## Task 3 - 330843

At one corner of a cube with an edge length of $1 \mathrm{~cm}$, there is an ant. Along each edge of the cube, $1 \mathrm{~g}$ of honey is distributed. The ant is to reach the endpoint of the body diagonal whose starting point it occupies. To do this, it must travel a path of exactly $7 \mathrm{~cm}$ in length and consume exactly 7 grams of honey.

Determine the number of all possible paths under these conditions!
I. The corners of the cube are labeled as $A, B, C, D, E, F, G, H$ as shown in the figure.

The starting point of the paths to be counted is $A$, and the endpoint is $G$.

From $A$, the first edge can only be traveled to $B$, $D$, or $E$. It suffices to consider one of these possibilities, for example, the one to $B$, and multiply the number of paths starting with this by 3.

II. From $B$, the second edge can only follow to $C$ or $F$. It suffices to multiply the number of paths found for the continuation to $C$ by 2. From $C$, there are again exactly two continuation possibilities:

III. If one continues from $C$ to $G$, one is already at the destination after a path of $3 \mathrm{~cm}$. Therefore, one must still add a path of $4 \mathrm{~cm}$ from $G$ to $G$. This is only possible by circling a side face. The only side face containing $G$ where this is possible without repeating an already traversed edge is GHEF.

Thus, one arrives (after starting via $C, G$) at exactly the 2 paths $A B C G H E F G$ and $A B C G F E H G$.

IV. From $C$ to $D$, there are exactly the two continuations:

a) Initial path $A B C D H$.

To go directly from $H$ to $G$ would not be possible, as one would then need to reach $G$ again on a path of $2 \mathrm{~cm}$, which is not possible without repeating an edge. Therefore, only the path $A B C D H E F G$ remains.

b) Initial path $A B C D A$.

From $A$, it must go uniquely to $E$. From there, one must reach $G$ on a path of $2 \mathrm{~cm}$. There are exactly two possibilities for this; thus, one obtains the two paths $A B C D A E F G$ and $A B C D A E H G$.

The number of the five paths found in III. and IV. is to be multiplied by 3 and 2 according to I. and II. Thus, the number of paths that are possible under the conditions of the problem is $5 \cdot 3 \cdot 2 = 30$.