A circular billiard table with a diameter of 2 meters has a billiard ball at point $P$, which is $\frac{1}{2}$ meter away from the center $O$ of the table. The ball needs to be hit in such a way that after two bounces, it passes through $P$ again. What angle does the direction of the hit make with the $P O$ direction in this case?
I. Solution: Let the points of reflection be denoted by $Q$ and $R$ (Fig. 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_7ece04176462bd38c501g-1.jpg?height=718&width=653&top_left_y=218&top_left_x=714)

The $\triangle PQR$ is isosceles, and by symmetry, $PO \perp QR$. The normal at the point of incidence (the line perpendicular to the tangent at the point of reflection) in this case is the radius $QO$. Let $\angle OQP = \angle OQR = \alpha$, the intersection of the line $PO$ with $QR$ be $S$, and the desired $\angle OPQ$ be denoted by $x$. Then $OS = \sin \alpha$, and according to the problem, $OQ = 1$, and $OP = \frac{1}{2}$.

Using the Angle Bisector Theorem,

$$
\frac{OS}{OP} = \frac{QS}{QP}
$$

which gives

$$
2 \sin \alpha = \cos 2\alpha
$$

But $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2 \sin^2 \alpha$. Substituting this value on the right-hand side and rearranging, we get the trigonometric equation

$$
2 \sin^2 \alpha + 2 \sin \alpha - 1 = 0
$$

One of the roots of this equation has an absolute value greater than 1, and the usable root is

$$
\sin \alpha = \frac{\sqrt{3} - 1}{2} \approx 0.3660, \quad \text{and thus} \quad \alpha = 21^\circ 28'
$$

(The obtuse angle solution is again not considered.)

Therefore, the desired angle is

$$
x = 90^\circ - 2\alpha = 90^\circ - 42^\circ 56' = 47^\circ 4'
$$

II. Solution: We can arrive at equation (1) in several ways without using the Angle Bisector Theorem. The simplest way is to apply the Law of Sines to the triangle $\triangle OPQ$:

$$
\sin x : \sin \alpha = 1 : \frac{1}{2}
$$

which gives

$$
\sin x = 2 \sin \alpha
$$

But $\sin x = \sin (90 - 2\alpha) = \cos 2\alpha = 1 - \sin^2 \alpha$. Substituting this value into (2), we obtain equation (1).

Remark: We do not need to consider the trivial case where the ball is shot in the direction of $PO$ and passes through point $P$ again after one (or any number of) reflections, as this case is essentially excluded by the problem statement.