A circular billiard table has a diameter of 2 meters. A perfectly elastic (point-like) ball placed at a point $P$, which is $40 \mathrm{~cm}$ away from the center, is hit such that after two bounces, it passes through $P$ again. Continuing the ball's path, how far will the ball be from $P$ after the tenth bounce?
Let $A$ and $B$ be the points on the perimeter of the table where the ball bounces consecutively, and let $O$ be the center of the circle (the table). The order of $A$ and $B$ can be switched, since in the sequence $P, A, B, P$, the law of elastic reflection implies that the line $AB$ is the reflection of $AP$ over the symmetry axis $AO$ of the circle, and $BP$ is the reflection of $BA$ over $BO$. Therefore, the path will be the same if the ball is launched from $P$ towards $B$. It is also evident that the ball's path is symmetric with respect to the axis $PO$, so $AB \perp PO$.

# $1985-04-156-1 . \mathrm{eps}$ 

Figure 1

Let the midpoint of $AB$ be $F$, and let $OP = p$, $OF = x$. According to the problem, $OA = r = 1$ (meter) and $\angle PAO = \angle OAB = \alpha$, $\angle OPA = 90^\circ - 2\alpha$.

In the triangle $OPA$, by the Law of Sines,

$$
\begin{array}{ll}
OA \sin \alpha = OP \sin \angle OPA, & r \sin \alpha = p \cos 2\alpha = p(1 - 2 \sin^2 \alpha) \\
2p \sin^2 \alpha + r \sin \alpha - p = 0, & \sin \alpha = \frac{1}{4p} \left(-r \pm \sqrt{r^2 + 8p^2}\right) = \frac{OF}{r}
\end{array}
$$

With our data, the positive root of the equation is $\sin \alpha = (\sqrt{57} - 5) / 8$. The other root is negative because the constant term has the opposite sign of the coefficient of $\sin^2 \alpha$, and it does not represent a solution for us. The positive root is valid only if $\alpha$ is an acute angle, so $\alpha = 18^\circ 35' 10''$ and $OF = 31.87 \, \text{cm}$.

For the further path of the ball, let the points of reflection be $A_3, A_4, \ldots, A_{10}, A_{11}$, with $A_1 = A$ and $A_2 = B$. We characterize their positions by the angle of rotation from the reference direction $OP$. For $OA_1$, $\varphi_1 = 90^\circ + \alpha$, and for each subsequent ray, the increase in $\varphi$ is $\angle A_iOA_{i+1} = \angle AOB = 180^\circ - 2\alpha$, so

$$
\varphi_i = (2i-1) \cdot 90^\circ - (2i-3) \alpha
$$

Subtracting the appropriate multiple of $360^\circ$,

$$
\varphi_{10} = -45^\circ 57' 49'' \quad \varphi_{11} = 96^\circ 51' 50''
$$

($OA_{11}$ is about $12^\circ$ behind $OA_1$ (Figure 2), and $OA_6$ is about $6^\circ$ behind; the ball roughly traces the perimeter of a regular star polygon.)

## $1985-04-156-2$. eps

Figure 2

Let the intersection of $OP$ and $A_{10}A_{11}$ be $M$, and the projection of $P$ onto the segment $A_{10}A_{11}$ be $P'$. Then $PP' = PM \cos \gamma$, where $\gamma = (\varphi_{10} + \varphi_{11}) / 2 = 25^\circ 27'$ and $PM = OP - OM$. We can calculate the subtrahend from the triangle $MOA_{11}$:

$$
OM = OA_{11} \cdot \frac{\sin 18^\circ 35' 10''}{\sin (90^\circ - \gamma)} = 0.353
$$

since $\angle MA_{11}O = \alpha$. Therefore,

$PP' = (0.4 - 0.353) \cos \gamma = 0.0425 \, \text{m}$, or $PP' = 4.25 \, \text{cm}$.

Remarks. 1. Some considered the trivial and meaningless statement that the ball must be launched along the diameter passing through $P$. In this case, after the first bounce, it will pass through $P$ again, and in every subsequent path, and the distance in question is zero in any bounce. We do not argue with this interpretation.

2. Some started from the assumption that the path must be a regular star polygon, and thus the data $OP = 0.4 \, \text{m}$ is "incorrect". The truth is that the data comes from rounding the number $\cos 72^\circ / \cos 36^\circ = (3 - \sqrt{5}) / 2 = 0.382 \ldots$—this is the distance from the center to the intersection of two sides of a regular star polygon. We deliberately "slightly distorted" the regularity with the rounding. The world is not made up of only regular things, and most things are far from any kind of idealized regularity.