Game of pool? (own)

[nayel]

A frictionless rectangular table $ABCD$, where $AB=2BC$, has holes at $A,B,C,D$, and at the midpoints of $AB$ and $CD$. A ball on the table is hit from $A$ at an angle $\theta$ to $AB$; it can reflect off the boundary of $ABCD$. Find all angles $\theta$ for which the ball will eventually fall into a hole. (Think of the ball and the holes as points)

[Zzzz]

I will explain my solution later (Sorry... actually I don't have enough time to write it right now )

I just want to be sure about the solution. Is it

$\theta \in \{ \alpha : tan\ \alpha$ is rational $\}$

?

[nayel]

Yep, that's right

[*Mahi*]

Let us reflect the pool table along it's sides as many times needed(as I showed in the diagram).Then a ball ricocheted from the sides will be passing through the sides to a reflected pool table.
Now a ball falling in one of the holes will be a ball shot to a hole in some reflected table in the new diagram.
Now,let the side length of $AB=l$ and $BC=2l$
Let the ball fall through a hole in side $BC$ or $AD$ .Then the ball will be falling through a hole after $k$ reflections of $BC$ and $m$ reflections of $CD$.
Then the length traveled towards the '$AD$' axis is $2lm- \frac{2l} 2$
And the length traveled towards the $AB$ axis will be $lk$
So if it was shot at angle $\theta$ then $tan \theta =\frac{m-\frac 1 2} k=\frac {2m-1} {2k}$
Again let it fall through a hole in side $AB$ or $CD$ and just like above arguments let the ball be falling through a hole after $k$ reflections of $BC$ and $m$ reflections of $CD$.
Then if the angle was $\theta$ then $tan \theta = \frac {2lm} {lk- \frac l 2}=\frac {2m} {k- \frac 1 2}=\frac {4m} {2k-1}$
So if $tan \theta $ can be expressed in one of the ways shown above for natural $k,m$ then the ball shot at angle $\theta $ will fall at a hole.

(Please let me know if the solution is correct )

[Zzzz]

@Mahi.. Thanks for the diagram. I did it in same way. I skipped the calculation part of your solution. You assumed that $2AB=BC$ but the problem stated $AB=2BC$. That's not a problem. But probably you have considered holes in the middle of shorter sides. But there are holes in the middle of longer sides only.

As you said, we will reflect the table each time the ball reflects. So, we can consider all the holes as lattice points of a graph. The ball will go through a lattice point iff $tan\ \theta$ is rational.

[*Mahi*]

@Mahi.. Thanks for the diagram. I did it in same way. I skipped the calculation part of your solution. You assumed that $2AB=BC$ but the problem stated $AB=2BC$. That's not a problem. But probably you have considered holes in the middle of shorter sides. But there are holes in the middle of longer sides only.

As you said, we will reflect the table each time the ball reflects. So, we can consider all the holes as lattice points of a graph. The ball will go through a lattice point iff $tan\ \theta$ is rational.

Oh shit.....I think I adapted the problem as I wish.......You are right.....

[nayel]

Good job guys I think this is a good way of writing up the solution formally:

We can instead think of the ball moving forever in the same direction and the table being reflected every time the ball hits one of its walls. Now as the table is reflected the holes form a square grid, which we can think of as lattice points in the $xy$-plane. Then the problem is equivalent to finding all angles $\theta$ such that if we hit the ball in the $xy$-plane from the origin at angle $\theta$ to the $x$-axis, the trajectory of the ball will pass through another lattice point. *Now prove that this is possible iff $\tan\theta$ is rational.*