A rectangular $A B C D$ billiard table, from the vertex $A$, we shoot a ball along the angle bisector direction, which, after bouncing off the $C D, B C$ and then $A B$ sides, precisely hits the ball standing in the center of the rectangle. In what other direction can we shoot the ball from point $A$ so that after bouncing off three different sides, it also precisely hits the ball standing in the center?
Solution. Consider the rectangular $ABCD$ representing the billiard table, which we repeatedly reflect onto the side where the collision occurs. Since the angle of incidence equals the angle of reflection, the path of the ball can be straightened (Figure 1). Let the side $AD$ of the rectangle $ABCD$ be $2a$ and the side $AB$ be $2b$ in length. In the isosceles triangle $ATK_3$, due to the reflections, $AT = 5a$ and $TK_3 = 3b$, so $5a = 3b$, from which $b = \frac{5}{3}a$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b4f82e4f0d793f63ed58g-1.jpg?height=515&width=585&top_left_y=388&top_left_x=759)

Figure 1

Since the ball collides with the sides three times, it must travel across 4 tables on the straightened path to the collision.

As shown in Figure 2, this can happen in 4 different ways. In the cases of "", and "", the ball bounces off only two different sides, so we exclude these. The case of "," is given as an example in the problem. Thus, only the case of ",c" remains. In the triangle $AKT$, $AT = 5b$ and $KT = 3a$. From the first case, we know that $b = \frac{5}{3}a$, so

$$
\operatorname{tg} \alpha = \frac{KT}{AT} = \frac{3a}{5b} = \frac{9}{25}, \quad \text{from which} \quad \alpha = 19.798^\circ \approx 19.8^\circ
$$

Therefore, the ball must be struck at an angle of 19.8 degrees relative to the side $AB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b4f82e4f0d793f63ed58g-1.jpg?height=851&width=1106&top_left_y=1428&top_left_x=499)