[ Completing the tetrahedron to a parallelepiped ] A sphere circumscribed around a tetrahedron $\quad]$

The lateral edges of a triangular pyramid are pairwise perpendicular and equal to $a, b$ and $c$. Find the radius of the circumscribed sphere.
Let $A B C D$ be a given triangular pyramid with vertex $D$, and $D A \perp A B, D A \perp A C$ and $A B \perp A C, D A=a, A B$ $=c$ and $A C=b$. Then the line $A D$ is perpendicular to two intersecting lines in the plane $A B C$, so the line $A D$ is perpendicular to the plane $A B C$, i.e., $D A$ is the height of the pyramid. We can complete the right triangle $A B C$ to a rectangle $A B M C$, and draw perpendiculars from vertices $B, C$ and $M$ to the plane $A B C$, and through vertex $D$ draw a plane parallel to the plane $A B C$. We obtain a rectangular parallelepiped $A B M C D K L N(A D\|C N\| M L \| B K)$. The sphere circumscribed around this parallelepiped passes through points $A, B, C, D$, so it is also circumscribed around the given pyramid $A B C D$. Its center $O$ is the point of intersection of the diagonals of the parallelepiped, and the radius $R$ is half the diagonal of the rectangular parallelepiped.

Therefore,

$$
R=\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}}
$$

## Answer

$\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}}$

The opposite edges of the tetrahedron are pairwise equal. The base of the tetrahedron is a triangle with sides $a, b, c$. Find the volume of the tetrahedron.

## Solution

We complete the given tetrahedron $A B C D$ to a parallelepiped $A K B L N D M C$ ( $A N\|K D\| B M \| L C$ ), by drawing through its opposite edges pairs of parallel planes. Let $A B=C D=a, A C=B D=b$ and $A D=B C=c$. Since $K L \| C D$ and $K D \| L C$, then $K L=A B$, so the parallelogram $A K B L$ is a rectangle. Similarly, all six faces of the parallelepiped $A K B L N D M C$ are rectangles. Therefore, the parallelepiped $A K B L N D M C$ is a rectangular parallelepiped. Let $A K=x, A L=y, A N=z$. Then

$$
x 2+y 2=a 2, y 2+z 2=b 2, x 2+z 2=c 2
$$

We add the first and third equations term by term and subtract the second from the result. We get

$$
x 2=\frac{1}{2}(a 2+c 2-b 2)
$$

Similarly,

$$
y 2=\frac{1}{2}(a 2+b 2-c 2), z 2=\frac{1}{2}(b 2+c 2-a 2)
$$

If $V$ is the volume of the parallelepiped, then

$$
V=x y z=\sqrt{\frac{a^{2}+a^{2}-b^{2}}{2}} \cdot \sqrt{\frac{a^{2}+b^{2}-a^{2}}{2}} \cdot \sqrt{\frac{b^{2}+a^{2}-a^{2}}{2}}
$$

and since the volume $V 1$ of the tetrahedron $A B C D$ is one-third of the volume of the parallelepiped $A K B L N D C$, then

$$
\begin{aligned}
V 1 & =\frac{1}{3} \sqrt{\frac{a^{2}+r^{2}-b^{2}}{2}} \sqrt{\frac{a^{2}+b^{2}-r^{2}}{2}} \sqrt{\frac{b^{2}+r^{2}-a^{2}}{2}}= \\
& =\frac{1}{6} \sqrt{\frac{\left(a^{2}+r^{2}-b^{2}\right)\left(a^{2}+b^{2}-r^{2}\right)\left(b^{2}+r^{2}-a^{2}\right)}{2}}
\end{aligned}
$$

## Answer

$\frac{1}{6} \sqrt{\frac{\left(a^{2}+r^{2}-b^{2}\right)\left(a^{2}+b^{2}-r^{2}\right)\left(b^{2}+r^{2}-a^{2}\right)}{2}}$