## Task 3 - 250933

Let the volume of a regular tetrahedron be $V_{T}$, and the volume of its circumsphere be $V_{K}$.

Calculate the ratio $V_{K}: V_{T}$ and round it to the nearest integer (i.e., determine the integer closest to $V_{K}: V_{T}$)!

You may use the approximate values $\sqrt{3} \approx$ 1.7 and $\pi \approx 3.14$ (accurate to one or two decimal places).
In a regular tetrahedron with edge length $a$, all side faces are equilateral triangles with edge length $a$. The median $s$ in such a triangle runs through the midpoint of the corresponding side and is orthogonal to it, thus it has the length $s=\sqrt{a^{2}-\left(\frac{a}{2}\right)^{2}}=\frac{\sqrt{3}}{2} a$ and the side face has an area of $A=\frac{1}{2} \cdot a \cdot s=\frac{\sqrt{3}}{4} a^{2}$.

The centroid of a triangle divides its median in the ratio 2:1, so the distance from a vertex to the centroid of the corresponding side face is exactly $\frac{\sqrt{3}}{3} a$.

In a regular tetrahedron, the median (i.e., the line segment connecting a vertex to the centroid of the opposite side face) is perpendicular to the corresponding side face, so it has a length of $h=\sqrt{a^{2}-\left(\frac{\sqrt{3}}{3} a\right)^{2}}=\sqrt{\frac{2}{3}} a=\frac{\sqrt{6}}{3} a$.

Thus, the tetrahedron has a volume of

$$
V_{T}=\frac{1}{3} \cdot A \cdot h=\frac{\sqrt{3} \cdot \sqrt{6}}{3 \cdot 4 \cdot 3} a^{3}=\frac{\sqrt{2}}{12} a^{3}
$$

The circumcenter of a regular tetrahedron coincides with its centroid. This centroid divides the medians in the ratio 3:1, so the circumradius $r$ is given by $r=\frac{3}{4} \cdot h=\frac{\sqrt{6}}{4} a$. Thus, the circumcircle has a volume of

$$
V_{K}=\frac{4}{3} \cdot \pi \cdot r^{3}=\frac{4 \cdot 6 \sqrt{6}}{3 \cdot 4^{3}} \cdot \pi \cdot a^{3}=\frac{\sqrt{6}}{8} \cdot \pi \cdot a^{3}
$$

so that

$$
V_{K}: V_{T}=\frac{\sqrt{6} \cdot 12}{8 \cdot \sqrt{2}} \cdot \pi=\frac{\sqrt{3} \cdot 3}{2} \cdot \pi \approx \frac{1.7 \cdot 3}{2} \cdot \pi=2.55 \cdot \pi \approx 2.55 \cdot 3.14=8.007 \approx 8
$$