Problem source

The lengths of the sides of a rectangular billiards table $ABCD$ are given by  $AB = DC = a$ and $AD=BC = 2b$. There are small pockets at the midpoints $M$ and $N$ of the sides $AD$ and $BC$, respectively. The sides of the table may be taken as smooth vertical walls. A small ball is projected along the table from the corner $A$. It strikes the side $BC$ at $X$, then the side $DC$ at $Y$ and then goes directly into the pocket at $M$.
The angles $BAX$, $CXY$ and $DY\!M$ are $\alpha$, $\beta$ and $\gamma$ respectively. On each stage of its path, the ball moves with constant speed in a straight line, the speeds being $u$, $v$ and $w$ respectively.  The coefficient of restitution between the ball and the sides is $e$, where $e>0$.
\begin{questionparts}
\item Show that $\tan\alpha \tan \beta = e$ and find $\gamma$ in terms of $\alpha$.
\item Show that $\displaystyle \tan\alpha = \frac {(1+2e)b} {(1+e)a}$ and deduce that the shot is possible whatever the value of $e$.
 
\item Find an expression in terms of $e$ for the fraction of the kinetic energy of the ball that is lost during the motion.
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\a{2};
        \def\b{1.618};
        \def\e{.8};
        \def\ta{((1+2*\e)*\b/(1+\e)/\a)};
        \coordinate (A) at (0,0);
        \coordinate (B) at (\a,0);
        \coordinate (C) at (\a,{2*\b});
        \coordinate (D) at (0,{2*\b});
        \coordinate (M) at (0,{\b});
        \coordinate (N) at (\a,{\b});

        \coordinate (X) at ({\a},{\ta*\a});
        \coordinate (Y) at ({\b/\ta},{2*\b});
        
        \draw (A) -- (B) -- (C) -- (D) -- cycle;

        \node[left] at (A) {$A$};
        \node[right] at (B) {$B$};
        \node[right] at (C) {$C$};
        \node[left] at (D) {$D$};
        \node[left] at (M) {$M$};
        \node[right] at (N) {$N$};
        \node[right] at (X) {$X$};
        \node[above] at (Y) {$Y$};

        \draw (A) -- (X) -- (Y) -- (M);

        \pic [draw, angle radius=.8cm, angle eccentricity=1.5, "$\alpha$"] {angle = B--A--X};
        \pic [draw, angle radius=.9cm, angle eccentricity=1.5, "$\beta$"] {angle = C--X--Y};
        \pic [draw, angle radius=.9cm, angle eccentricity=1.5, "$\gamma$"] {angle = D--Y--M};
        
    \end{tikzpicture}
\end{center}

\begin{questionparts}
\item The initial velocity is $u = \binom{u \cos\alpha}{u \sin \alpha}$, therefore $v = \binom{v_x}{u \sin \alpha}$. Newton's experimental law tells us $v_x = -e u_x = -eu \cos\alpha$, therefore $v = \binom{-eu \cos \alpha}{u \sin \alpha} = \binom{-v \sin \beta}{v\cos \beta} \Rightarrow -\tan \beta = -e \cot \alpha \Rightarrow \tan \alpha \tan \beta = e$.

There is nothing special about the result here, and so it must also be the case that $\tan \beta \tan \gamma = e \Rightarrow \tan \gamma = \tan \alpha$

\item $\tan \alpha = \frac{XB}{BA}$ so the point $X$ is at $(a, \tan \alpha a)$.  
The point $Y$ satisfies $\tan \beta = \frac{CY}{CX} = \frac{CY}{2b - \tan \alpha a}$ so the point $Y$ is $(a-(2b - a \tan \alpha)\tan \beta,2b) = (a - 2b\tan \beta + ea, 2b) = ((1+e)a-2b\tan \beta, 2b)$. 
Finally, the point $M$ is the midpoint, so $\tan \gamma = \frac{DM}{DY}$ so $M$ is the point $(0, 2b - ((1+e)a-2b\tan \beta)\tan \gamma) = (0, 2b - (1+e)a \tan \gamma - 2b e) = (0, (2b(1-e) - (1+e)a \tan \gamma)$, but $M$ is the point $(0, b)$, ie

\begin{align*}
&& b &= 2b(1-e) - (1+e)a \tan \gamma \\
\Rightarrow && b+2eb &= (1+e)a \tan \gamma \\
\Rightarrow && \tan \gamma &= \frac{(1+2e)b}{(1+e)a} \\
\Rightarrow && \tan \alpha &= \frac{(1+2e)b}{(1+e)a}
\end{align*}

Since $ \frac{(1+2e)b}{(1+e)a} = \frac{b}{a} + \frac{e}{1+e}b$ which is clearly an increasing function of $e$ on $[0,1]$, so $\tan \alpha \in \left [\frac{b}{a}, \frac{3b}{2a} \right]$ which are all all angles which place $X$ in sensible places, therefore we can always hit the middle pocket. (Except $e = 0$, where we would put the ball in $N$, but we are given $e > 0$).

\item After the first collision the velocity is $\binom{-eu \cos \alpha}{u \sin \alpha}$ after the second collision the velocity is $\binom{-eu \cos \alpha}{-eu \sin \alpha}$. Initial kinetic energy is therefore $\frac12 m u^2$ and final kinetic energy is $\frac12 m e^2u^2$ therefore the fraction lost is $\frac{\frac12 m u^2 - \frac12 m e^2u^2}{\frac12 m u^2} = 1-e^2$

\end{questionparts}