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\title{%
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Irish Intervarsity Mathematics Competition 1992}
\author{Trinity College Dublin}
\date{9.30--12.30 15\textsuperscript{th} February}
\maketitle
\thispagestyle{empty}
\begin{quote}
\sl
Answer all questions.\\
Calculators permitted.
\end{quote}
\begin{enumerate}
\item Solve the equation
\[
(x-2)(x-3)(x+4)(x+5) = 44.
\]
\begin{answer}
Let $x-2=y-3$, ie $y=x+1$.
Then the equation becomes
\[
(y-3)(y-4)(y+3)(y+4) = 44.
\]
In other words,
\[
(y^2-9)(y^2-16) = 44.
\]
Let $z = y^2$.
We have a quadratic
\[
(z-9)(z-16) = 44.
\]
This gives
\[
z^2 - 25z + 100 = 0.
\]
Let $z = 5u$.
Then
\begin{eqnarray*}
25u^2 - 125u + 100 = 0 &\implies& u^2 - 5u + 4 = 0\\
&\implies& (u-4)(u-1) = 0\\
&\implies& u = 1 \mbox{ or } 4\\
&\implies& z = 5 \mbox{ or } 20\\
&\implies& y = \pm\sqrt5 \mbox{ or } \pm2\sqrt5\\
&\implies& x = \pm\sqrt5 - 1 \mbox{ or } \pm2\sqrt5 - 1.
\end{eqnarray*}
\end{answer}
\item Find the greatest value of
\[
\frac{x+2}{2 x^2 + 3x + 6}.
\]
\begin{answer}
To {\em maximize}
\[
f(x) = \frac{x+2}{2 x^2 + 3x + 6},
\]
it is sufficient to {\em minimize}
\[
g(x) = \frac{2 x^2 + 3x + 6}{x+2}.
\]
Setting $x+2 = y$,
\begin{eqnarray*}
g(x) &=& \frac{2(y-2)^2 + 3(y-2) + 6}{y}\\
&=& \frac{2y^2 - 5y + 8}{y}\\
&=& 2y - 5 + \frac{8}{y}\\
&=& 2\left(\sqrt y - \frac{2}{\sqrt y}\right)^2 + 3.
\end{eqnarray*}
Thus
\[
g_{\min} = 3,
\]
the minimum being attained when $y = 2$.
Hence
\[
f_{\max} = \frac{1}{3},
\]
the minimum being attained when $x = 0$.
\end{answer}
\item Writing numbers to the base 8, show that there are infinitely
many numbers which are doubled by reversing their `digits'.
\begin{answer}
Consider 2 \lq digit' numbers $x = ab$, ie
\[
x = 8a +b \qquad (0\le a,b\le 7).
\]
This is doubled on reversal if
\begin{eqnarray*}
8b + a = 2(8a + b) &\implies& 15a = 6b\\
&\implies& 5a = 2b.
\end{eqnarray*}
This has the solution $a = 2, b = 5$, ie
\[
2\cdot25_8 = 52_8.
\]
Now we see that any number of the form
\[
2525\cdots25_8
\]
has the required property.
\end{answer}
\item Show that for any positive real numbers $a$, $b$ with $a > b$,
\[
a^n - b^n > n (a-b) (ab)^{(n-1)/2}.
\]
\begin{answer}
We have
\[
\frac{a^n - b^n}{a-b} = a^{n-1} + ba^{n-2} + \cdots + b^{n-1}.
\]
Thus
\[
\frac{a^n - b^n}{n(a-b)} =
\frac{1}{n}\left(a^{n-1} + ba^{n-2} + \cdots + b^{n-1}\right).
\]
The quantity on the right is the arithmetic mean
of $a^{n-1}, ba^{n-2}, \dots, b^{n-1}$.
Using the result that {\em Arithmetic Mean $\ge$ Geometric Mean},
with equality only when all terms are equal,
we deduce that
\begin{eqnarray*}
\frac{a^n - b^n}{n(a-b)} &>&
\left(a^{n-1}\cdot ba^{n-2}\cdot \cdots \cdot b^{n-1}\right)^{\frac{1}{n}}\\
&=& (ab)^{\frac{n-1}{2}}.
\end{eqnarray*}
\end{answer}
\item Describe geometrically the points $P$, $Q$ in an arbitrary triangle
$ABC$ that minimize
\begin{enumerate}
\item $ AP + BP + CP $
\item $ AQ^2 + BQ^2 + CQ^2 $.
\end{enumerate}
\begin{answer}
(a) The point $P$ minimizing $AP+BP+CP$ is the unique point $P$
such that
\[
A\hat{P}B = B\hat{P}C = C\hat{P}A = \frac{2\pi}{3} (= 120^\circ).
\]
(To construct this point, draw a circular arc on $AB$
such that $A\hat{P}B = 2\pi/3$ for all points $P$ on the arc;
and a similar circular arc on $BC$ such that $B\hat{P}C = 2\pi/3$
for all point on the arc.
Then the required point $P$ is the 2nd point---%
apart from the point $B$---%
where these 2 arcs meet.)
Suppose that $P$ minimizes $AP+BP+CP$.
Keep $CP$ fixed,
allowing $P$ to vary on a circle centred on $C$.
Draw the tangent $t$ to this circle at $P$.
A small variation $dP$ of $P$ along the tangent $t$
must leave $AP+BP$ unchanged, to the first order in $dP$.
It follows that $P$ must be the point on $t$ minimizing $AP+BP$.
But by the Law of Reflection,
this will occur when $AP$ and $BP$ make the same angle $\theta$ with $t$.
(To see this, let $B'$ be the reflection of $B$ in $t$.
Then
\[
AP + BP = AP + PB'
\]
will be minimized when $APB'$ is a straight line.)
But now
\[
A\hat{P}C = \theta + \frac{\pi}{2} = B\hat{P}C.
\]
Thus at the minimizing point,
\[
A\hat{P}C = B\hat{P}C.
\]
By the same argument with $B$ in place of $C$,
\[
A\hat{P}B = C\hat{P}B.
\]
Hence all 3 angles $A\hat{P}B, B\hat{P}C, C\hat{P}A$ are equal;
and since they add up to $2\pi$ ($360^\circ$),
each must be equal to $2\pi/3$ ($120^\circ$).
(b) The point minimizing $AP^2+BP^2+CP^2$ is the centroid $G$ of $ABC$.
For suppose $Q$ is the minimal point.
Let $A'$ be mid-point of $BC$.
Then
\[
AQ^2 + BQ^2 = 2A'Q^2 + 2A'B^2.
\]
So we have to minimize
\[
2A'Q^2 + QA^2.
\]
Evidently this will be minimal when $AQA'$ lie in a straight line.
Thus $Q$ must lie on the median $AA'$.
Similarly it must lie on the other 2 medians.
Hence it lies at the centroid $G$ of $ABC$.
\end{answer}
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\item For $m$ a positive integer, let $k(m)$ denote the largest integer
$k$ such that $2^k$ divides $m!$. Let $c(m)$ denote the number of 1's
in the binary representation of $m$. Show that $k(m) = m - c(m)$.
\begin{answer}
To compute $k(m)$,
consider how many of the numbers $1,2,\dots,m$
are divisible by $2^i$.
\[
\begin{tabular}{l}
$\left[\frac{m}{2}\right]$ numbers are divisible by 2.\\
$\left[\frac{m}{4}\right]$ numbers are divisible by 4.\\
\dots
\end{tabular}
\]
We deduce that
\[
k(m) = \left[\frac{m}{2}\right] + \left[\frac{m}{4}\right] + \cdots.
\]
Note that
\[
m = \frac{m}{2} + \frac{m}{4} + \cdots.
\]
Consider repeated division of $m$ by 2.
At the first step we obtain $[\frac{m}{2}]$,
with remainder 1 if $m$ is odd, ie if the last bit of $m$ is 1.
At the second step we obtain $[\frac{m}{4}]$,
with remainder 1 if the second last bit of $m$ is 1.
And so on.
We conclude that the sum of the remainders is equal to $c(m)$,
the number of bits of $m$ equal to 1.
Thus
\[
k(m) = m - c(m).
\]
\end{answer}
\item If $x_1$, $x_2$, \dots , $x_n$ are positive numbers and $s$ is their
sum, prove that
\[
(1+x_1)(1+x_2) \cdots (1+x_n) \leq 1 + s + \frac{s^2}{2!} + \cdots
+ \frac{s^n}{n!}.
\]
\begin{answer}
We have
\[
e^x = 1 + x + \frac{x^2}{2!} + \cdots .
\]
Thus
\[
(1+x_1)\cdots(1+x_n) \le e^{x_1}\cdots x^{x_n}
= e^s = 1 + s + \frac{s^2}{2!} + \cdots.
\]
But we can go further.
The left hand side only contains terms
of degree $\le n$ in $x_1,\dots,x_n$.
Therefore we need only include terms on the right of degree $\le n$.
But $s^{n+1}, s^{n+2},\dots$
only contain terms of degree $>n$, and can therefore be omitted.
Thus
\[
(1+x_1)(1+x_2) \cdots (1+x_n) \leq 1 + s + \frac{s^2}{2!} + \cdots
+ \frac{s^n}{n!}.
\]
\end{answer}
\item A table tennis club with 20 members organises 14 singles games
(2-player games) one Saturday morning in such a way that each member
plays at least once. Show that there must be 6 games involving 12
different players.
\begin{answer}
Call the players $1,\dots,20$.
Let player $i$ play $1+n_i$ games.
Then the total number of games is
\[
\frac{1}{2}\sum (1 + n_i) = 10 + \frac{1}{2}\sum n_i.
\]
But we know that there are 14 games. Hence
\[
\sum n_i = 8.
\]
Now let us go through the 20 players.
For each player with $n_i > 0$,
delete $n_i$ of his games.
This leaves at least
\[
\frac{1}{2}(20-8) = 6 \mbox{ games.}
\]
By construction, no player plays in more than 1 of these 6 games.
\end{answer}
\item Let $a_1$, $a_2$, \dots be the sequence of all positive integers
with no 9's in their decimal representation. Show that the series
\[
\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \cdots% \leq 30.
\]
converges.
\begin{answer}
Consider numbers with exactly $k$ digits.
Each such number is
\[
\ge10\cdots0 = 10^{k-1}.
\]
There are $10^k - 10^{k-1}$ k-digit numbers,
of which $9^k - 9^{k-1}$
do not contain the digit 9.
Thus the k-digit numbers not containing 9 contribute
\begin{eqnarray*}
&\le& \left(9^k - 9^{k-1}\right)10^{-(k-1)}\\
&\le& 10\left(\frac{9}{10}\right)^k
\end{eqnarray*}
to the sum.
Hence
\[
\sum \frac{1}{a_i} \le 10\sum \left(\frac{9}{10}\right)^k = 100.
\]
In particular the series converges.
\end{answer}
\item In a convex quadrilateral $ABCD$, let $E$ and $F$ be the
midpoints of the sides $BC$ and $DA$ (respectively). Show that the sum
of the areas of the triangles $EDA$ and $FBC$ is equal to the area of the
quadrilateral.
\begin{answer}
Writing $|XYZ|$ for the area of the triangle $XYZ$,
\[
|EDA| = \frac{1}{2} |DA| \times p,
\]
where $p$ is the perpendicular distance from $E$ to $DA$.
But
\[
p = \frac{1}{2}(p_1 + p_2).
\]
where $p_1,p_2$ are the perpendicular distances from $B,C$ to $DA$,
respectively.
Hence
\begin{eqnarray*}
|EDA| &=& \frac{1}{4} |DA| \times p_1 + \frac{1}{4} |DA| \times p_2\\
&=& \frac{1}{2} \left(|BDA| + |CDA|\right).
\end{eqnarray*}
Similarly
\[
|FBC| = \frac{1}{2} \left(|ABC| + |DBC|\right).
\]
Hence
\begin{eqnarray*}
|EDA| + |FBC| &=& \frac{1}{2} \left(|BDA| + |DBC| + |CDA| + |ABC|\right)\\
&=& \frac{1}{2} \left(|ABCD| + |ABCD|\right)\\
&=& |ABCD|
\end{eqnarray*}
\end{answer}
\end{enumerate}
\end{document}