tangents $PA$
and $PB$ are drawn. $PQR$ is any secant\footnote{i.e. a straight
line through $P$ which intersects the circle twice}, with $Q$ and $R$
on the circumference. Chord $BS$ is parallel to $PQR$. 
Prove that $SA$ bisects $QR$. \newline
\noindent {\bf Solution } {\em Let $C$ be the intersection of
$AS$ and $PQR$. Then $\angle ACP = \angle SCR
= \angle CSB = \angle ABP$ (tangent-chord theorem). Therefore
$ACBP$ is a cyclic quadrilateral. Also $\angle SBC =
\angle BCP = \angle BAP = \angle ABP = \angle BSC$ (since
$PA = PB$) and thus $\Delta SCB$ is isosceles.

It follows that the perpendicular bisector of $BS$ passes through $C$.
But oeroendicular bisectors of chords pass through the centre and
$BS \parallel QR$, so $BS$ and $QR$ must share perpendicular bisectors. Thus $C$
lies on the perpendicular bisector of $QR$, and therefore it is the midpoint
of $QR$.}
\item[6.] A company has five directors. The regulations of the company
require that any majority (three or more) of the directors
should be able to open the strongroom, but any minority
(two or fewer) of the directors should not be able to do so.
It is proposed to equip the strongroom with ten locks, so that
it can only be opened when keys to all ten locks are available,
and to give each director a set of keys to $n$ different locks. 
Find all values of $n$ for which there is a way to allocate
the keys according to the regulations of the company. \newline
\noindent {\bf Solution } {\em Directors $A$ and $B$ cannot open all
the locks together, so there is a lock (say number 1), which neither can open,
but all other directors must have key number 1. The same reasoning
applies to all $(^5_2) = 10$ unordered pairs of directors; each pair will
be missing a particular key, and all other directors will own that key.
This tells us how the 10 keys must be distributed to have the desired effect
(the only choice being the particular labelling of the keys). Each director
is missing a different key for each of the 4 directors with whom he or she
can form a pair, so $n = 6$.}
\end{enumerate}
\end{document}

