A smart ant, which walks on the surface of the cube below, always takes the shortest path between two points. The cube has edges of size $1 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-03.jpg?height=571&width=662&top_left_y=1733&top_left_x=800)

What distance will the smart ant travel if it goes:

a) From vertex $A$ to vertex $B$?

b) From point $M$ to point $N$?

c) From vertex $A$ to vertex $D$?
Solution

a) Since the cube has edges of size 1, the distance between $A$ and $B$ is equal to $1 \mathrm{~cm}$.

b) The shortest path between $M$ and $N$ is made by going in a straight line from $M$ to $L$, and then, again in a straight line, from $L$ to $N$. See the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-25.jpg?height=569&width=628&top_left_y=1189&top_left_x=817)

The reason for this is simple: if we place the faces $A B H F$ and $B C D H$ in the same plane, the path $M \rightarrow L \rightarrow N$ is a straight line! Therefore, it is the shortest path between two points. See the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-25.jpg?height=462&width=802&top_left_y=2005&top_left_x=730)

Thus, the distance to be traveled in this case is $2 \mathrm{~cm}$.

c) Repeating the idea from the previous item, we will place the faces $A B H F$ and $F H D E$ in the same plane, see the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-26.jpg?height=782&width=486&top_left_y=226&top_left_x=685)

Therefore, the shortest distance between $A$ and $D$ will be given by the Pythagorean Theorem:

$$
\begin{aligned}
& \overline{A D}^{2}=1^{2}+(1+1)^{2} \\
& \Longrightarrow \overline{A D}^{2}=5 \\
& \Longrightarrow \overline{A D}=\sqrt{5}
\end{aligned}
$$

Since the ant is smart, it will take this shortest path and travel $\sqrt{5} \mathrm{~cm}$!