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\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip
\centerline{\large \bf Christchurch, 1998}
\bigskip
\medskip
\centerline{\Large \bf TEAM PROBLEMS NO. 2}
\vspace{2cm}
\begin{enumerate}
\item Let $S$ be a finite set of points in the plane with the following
property:
any three points $A$, $B$, $C$ of $S$ are not collinear and the
orthocentre
of the
triangle $ABC$ also belongs to $S$. Prove that $S$ contains $4$ points.
\item Prove that for any three points $A$, $B$, $C$ of
the hyperbola
$y={\lambda /x}$ the orthocentre $H$ of the triangle $ABC$ is defined
and is also a point of this hyperbola.
\item An infinite grid divides the plane into squares by vertical and
horizontal
lines. Prove that there exists an infinite set $S$ of grid-points with
the
following
property: any three points $A$, $B$, $C$ of $S$ are not collinear and
the
orthocentre of the triangle $ABC$ also belongs to $S$.
\end{enumerate}
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\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip
\centerline{\large \bf Christchurch, 1997}
\bigskip
\medskip
\centerline{\Large \bf TEAM PROBLEMS NO. 2}
\vspace{1cm}
\centerline{\Large \bf Solutions}
\vspace{1cm}
{\bf 1.} Let $A_1A_2\dots A_k$ be the convex hull of $S$. If $k\ge 5$
then
one of
the angles of this polygon is obtuse. Suppose, for example,
$\angle A_iA_{i+1}A_{i+2}$ is obtuse, then the orthocentre of the
triangle $
A_iA_{i+1}A_{i+2}$ is outside the polygon, which is a contradiction.
For the other two cases we need the following
\medskip
{\bf Lemma:} Let $ABC$ be a triangle with the orthocentre $H$. Then
$\angle
AHB=\pi -\angle C$.
{\it Proof} is by a simple calculation.
\medskip
Suppose $k\ge 4$. Then the Lemma above shows that the convex hull
$A_1A_2A_3A_4$
must be a rectangle. These four points satisfy the condition. Suppose
that $S$
contains a fifth point $P$. Without loss of generality we may assume
that $P$ is
inside the triangle $A_1A_2A_3$. Then the angle $A_1PA_3$ is obtuse,
thus
the angle
$A_1HA_3$, where $H$ is the orthocentre of $A_1PA_3$, is acute. Hence
$H$ is
outside the convex hull, the contradiction.
\medskip
Suppose $k\ge 3$. Then the convex hull is the triangle $A_1A_2A_3$; let
us
denote
its orthocentre by $H$. Of course, $H$ is inside $A_1A_2A_3$. If we had
a fifth
point $P$, then without loss of generality we may assume that $P$ lies
inside
$A_1HA_3$. Let us denote the orthocentre of $A_1HA_3$ by $Q$. Then
$\angle
A_1PA_3>
\angle A_1HA_3$, then by the lemma $\angle A_1QA_3<
\angle A_1A_2A_3$, which means that $Q$ lies outside the circumcircle of
$A_1A_2A_3$, the contradiction.
\bigskip
{\bf 2.} We will prove that for any
three points
$A_1$,
$A_2$,
$A_3$ of the hyperbola
$y={\lambda /x}$ the orthocentre $A_4$ of the triangle $A_1A_2A_3$ is
defined
and also a point of this hyperbola. Moreover, if
$$
A_1=(a_1,{\lambda \over a_1}),\ A_2=(a_2,{\lambda \over a_2}),\
A_3=(a_3,{\lambda
\over a_3}),
$$
then
$$
A_4=(a_4,{\lambda \over a_4})\quad \mbox{with}\ a_4=-{\lambda^2 \over
a_1a_2a_3}.
$$
Indeed, since a line can have only two points of
intersection with a hyperbola, the points $A_1$, $A_2$, $A_3$ are not
collinear.
For $A_4$ being the orthocentre it is necessary and sufficient
that the
following two
dot products are zero:
$$
\overrightarrow{A_1A_2}\cdot \overrightarrow{A_3A_4}=0,\qquad
\overrightarrow{A_1A_3}\cdot \overrightarrow{A_2A_4}=0.
$$
(the third will follow) and setting $A_4=(x,y)$ we obtain two linear
equations from
which
$$
x=-{\lambda^2\over a_1a_2a_3}, \qquad y=-{a_1a_2a_3\over \lambda }.
$$
{\bf 3.} The hyperbola $y={\lambda
/x}$ possesses the property we need but it does not have enough
points with integer coordinates on it. The idea is to consider rotations
of it
such that the orthocentre of any triangle, whose vertices are on the
hyperbola
and have integer coordinates, also has integer coordinates. Then we will
try to
secure that the rotated hyperbola has an infinite number of points with
integer
coordinates.
Let us rotate the hyperbola about the origin through the angle
$\alpha$ in clockwise direction and denote the coordinates of the point
$(x,y)$ in
the new coordinate system by $(X,Y)$. Then
\begin{eqnarray*}
X&=&x\cos \alpha -y\sin \alpha\\
Y&=&x\sin \alpha +y\cos \alpha
\end{eqnarray*}
and
\begin{eqnarray*}
x&=&\ \ X\cos \alpha +Y\sin \alpha\\
y&=&-X\sin \alpha +Y\cos \alpha
\end{eqnarray*}
The equation of the hyperbola will take the form
\begin{equation}
(Y^2-X^2)\,\frac{\sin \alpha \cos \alpha }{\lambda}
+XY\,\frac{ (\cos^2 \alpha -\sin^2 \alpha )}{\lambda} =1.
\end{equation}
Let us compute now the orthocentre $(X_4,Y_4)$ of the triangle with
vertices at
$(X_1,Y_1)$, $(X_2,Y_2)$, $(X_3,Y_3)$ in new coordinates. The conditions
$$
x_4=-\frac{y_1y_2y_3}{\lambda},\qquad y_4=-\frac{x_1x_2x_3}{\lambda}
$$
become
\begin{eqnarray*}
X_4\cos \alpha +Y_4\sin \alpha&=&-{1\over \lambda }\Pi^3_{i=1}
(-X_i\sin \alpha +Y_i\cos \alpha),\\
-X_4\sin \alpha +Y_4\cos \alpha&=&-{1\over \lambda }\Pi^3_{i=1}
(\ \ X_i\cos \alpha +Y_i\sin \alpha)
\end{eqnarray*}
Solving this system of linear equations we get
\begin{eqnarray*}
X_4&=&-\left[K\left(\frac{ (\cos^2 \alpha -\sin^2 \alpha
)}{\lambda}\right)
-L\left(\frac{\sin \alpha \cos \alpha
}{\lambda}\right)\right]\\
Y_4&=&-\left[M\left(\frac{ (\cos^2 \alpha -\sin^2 \alpha
)}{\lambda}\right)
+N\left(\frac{\sin \alpha \cos \alpha
}{\lambda}\right)\right]
\end{eqnarray*}
where $K=Y_1Y_2Y_3$, $L=X_1X_2X_3+Y_1Y_2X_3+Y_1X_2Y_3+X_1Y_2Y_3$,
$M=X_1X_2X_3$, $N=Y_1Y_2Y_3+X_1X_2Y_3+X_1Y_2X_3+Y_1X_2X_3$. If we
suppose
that all
$X_i$'s and $Y_i$'s are integers, then $K,L,M,N$ are integers too. In
order to
have $X_4$ and $Y_4$ integers it is enough to secure that the numbers
$$
\frac{ \cos^2 \alpha -\sin^2 \alpha }{\lambda}\quad \mbox{and}\quad
\frac{\sin \alpha \cos \alpha
}{\lambda}
$$
are integers. This is the same as saying that the coefficients of (1)
are
integers.
Let us show that, given any two integers $C$ and $D$ which are not both
zero, by a
rotation and a proper selection of
$\lambda$ we can ensure that the hyperbola $y=\lambda /x$ in new
coordinates will
be written as
\begin{equation}
C(Y^2-X^2)+DXY=1.
\end{equation}
Indeed, the equation (1) can be rewritten as
$$
(Y^2-X^2)\,\frac{\sin 2\alpha }{2\lambda}
+XY\,\frac{ \cos 2\alpha }{\lambda} =1.
$$
It is clear that for $\lambda =\sqrt{4C^2+D^2}$ the angle $\alpha$ with
the
property
$$
\frac{\sin 2\alpha }{2\lambda}=C,\qquad \frac{\cos 2\alpha }{\lambda}=D
$$
exists.
What is left is to find $C$ and $D$ for whichequation (2) has infinite
integer solutions. This search is easy as $C=D=1$ do the job. In this
case
we have
the equation
\begin{equation}
(Y^2-X^2)+XY=1
\end{equation}
for which we can prove that if $(X,Y)$ is a solution, then $(2X+Y,X+Y)$
is
another
solution. If we start from the solution $(0,1)$ then all solutions
obtained
this way
will be different as their $X$-coordinates grow.
\end{document}