## Task 6B - 201246B

If $T = ABCD$ is a tetrahedron, let $s$ be the sum of all edge lengths of $T$.

In this task, each edge length (measurable in centimeters) is given only by its numerical value.

Determine whether there exists a tetrahedron $T$ among all tetrahedra $T$ with the following properties (1), (2), (3) for which $s$ attains its maximum value.

If this is the case, determine this maximum value of $s$.

The required properties are:

(1) $\angle BDC = \angle CDA = \angle ADB = 90^{\circ}$.

(2) All edge lengths of $T$ are not less than $\frac{1}{6}$.

(3) The volume of $T$ is equal to $\frac{1}{6}$.
We consider the set of all tetrahedra that possess properties (1), (2), and (3). Setting \( A D = x \), \( B D = y \), and \( C D = z \), we have, according to (1) and using the Pythagorean theorem, for the sum \( s \) to be investigated:

\[
s = x + y + z + \sqrt{x^2 + y^2} + \sqrt{y^2 + z^2} + \sqrt{z^2 + x^2}
\]

Due to (1), the volume is \(\frac{1}{6} x y z\), so according to (3), we have: \( x y z = 1 \).

For a fixed \( z \geq \frac{1}{6} \) (see (2)), we now investigate the sum \( s \) as a function of the quotient \( k = \frac{x}{y} \), where we can assume without loss of generality that \( k \geq 1 \).

From \( x y = \frac{1}{z} \) and \( \frac{x}{y} = k \), it follows that \( x^2 = \frac{z}{k} \) and \( y^2 = \frac{1}{k \cdot z} \). Substituting these into (4), we get:

\[
\begin{aligned}
& s(k) = \sqrt{\frac{k}{z}} + \sqrt{\frac{1}{k \cdot z}} + z + \sqrt{\frac{k}{z} + \frac{1}{k z}} + \sqrt{\frac{1}{k z} + z^2} + \sqrt{z^2 + \frac{k}{z}} \\
& = z + \frac{1}{\sqrt{z}} \left( \sqrt{k} + \frac{1}{\sqrt{k}} \right) + \sqrt{\frac{1}{z} \left( k + \frac{1}{k} \right)} + \sqrt{\left( \sqrt{z^2 + \frac{1}{k z}} + \sqrt{z^2 + \frac{k}{z}} \right)^2} \\
& = z + \frac{1}{\sqrt{z}} \left( \sqrt{k} + \frac{1}{\sqrt{k}} \right) + \sqrt{\frac{1}{z} \left( k + \frac{1}{k} \right)} + \sqrt{2 z^2 + \frac{1}{z} \left( k + \frac{1}{k} \right) + 2 \sqrt{z^4 + \frac{1}{z^2} + z \left( k + \frac{1}{k} \right)}}
\end{aligned}
\]

The functions \(\sqrt{x}\) for \( x \geq 0 \) and \( f(t) = t + \frac{1}{t} \) for \( t \geq 1 \) are monotonically increasing \(\left[ f'(t) = 1 - \frac{1}{t^2} \geq 0 \right]\), so \( s(k) \) is monotonically increasing.

For a fixed \( z \), the largest ratio of \( k = \frac{1}{y^2 z} \) is achieved for \( y = \frac{1}{6} \). Therefore, \( s \) (for \( k \)) takes its maximum for \( y = \frac{1}{6} \) and \( x = \frac{6}{z} \). Thus, we only need to investigate whether and, if so, for which \( z \) the sum \( s \) is maximal.

Holding \( y = \frac{1}{6} \) fixed and assuming without loss of generality that \( x \geq z \), it follows from the same reasoning that there is a tetrahedron for which \( s \) attains a maximum, and this is for \( z = \frac{1}{6} \) and \( x = 36 \). The maximum value of \( s \) is thus achieved for \( x = 36 \), \( y = z = \frac{1}{6} \) and is

\[
s = 36 + \frac{2 + \sqrt{2}}{6} + 2 \sqrt{36^2 + \frac{36}{1}}
\]

Taken from [3]

### 9.23 21st Olympiad 1981

### 9.23.1 First Round 1981, Class 12