Consider  a sphere $S$ of radius  $R$ tangent to a plane $\pi$. Eight other spheres of the same radius $r$ are tangent to $\pi$ and tangent to $S$, they form a "collar" in which each is tangent to its two neighbors. Express $r$ in terms of $R$.
1. **Understanding the Geometry**:
   - We have a large sphere \( S \) of radius \( R \) tangent to a plane \( \pi \).
   - Eight smaller spheres, each of radius \( r \), are tangent to \( \pi \) and to \( S \). These smaller spheres form a "collar" around \( S \) and are tangent to each other.

2. **Analyzing the Circle of Tangency**:
   - When viewed from above, the points of tangency of the smaller spheres with the plane \( \pi \) form a circle \( C \) of radius \( r' \).
   - The centers of the smaller spheres lie on this circle \( C \).

3. **Angle Subtended by Adjacent Spheres**:
   - Since there are 8 smaller spheres, the angle subtended by the centers of two adjacent smaller spheres at the center of \( C \) is \( \frac{\pi}{4} \).

4. **Radius of Circle \( C \)**:
   - The radius \( r' \) of circle \( C \) can be expressed in terms of \( r \) using the cosecant function:
     \[
     r' = r \csc\left(\frac{\pi}{8}\right)
     \]

5. **Cross-Section Analysis**:
   - Consider a cross-section through the center of \( S \) and one of the smaller spheres.
   - The distance between the centers of \( S \) and a smaller sphere is \( R + r \).
   - The length of the common tangent between \( S \) and a smaller sphere in this cross-section is \( r' \).

6. **Using the Pythagorean Theorem**:
   - The length of the common tangent \( r' \) can also be found using the Pythagorean theorem:
     \[
     r' = \sqrt{(R + r)^2 - (R - r)^2}
     \]
   - Simplifying the expression inside the square root:
     \[
     r' = \sqrt{(R + r)^2 - (R - r)^2} = \sqrt{(R + r)^2 - (R^2 - 2Rr + r^2)} = \sqrt{4Rr}
     \]
     \[
     r' = 2\sqrt{Rr}
     \]

7. **Equating the Two Expressions for \( r' \)**:
   - We have two expressions for \( r' \):
     \[
     r' = r \csc\left(\frac{\pi}{8}\right) \quad \text{and} \quad r' = 2\sqrt{Rr}
     \]
   - Equating these:
     \[
     r \csc\left(\frac{\pi}{8}\right) = 2\sqrt{Rr}
     \]

8. **Solving for \( r \)**:
   - Square both sides to eliminate the square root:
     \[
     r^2 \csc^2\left(\frac{\pi}{8}\right) = 4Rr
     \]
   - Divide both sides by \( r \) (assuming \( r \neq 0 \)):
     \[
     r \csc^2\left(\frac{\pi}{8}\right) = 4R
     \]
   - Solve for \( r \):
     \[
     r = \frac{4R}{\csc^2\left(\frac{\pi}{8}\right)}
     \]

9. **Simplifying the Expression**:
   - Recall that \( \csc(x) = \frac{1}{\sin(x)} \), so:
     \[
     \csc^2\left(\frac{\pi}{8}\right) = \frac{1}{\sin^2\left(\frac{\pi}{8}\right)}
     \]
   - Therefore:
     \[
     r = 4R \sin^2\left(\frac{\pi}{8}\right)
     \]

10. **Using Trigonometric Identity**:
    - Using the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
      \[
      \sin^2\left(\frac{\pi}{8}\right) = \frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}
      \]
    - Since \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \):
      \[
      \sin^2\left(\frac{\pi}{8}\right) = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}
      \]
    - Thus:
      \[
      r = 4R \cdot \frac{2 - \sqrt{2}}{4} = R(2 - \sqrt{2})
      \]

The final answer is \( \boxed{ r = R(2 - \sqrt{2}) } \).