13.410 Along the sides of a right angle, towards the vertex, two spheres with radii of 2 and 3 cm are moving, with the centers of these spheres moving along the sides of the angle at unequal but constant speeds. At a certain moment, the center of the smaller sphere is 6 cm from the vertex, and the center of the larger sphere is 16 cm away. After 1 second, the distance between the centers became 13 cm, and after another 2 seconds, the spheres collided without reaching the vertex. Find the speeds of the spheres.
Solution. Let $x$ and $y$ be the speeds of the balls. After 1 second, the distances from the centers of the balls to the corner are $6-x$ and $16-y$, and after 3 seconds, they are $6-3x$ and $16-3y$. By the Pythagorean theorem, we have: $\left\{\begin{array}{l}(6-x)^{2}+(16-y)^{2}=13^{2}, \\ (6-3 x)^{2}+(16-3 y)^{2}=5^{2},\end{array}\right.$ (since at the moment of impact, the distance between the centers of the balls is 5 cm).

From this, $\left\{\begin{array}{l}x^{2}+y^{2}=32 y+12 x-123, \\ 3\left(x^{2}+y^{2}\right)=32 y+12 x-89\end{array} \Rightarrow\left\{\begin{array}{l}3(32 y+12 x-123)=32 y+12 x-89, \\ 2\left(x^{2}+y^{2}\right)=34\end{array} \Rightarrow\right.\right.$ $\Rightarrow\left\{\begin{array}{l}3 x+8 y=35, \\ x^{2}+y^{2}=17\end{array} \Rightarrow\left\{\begin{array}{l}x=1, \\ y=4 .\end{array}\right.\right.$

Answer: 1 cm/s, 4 cm/s.