\documentclass[10pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{wrapfig}
\usepackage{tools/multicol}
\usepackage{textcomp}
\usepackage{fouriernc}
\usepackage{calc}
\usepackage{amsmath}
\usepackage{flexisym}
\usepackage{breqn}
\usepackage{ifthen}
\usepackage{multicol}
\usepackage{amsthm}
% \usepackage[T1]{fontenc}
% \usepackage[sc]{mathpazo}
% \linespread{1.05} % Palatino needs more leading (space between lines)
\pagestyle{empty}
\textwidth = 7.0 in
\textheight = 9.5 in
\oddsidemargin = -0.25 in
\evensidemargin = -0.6 in
\topmargin =-0.3 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.1in
\parindent = 0.0in
\columnsep = 0.25in
\newcommand{\blank}{\underline{\hspace{1.5cm}}}
\setlength{\linewidth}{(\textwidth-\columnsep)/2}
\newcounter{prob}
\setcounter{prob}{1}
\newcounter{test}
\setcounter{test}{1}
\newcommand{\sol}{\underline{Solution.} }
\newcommand\solutionmode{nosolutions}
\newcommand\corrhide[1]{\ifthenelse{\equal{\solutionmode}{solutions}}{#1}{}}
\newcommand{\solution}[1]{\ifthenelse{\equal{\solutionmode}{solutions}}{\sol
#1}{}}
\newcommand{\itm}{\nopagebreak \vspace{0.5mm} \newline (\alph{prob}) \stepcounter{prob}\nopagebreak}
\newcommand{\itmn}{\nopagebreak \vspace{0.5mm} (\alph{prob}) \stepcounter{prob}\nopagebreak}
\newcommand{\testitem}{\hspace{-3.5mm}\textbf{\thetest.}\ \stepcounter{test}\setcounter{prob}{1}\nopagebreak}
\newcommand{\nota}{None of the above}
\newcommand{\gcf}{\mathop{\mathrm{gcf}}\nolimits}
\renewcommand{\labelenumi}{(\textit{\roman{enumi}})}
\begin{document}
\begin{center}
\Large \textbf{University of Mississippi} \\ \vspace{1mm} \normalsize 4\raisebox{3.6pt}{\scriptsize{th}} Annual High School Mathematics Contest \\
Team Competition \corrhide{Solutions} \\ \input{date.tex}
\end{center}
\begin{multicols}{2}
\testitem Find $z$ given that
\begin{align*}
2x+4y+z&=27 \\
\frac{x}{6}+\frac{y}{3}-z&=16
\end{align*}
\solution{Multiply the bottom equation by $-12$ and add the two equations to
obtain $13z=-165$ or \fbox{$z=-165/13$}.}
\testitem Prove that an 85$\times$16 grid cannot be tiled (covered without
overlap) using tiles of the following shape:
\begin{center}
\includegraphics[width=0.2\textwidth]{images/tile.mps}
\end{center}
\solution{The area of one tile is 6, so the area of any region which may be
tiled using these tiles is a multiple of six. However,
$85\cdot16=2^4\cdot5\cdot17$ is not divisible by $6$. \hfill $\square$ }
\testitem Prove that for all real numbers $a,\,b,$ and $c$, we have
\[2a^2+b^2+c^2\geq2a(b+c).\]
\solution{Move everything to the left hand side, distribute the $2a$ and
write $2a^2$ as $a^2+a^2$:
\[a^2+b^2-2ab+a^2+c^2-2ac\geq0.\] Now each trinomial may be factored, and
we have shown that the original inequality is equivalent to
$(a-b)^2+(a-c)^2\geq0$, which is true because squares of real numbers are
nonnegative. \hfill $\square$ }
\testitem An integer $n$ is called \emph{perfect} if it equals the sum
of all its divisors $d$ with $1 \leq d < n$. For example, $28$ is
perfect because $1+2+4+7+14 = 28$. Let $a$ be a positive integer. Prove
that if $2^a-1$ is prime, then $2^{a-1}(2^a-1)$ is perfect.
\solution{
Define $p=2^a-1$ and $n=2^{a-1}(2^a-1)$. Since $p$ is an odd
prime, $2^{a-1}p$ is the prime factorization of $n$. Therefore, the
proper factors of $n$ are $1,\,2,\,2^2,\,\ldots,\,2^{a-1}$, and
$p,\,2p,\,2^2p,\,\ldots,\,2^{a-2}p$. Summing these geometric series and
subtituting $2^a-1=p$
gives \[ \quad \sum_{k=0}^{a-1}2^k+\sum_{k=0}^{a-2}2^kp=\frac{2^a-1}{2-1}+p\frac{2^{a-1}-1}{2-1}=2^{a-1}p=n.
\quad \square \]
}
\testitem Prove that every integer which can be written as the sum of two
cubes may also be written in the form $m(m^2-3n)$ for integers $m$ and $n$.
\solution{Write $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)\left[(a+b)^2-3ab\right]$,
and define $m=a+b$ and $n=ab$. \hfill $\square$}
\testitem A cone whose diameter is equal to its height is inscribed in a
sphere of radius 1. What is the volume of the cone?
\solution{Let $r$ denote the radius of the cone, $C$ the center of the
sphere, $D$ the center of the cone's base and $B$ a point on the edge of
the base of the cone. Looking at triangle $CDB$, we can see that the
height of the cone is $1+\sqrt{1-r^2}$. Setting the height equal to the
diameter, we get $1+\sqrt{1-r^2}=2r \Rightarrow r=4/5$, so that $V=\pi r^2
h/3=\pi (4/5)^2\cdot(8/5)/3=\fbox{$128\pi/375$}$.}
\testitem A ship located at point $P$ is known to be 20 nautical miles from
point $C$. Point $A$ is 5 nautical miles west of $C$, and $B$ is $5$
nautical miles east of $C$. Based on radio signals from stations based at
point $A$ and $B$, we know that the the distance $AP$ is 6 nautical miles
larger than the distance $BP$. Find the possible locations of the ship,
expressed as a number of nautical miles east/west and a number of nautical
miles north/south of $C$.
\solution{Define a coordinate plane with $C$ at the origin and the $x$-axis
passing through $A$ and $B$. Let $x$ and $y$ be the coordinates of $P$.
Then we know $x^2+y^2=20^2$, and because $AP-BP=6$ we know that $P$ is on
a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with foci at $(\pm
c,0)=(\pm5,0)$ and with $a=6/2=3$. Then we calculate
$b=\sqrt{c^2-b^2}=4$, so that $\frac{x^2}{9}-\frac{y^2}{16}=1$.
Multiplying this equation by $-9$ and adding it to $x^2+y^2=400$ gives
\fbox{$y=\displaystyle{\pm\frac{4 \sqrt{391}}{5}}$} and \fbox{$x=\displaystyle{\frac{12\sqrt{26}}{5}}$}.
\underline{Alternate Solution.} Use the coordinate representation in the
first solution, and write $\sqrt{y^2+(x+5)^2}-\sqrt{y^2+(x-5)^2}=6$ from
the Pythagorean theorem. Squaring and substituting the $x^2+y^2=400$
into this equation, we get
\[\sqrt{425+10x}-\sqrt{425-10x}=6,\]
which we may square and rearrange to get
\[407=\sqrt{425^2-100x^2}.\] This gives $x=\sqrt{425^2-407^2}/10$ which may
be simplified using difference of squares to obtain the same answer as
above.}
\testitem Consider the following sequence of strings of digits:
\[1,\, 2,\, 4,\, 13,\, 31,\, 112,\, 224,\, 1003, \ldots\] Describe a rule
for finding the terms of this sequence, and use it to find the next three
terms of the sequence.
\solution{Noticing that the numbers use only the digits 0, 1, 2, 3, and 4, we
convert the sequence to base 5, which gives
1, 2, 4, 8, 16, 32, 64, 128, 256, $\ldots$. Therefore, the next three terms are
the base 5 representations of $512$, $1024$, and $2048$, i.e.\
\fbox{4022, 13044, and 31143}.}
\testitem Mark is taking a long test, and he knows that if he works at a
rate of $r$ questions per minute, his accuracy is $1-r/5$ (e.g. he will get
80\% of the questions that he works at a rate of 1 question per minute
correct). There is no partial credit, and he gets no credit for problems
he doesn't work completely. With 25 minutes left, he has $Q$ questions
remaining. He faces a decision between (a) losing one minute of test time
to calculate his optimum rate for the remaining 24 minutes, or (b)
continuing at a rate of $Q/25$ questions per minute so that he finishes the
test. Determine the values of $Q$ for which strategy (a) will result in a
better score than (b).
\solution{For
real $x$, let $\lfloor x\rfloor$ denote the greatest integer that does
not exceed $x$. Working at a rate of $r$ \begin{wrapfigure}{r}{6cm}
\includegraphics[width=6cm]{images/markstest.pdf}
\end{wrapfigure} questions per minute for $n$ minutes will allow Mark
to work $nr$ total questions (as long as $nr$ is not more than the
number of questions remaining), and therefore he will get $nr(1-r/5)$
questions correct. So for choice (b), Mark will get $\lfloor
Q(1-Q/125)\rfloor$ questions correct. Now to determine how many
questions Mark will get with choice (a), we first optimize the function
$C(Q,r)=24r(1-r/5)$ for large $Q$. As a function of $r$, this is a
concave parabola, so its maximum occurs halfway between its two roots
$r=0$ and $r=5$, i.e.\ at $r=5/2$. Therefore, the maximum of $C(Q,r)$
is $30$ for $Q\geq24(5/2)=60$. For $Q<60$, it is best to decrease the
rate from 2.5 questions per minute to $Q/24$, since he has time to
finish the questions, and he increases his accuracy by slowing down.
This gives him $\lfloor Q(1-Q/120)\rfloor$ questions correct for
$Q<60$. Plotting these two functions together and solving
$30=Q(1-Q/125)$, we see that (a) is better than (b) for
\fbox{$Q>75$}. }
% \testitem For postive integers $n$, define the function
% \begin{dmath*}
% D(n)=\text{the sum of digits in the decimal representation of $n$}
% - \text{ the number of digits in the decimal representation of $n$}.
% \end{dmath*}
% For example, $D(427)=13-3=10$. Find with proof the minimum positive
% integer $n$ such that $D(2n)=2D(n)$.
% \solution{Define $d(n)$ to be the number of digits of $n$ and $s(n)$ to be
% the sum of $n$'s digits. Notice that if all of the digits of $n$ are
% less than 5 (so that adding $n+n$ does not involve carrying),
% $D(2n)=s(2n)-d(2n)=2s(n)-d(n)\neq2s(n)-2d(n)=2D(n)$. Also, if any of the
% digits of $n$ are greater than 4, then $s(2n)~~0$ and $R>0$, and equally space $2n+1$ points with
angular separation $\alpha$ on the circumference of a circle of radius
$R$. Then project all the points onto the diameter perpendicular to the
radius drawn to the middle point. We require that
$n\alpha\leq90^\circ$ so that all the points lie on the same
semicircle.
\vspace{-.4cm}
\begin{center}
\includegraphics[width=6.5cm]{images/parthenonp.mps}
\end{center}
\item Ensure that ratios between adjacent column spacings are equal. That
is, choose a real number $r>1$ and place $C_1,C_2,\ldots,C_{2n+1}$ so that
$rC_kC_{k+1}=C_{k+1}C_{k+2}$ for all $1\leq k < n$,
$C_nC_{n+1}=C_{n+1}C_{n+2}$, and $C_kC_{k+1}=rC_{k+1}C_{k+2}$ for all $n
< k < 2n$. (Here $C_{i}C_{j}$ denotes the distance from $C_i$ to $C_j$.)
\vspace{-0.5cm}
\begin{center}
\hspace*{-0.1cm}\includegraphics[width=8cm]{images/parthenon2p.mps}
\end{center}
\end{enumerate}
Prove that (\textit{i}) and (\textit{ii}) represent fundamentally
different ways of determining column placement. In particular, prove
that for all $n\geq3$, there do not exist $r$, $R$, and $\alpha$ so that
the points obtained from construction (\textit{i}) coincide with the
points obtained from construction (\textit{ii}). (Note: the distance
$C_1C_{2n+1}$ between the first and last points is assumed to be fixed.)
\solution{First, from the triangles in the upper figure we see that the
distances $C_nC_{n+k}$ in construction (i) are $R\sin(k\alpha)$.
Therefore, if for some $n\geq3$ constructions (i) and (ii) coincide,
then we would have \[\frac{\sin2\alpha-\sin\alpha}{\sin\alpha-\sin0}=
\frac{\sin3\alpha-\sin2\alpha}{\sin2\alpha-\sin\alpha},\] by looking at
the ratios $C_{n+3}C_{n+2}/C_{n+2}C_{n+1}$ and
$C_{n+2}C_{n+1}/C_{n+1}C_{n}$. Now recall that for all $u$ and $v$,
$\sin u-\sin
v=2\sin\left(\frac{u-v}{2}\right)\cos\left(\frac{u+v}{2}\right)$.
Using this identity to simplify all the differences in the previous
equation gives
\[\frac{\cos(3\alpha/2)}{\cos(\alpha/2)}=\frac{\cos(5\alpha/2)}{\cos(3\alpha/2)}.\]
Abbreviating $\beta=3\alpha/2$, we can rewrite this equation as
\[\cos^2\beta=\cos(\beta-\alpha)\cos(\beta+\alpha),\]
and we can use the cosine sum-angle formulas:
$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$. We
get \[\cos^2\beta=\cos^2\beta\cos^2\alpha-\sin^2\beta\sin^2\alpha,\] and
moving everything to the left side and rewriting
$1-\cos^2\alpha=\sin^2\alpha$, we get
$\sin^2\alpha(\cos^2\beta+\sin^2\beta)=0$. This is a contradiction since
$\cos^2\beta+\sin^2\beta=1$ and $0<\alpha\leq \pi/6 \Rightarrow
\sin^2\alpha\neq0$. \hfill $\square$}
\end{multicols}
\end{document}
~~