## Task 4 - 160914

Determine whether there exist bodies for which the following applies!

|  | Edge length or Diameter in $\mathrm{cm}$ | Surface area in $\mathrm{cm}^{2}$ | Volume in $\mathrm{cm}^{3}$ |
| :--- | :---: | :---: | :---: |
| Cube | $a$ | $b$ | $b$ |
| Sphere | $c$ | $d$ | $d$ |
| Regular Tetrahedron | $e$ | $f$ | $f$ |

Determine all real numbers $a, c, e$ that satisfy these conditions! Here, the same letters denote the same real numbers, while different letters do not necessarily denote different numbers.
Assuming there are such bodies, then for the cube it must hold that

$$
\begin{aligned}
b & =6 a^{2} \\
b & =a^{3} \quad \text { and thus } \quad a^{3}=6 a^{2} \\
a & =6
\end{aligned}
$$

Indeed, a cube with an edge length of $6 \mathrm{~cm}$ has a surface area of $216 \mathrm{~cm}^{2}$ and a volume of $216 \mathrm{~cm}^{3}$.

2) for the sphere

$$
\begin{aligned}
& d=\frac{\pi}{6} c^{3} \\
& d=\pi c^{2} \quad \text { and thus } \quad \frac{\pi}{6} c^{3}=\pi c^{2} \\
& c=6
\end{aligned}
$$

Indeed, a sphere with a diameter of $6 \mathrm{~cm}$ has a surface area of $36 \mathrm{~cm}^{2}$ and a volume of $36 \mathrm{~cm}^{3}$.

3) for the regular tetrahedron

$$
\begin{aligned}
f & =e^{2} \sqrt{3} \\
f & =\frac{e^{3}}{12} \sqrt{2} \quad \text { and thus } \quad \frac{e^{3}}{12} \sqrt{2}=e^{2} \sqrt{3} \\
e & =\frac{12 \sqrt{3}}{2} \\
e & =6 \sqrt{6}
\end{aligned}
$$

Indeed, a regular tetrahedron with an edge length of $6 \sqrt{6} \mathrm{~cm}$ has a surface area of $216 \sqrt{3}$ $\mathrm{cm}^{2}$ and a volume of $216 \sqrt{3} \mathrm{~cm}^{3}$.

Thus, exactly the following numbers satisfy the given conditions:

$$
a=6, \quad c=6 \quad \text { and } \quad e=6 \sqrt{6}
$$

Solutions of the I. Round 1976 taken from [5]

### 6.18.2 II. Stage 1976, Class 9