Example 3.8.2 (Steinhaus Problem) Can a tetrahedron be made using four identical triangles with side lengths $a$, $b$, and $c$? If it can be made, find its volume.
Analyzing the method of constructing a figure during the motion process to answer "whether a tetrahedron can be formed" is also a strategy that varies with the problem.
The answer to Method 1 is that a tetrahedron can be formed when \(a, b, c\) form an acute triangle.
An ingenious intuitive proof is to take two acute triangles with side lengths \(a, b, c\), and overlap their \(c\) sides. Place them in the same plane, with the third vertices not coinciding and on the same side (1) and on opposite sides (2) of the \(c\) side.
In figure (1), \(A B C_{2} C_{1}\) is an isosceles trapezoid, \(\angle C_{1} A B\) and \(\angle C_{2} B A\) are acute angles,
\[
\therefore d_{1}=C_{1} C_{2}0 .
\end{array}
\]
Since \(a, b, c\) form an acute triangle, we also have \(b^{2}+c^{2}-a^{2}>0, c^{2}+a^{2}-b^{2}>0\).
Solving the system of equations, we get: \(x=\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2}}, y=\sqrt{\frac{c^{2}+a^{2}-b^{2}}{2}}, z=\sqrt{\frac{a^{2}+b^{2}-c^{2}}{2}}\)
Thus, using a rectangular prism with length, width, and height \(x, y, z\), we can construct a tetrahedron that meets the requirements of the problem. And we can calculate
\[
\begin{aligned}
V & =x y z-4\left(\frac{1}{3} \cdot \frac{1}{2} \cdot x y z\right)=\frac{1}{3} x y z \\
& =\frac{1}{6 \sqrt{2}} \sqrt{\left(a^{2}+b^{2}-c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\left(c^{2}+a^{2}-b^{2}\right)} .
\end{aligned}
\]