Given a triangular pyramid $A B C D$. The skew edges $A C$ and $B D$ of this pyramid are perpendicular. The skew edges $A D$ and $B C$ are also perpendicular, and $A B=C D$. All edges of this pyramid are tangent to a sphere of radius $r$. Find the area of the face $A B C$.
Let's prove that the opposite edges $AB$ and $CD$ of the tetrahedron $ABCD$ (Fig.1) are also perpendicular. Extend the given tetrahedron $ABCD$ to a parallelepiped $AKBLNDMC$ ($AN \| KD \| BM \| LC$) by drawing through its opposite edges pairs of parallel planes. Since $AC \perp BD$ and $LN \| BD$, then $LN \perp AC$, so the parallelogram $ALCN$ is a rhombus. Then the equal parallelogram $KMBD$ is also a rhombus. Similarly, the faces $AKDN$ and $LBCM$ are rhombuses. Therefore, $AK = AN = AL$, so the remaining faces $AKBL$ and $NDMC$ are also rhombuses.

Thus, $AB \perp CD$. Therefore, all edges of the parallelepiped are equal. Let their lengths be denoted by $x$. By the theorem on the sum of the squares of the diagonals of a parallelogram,

\[
AB^2 + CD^2 = 4 \times 2, \quad AC^2 + BD^2 = 4 \times 2, \quad AD^2 + BC^2 = 4 \times 2
\]

Therefore,

\[
AB^2 + CD^2 = AC^2 + BD^2 = AD^2 + BC^2
\]

Let $AB = CD = a$. Suppose a sphere of radius $r$ touches the edges $AB, CD, AD$, and $BC$ of the tetrahedron $ABCD$ at points $E$, $F$, $G$, and $H$ respectively (Fig. 2). Then

\[
\begin{gathered}
AB + CD = (AE + BE) + (CF + DF) = (AG + BH) + (CH + DG) = \\
= (AG + DG) + (CH + BH) = AD + BC
\end{gathered}
\]

Similarly, $AC + BD = AB + CD$, and since

\[
AB^2 + CD^2 = 2a^2 \quad \text{and} \quad AB + CD = 2a
\]

then

\[
AC^2 + BD^2 = 2a^2 \quad \text{and} \quad AC + BD = 2a
\]

Square both sides of the second equation and add the resulting equation to the first. We have the system of equations

\[
\left\{\begin{array}{l}
AC + BD = 2a \\
AC \cdot BD = a^2
\end{array}\right.
\]

from which we find that $AC = BD = a$. Similarly, $AD = BC = a$. Therefore, $ABCD$ is a regular tetrahedron with edge $a$, and the parallelepiped $AKBLNDMC$ is a cube, the diagonal of whose face is $a$, and the edge is $\frac{a}{\sqrt{2}}$. The given sphere is inscribed in this cube, so the edge of the cube is equal to the diameter of the sphere, i.e., $2r$. Therefore, $a = 2r\sqrt{2}$, and the area of each face of the tetrahedron is $\frac{1}{4}a^2\sqrt{3}$. Thus,

\[
S_{\triangle ABC} = \frac{1}{4}a^2\sqrt{3} = \frac{1}{4} \cdot 8r^2\sqrt{3} = 2r^2\sqrt{3}
\]

## Answer

$2r^2\sqrt{3}$.