Bruno and Pedro play billiards on a strange table. It contains two walls that meet forming an angle of $20^{\circ}$. They observed that the reflections of the ball against the walls are perfect, that is, if the trajectory of incidence makes an angle of $x^{\circ}$ with the wall, then the angle that the reflected trajectory makes with the wall will also be equal to $x^{\circ}$. The figure below illustrates this reflection rule.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-19.jpg?height=271&width=697&top_left_y=658&top_left_x=788)

Playing with the ball, they noticed that it is possible that, after reflecting several times off the wall of the table, the ball's trajectory intersects itself. For example, Bruno launched a ball in such a way that, after 3 reflections, its trajectory intersected itself, as illustrated in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-19.jpg?height=280&width=798&top_left_y=1205&top_left_x=729)

After Bruno, Pedro launched the ball as shown in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-19.jpg?height=269&width=768&top_left_y=1616&top_left_x=747)

Say how many reflections the ball sent by Pedro suffered before its trajectory intersected itself.
Solution

After the third bounce, we will observe the following drawing:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-48.jpg?height=406&width=1145&top_left_y=1476&top_left_x=358)

From the drawing, we see that $\measuredangle B A O=20^{\circ}$, since the bounces are perfect.

Let's recall that the sum of the interior angles of a triangle is $180^{\circ}$. Consequently, the sum of the interior angles at two vertices of a triangle coincides with the exterior angle at the third vertex. We will use this fact systematically.

Applying this fact to triangle $A B O$, we have $\angle A B P=\measuredangle B A O+\measuredangle A O B=40^{\circ}$. Using the fact that the bounce is perfect, we conclude that $\measuredangle C B D=\measuredangle A B P=40^{\circ}$.

We can now use the previously mentioned fact in triangle $B C O$, and thus conclude that $\measuredangle B C A=\measuredangle C B O+\measuredangle B O C=40^{\circ}+20^{\circ}=60^{\circ}$. Knowing that the bounce is perfect, we have $\measuredangle B C A=\measuredangle D C O=60^{\circ}$. Therefore, in triangle $C D O$, we have $\measuredangle C D B=$ $\measuredangle D C O+\measuredangle C O D=60^{\circ}+20^{\circ}=80^{\circ}$. The fourth bounce is perfect, so $\measuredangle E D O=$ $\measuredangle C D B=80^{\circ}$, where $E$ is the point shown in the following figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-49.jpg?height=408&width=1139&top_left_y=236&top_left_x=561)

Since the sum of the interior angles in triangle $E D O$ must be $180^{\circ}$, then $\measuredangle D E O=$ $80^{\circ}$. It follows that after the fifth bounce (at point $E$), the trajectory will return and intersect itself.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-49.jpg?height=403&width=1134&top_left_y=878&top_left_x=561)

Therefore, the answer is 5 bounces.