10,11

The centers of four spheres of radius $r\left(r<\frac{1}{\sqrt{2}}\right)$ are located at the vertices of an isosceles right triangle with legs of length 2, and at the midpoint of its hypotenuse. Find the radius of the sphere that is tangent to these four spheres.
Let $ABC$ be the given triangle with sides $AC = BC = 2$ and a right angle at vertex $C$, $M$ be the midpoint of the hypotenuse $AB$, $O$ be the center of the desired sphere, and $R$ be its radius. It is clear that the desired sphere cannot touch all the given spheres externally or internally. We will find $R$ under the condition that the desired sphere touches the spheres with centers at $A$, $B$, and $C$ externally, and the sphere with center $M$ internally (Fig. 1). In this case,

$$
OA = OB = OC = r + R, \quad OM = |R - r|.
$$

From the right triangle $OAM$, we have

$$
OA^2 = OM^2 + AM^2, \text{ or } (R + r)^2 = (R - r)^2 + 2,
$$

from which $R = \frac{1}{2\varphi}$. Suppose the sphere with center $O$ touches the spheres with centers at $A$, $C$, and $M$ internally, and the sphere with center $B$ externally (Fig. 2). In this case,

$$
OA = OC = OM = |R - r|, \quad OB = R + r.
$$

Since $OA = OC = OM$, the orthogonal projection $K$ of point $O$ onto the plane of triangle $ABC$ is equidistant from the vertices of the right triangle $ACM$. Therefore, $K$ is the midpoint of $AC$. From the right triangle $OKB$, we have

$$
OB^2 = OK^2 + KB^2 = OA^2 - AK^2 + KB^2,
$$

or

$$
(R + r)^2 = (R - r)^2 - 1 + 5
$$

from which $R = \frac{1}{r}$. Similarly, for the case where the sphere with center $O$ touches the spheres with centers at $B$, $C$, and $M$ internally, and the sphere with center $A$ externally. Suppose the sphere with center $O$ touches the spheres with centers at $C$ and $M$ internally, and the spheres with centers at $A$ and $B$ externally (Fig. 3). In this case,

$$
OA = OB = R + r, \quad OC = OM = |R - r|.
$$

Since $OA = OB$ and $OC = OM$, point $O$ lies on the line passing through the midpoint $P$ of segment $CM$ and perpendicular to the plane of triangle $ABC$. From the right triangles $OAP$ and $OCP$, we have

$$
OP^2 = OA^2 - AP^2 = OC^2 - CP^2,
$$

or

$$
(R + r)^2 - \frac{5}{2} = (R - r)^2 - \frac{1}{2}
$$

from which $R = \frac{1}{24}$. No other cases of tangency are possible.