\documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage[dvips]{graphicx}
\usepackage{verbatim}
\ifx\answers\undefined
\newenvironment{answer}{\comment}{\endcomment}
\else
\newenvironment{answer}{\textbf{Answer:}\em}{}
\fi
\let\divides=\mid
\def\notdivides{\not\,\mid}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\A}{\mathcal{A}}
\newcommand{\B}{\mathcal{B}}
\newcommand{\abs}[1]{\lvert#1\rvert}
%\addtolength{\topmargin}{-2.5cm}
\addtolength{\textheight}{18pt}
\pagestyle{empty}
\begin{document}
\title{%
\includegraphics[width=4cm]{tcdarms.ps}\\[5mm]
Irish Intervarsity Mathematics Competition 2001}
\author{Trinity College Dublin}
\date{9.30--12.30 Saturday 3\textsuperscript{rd} March 2001}
\maketitle
\thispagestyle{empty}
\enlargethispage*{13mm}
\begin{center}
\sl Answer as many questions as you can;
all carry the same mark.
Give reasons in all cases.
Tables and calculators are not allowed.
\end{center}
\begin{enumerate}
\item
For which integers $a$ does the equation
\[
x^3 - 3x + a
\]
have three integer roots?
\begin{answer}
Suppose the roots are $u,v,w$.
Then
\[
u + v + w = 0,\; uv + uw + vw = -3, uvw = -a.
\]
Setting
\[
w = -(u + v),
\]
the second equation becomes
\begin{gather*}
uv + uw + vw = -3,\\
\intertext{ie}
uv - (u+v)^2 = -3,\\
\intertext{ie}
u^2 +uv + v^2 = 3.
\end{gather*}
We have to find solutions $u,v \in \Z$ of this equation.
We can write the equation in the form
\[
(u + v/2)^2 + 3v^2/4 = 3.
\]
Thus
\begin{gather*}
3v^2/4 \le 3,\\
\intertext{ie}
v^2 \le 4,\\
\intertext{ie}
\abs{v} \le 2.
\end{gather*}
Similarly
\[
\abs{u} \le 2.
\]
On going through the possibilities
\[
u,v \in \{-2,-1,0,1,2\}
\]
we see that the only integer solutions are
\[
(u,v) = (-2,1), (-1,-1), (1,1), (2,-1).
\]
Substituting for $w$, the complete solutions are
\[
(u,v,w) = (-2,1,1), (-1,-1,2), (1,1,-2), (2,-1,-1).
\]
Thus there are really just two solutions (up to order) namely
\[
\pm (1,1,-2)
\]
giving
\[
a = -uvw = \pm 2.
\]
\end{answer}
\item
Find all solutions in integers $a,b,c$ of
\[
a^2 + b^2 + c^2 = a^2b^2.
\]
\begin{answer}
One solution is evidently
\[
a = b = c = 0.
\]
Suppose we have a different solution.
If $x \in \Z$ then
\[
x^2 \equiv 0 \text{ or } 1 \bmod 4
\]
according as $x$ is even or odd.
Suppose $a,b$ are both odd.
Then
\[
a^2b^2 \equiv 1 \bmod 4,
\]
while
\[
a^2 + b^2 + c^2 \equiv 2 \text{ or } 3 \bmod 4.
\]
Thus one (at least) of $a,b$ is even.
Hence
\[
a^2b^2 \equiv 0 \bmod 4,
\]
and so
\[
a^2 + b^2 + c^2 \equiv 0 \bmod 4,
\]
which is only possible if $a,b,c$ are all even.
Let $2^r$ be the highest power of 2 dividing $a,b,c$.
Then $r \ge 1$ from above.
Suppose
\[
a = 2^r a',\; b = 2^r b',\; c = 2^r c'.
\]
Then
\[
{a'}^2 + {b'}^2 + {c'}^2 = 2^{2r} {a'}^2 {b'}^2.
\]
Hence
\[
{a'}^2 + {b'}^2 + {c'}^2 \equiv 0 \bmod 4,
\]
which as we have seen implies that $a',b',c'$ are all even,
contrary to our choice of $r$ as the highest power of 2
dividing $a,b,c$.
We conclude that
\[
a = b = c = 0
\]
is the only solution.
\end{answer}
\item
Find all polynomials $P(x)$ satisfying
\[
P(x^2) = P(x) P(x-1).
\]
\begin{answer}
Suppose the roots of $P(x)$ are
\[
\alpha_1, \dots \alpha_n.
\]
Then the roots of $P(x^2)$ are
\[
\pm\sqrt{\alpha_1}, \dots \pm\sqrt{\alpha_n},
\]
while the roots of $P(x)P(x-1)$ are
\[
\alpha_1, \alpha_1 - 1, \dots, \alpha_n, \alpha_n - 1.
\]
These must be the same, up to order.
Let
\[
\max \abs{\alpha_i} = M.
\]
Then
\[
\max \abs{\sqrt{\alpha_i}} = \sqrt{M}.
\]
Thus if $M > 1$ there is no way of matching the roots.
Similarly, let
\[
\min_{\alpha_i \neq 0}\; \abs{\alpha_i} = m.
\]
Then
\[
\min_{\alpha_i \neq 0}\; \abs{\sqrt{\alpha_i}} = \sqrt{m}.
\]
Thus if $m < 1$ there is no way of matching the roots.
We conclude that the non-zero roots must all lie on the unit circle:
\[
\alpha_i \neq 0 \implies \abs{\alpha_i} = 1.
\]
Now suppose $\alpha \neq 0$ is a root of $P(x)$.
Then $\alpha + 1$ is a root of $P(x-1)$.
It follows that either $\alpha = -1$ or else
\[
\abs{\alpha} = 1,\; \abs{\alpha + 1} = 1.
\]
These two equations represent the two circles of radius 1
centred on $0$ and $-1$,
meeting in the points $\omega,\omega^2$
(where $\omega = \exp(2\pi i/3)$).
We conclude that if $\alpha$ is a roots of $P(x)$ then
\[
\alpha \in \{0,-1,\omega,\omega^2\}.
\]
If $0$ is a root of $P(x)$ then $-1$ is a root of $P(x-1)$,
and so $\pm i$ are roots of $P(x^2)$.
Hence $i$ (for example) is a root of $P(x)$ or $P(x-1)$,
ie $i$ or $i+1$ is a root of $P(x)$,
both of which we have already excluded.
Similarly if $-1$ is a root of $P(x)$ then $\pm i$ are roots of $P(x^2)$,
which we have just seen is impossible.
It follows that the only possible roots of $P(x)$ are
\[
\alpha = \omega \text{ or } \omega^2.
\]
Moreover these roots must occur equally often.
For
\[
\omega + 1 = -\omega^2,\; \omega^2 + 1 = -\omega;
\]
and
\[
\sqrt{\omega} = \pm \omega^2,\; \sqrt{\omega^2} = \pm \omega.
\]
Thus if $\omega$ occurs $r$ times
and $\omega^2$ occurs $s$ times
as roots of $P(x)$
then $\omega, -\omega, \omega^2, -\omega^2$
occur with multiplicities $s,s,r,r$ in $P(x^2)$,
and with multiplicies $r,s,r,s$ in $P(x)P(x-1)$.
Since
\[
(x - \omega)(x - \omega^2) = x^2 + x + 1,
\]
we have shown that the only solutions of the given relation are
\[
P(x) = (x^2 + x + 1)^n
\]
for $n = 0,1,2,\dots$
(to which should be added the trivial solution $P(x) = 0$).
\end{answer}
\item
If the plane is partitioned into two disjoint subsets,
show that one of the subsets contains three points
forming the vertices of an equilateral triangle.
\begin{answer}
There are probably many ways of doing this.
Let us call the two sets $\A$ and $\B$.
If $\A$ and $\B$ are empty the result is trivial;
so we may assume that there are points $A \in \A,\; B \in \B$.
Consider the line $AB$, and equidistant points
\[
\dots,P_{-2},P_{-1},A,B,P_1,P_2,\dots
\]
on the line.
There are two possibilities.
Either these points are alternately in $\A$ and $\B$,
\[
\dots,A_{-1},B_{-1},A_0,B_0,A_1,B_1,\dots;
\]
or we can find three equidistant points $X,Y,Z$ with
the central point $X$ in the same set as just one of the end-points $X,Z$, eg
\[
A_1A_2B_1.
\]
In the first case, consider the equilateral triangles
\[
A_iX_iA_{i+1} \text{ and } A_iY_iA_{i+1},
\]
with all the points $X_i$ on the same side of the line.
Then the points $X_i,Y_i$ must all lie in $\B$;
and so
\[
X_1Y_2X_3
\]
is an equilateral triangle in $\B$.
In the second case we may assume the three equidistant points are
\[
A_1A_2B_1.
\]
Consider the hexagon
\[
A_1P_1P_2B_1P_3P_4
\]
with centre $A_2$ and diagonal $A_1B_1$.
Then $P_1 \in \B$, or $A_1P_1A_2$ would be an equilateral triangle in $\A$.
Similarly $P_4 \in \B$.
But then $P_1B_1P_4$ is an equilateral triangle in $\B$.
\end{answer}
\bigskip
\bigskip
\bigskip
\begin{flushright}
\emph{\small More questions overleaf!}
\end{flushright}
\newpage
\item
For which real numbers $x$ is
\[
\sin(\cos x) = \cos(\sin x)?
\]
\begin{answer}
If
\[
\sin(\cos x) = \cos(\sin x),
\]
then
\[
\sin(\cos x) = \sin(\pi/2 - \sin x).
\]
But
\[
\sin A = \sin B \iff A = B + n\pi,
\]
where $n \in \Z$.
Thus
\[
\cos x = n\pi + \pi/2 - \sin x.
\]
Since
\[
\abs{\cos x}, \abs{\sin x} \le 1,
\]
this is only possible if
\[
n = 0 \text{ or } -1.
\]
Thus
\[
\cos x + \sin x = \pm\frac{\pi}{2}.
\]
Squaring both sides,
\[
\cos^2x + 2\cos x \sin x + \sin^2 x = \frac{\pi^2}{4}.
\]
Since
\[
\cos^2x + \sin^2x = 1,\quad 2\cos x \sin x = \sin2x,
\]
our equation gives
\[
\sin2x = \frac{\pi^2}{4} - 1.
\]
But $\pi^2 > 9$, and so
\[
\frac{\pi^2}{4} - 1 > \frac{9}{4} - 1 > 1.
\]
Since $\abs{\sin2x} \le 1$ for all $x$,
we conclude that the original equation has no solution.
\end{answer}
\item
For which real numbers $c \neq 0$ does the sequence defined by
\[
a_0 = c,\quad a_{n+1} = \frac{1}{2}\left(a_n + \frac{1}{a_n}\right)
\]
converge?
\begin{answer}
Suppose the sequence converges to $\ell$.
Then
\[
\ell = \frac{1}{2}\left(\ell + \frac{1}{\ell}\right)
\]
Hence
\begin{gather*}
\ell = \frac{1}{\ell},\\
\intertext{ie}
\ell = \pm 1.
\end{gather*}
If $c$ is replaced by $-c$ then $a_n$ is replaced by $-a_n$ for all $n$.
Thus it is sufficient to consider the case $c > 0$;
if $a_n \to \ell$ when $a_0 = c > 0$ then $a_n \to -\ell$ when $a_0 = -c$.
If $c > 0$ then $a_n > 0$ for all $n$.
Thus if $a_n$ is convergent then $a_n \to 1$.
By the inequality between the arithmetic and geometric means
(or by a simple calculation) if $a_n > 0$ then
\[
\frac{1}{2}\left(a_n + \frac{1}{a_n}\right) \ge 1.
\]
Thus
\[
a_n \ge 1
\]
for $n \ge 1$.
Also
\begin{align*}
a_{n+1} \le a_n &\iff \frac{1}{a_n} \le a_n\\
&\iff a_n \ge 1
\end{align*}
Hence the sequence $a_n$ is decreasing for $n \ge 1$,
and so must converge to a limit, which we have seen must be 1.
We conclude that the sequence always converges if $c \neq 0$:
\[
a_n \to
\begin{cases}
1 & \text{ if $c > 0$},\\
-1 & \text{ if $c < 0$}.
\end{cases}
\]
\end{answer}
\item
Show that for any sequence $a_n$ of strictly positive real numbers
one (or both) of the series
\[
\sum \frac{a_n}{n^2} \quad \text{ and } \quad \sum \frac{1}{a_n}
\]
must diverge.
\begin{answer}
Suppose the two series both converge.
Then
\[
\frac{1}{a_n} \to 0 \implies a_n \to \infty.
\]
Since $a_n > 0$ we can re-arrange the series
without affecting their convergence or divergence.
Thus we may assume that the $a_n$ are increasing:
\[
a_{n+1} \ge a_n.
\]
Let $\pi(m)$ denote the number of $a_n$ which are $\le m$:
\[
\pi(m) = \abs{\{n: a_n \le m\}}.
\]
The number of $a_n$ in the range $[m,2m)$ is
\[
\pi(2m) - \pi(m).
\]
The $a_n$ in this range contribute
\[
\ge \frac{\pi(2m) - \pi(m)}{2m}
\]
to the second series $\sum 1/a_n$.
It follows that
\[
\frac{\pi(2m) - \pi(m)}{2m} \to 0
\]
as $m \to \infty$.
In particular there is an $N > 0$ such that
\begin{gather*}
\frac{\pi(2m) - \pi(m)}{2m} < \frac{1}{8}\\
\intertext{ie}
\pi(2m) - \pi(m) < \frac{1}{4}m
\end{gather*}
for $m \ge N$.
Thus
\begin{gather*}
\pi(2N) - \pi(N) \le \frac{1}{4}N,\\
\pi(4N) - \pi(2N) \le \frac{1}{4}2N,\\
\dots\\
\pi(2^{r+1}N) - \pi(2^r2N) \le \frac{1}{4}2^rN.
\end{gather*}
Adding,
\begin{align*}
\pi(2^{r+1}N) - \pi(N)
&\le \frac{1}{4} \left(1 + 2 + 2^2 + \cdots + 2^r\right)N\\
&\le \frac{1}{4} 2^{r+1}N
\end{align*}
Thus if
\[
2^rN \le n < 2^{r+1}N
\]
then
\begin{align*}
\pi(n) &\le \pi(2^{r+1}N)\\
&\le \pi(N) + \frac{1}{2}{2^rN}\\
&\le \pi(N) + \frac{1}{2}n.
\end{align*}
We conclude that, for some $N_1 \ge N$,
\[
\pi(n) \le n
\]
for $n \ge N_1$.
But
\[
\pi(n) \le n \implies a_n \ge n.
\]
Thus
\[
\frac{a_n}{n^2} \ge \frac{1}{n}
\]
for $n \ge N_1$;
and so the first series
\[
\sum \frac{a_n}{n^2}
\]
diverges by comparison with
\[
\sum \frac{1}{n}.
\]
\end{answer}
\item
The point $A$ lies inside a circle centre $O$.
At what point $P$ on the circumference of the circle
is the angle $OPA$ maximised?
\begin{answer}
It is evident that
\[
O\hat{P}A < \pi/2,
\]
eg by drawing the diameter $D_1OAD_2$ through $A$,
and comparing $O\hat{P}A$ with $D_1PD_2 = \pi/2$.
Applying the sine law to the triangle $OPA$,
\begin{gather*}
\frac{\sin O\hat{P}A}{OA} = \frac{\sin O\hat{A}P}{OP}\\
\intertext{ie}
\sin O\hat{P}A = \frac{OA}{OP}\sin O\hat{A}P.
\end{gather*}
Now $O\hat{P}A$ is maximised when $\sin(O\hat{P}A)$ is maximised.
Since $OA$ and $OP$ are fixed (with $OA < AP$)
this will occur when $\sin(O\hat{A}P)$ is maximised,
ie when $O\hat{A}P = \pi/2$.
We conclude that $O\hat{P}A$ is maximised
when $P$ is one of the two points where the line through $A$
perpendicular to $OA$ meets the circle.
\end{answer}
\item
What is the maximum volume of a cylinder contained within a sphere of radius 1?
\begin{answer}
This is straightforward.
Clearly the largest cylinder will have its centre
at the centre of the sphere.
Suppose the radius of the cylinder is $r$,
and the height $h$.
Then
\[
r^2 + h^2 = 1.
\]
The volume of the cylinder is
\begin{align*}
V &= \pi r^2h\\
&= \pi h(1-h^2).
\end{align*}
We have to maximise this function,
with the restriction that $0 < h < 1$.
Let
\[
f(x) = x(1 - x^2) = x - x^3.
\]
Then $f(0) = f(1) = 0$ and
\[
f'(x) = 1 - 3x^2.
\]
Thus $f(x)$ attains its maximum when
\[
x = \frac{1}{\sqrt{3}};
\]
and
\[
V_{\text{max}} = \frac{\pi}{\sqrt{3}}.
\]
\end{answer}
\item
The function $f(x)$ is twice-differentiable for all $x$,
and both $f(x)$ and $f''(x)$ are bounded.
Show that $f'(x)$ is also bounded.
\begin{answer}
Suppose
\[
\abs{f(x)} \le C,\quad \abs{f''(x)} \le C.
\]
Suppose $f'(x)$ is large at some point, say $x_0$.
On taking $-f(x)$ in place of $f(x)$ if necessary,
we may suppose that
\[
f'(x_0) = X > 0.
\]
Since $f''(x)$ is bounded, $f'(x)$ must remain large
for a considerable interval beyond $x_0$.
More precisely, suppose
\[
f'(x) < X/2
\]
for some $x > x_0$.
Then
\[
f'(x_0) - f'(x) > X/2.
\]
But we know, from Rolle's Theorem, that
\[
\frac{f'(x) - f'(x_0)}{x - x_0} = f''(\xi)
\]
for some $\xi \in (x_0,x)$.
Thus
\begin{align*}
x - x_0 &= \frac{f'(x) - f'(x_0)}{f''(\xi)}\\
\ge \frac{X}{2C}.
\end{align*}
In other words,
\[
f'(x) \ge X/2
\]
for $x \in [x_0, x_1]$,
where
\[
x_1 = x_0 + \frac{X}{2C}.
\]
But this in turn means that $f(x)$ is increasing steadily over this interval.
More precisely,
for some $\eta \in (x_0,x_1)$,
\begin{align*}
f(x_1) - f(x_0) &= (x_1 - x_0) f'(\eta)\\
&\ge (x_1 - x_0) \frac{X}{2}.
\end{align*}
Thus
\[
f(x_1) - f(x_0) \ge \frac{X^2}{4C}.
\]
But since
\[
\abs{f(x) \le C
\]
it follows that
\[
\abs{f(x_1 - f(x_0)} \le \abs{f(x_1}} + \abs{f(x_0)} \le 2C.
\]
Hence
\begin{gather*}
\frac{X^2}{4C} \le 2C,
\intertext{ie}
X < 2^{3/2} C.
\end{gather*}
So we have shown that
\[
\abs{f'(x)} \le 2^{3/2} C
\]
for all $x$.
\end{answer}
%\item
%If the continuous function $f(x)$ on $\R$ satisfies
%\[
%f(x) + f(2x) + f(3x) = 0
%\]
%does it follow that $f(x) = 0$ for all $x$?
\end{enumerate}
\end{document}