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\title{TJ USAMO Practice 6 - Cyclic Quadrilaterals}
\author{PDiao05}
\date{\today}
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\maketitle

\par Cyclic quadrilaterals come up very often so it is instructive to learn as 
much as possible about them.  They are most often useful for angle chasing 
because as you will see, they involved a lot of angles.  It is important to 
realize that very few problems will only require noticing a cyclic quadrilateral
, more often, it could be one critical observation to get us closer to our goal.
Before I tell you anything, why don't we make some observations about cyclic 
quadrilaterals?

\section{What does a Cyclic Quadrilateral look like?}

\par To answer this question why don't we try drawing one?  Take a circle $\omega$ with
center O.  Inscribe a cyclic quadrilateral ABCD into it.  Now draw the diagonals AC and BD.
Let us make some observations.  The first thing to remember is that the measure of an inscribed
angle is half the measure of the central angle.  Also, the measures of two inscribed angles that
cut off the same arc are equal.  Using these two facts combined with the knowledge that a whole
circle constitutes $360^{\circ}$, we make two observations about the inscribed quadrilateral.  They 
turn out being if and only if statements:

\begin{enumerate}

\item Convex quadrilateral ABCD is cyclic iff $\angle ABD = \angle ACD$.
\item Convex quadrilateral ABCD is cyclic iff $\angle ABC + \angle CDA = 180^{\circ}$.

\end{enumerate}

\section{Some Theoroms}

\par Usually we will want to use cyclic quadrilaterals to do angle chasing since they give us a lot 
of information about angles within the quadrilateral.  To get information about other things there are
a couple of other theoroms you should know.

\begin{itemize}

\item \textbf{Power of a Point} - Take a circle $\omega$ and a point P.  Take any line through P that 
intersects the circle.  The product of the distance from the point P to the first intersection and the distance
from P to the other intersection is independant of the choice of line.  If the line is tangent, we just take the
tangent squared.

\item \textbf{Ptolemy's Inequality} - Take a quadrilateral ABCD.  We know that: $$(AB)(CD) + (BC)(AD) \ge (AC)(BD)$$
With equality iff ABCD is cyclic.

\item \textbf{Brahmagupta's Theorom} - Given a cyclic quadrilateral ABCD with side lengths a, b, c, and d.
$$K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$ where s is the semiperimeter and K is the area of the quadrilateral.

\end{itemize}

\section{Practice}

\begin{enumerate}

\item Prove Power of a Point.
\item (Honsberger) Let ABC be an equilateral triangle inscribed inside of a circle. Then choose a point P on the circumscribed
circle.  Prove that the sum of the shorter two of PA, PB, and PC equals the third.
\item Prove the equality part of Ptolemy's inequality.
\item (MOSP) In triangle ABC, points S and T lie on AB, such that $\angle ACT = \angle BCS$.  Points K and L are drawn such that
AK is perpendicular to CK and BL is perpendicular to CL.  CH is an altitude of the triangle and M is the midpoint of side AB.  Prove
that points H, M, K, and L are concyclic.

\end{enumerate}
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