## Task 2 - 090922

Given is a cube with edge length $a_{1}$ and volume $V_{1}$, and a regular tetrahedron with edge length $a_{2}$ and volume $V_{2}$. For the edge lengths, it holds that $a_{1}: a_{2}=1: \sqrt{2}$.

Calculate the ratio $V_{1}: V_{2}$!
It is

$$
V_{1}=a_{1}^{3} \quad \text{and} \quad V_{2}=\frac{1}{12} \cdot \sqrt{2} \cdot a_{2}^{3}=\frac{1}{12} \cdot \sqrt{2} \cdot\left(\sqrt{2} a_{1}\right)^{3}=\frac{1}{3} a_{1}^{3}=\frac{1}{3} V_{1}
$$

so that the ratio $V_{1}: V_{2}$ is exactly $3: 1$.

Remark:

The volume $V$ of a regular tetrahedron with edge length $a$ is (as for any pyramid) given by $V=\frac{1}{3} \cdot A_{G} \cdot h$, where $A_{G}$ is the area of the base and $h$ is the length of the corresponding height. The base is an equilateral triangle with edge length $a$. The area of this triangle can be calculated (as in any triangle) via $A_{G}=\frac{1}{2} \cdot a \cdot h_{g}$, where $h_{g}$ is the length of a height in the triangle.

Since the heights in an equilateral triangle coincide with the medians, such a height divides the equilateral triangle into two right triangles, where one of the legs of such a triangle is the height $h_{g}$, the second leg is half the base, and the hypotenuse is the uncut side of the triangle.

By the Pythagorean theorem, we get $h_{g}^{2}+\left(\frac{a}{2}\right)^{2}=a^{2}$, thus $h_{g}=\frac{\sqrt{3}}{2} a$ and therefore $A_{G}=\frac{\sqrt{3}}{4} a^{2}$.

For the height of the regular tetrahedron, note that it coincides with the corresponding median (the line connecting a vertex to the centroid of the opposite face). This results in a right triangle with the "apex of the tetrahedron", the "centroid of the base", and a vertex of the base as vertices.

The hypotenuse of this right triangle is an edge of the regular tetrahedron, one leg is the height $h$, and the second leg is the segment of the median in the base between the centroid and the vertex. Since the centroid divides the medians in the ratio 2:1, with the segment between the vertex and the centroid being the longer one, and since in an equilateral triangle the heights and medians coincide, this segment is $\frac{2}{3} \cdot h_{g}=\frac{\sqrt{3}}{3} a$ long.

For the right triangle considered between the apex, the centroid, and the vertex, we now get by the Pythagorean theorem $h^{2}+\left(\frac{\sqrt{3}}{3} a\right)^{2}=a^{2}$, thus $h^{2}+\frac{3}{9} a^{2}=a^{2}$ or $h=\sqrt{\frac{2}{3}} a=\frac{\sqrt{2} \cdot \sqrt{3}}{3} a$.

Together with the base area calculated previously, we now get

$$
V=\frac{1}{3} \cdot A_{G} \cdot h=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} \cdot \frac{\sqrt{2} \cdot \sqrt{3}}{3} a=\frac{\sqrt{2}}{3 \cdot 4} a^{3}
$$

which we used above.