Bouncing Ball


A ball is projected from the bottom left corner of unit square $ABCD$ into its interior. We shall assume that the speed of the ball remains constant and it will continue bouncing off the edges until it arrives at a corner. For example, if the ball strikes $\frac{2}{3}$ of the way from $D$ to $C$ it will terminate at $D$.

Where must the ball strike on $DC$ to finish at $A$?


We shall solve this problem in two ways.

Method 1

By allowing the square to be reflected we can consider the reflection to be a continuous straight path. In the example below, the reflected path on the left diagram can be represented by the continuous path in the diagram on the right.

Extending this across the entire plane we can place the original unit square $ABCD$ on co-ordinate axes with $A$ at the origin.

It can be seen that for the ball to finish at $A$ it must terminate at one of the image points of $A$, represented by $A'$. As we are dealing with a unit square these points would be represented by the co-ordinate $(x,y)$ where both $x$ and $y$ are even. Let $x = 2a$ and $y = 2b$.

However, if a line passes through the point $(2a,2b)$ then it must first have passed through the point $(a,b)$, which means that it is impossible for the ball to return to its corner of origin before first arriving at one of the other corners.

Method 2

First we shall consider rectangle $ABCD$. Suppose the ball is projected from $A$ towards $E$, a point on $DC$, such that the length $DE$ splits $DC$ into equal sections. In the example below, $DE = \frac{1}{4}DC$.

It can be seen that the ball will terminate at $B$ if the number of sections is even and it will terminate at $C$ if the number of sections is odd.

We shall now approach the problem in a similar manner to the previous method. We shall allow the ball to bounce off the top and bottom edges: DC and $AB$, but instead of allowing the ball to bounce off the vertical edges: $DA$ and $CB$, we shall repeatedly reflect the square in a horizontal direction to consider the path of the ball to be a continuous path.

For example, in the unit square $ABCD$ let $DE = \frac{2}{3}$. As $3 \times \frac{2}{3} = 2$, it will be necessary to have two complete squares for the ball to reach a corner; in this case the ball would terminate at $D$.

In general, if $DE = \frac{p}{q}$, where $GCD(p,q) = 1$, then we shall have $q \times \frac{p}{q} = p$ complete squares and we shall generate $q$ sections.

We have already established that it is necessary for there to be an even number of sections in a rectangle for the ball to terminate in the bottom corner, so $q$ must be even. However, it can be seen that as the square is repeatedly reflected the bottom right corner alternates between the images of $A$ and $B$ Hence there must be an even number of squares for the image of $A$ to be in the bottom right corner.

This would require both $p$ and $q$ to be even which is a contradiction. Hence it is impossible for the ball to terminate at $A$.

Given that $DE = \frac{p}{q}$ where $GCD(p,q) = 1$, can you deduce which corner the ball will terminate in?
What would happen if $DE$ were irrational?

Problem ID: 363 (28 Oct 2009)     Difficulty: 3 Star

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