10,11

Two spheres with centers $O 1$ and $O 2$ are intersected by a plane $P$, which is perpendicular to the segment $O 1 O 2$ and passes through its midpoint. The plane $P$ divides the surface area of the first sphere in the ratio $m: 1$, and the surface area of the second sphere in the ratio $n: 1 (m>1, n>1)$. Find the ratio of the radii of these spheres.
Let $R_1$ and $R_2$ be the radii of spheres with centers $O_1$ and $O_2$ respectively, $M$ be the midpoint of the segment $O_1 O_2$, $A$ and $B$ be the points of intersection of the first and second spheres with the segment $O_1 O_2$. Denote $A M = h_1, B M = h_2$. Applying the formula for the surface area of a spherical cap, we get:

$$
\frac{2 \pi R_1 h_1}{2 \pi R_1 (2 R_1 - h_1)} = \frac{1}{m}, \quad \frac{2 \pi R_2 h_2}{2 \pi R_2 (2 R_2 - h_2)} = \frac{1}{m_2}
$$

from which

$$
h_1 = \frac{2 R_1}{m + 1}, \quad h_2 = \frac{2 R_2}{n + 1}
$$

Since $O_1 M = O_1 A - A M = R_1 - h_1$ and $O_2 M = O_2 B - B M = R_2 - h_2$, and $M$ is the midpoint of $O_1 O_2$, then $R_1 - h_1 = R_2 - h_2$, or $h_1 - h_2 = R_1 - R_2$. Substituting the previously found expressions for $h_1$ and $h_2$ into this equation, we get:

$$
\frac{2 R_1}{m + 1} - \frac{2 R_2}{n + 1} = R_1 - R_2
$$

Dividing both sides of this equation by $R_1$ and denoting $\frac{R_2}{R_1} = t$, we get the equation:

$$
\frac{2}{m + 1} - \frac{2 t}{n + 1} = 1 - t,
$$

from which we find that $\frac{R_2}{R_1} = t = \frac{(n + 1)(m - 1)}{(m + 1)(n - 1)}$.

## Answer

$(n + 1)(m - 1)$

$(m + 1)(n - 1)$.