On a right-angled triangular billiard table (small), we shoot a ball from the vertex of one of the acute angles along the angle bisector. The ball, after being reflected from the third side, returns to the starting vertex along the path it has traveled. What are the acute angles of the table? (The law of elastic collision geometrically is the same as the reflection of light.)
Let's launch the ball from the vertex of the acute angle at $A$ of the right-angled triangle $ABC \left(\angle ACB = 90^\circ\right)$, and denote the successive points of reflection on its path by $A_1, C_1, B_1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d79a7926449a53cd5a15g-1.jpg?height=488&width=466&top_left_y=263&top_left_x=800)

$A_1$ is clearly on $BC$; $C_1$ is on $AB$ because in the right-angled triangle $AA_1C$, $\angle CA_1A$ is an acute angle, so by the law of reflection, $\angle BA_1C_1$ is also an acute angle, and therefore $\angle CA_1C_1$ is an obtuse angle. Finally, $B_1$ is a point on the third side, i.e., $AC$. The reflection at $B_1$ causes $C_1B_1$ to be perpendicular to $AC$, so the triangle $AB_1C_1$ is right-angled.

With the notation $\angle A_1AC = \delta$, we have $\angle CAB = 2\delta$, $\angle ABC = 90^\circ - 2\delta$, and $\angle CA_1A = \angle BA_1C_1 = 90^\circ - \delta$. In the triangle $A_1BC_1$, $\angle BC_1A_1 = 180^\circ - (90^\circ - \delta) - (90^\circ - 2\delta) = 3\delta$, and the angle $\angle AC_1B_1$ is also $3\delta$. The angles at $A$ and $C_1$ in the triangle $AB_1C_1$ are complementary: $2\delta + 3\delta = 90^\circ$, and from this, $\delta = 18^\circ$. The acute angles of the table are $36^\circ$ and $54^\circ$.

Ferenc Szalay (Tatabánya, Árpád g. I. o. t.)

Second solution: After we see that the first two reflections occur on the sides $BC$ and $AB$, reflect the triangle $ABC$ over the side $BC$, and then reflect the resulting triangle $A'BC$ over the side $A'B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d79a7926449a53cd5a15g-1.jpg?height=442&width=767&top_left_y=1275&top_left_x=663)

Figure 2

Thus, the path $AA_1C_1B_1$ has the segment $AA_1$ and the segment $A_1C_1$ as the image of $A_1C_1'$ and the segment $C_1B_1$ as the image of $C_1'B_1''$, all lying on a straight line, and this line is perpendicular to $A'C''$. In the right-angled triangle $AA'B_1''$, $\angle AA'B_1'' = 2 \angle CA'B = 2 \angle CAB = 4\delta$, which is complementary to the angle $\angle CAA_1$, so $\delta + 4\delta = 90^\circ$ and $\delta = 18^\circ$.

László Haris (Bp., VIII. Vörösmarty M. g. II. o. t.)