The bouncing ball

Consider the rectangular lattice generated by $(m,0)$ and $(0,n)$ . The line $y = x$ represents the path of the ball on a series of (possibly reflected) images of the pool table. Bouncing off a wall corresponds to crossing over one of the sides of one of the fundamental rectangles.

The ball goes into the hole at $(ma,nb)$ where $ma=nb$ and $a$ and $b$ are the minimum positive solutions to this equation. Clearly $a = n/d, b = m/d$ where $d = {\rm gcd}(m,n).$ The ball bounces $a+b-2$ times. In order to maximize this, we should have $d = 1$ and so $M = 2013$ . The answer is the number of $(m,n)$ such that $m + n = 2015$ and ${\rm gcd}(m,n) = 1$ , which is equivalent to ${\rm gcd}(m,2015) = 1$ , so it's $\varphi(2015) = \fbox{1440}$ .

Imagine the ball lies in any corner and is hit at an angle of $\SI{45}{\degree}$ . After some distance it will reach a wall and bounce back. Now imagine you mirrored the pool table on the side where it bounced back - on the mirrored table, the ball follows the same line it had been on before (also see Vu Van Luan's solution)!

Notice: By mirroring the table along the side where a bounce occurs, the ball rolls along a line of the form $(x,\:y)=t(\pm1,\:\pm1),\quad t\in\mathbb{R}$ (signs depend on the initial direction the ball was hit to).

Notice: The corners of the mirrored $m\times n$ -table lie at $(km,\:ln),\quad k,\:l\in\mathbb{Z}$

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Now let's find the number of bounces $B\leq M$ . The ball reaches a hole of the (mirrored) table if $\begin{aligned} (x,\:y)&\overset{!}{=}(km,\:ln)&\Rightarrow&&\pm km&\overset{!}{=}\pm ln&\Rightarrow &&(km,\:ln)&=z\operatorname{lcm}(m,\:n)\cdot (\pm1,\:\pm1),&z&\in\mathbb{Z}&&(*) \end{aligned}$ We need the solution where the ball reaches a hole for the first time , i.e. $z=1$ . The number of bounces against the horizontal and vertical walls until the ball reaches a hole are $l-1,\:k-1$ respectively. We need to subtract one as there is no bouncing back when the ball reaches the hole: $\begin{aligned} B&=l+k-2\underset{(*)}{=}\frac{\operatorname{lcm}(m,\:n)}{m}+\frac{\operatorname{lcm}(m,\:n)}{n}-2&&\left|\operatorname{lcm}(m,\:n)\cdot \gcd(m,\:n)=mn\right.\\ &=\frac{m+n}{\gcd(m,\:n)}-2=\frac{N}{\gcd(m,\:n)}-2\leq N-2&&\left|m+n=N:=2015\right. \end{aligned}$ We really can get the maximum number of bounces (consider an $1\times(N-1)$ -table), so the maximum number of bounces is $M=N-2$ . Using $m+n=N$ , we find $\gcd(m,\:n)=\gcd(m,\:N-m)=\gcd(m,\:N)$

Notice: We get the maximum number of bounces if (and only if) $\gcd(m,\:n)=\gcd(m,\:N)=1$

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We can count the number of tables with maximum bounce by counting the numbers $1\leq m\leq N$ with $\gcd(m,\:N)=1$ . That number is given by $\varphi(N)=4\cdot 12\cdot 30=\boxed{1440}$

Let $m$ be on $x$ -axis and $n$ on $y$ -axis.

Then we can see that ball has to travel $\frac{LCM(m,n)}{n}$ on $x$ -axis and $\frac{LCM(m,n)}{m}$ on $y$ -axis

Then the number of bounces on side $m = \frac{LCM(m,n)}{n} - 1$ .

Similarly, number of bounces on side $n = \frac{LCM(m,n)}{m} - 1$ .

$\therefore$ Total number of bounces = $(\frac{LCM(m,n)}{n} -1)+(\frac{LCM(m,n)}{m}-1) = LCM(m,n)(\frac{1}{m} + \frac{1}{n})-2$

$\implies$ Total Bounces $= \frac{mn(m+n)}{HCF(m,n)mn}-2$

$\implies$ Total Bounces $= \frac{2015}{HCF(m,n)} -2$

$\therefore$ For maximum bounces $HCF(m,n)=1$ and $M=2013$ .

$1=HCF(m,n)=HCF(m,2015-m)=HCF (m,2015)$

$\implies \# m = \phi(2015) = 1440$

$\therefore$ Number of solutions $= 1440$ .