\begin{problem}{2}

TA's Paul and Zardosht are playing a game a chicken by racing on bicycles
toward each other (to see who turns aside first).  The initial distance
between them is $d$ and they each ride at $p$ mph.  Eric tries to mediate
the race by traveling back and forth between Paul and Zardosht on a
motorcycle.  Eric starts at the same initial position as Paul and drives
back and forth at a constant $m$ mph, where $m > p$.

\begin{problemparts}{1}

\problempart Write a simple expression using $d,p,z,m$ for
\begin{enumerate}

\item the total time it takes Paul and Zardosht to meet, and
\item the total distance back and forth Eric travels until Paul and
Zardosht meet.

\end{enumerate}

\solution{Paul and Zardosht close at a rate of $2p$ mph, so the time is
$h \eqdef d/2p$ hours.  Eric goes $md/2p$ miles in this time.}

\ppart Find a geometric series describing the total distance that Eric
travels.

\solution{If the bikes are a distance $d$ from each other when Eric leaves
one TA to head toward the other, he reaches the other TA after $t \eqdef
d/(m+p)$ hours having traveled $dm/(m+p)$ miles.  The distance between the
TA's at this time is $2pd/(m+p)$


leaves one player to meet up with the other player, Eric arrives at the
other player in time $d/60$ because Eric rides towards the other bike at
40 mph, and the other bike travels towards Eric at 20 mph, for a total of
60 mph of closing speed.  Then Eric travels $2d/3$ miles on that
journey. Thus, his first journey is 80 miles long.

We now try to describe how far Eric travels on a given journey
between the two players.  Say that on the previous journey, Eric
traveled $d$ miles, by the previous observation that must mean
the two bikes were $3d/2$ apart when Eric set out.

The total time that Eric was traveling is: $d/40$.  In that time, the two
bikes both traveled at 20 mph towards each other for that much time, and
the distance between bikes closes by $40 \cdot x/40 = x$. Thus, the two bikes
are now $3x/2-x=x/2$ miles apart.  Eric meets the other bike in $(x/2)/60$
hours, and travels $(x/2)/60 \cdot 40=x/3$ miles.

So, on every journey Eric travels a third as far as on his previous
journey.  This yields the infinite geometric series:
\begin{equation}\label{s}
80  \sum_{i=0}^\infty \paren{\frac{1}{3}}^i.
\end{equation}
Using the formula for the sum of a geometric series, we conclude that~\eqref{s}
equals $80(3/2) = 120$ miles.
}

\ppart Use the formula for the sum of a geometric sum to verify that the
value of the sum agrees with the answer to part~(a).

\end{problemparts}
\end{problem}
