A point starts from the vertex $A$ of the triangle $ABC$, moves in a straight line, and after being reflected from the two sides of the triangle, it reaches the vertex $B$. Let us construct the path of the point and determine the conditions for solvability.
Consider the task as solved. Let the path of the moving point be $A P Q B$, where $P$ is on side $B C$ and $Q$ is on side $C A$; let the reflection of $A$ over $B C$ be $A^{\prime}$, and the reflection of $B$ over $C A$ be $B^{\prime}$. According to the law of reflection, $\angle Q P C = \angle A P B = \angle A^{\prime} P B$ and $\angle P Q C = \angle B Q A = \angle B^{\prime} Q A$, so $A^{\prime}, P, Q, B^{\prime}$ lie on the same line. Based on this, the path of the point can be constructed.

![](https://cdn.mathpix.com/cropped/2024_05_02_381ea95128926b94589dg-1.jpg?height=572&width=727&top_left_y=359&top_left_x=688)

The solvability condition is that $A^{\prime} B^{\prime}$ intersects both $B C$ and $C A$ (see the figure).

Denoting the angles of the triangle by $\alpha, \beta, \gamma$, the first condition is that $\angle A^{\prime} C B^{\prime} = 3 \gamma - \beta$, or in the case of $C A \beta$,

$$
\alpha < 90^{\circ} + \gamma
$$

In the limiting cases where $\gamma = 60^{\circ}$ and $\gamma < 45^{\circ}, \alpha = 90^{\circ} + \gamma$, the required reflections of the ball as stated in the problem cannot occur.

László Tóth (Miskolc, Vill. kp. t. III. o. t.)