\documentclass[a4paper,12pt,twoside]{amsart}

\title{Math 304, 2009/10, second examination \emph{solutions}}
\author{David Pierce}
\email{dpierce@metu.edu.tr}
\address{Mathematics Dept., Middle East Technical University, Ankara
06531, Turkey} 
\urladdr{metu.edu.tr/~dpierce/}
\date{Tuesday, May 18, 2010}

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\newtheorem*{solution}{Solution}

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\begin{document}

\maketitle
\thispagestyle{empty}

\begin{problem}\mbox{}
The ellipse $AEB$ is
determined as follows. 
Triangle $ABC$ is given, the angle at $A$ being right.  If a point $D$
is chosen at random on $AB$, and $DE$ is erected at right angles to
$AB$, then $E$ lies on the ellipse if (and only if) the square on $DE$
is equal to the rectangle $ADFG$ (which is formed by letting $ED$,
extended as necessary, meet $BC$ at $F$).  Let also the circle $AHB$
with diameter $AB$ be given.

Find $h$ (in terms of the given straight lines) such that $h$ is
to $AB$ as the ellipse is to the circle.
Prove that your answer is correct, using Newton's lemmas as needed.
\end{problem}
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\begin{remark}
  The ellipse appears to result from contracting the circle in one
  direction.  If this is so, then by Newton's Lemma 4, the ratio of
  ellipse to circle is the factor of contraction, which should be
  $DE/DH$.  So one should find this ratio and check that it is indeed
  independent of the choice of $D$.

Two students solved this problem perfectly.  Five others used without
proof a rule for the area of an ellipse; but we do not officially have
such a rule, and in fact the point of this problem is to establish
this rule.
\end{remark}

  \begin{solution}
    By construction and the similarity of the triangles $BDF$ and $BAC$,
    \begin{equation*}
      DE^2=ADFG=AD\times DF=AD\times DB\times\frac{AC}{AB}.
    \end{equation*}
In the circle,
\begin{equation*}
  DH^2=AD\times DB.
\end{equation*}
Let $h$ be a mean proportional of $AB$ and $AC$, so
\begin{align*}
  h^2&=AB\times AC,&
\frac{AC}{AB}=\frac{h^2}{AB^2}.
\end{align*}
Then
\begin{align*}
  \frac{DE^2}{DH^2}&=\frac{AC}{AB},&
\frac{DE}{DH}=\frac h{AB}.
\end{align*}
If we inscribe series of parallelograms in the ellipse and circle, all
of the same breadth, then corresponding parallelograms will be to each
other as $DE$ to $DH$, that is, $h$ to $AB$.  Therefore this is the
ratio of the ellipse to the circle [by Newton's Lemma 4].
  \end{solution}


  \begin{problem}
We have used without proof Propositions I.33 and 49 of the
\emph{Conics} of Apollonius.  This problem is an opportunity to prove
those propositions, using the techniques of Descartes and Newton as
appropriate.

    A straight line $\ell$ (not shown), a curved line
    $ABE$, and a straight line
    $AC$ are given such that, whenever a point $B$ is chosen at random
    on $ABE$, and straight line $BC$ is dropped
    perpendicular to $AC$, then the square on $BC$ is equal to the
    rectangle bounded by $\ell$ and $AC$.  So $ABE$ is a parabola with
    axis $AC$.

Let $B$ now be fixed; so we may write $BC=a$ and $AC=b$.  Extend $CA$ to $D$
so that $AD=AC$.  Draw straight line $DBK$, and let $c=BD$.

Let a point $E$ be chosen at random on the parabola $ABE$.  Draw
straight lines $BF$ parallel to $AC$, and $EF$ parallel to $BD$.
\begin{enumerate}
\item
Show that the parabola
$ABE$ must indeed lie all on one side of $DBK$.
\item 
Show that the square on $EF$ varies as $BF$, and find $m$ (in
terms of $a$, $b$, and $c$ only) such that $m\times BF$ is equal to the
square on $EF$.  For your computations, let $x=EF$ and $y=BF$.
\item
Explain why $BD$ is tangent to the parabola at $B$.
\end{enumerate}
  \end{problem}

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  \end{center}

  \begin{remark}
    One approach to (a) is showing that $E$ lies above $K$.  The
    height of $E$ above $D$ is the length of $DH$; by similarity of
    triangles, the height of $K$ above $D$ is $2b/a$ times $EH$.  The
    point of using $DH$ and $EH$ is that we know how their lengths are
    related.  Two students solved this problem perfectly; one other
    was partially successful.

In (b), we want to find $x^2/y$ in terms of fixed magnitudes.  We have
one equation, $EH^2=\ell\times AH$, and we can write this in terms of
$x$ and $y$ (and fixed magnitudes) by using the similar triangles
$BCD$ and $EGF$.  Three students solved this problem completely; two
others got halfway there.

For (c), one student showed that $DB$ is the only straight line
passing through $B$ and meeting $AD$ that meets the parabola exactly
once.  A number of students observed that $DB$ does meet the parabola
just once; but this is not enough to establish that $DB$ is a
tangent.  Note also that $BG$ also meets the parabola exactly once,
but is not a tangent.
  \end{remark}

  \begin{solution}
    \begin{enumerate}
    \item 
Assuming $KE$ is parallel to $AC$, drop a perpendicular $KL$ to $AC$.
We want to show $DH\geq DL$ or $AH\geq AL$.  We have
\begin{align*}
  AH&=\frac{EH^2}{\ell},& DL&=LK\times\frac{2b}a=EH\times\frac{2b}a,
\end{align*}
so
\begin{equation*}
  \ell\times(DH-DL)=EH^2+b\ell-EH\times\frac{2b\ell}a
=EH^2+a^2-EH\times2a
=(EH-a)^2,
\end{equation*}
which is positive when $E$ is not $B$; so $DH>DL$.
\item
We have $EH^2=\ell\times AH$.
Since $EG=ax/c$ and $GF=2bx/c$, this means
\begin{gather*}
  \Bigl(a+\frac{ax}c\Bigr)^2=\ell\Bigl(y+\frac{2bx}c+b\Bigr),\\
a^2+\frac{2a^2x}c+\frac{a^2x^2}{c^2}=\ell y+\frac{2b\ell x}c+b\ell,
\end{gather*}
and since $a^2=\ell b$, we have
\begin{align*}
  \frac{a^2x^2}{c^2}&=\ell y,& x^2&=\frac{c^2}{a^2}\ell y,&
m&=\frac{c^2}b.
\end{align*}
\item
In the figure, as $E$ approaches $B$, $EK$ varies as $BK^2$.
Therefore $EK/BK$ varies as $BK$, so the angle $EBK$ ultimately vanishes.
    \end{enumerate}
  \end{solution}

\end{document}
