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\title{HMO2 Solutions by }
\author{G.~C.~Smith, Universities of Bath and Paris VI}
\date{14/1/2001}
\begin{document}
\maketitle

\begin{enumerate}
\item {\em The graphs of $xy = 1$ and $xy = -1$ are given in the same
co-ordinate system. We have a circle around the origin with radius $R$.
The common points of the circle and the two hyperbolas are the vertices of
a regular $n$-gon. Show that the area of the $n$-gon is $R^4$.}

\noindent{\bf Solution } The union of the graphs of the hyperbolas
has equation $x^2y^2 =1$. Solving this against $x^2 + y^2 = R^2$
we obtain $x^2 = (R^2 \pm \sqrt{R^4 -2})/2.$ Thus there are either 
0, 4 or 8 points of intersection of the circle with the union of the
hyperbolas. We investigate for which values of $R$ these points form the
vertices of a square or a regular octagon.

The square is obtained when $R = \sqrt 2$, and then $x^2 =1$.
The area of the square is $4x^2 = 4 = r^4.$

Let $z = \tan \pi/8$ so 
\[ 1 = \tan \pi/4 = \frac{2z}{1-z^2}\]
so $z= \sqrt 2-1$. The condition that the vertices form a regular
octagon amounts to the condition that the circle, the hyperbola $xy =1$
and the ray $y = zx$ $(x > 0)$ should have a point of intersection.
Solving $y = zx$ and the hyperbola
yields $x^2 = \sqrt 2 + 1$ so $x = \sqrt{\sqrt{2} + 1}$
and $y = \sqrt{\sqrt{2}-1}$. Now $R^2 = x^2 + y^2 = 2\sqrt 2$
so $R^4 = 8$ but using ``half height times base'', the area of
the octagon is \[ 16 \cdot x \cdot y =
16 \cdot \frac{1}{2} \cdot \sqrt{\sqrt{2}+1}\sqrt{\sqrt{2}-1} = 8 = R^4.\] 
\item {\em Find $x,y,z,t \in \mathbb R \setminus \{0, -1\}$ for which
both (a) and (b) hold:
\begin{enumerate}
\item[(a)] 
$x + y + z = 3/2$
\item[(b)] $\sqrt{4x -1} + \sqrt{4y -1} + \sqrt{4z-1} \geq 2 + 3^{\sqrt{t-2}}.$
\end{enumerate}}
\noindent{\bf Solution } Notice that $x,y,z$ must all be at least 1/4 so that
the left-hand side will make sense. Apply the Cauchy-Schwarz inequality:
\[ (1,1,1) \cdot (\sqrt{4x-1}, \sqrt{4y-1}, \sqrt{4z-1}) \leq \sqrt 3
\cdot \sqrt{4(x+y+z)-3}= 3.\] Thus the given condition holds if and only
if $t=2$, and the vectors in the dot product are parallel,
so $x = y =z =
1/2$. 
{\bf Comment: This method is obvious. I was amazed to see other (more complicated)
methods in Hungary (Jensen indeed!).}

\item {\em $f(x)$ is defined on $\mathbb R \setminus \{0, -1\}$
and takes real values. Determine $f(x)$ if for every $x\ (\not = 0,-1)$
\[ f(x) = kx^2 f\left(\frac{1}{x}\right) = \frac{x}{x+1}.\]
(here $k$ is a constant, $0 < k^2 \not = 1)$. Find those values
of $x$ for which $f(x) = 0$.}\newline
\noindent{\bf Solution} Replace $x$ by $1/x$ and solve the linear equations.
In the `unknowns' $f(x)$ and $f(1/x)$ and deduce that
\[f(x) = \frac{x(1-kx)}{(x+1)(1-k^2)}.\] Notice how the side conditions
keep the denominator under control. 
To obtain full credit, you must substitute back into the original
equation to verify that this is indeed a solution.
Now $f(x) = 0$ exactly when $x = 1/k$ (a sucker will allow $x=0$
but this violates a side condition).

\item {\em In triangle $ABC$ the bisector of $\angle BAC$ meets the incircle
first at $O_A.$ 
($O_A$ is closer to $A$ than the other common point of the bisector and the   
incircle.)
Let $k_A$ be the circle with centre $O_A$ which nis tangent to
$AB$ and $AC$. We get $k_B$ and $k_C$ similarly.
The external common tangent of $k_B, k_C$ which is not a side of
$\Delta ABC$ is $t_A$. We get $t_B, t_C$ similarly. Prove that
$t_A, t_B, t_C$ are concurrent.}\newline
\noindent{\bf Solution (outline)} Recall the following important fact
about the orthocentre and circumcircle of a triangle:
{\em the reflection of the orthocentre in any side of a triangle
is on the circumcircle.} A short proof is available by joining
the orthocentre to one of the vertices on the reflecting side,
and chasing angles. Now to the question. The point of concurrency turns
out to be the orthocentre $H$ of $\Delta O_AO_BO_C$. In order to verify this
fact, it suffices to do the following: drop the perpendicular $IF_A$
to $BC$. Draw $O_AH$ and $HF_A$. It suffices to show that $O_BO_C
\perp HF_A$ by the important fact which we recalled earlier, since
then $H$ will sit on one (and by symmetry all) of the internal
common tangents mentioned in the question.
Thus we have reduced the question to a single perpendicularity problem.
Now draw another diagram which is less littered (omit the smaller circles
and the internal common tangents). Let the angles of $\Delta ABC$ be
$2\alpha, 2\beta, 2\gamma$ respectively, then chase angles (nailing down
the angles around $I$ is helpful), and soon you will be done. 
\item 
{\em In a folk dance the dancers are standing in two rows,
9 boys facing 9 girls. Each dancer gives his/her left hand to
the person opposite, to his/her left neighbour, or to the person
opposite his/her left neighbour. The analogous rule applies
to right hands. Nobody gives both hands to the same person.
\begin{enumerate}
\item[(a)] Find the number of possible configurations.
\item[(b)] Is 2002 a divisor of the number of configurations
if there are 2002 people (rather than 18).
\end{enumerate}}
\noindent{\bf Solution } \begin{enumerate}
\item[(a)] Replace 9 by $n$ and induct on it (inducting on 
9 itself may confuse the casual reader). Let the number of configurations
with $n$ couples be $u_n$ so $u_1 = 0$ and $u_2 = 2$ (the square or the infinity
symbol). An induction argument for $n \geq 3$ goes as follows: either
couple $n-1$ hold hands or they don't. The first possibility gives rise to
$2u_{n-2}$ configurations (letting couples $n-1$ and $n$ do their
own thing in two ways) and the second to $2u_{n-1}$ configurations
(by slicing off the arms linking couples
$n-1$ and $n$, and then jamming the blokes and the blokesses together
to avoid unsightly embarrassment and to create a replacement couple $n-1$).
Thus $u_n = 2u_{n-1} + 2u_{n-2}$ for every $n \geq 3$. Thus the sequence
goes
\[ 0, 2, 4, 12, 32, 88, 240, 656, 1792\]
and the number of configurations with 9 couples is 1792 (which has a certain 
{\em je ne sais quoi}).
{\bf Note that if you are honest enough to allow
$n=0$ then there is a unique empty configuration for them, $u_0 = 1$
and the induction formula still works; such is the power of careful reasoning.}
\item[(b)] Note that $2002 = 2 \times 7 \times 11 \times 13$ {\bf A fact which should
be inscribed upon your soul.} 
Consider the sequence $u_n$ (beginning $n=1$) modulo each of the primes involved
in the factorization of 2002. Of course all terms $u_n$ are even 
so there is no problem there. You discover that
the sequence is
\[0,2,4,5,4,4,2,5,0,3,\ldots \mbox{ mod } 7\]
and since $2 \times 5 = 3 \mbox{ mod } 7$ we can write this as
\[0,2,4,5,4,4,2,5, 5\times (0,2,4,5,4,4,2,5, \ldots)\]
which is an ostentatious way of showing that $u_n \equiv 0 \mbox{ mod }7$
if and only if $n \cong 1 \mbox{ mod } 8$, so 7 divides $u_{1001}.$
Now for 11. The series is
\[0,2,4,1,10,0,9,7,10,1,0,2, \ldots\]
and this time we haven't the cheek to repeat the trick since the series repeats
so quickly. Now $u_n \equiv 0 \mbox{ mod }11$ if and only if $n \equiv 1$ mod $5$
so $11$ divides $u_{1001}$. Next work modulo 13. The series is
\[0,2,4,12,6,10,6,6,11,8,12,1,0,2 \ldots\]
and the series repeats. Thus $u_n \equiv 0 \mbox{ mod }13$ if and only if
$n \equiv 1$ mod $12$. But 12 is not a divisor of 1000, so
$u_{1001}$ is not divisible by 13 and therefore not divisible
by 2002.
\end{enumerate} 
\end{enumerate}
\end{document}