Problem 11.7. The centers of six spheres of radius 1 are located at the vertices of a regular hexagon with side 2. These spheres are internally tangent to a larger sphere $S$ with its center at the center of the hexagon. The sphere $P$ is externally tangent to the six spheres and internally tangent to the sphere $S$. What is the radius of the sphere $P$?
Answer: 1.5.

Solution. Let the centers of the first six spheres be $A, B, \ldots, F$ (in the order in which they form a hexagon), and the plane of the hexagon be $\alpha$; the center of sphere $S$ (i.e., the center of the hexagon) be $O$, and its radius be $r$; the center of sphere $P$ be $Z$, and its desired radius be $z$.

First, we find the radius of sphere $S$. For this, consider the cross-section of the construction by the plane $\alpha$ (Fig. 16a). The point of tangency of $S$ with the sphere centered at $A$ - let's denote this point as $A_{1}$ - lies on the line $O A$. Since $A A_{1}=1$ and $O A_{1}=r$, then $O A=r-1$. On the other hand, from the properties of a regular hexagon, it is clear that triangles $A O B, B O C, \ldots, F O A$ are equilateral, i.e., $O A=A B=2$. We get $r=3$.

Now consider sphere $P$. From its tangency with the six spheres, it follows that the distances from $Z$ to points $A, B, \ldots, F$ are all $1+z$. This means that $Z$ lies on the perpendicular to $\alpha$ passing through $O$ (indeed, from the equality of distances, it follows that points $Z$ and $O$ lie on the perpendicular bisectors of segments $A D$ and $B E$, so the line $Z O$ is perpendicular to lines $A D$ and $B E$, and hence to the plane containing them). This perpendicular is the line of centers of spheres $S$ and $P$, so their point of tangency - let's denote it as $X$ - also lies on it. Note that from the internal tangency of $S$ and $P$, it follows that $O Z=O X-X Z=3-z$.

![](https://cdn.mathpix.com/cropped/2024_05_06_28b5ece3789ad9441c8ag-5.jpg?height=609&width=608&top_left_y=97&top_left_x=124)

(a)

![](https://cdn.mathpix.com/cropped/2024_05_06_28b5ece3789ad9441c8ag-5.jpg?height=605&width=575&top_left_y=91&top_left_x=750)

(b)

Fig. 16: to the solution of problem 11.7

Consider triangle $A O Z$ (Fig. 16b). It is a right triangle because $O Z \perp \alpha$ and $\alpha \ni O A$. We know the expressions for all its sides in terms of $z$; let's write the Pythagorean theorem:

$$
\begin{gathered}
Z A^{2}=O A^{2}+O Z^{2} \Rightarrow \\
(1+z)^{2}=2^{2}+(3-z)^{2} \Rightarrow \\
1+2 z+z^{2}=4+9-6 z+z^{2} \Rightarrow \\
8 z=12
\end{gathered}
$$

From this, $z=1.5$.