3. In the middle of one of the side edges of a cube, there is an ant. If the ant moves along the side faces at a speed of 6, and along the bases at $5 \sqrt{2}$, find the number of different paths (up to symmetry) for the ant to reach the middle of the opposite edge as quickly as possible.
Solution. Let the ant be at the midpoint $M$ of edge $A A_{1}$, and it needs to reach the midpoint $R$ of edge $C C_{1}$ (see diagram). Clearly, one of the shortest paths from $M$ to $R$ is the path $M N R$, where $N$ is the midpoint of edge $B B_{1}$. By this path, the ant will reach $R$ in time $t=\frac{2 a}{6}=\frac{a}{3}$.

However, the ant can also move along the bases. Let $M P Q R$ be such a path

![](https://cdn.mathpix.com/cropped/2024_06_05_d61831f2f852c6e23b39g-7.jpg?height=466&width=488&top_left_y=1476&top_left_x=991)
(see diagram). Due to symmetry, we have

$$
\overline{M P}=\overline{Q R} \text { and } \overline{P B_{1}}=\overline{B_{1} Q}
$$

Let $x=\overline{A_{1} P}$; then we have:

$$
\overline{M P}=\sqrt{x^{2}+\frac{a^{2}}{4}}, \overline{P Q}=(a-x) \sqrt{2}
$$

If $t_{2}$ is the time it takes for the ant to travel the path $M P Q R$, then

$$
t_{2}=\frac{\overline{M P}}{6}+\frac{\overline{P Q}}{5 \sqrt{2}}+\frac{\overline{Q R}}{6}=\frac{\sqrt{4 x^{2}+a^{2}}}{6}+\frac{a-x}{5}
$$

From $t_{2} \leq t_{1}$ we get

$$
\begin{aligned}
& \frac{\sqrt{4 x^{2}+a^{2}}}{6}+\frac{a-x}{5} \leq \frac{a}{3} \\
& 5 \sqrt{4 x^{2}+a^{2}} \leq 4 a+6 x \\
& (8 x-3 a)^{2} \leq 0
\end{aligned}
$$

i.e., $8 x-3 a=0$, from which we get $x=\frac{3 a}{8}$. The corresponding time is $t_{2}=t_{1}=\frac{a}{3}$.

Therefore, there are (up to symmetry) two such paths.