53. The lateral edges of a triangular pyramid are pairwise perpendicular, and one of them is equal to $a$ and is equal to the sum of the other two. Find the radius of the sphere that touches the base of the pyramid and the extensions of its lateral faces.
53. Let $ABCD$ be a given pyramid, the lateral edges of which $|DA|=a$, $|DB|=x$, $|DC|=y$; by condition, these edges are perpendicular and $x+y \doteq a$. It is not difficult to find that

$$
S_{ABC}=\frac{1}{2} \sqrt{a^{2}\left(x^{2}+y^{2}\right)+x^{2} y^{2}}, \quad V_{ABCD}=\frac{1}{6} a x y
$$

On the other hand, if $R$ is the radius of the desired sphere; then

$$
\begin{aligned}
V_{ABCD}=\frac{R}{3}\left(S_{DAB}+\right. & \left.S_{DBC}+S_{DCA}-S_{ABC}\right)= \\
=\frac{R}{6}[ax+ & \left.ay+xy-\sqrt{a^{2}\left(x^{2}+y^{2}\right)+x^{2} y^{2}}\right]= \\
& =\frac{R}{6}\left(a^{2}+xy-\sqrt{\left.a^{4}-2xya^{2}+x^{2} y^{2}\right)}=\frac{R}{3}xy\right.
\end{aligned}
$$

By equating the two expressions for $V_{ABCD}$, we find $R=\frac{a}{2}$.