## Task 6 - 130936

Given a regular tetrahedron with vertices $A, B, C, D$ and edge length $a$. A point $D^{\prime}$ should have the following properties:

a) The tetrahedron with vertices $A, B, C, D^{\prime}$ is of equal volume to the given tetrahedron, b) $B D^{\prime}=C D^{\prime}=a$, c) $A D^{\prime} \neq a$.

Investigate whether such points $D^{\prime}$ exist, and determine for each such $D^{\prime}$ the length of the edge $A D^{\prime}$.
If $h$ is the height of the tetrahedron $ABCD$, it follows from property a) that $D'$ lies in a plane $\epsilon$ parallel to the base $ABC$ at a distance $h$. In principle, two possibilities are conceivable: either $\epsilon$ lies in the same half-space generated by the base as $D$ (then $D$ lies on $\epsilon$), or the plane $\epsilon$ lies in the corresponding other half-space. Since the further conditions do not depend on which of the two cases occurs, symmetrical structures arise, so we can assume w.l.o.g. that $\epsilon$ lies in the upper half-space together with $D$, so that both $D$ and $D'$ are points on $\epsilon$.

According to condition b), $D'$ lies on the surface of the two spheres centered at points $B$ and $C$ with radius $a$. Since $h < a$, these two spheres are intersected by $\epsilon$ (and not just touched), resulting in two circles $k_B; k_C$ as intersection figures in the plane $\epsilon$. These two circles intersect in up to two points, one of which is $D$ (since $|BD| = |CD| = a$ also holds) and the other (if it exists) is $D'$.

A second intersection point would only not exist if the circles only touched at $D$. Since the centers $M_B$ and $M_C$ of the two circles, like the sphere centers $B$ and $C$, have the distance $a$ (since $\epsilon$ is parallel to the base), their equal radii would thus add up to $a$ and in the right triangle $\triangle BM_BD$ the Pythagorean theorem would yield the relationship $\left|BM_B\right|^2 + \left|M_BD\right|^2 = |BD|^2$ or $h^2 + \frac{a^2}{4} = a^2$, thus $h = \frac{\sqrt{3}}{2} a$. However, in a regular tetrahedron, $h = \frac{\sqrt{6}}{3} a$, so the circles do not just touch and there is therefore another point $D'$ besides $D$ that satisfies conditions a) and b).

Since $|AD| = a$, condition c) also rules out $D = D'$. Let's now calculate the distance from $A$ to the second intersection point $D'$ of the two circles:

The two intersection points of two circles are symmetric with respect to the line connecting the two centers. Since the plane $\epsilon_1$, which is perpendicular to the base and passes through the line $BC$, is also perpendicular to $\epsilon$ and there passes through the line $M_B M_C$, $D$ is thus transformed into $D'$ by reflection at $\epsilon_1$. Due to the orthogonality of $\epsilon_1$ to both $\epsilon$ and the base, the foot of the perpendicular $L_D$ from $D$ to the base is also transformed into the foot of the perpendicular $L_{D'}$ from $D'$ to the base by reflection at $\epsilon_1$: The perpendiculars are parallel to the reflection plane. When considering only in the base, $L_D$ is thus transformed into $L_{D'}$ by reflection at the line $BC$.

Since in the equilateral triangle $\triangle ABC$ the medians and altitudes coincide and $L_D$ is the centroid of this triangle, the line $AL_D$ as a median and at the same time an altitude is perpendicular to $BC$ and passes through its midpoint $M$. In particular, the reflection point $L_{D'}$ also lies on this line and it holds that

$$
\left|AL_{D'}\right| = \left|AL_D\right| + 2 \cdot \left|L_D M\right| = \frac{2}{3} s + 2 \cdot \frac{1}{3} s = \frac{4}{3} s
$$

where $s$ is the length of the median in the equilateral triangle with side length $a$. For this, in the right triangle $\triangle ABM$ the relationship $s^2 + \frac{a^2}{4} = a^2$ holds and thus $s^2 = \frac{3}{4} a^2$. Substitution yields
$\left|AL_{D'}\right|^2 = \frac{16}{9} s^2 = \frac{4}{3} a^2$.

In the right triangle $\triangle AL_{D'}D'$, we finally have

$$
\left|AD'\right|^2 = \left|AL_{D'}\right|^2 + \left|L_{D'}D'\right|^2 = \frac{4}{3} a^2 + h^2 = \frac{4}{3} a^2 + \frac{2}{3} a^2 = 2 a^2
$$

thus $\left|AD'\right| = \sqrt{2} a$.