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{\Large \sc Brutal Force \\ \mbox{}\\}
{\large \sc Po  ---  MOP '98 \\ \mbox{}\\ \mbox{}\\}
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This is probably going to be my final lecture at MOP (giving or receiving), so here's
what I'm leaving for the rest of you.

I'm showing how to use \emph{brutal force}\/ to solve geometry problems.  \emph{Brutal
force}\/ is just what I'm calling my method of twisting/smushing because hey,
twisting/smuching a diagram is exerting brutal force on it.

Now that my neat little introduction is over, let's get to the point.  What exactly is
this \emph{brutal force}\/ and how do I use it?

\vspace{0.25in}

There are many ways to deform a diagram.  Here are the basic tools:

\begin{description}

\item[Reflection] This is one of the most basic operations.  It involves finding a
line in the plane and reflecting [part] of the problem around it, possibly
superimposing the two images.  (Duh)

\item[Rotation] This is another basic operation.  Very useful in circles (for the
obvious reasons).

\item[Homothety] I hope you know what this is.  This is fun because stuff stays
similar and that's cool.

\item[Spiral Similarilty] This is a useful tool later, when we do much more drastic
stuff.  But what it really does is rotate something, and then homothety it around
somewhere.

\end{description}



All right!  Now that you know that, let's find out how to do some more advanced stuff.



\begin{description}

\item[Look for spinning stuff] In a picture, sometimes there's a hidden
spirally-similar figure.  Look closely, and you may be able to
rotate/homothety/spiral-similar something.

\item[Stretch] Here we go.  This is more interesting.  Stretching stuff is (kinda)
like homothety-ing something in one dimension.  A nice fact is that stuff always
travels in one direction, emabling you to do lots of cool stuff.  Important:
everything moves by the same ratio.

\item[Twist] This involves moving a part of the diagram along an arc of a circle.  The
neat thing that it does is preserve angles.

\item[Pull] This involves moving one part of the diagram along a ray, and then
adjusting the rest of the picture to abide by that deformation.

\end{description}



Those usually work.  But by far the most fun operations to perform are the things that
generalize facts.



\begin{description}

\item[``Induction'' on geometrical figures] This is the coolest (and also the most
difficult) algorithm to perform on a picture.  What you do is:

\begin{enumerate}
\item Find an easy (usually symmetric) case for the problem.  For example, if the
problem is about a triangle, show that it's true for an isosceles triangle.  This is
the base case.
\item Find a line/point/circle that the entire problem depends on.  Treat the problem
as a function:  the input is the position of that crucial line/point/circle and the
output is the status of the problem (T/F \ldots hopefully it's true).
\item Stretch/Twist/Pull that crucial element by an incremental amount ($\epsilon$,
$\theta$, $d$, etc.).
\item Superimpose the original and deformed figures (use at least two colors!).
\item Assuming the base case, show that the problem is still true for this deformed
figure.
\item By Po's Geometrical Induction Theorem (also the
\emph{If you can't see this you don't deserve to be reading this}\/ Theorem), the
problem is universally true. 
\end{enumerate}

\end{description}


\newpage

Now that you've got the tools, let's solve some problems!

\begin{enumerate}

\item (MOP Test \#3, Problem \#3) Let $\bigtriangleup ABC$ be inscribed in circle $O$. 
Let $X$ be the intersection of $BO$ and circle $O$.  Let $Y$ be the midpoint of $AC$. 
Choose an arbitrary $D$ on arc $AC$.  Construct equilateral triangles $CDF$ and $ADE$
outwards.  Let $M$ be the midpoint of $EF$.  Prove that $XY = XM$.

\item (Razvan, 6/19, \emph{Quadrilaterals}\/ \#6) Prove that if in a convex
quadrilateral two opposite angles are congruent, the bisectors of the other two angles
are parallel.

\item (Rasvan, 6/19, \emph{Quadrilaterals}\/ \#13) Prove that the interior bisectors of
the angles of a parallelogram form a rectangle whose diagonals are parallel to the
sides of the parallelogram.

\item (Elgin, 6/30, \emph{Complex Numbers}\/ \#4) Let $A$, $B$, $C$ be distinct points
on a line and $\alpha$ be a given angle with $0 < \alpha < 180$.  Isosceles triangles
$ABX$ and $BCY$, with $AX = BX$, $CY = BY$ and angle $AXB = \alpha =$ angle $BYC$ are
drawn so that their cyclic orientations $A$-$B$-$X$ and $B$-$C$-$Y$ are both
counterclockwise.  Then isosceles triangle $XYZ$ is drawn with $XZ = YZ$, angle $XZY =
\alpha$, and clockwise orientation $X$-$Y$-$Z$.  Prove that $Z$ is on the line
containing $A$, $B$, and $C$.

\item (MOP Test \#9, Problem \#5) Given a triangle $ABC$, construct the circle with
diameter $AB$.  Let the circle intersect $AC$ and $BC$ in $D$ and $E$.  Let the feet
of the perpendiculars from $D$ and $E$ to $AB$ be $F$ and $G$.  Let the intersection
of $DG$ and $EF$ be $M$.  Prove that $CM$ is perpendicular to $AB$.

\item (Rookie Contest, Po's Star Theorem) Given two congruent circles, $\omega_1$ and
$\omega_2$.  Let them intersect at $B$ and $C$.  Select a point $A$ on $\omega_1$. 
Let $AB$ and $AC$ intersect $\omega_2$ at $A_1$ and $A_2$.  Let $X$ be the midpoint of
$BC$.  Let $A_1X$ and $A_2X$ intersect $\omega_1$ at $P_1$ and $P_2$.  Prove that
$AP_1 = AP_2$.

\item (Razvan, 6/11, \emph{Triangles}\/ \#3) In the triangle $ABC$ we consider the
bisector $AA^\prime$.  The circumcircles of the triangles $ABA^\prime$ and
$ACA^\prime$ intersect $AC$ and $AB$ in $M$ and $N$.  Prove that $|BN| = |CM|$.

\item (Po, random problem) Let us have three concurrent circles $\omega_1$,
$\omega_2$, and $\omega_3$.  Let the point of concurrence be $P$.  Let the
intersection of $\omega_1$ and $\omega_2$ be $A$, the intersection of $\omega_2$ and
$\omega_3$ be $B$, and the other intersection be $C$.  Select a point $X$ on
$\omega_1$.  Let the intersection of $XA$ and $\omega_2$ be $Y$.  Let the intersection
of $YB$ and $XC$ be $Z$.  Prove that:

\begin{enumerate}
\item $Z$ lies on $\omega_3$.
\item $XY/XZ$ is a constant.
\end{enumerate}

\item (IMO Test \#3, Problem \#3) Let circle $\omega_1$, centered at $O_1$, and circle
$\omega_2$, centered at $O_2$, meet at $A$ and $B$.  Let $l$ be a line through $A$
meeting $\omega_1$ again at $Y$ and meeting $\omega_2$ again at $Z$.  Let $X$ be the
intersection of the tangent to $\omega_1$ at $Y$ and the tangent to $\omega_2$ at $Z$.
 Let $\omega$ be the circumcircle of $O_1O_2B$, and let $Q$ be the second intersection
of $\omega$ with $BX$.  Prove that the length of $XQ$ equals the diameter of $\omega$.

\item (APMO '98, Problem \#4) Let $ABC$ be a triangle and $D$ the foot of the altitude
from $A$.  Let $E$ and $F$ be on a line passing through $D$ such that $AE$ is
perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different
from $D$.  Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$,
respectively.  Prove that $AN$ is perpendicular to $NM$.

\item (Lots of other problems) Go look up any geometry problem and try to induct it.

\end{enumerate}














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