## Task A-1.3.

A billiard table has the shape of a rectangle $A B C D$ with dimensions $|A B|=2 \mathrm{~m}$ and $|B C|=1 \mathrm{~m}$. A billiard ball moves straight on the table until it reaches the edge of the rectangle, at which point it bounces off such that the path of the ball before and after the bounce forms equal angles with the edge. If the billiard ball starts moving from point $A$ and after bouncing off the sides $\overline{C D}, \overline{B C}$, and $\overline{A B}$ in sequence, it ends its movement at point $D$, determine the total distance the ball has traveled. The ball is considered as a material point.

![](https://cdn.mathpix.com/cropped/2024_05_30_bdf789747fbcdaaac460g-03.jpg?height=323&width=399&top_left_y=1455&top_left_x=1271)
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bdf789747fbcdaaac460g-03.jpg?height=529&width=973&top_left_y=1940&top_left_x=473)

Let $E, F,$ and $G$ be the points where the billiard ball bounces off the sides $\overline{C D}, \overline{B C},$ and $\overline{A B}$, respectively. The condition of the billiard ball's bounce means that the following equalities hold:

$$
\varangle A E D = \varangle C E F, \quad \varangle E F C = \varangle G F B, \quad \varangle F G B = \varangle D G A.
$$

Let $l = |A E| + |E F| + |F G| + |G D|$ be the length of the path traveled by the billiard ball.

Consider the triangles $A G D$ and $D E A$. These are right triangles with right angles at vertices $A$ and $D$, respectively, and they share the leg $\overline{A D}$. For the angles opposite this leg, we have:

$$
\varangle D G A = \varangle F G B = 90^{\circ} - \varangle G F B = 90^{\circ} - \varangle E F C = \varangle F E C = \varangle A E D
$$

Therefore, all the angles of these triangles match in pairs, so the triangles are congruent by the $\mathrm{K}-\mathrm{S}-\mathrm{K}$ (Angle-Side-Angle) congruence theorem.

From this congruence, it follows that $|A E| = |D G|$ and $|A G| = |D E|$, and thus $|B G| = |C E|$.

Consider the right triangles $B G F$ and $C E F$. All their angles match in pairs (due to $\varangle E F C = \varangle G F B$), and they have one pair of sides of equal length (due to $|B G| = |C E|$). Therefore, they are also congruent by the $\mathrm{K}-\mathrm{S}-\mathrm{K}$ (Angle-Side-Angle) congruence theorem.

We conclude that

$$
|C F| = |F B| = \frac{1}{2} |B C| = \frac{1}{2} \text{ m}
$$

and $|E F| = |F G|$.

Since $\varangle D G A = \varangle F G B$, the right triangles $D G A$ and $F G B$ are similar. Therefore, we have

$$
\frac{|D G|}{|F G|} = \frac{|A G|}{|G B|} = \frac{|F B|}{|D A|} = \frac{1}{2}
$$

Together with $|A G| + |G B| = 2 \text{ m}$, we get $|B G| = \frac{2}{3} \text{ m}$.

The length of the path traveled by the billiard ball is

$$
\begin{aligned}
l & = |A E| + |E F| + |F G| + |G D| \\
& = |G D| + |F G| + |F G| + |G D| \\
& = 2|F G| + |F G| + |F G| + 2|F G| \\
& = 6|F G|
\end{aligned}
$$

Applying the Pythagorean theorem to triangle $F G B$, we get

$$
|F G| = \sqrt{|F B|^2 + |G B|^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2} \text{ m} = \frac{5}{6} \text{ m}
$$

Thus, $l = 5 \text{ m}$.