An insect travels from vertex $A$ to vertex $B$ on the cube shown with edges of length 1 metre. The insect can travel in exactly one and only one of the following ways:

- crawl along the edges of the cube (it crawls at 5 metres per minute along edges)
- crawl along the faces of the cube (it crawls at 4 metres per minute along faces)
- fly through the interior of the cube (it flies at 3 metres per minute along any path through the interior of the cube)

What is the shortest possible time, in minutes, that the insect can take to get from $A$ to $B$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-2.jpg?height=474&width=390&top_left_y=2037&top_left_x=930)
To get from $A$ to $B$ along edges only, the insect must travel along at least 3 edges. Thus, the insect travels at least $3 \mathrm{~m}$, and so takes at least $\frac{3}{5}=0.6$ minutes.

To get from $A$ to $B$ along faces only, the insect must travel along at least 2 faces.

If we unfold the top and front faces to make a flat rectangle that is $1 \mathrm{~m}$ by $2 \mathrm{~m}$, the insect can travel from $A$ to $B$ along the diagonal of this rectangle. The distance from $A$ to $B$ in this rectangle is $\sqrt{2^{2}+1^{2}}=\sqrt{5} \mathrm{~m}$. Thus, the insect could travel $\sqrt{5} \mathrm{~m}$, which would take $\frac{\sqrt{5}}{4} \approx 0.56$ minutes.

To get from $A$ to $B$ through the interior of the cube, the shortest possible path is the diagonal from $A$ to $B$. Since the cube has edge length $1 \mathrm{~m}$, then the length of the diagonal is $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3} \mathrm{~m}$. Thus, the insect travels at least $\sqrt{3} \mathrm{~m}$ in this case, and so takes at least $\frac{\sqrt{3}}{3} \approx 0.58$ minutes.

Therefore, the shortest possible time that the insect can take is $\frac{\sqrt{5}}{4}$ minutes.