Problem Solving - Bouncing a billiards ball between two rays I am trying to understand how to do the following problem:
A billiard ball strikes the ray AX at point X, with angle of incidence α as shown. The billiard
ball continues its path, bouncing off rays AX and AR' according to the rule ”angle of incidence
equals angle of reflection.” If AX = AR', determine the number of times the ball will bounce
off the two line segments (including the first bounce, at C). Your answer will be a function of
α and β.
Now, I am trying to solve this in a way that takes advantage of the problem's symmetry. I know the answer to be:
$k=\left \lceil \frac{180-\alpha }{\beta } \right \rceil$
And the diagram that shows the symmetry looks like this:

I am having trouble trying to understand how that equation is derived from that diagram. If someone could help me understand this by explaining where this formula comes from, that would be excellent! Thanks
 A: I take it that $\angle R'AX = \beta$. Every time the ball hits a wall, we get a new line ($AX, AM, AN...$) which swings out by angle $\beta$. The ball will stop bouncing when the line swings out when the next line swings 'past parallel' to line $RX$, so that they don't intersect. As you can see in the diagram, this happens when the next line has rotated at least $180-\alpha$ degrees. Hence we seek the smallest $k$ such that $k \beta \geq 180 - \alpha$. This gives the formula.
A: I assume you mean for angle $\angle XAR'$ to have angular measure $\beta.$
Consider triangle $\triangle XAR.$
The angle $\angle AXR = \alpha.$
The angle $\angle XAR$ is subdivided by several repeatedly reflected copies of $\angle XAR'.$
Notice that $\angle XRA \leq \beta,$ since if $\angle XRA > \beta$
we could add another reflected copy of $\angle XAR'$ on the far side of ray $AR,$ and the additional reflected ray would intersect the line $XR$ somewhere on the side of $R$ opposite $X,$ contradicting the assumption that $R$ corresponds to the last "bounce" of the billiard ball.
And of course $\angle XRA > 0,$ otherwise we wouldn't have a triangle
where we have triangle $\triangle XAR.$
So if there are $k$ "bounces," including the initial bounce $X$ and the final bounce at a reflected image of $R,$ there are $k - 1$ copies of $\angle XAR'$ between the ray $AX$ and the ray $AR.$
Hence $\angle XAR = (k - 1)\beta.$
But the angles of a triangle sum to $180$ degrees, so
$$
\angle AXR + \angle XAR + \angle XRA
 = \alpha + (k - 1)\beta + \angle XRA = 180.
$$
Rearranging terms,
$$
 \angle XRA = 180 - \alpha - (k - 1)\beta. \tag1
$$
Since $0 < \angle XRA \leq \beta,$ the equation $(1)$ implies that
$$
0 < 180 - \alpha - (k - 1)\beta \leq \beta,
$$
and adding $(k - 1) \beta$ on both sides of each inequality it follows that
$$
(k - 1) \beta < 180 - \alpha \leq k\beta;
$$
dividing by $\beta$ (which is positive and therefore preserves the direction of inequalities),
$$
k - 1 < \frac{180 - \alpha}{\beta} \leq k. \tag2
$$
Since $k$ is an integer, equation $(2)$ tells us that
$$
\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = k.
$$
