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\title{Pan African Mathematical Olympiad 2000}
\author{Paper 1 Solutions}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item	Solve the trigonometric equation
\[ \sin^3 x (1 + \cot x)+ \cos^3 x (1 + \tan x) = \cos 2x.\] 
\noindent {\bf Solution } {\em
\[
\begin{array}{rl}
& \sin^3 x (1 + \cot x)+ \cos^3 x (1 + \tan x) = \cos 2x\\
\Rightarrow & \sin^2 x (\sin x + \cos x) + \cos^2 x(\cos x + \sin x)
= \cos^2 x - \sin^2 x\\
\Leftrightarrow & (\sin^2 x + \cos^2 x)(\sin x + \cos x)=
( \cos x + \sin x)(\cos x - \sin x)\\
\Leftrightarrow & 0 = (\cos x + \sin x)(\cos x - \sin x -1)\\ 
\Leftrightarrow & 0 = (\sin \pi/4 \cdot \cos x + \cos \pi/4 \cdot \sin x)
(\sin \pi/4 \cdot \cos x - \cos \pi/4 \cdot \sin x - \sqrt 2/2)\\
\Leftrightarrow &  0 = [\sin(\pi/4 + x)][\sin(\pi/4 -x) - \sqrt 2/2].
\end{array}\]
Thus $\sin(\pi/4 + x) = 0$ or $\sin(\pi/4 -x) = \sqrt 2/2$
The sine function takes the value 0 at and only at 
multiples of $\pi$, so the
solutions to the first case are of the form 
$k\pi - \pi/4$, for $k \in \mathbb Z$. These solutions give
$1 + \cot x = 0,$ $1 + \tan x = 0$
and $\cos 2x =0,$ so they clearly satisfy the original problem.

We now examine the other candidate solutions. 
Now $\sin x = \sqrt 2/2$ at and only at $\pi/4$ and $5\pi/4$ if
$x \in [0,2\pi]$.
Thus the  solutions of $\sin (\pi/4 -x) = \sqrt 2/2$ are of the form
$2k\pi$ and $2k\pi -\pi/2$ for $k \in \mathbb Z$.
None of these are solutions to the original problem,
since they give $\cot x$ or $\tan x$ as undefined.}

\item Let the polynomials $P_0, P_1, P_2, \ldots$ be defined by
\[ P_0(x) = x^3 + 213x^2 - 67x -2000\]
and
\[ P_n(x) = P_{n-1}(x-n) \mbox{ for }n =1, 2, 3, \ldots\]
What is the coefficient of $x$ in $P_{21}(x)$? \newline
\noindent {\bf Solution } {\em
\[ \begin{array}{rcl}
P_{21}(x) & = & P_{20}(x-21)\\
&=& P_{19}(x-21 -20)\\
&\vdots& \\
&=& P_0(x-21 -20 - \cdots -1)\\
&=&  P_0(x - 231)\\
&=& (x-231)^3 + 213(x-231)^2 -67(x-231) - 2000.
\end{array}\]
The coefficient of $x$ in each term can be found from the
binomial theorem, so the coeffiient of $x$ is
\[ 3 \cdot 231^2 - 213 \cdot 2 \cdot 213 -67 = 61610.\]}
\item Prove that if 
\[ \frac{p}{q} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}+ \ldots -\frac{1}{1334} + \frac{1}{1335}\] where 
$p$ and $q$ are natural numbers, then 2003 divides $p$.\newline
\noindent{\bf Solution } {\em Firstly note that 2003 is a prime
number (easily checked by trial division by all prime numbers
less than $\sqrt{2003}$, i.e. all prime number up to 43.

\[ \begin{array}{rcl}
1 - 1/2 + \cdots + 1/1335 & = &
1 + 1/2 + \cdots + 1/1335 -2(1/2 + 1/4 + \cdots + 1/1334)\\
& = & 1 + 1/2 + \cdots + 1/1335 - (1 + 1/2 + \cdots + 1/667)\\
& = & 1/668 + 1/669 + \cdots + 1/1335\\ 
& = & (1/668 + 1/1335) + (1/669 + 1/1334) + \cdots + (1/1001 + 1/1002)\\
& = & 2003( 1/(668 \times 1335) + 1/ (669 \times 1334) +
\cdots + 1/(1001 \times 1002)).
\end{array}\]
Now the denomominator of 

\[1/(668 \times 1335) + 1/ (669 \times 1334) +
\cdots + 1/(1001 \times 1002)\]
in simplest form
is coprime to 2003. This is because 2003 is prime and the numbers 
668 to 1335 are all smaller than 2003, so their product cannot
be a multiple of 2003. Thus the sum is $p/q$ in simplest form where
$2003 \mid p$.}
\end{enumerate}
\end{document}

