## Task 4 - 270824

A cube $W$ is divided into volume-equal smaller cubes. The surface area of the cube $W$ is $A$, and the sum of the surface areas of the separated smaller cubes is $S$. Determine the ratio $A: S$

(a) if the cube $W$ has an edge length of $14 \mathrm{~cm}$ and the number of smaller cubes is 8,

(b) for any edge length $a$ of the cube $W$ and the number of smaller cubes is 8,

(c) for any edge length $a$ of the cube $W$ and the number of smaller cubes is $n^{3}$, where $n$ is any natural number with $n \geq 2$!
(c) It holds that $A=6 a^{2}$. Furthermore, $W$ has the volume $a^{3}$; from this it follows:

Each sub-cube has the volume $\frac{a^{3}}{n^{3}}$, thus the edge length $\frac{a}{n}$ and consequently the surface area $6 \frac{a^{2}}{n^{2}}$. Therefore,

$$
S=n^{3} \cdot 6 \cdot \frac{a^{2}}{n^{2}}=6 n a^{2}
$$

Thus, we obtain

$$
A: S=6 a^{2}:\left(6 n a^{2}\right)=1: n
$$

If we choose $n=2$ here, we get (since $2^{3}=8$) as the result for both (a) and (b) the ratio $A: S=1: 2$.

Solutions of the 2nd Round 1987 adopted from [5]

# 5.29.3 3rd Round 1987, Class 8