Problem 11.4. On a horizontal floor, there are three volleyball balls with a radius of 18, each touching the other two. Above them, a tennis ball with a radius of 6 is placed, touching all three volleyball balls. Find the distance from the top point of the tennis ball to the floor. (All balls are spherical.)

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-03.jpg?height=385&width=521&top_left_y=84&top_left_x=466)
Answer: 36.

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-03.jpg?height=416&width=519&top_left_y=579&top_left_x=467)

Fig. 9: to the solution of problem 11.4

Solution. Let the centers of the volleyballs be denoted by $A, B, C$, and the center of the tennis ball by $D$; the top point of the tennis ball is denoted by $X$.

Consider the section of the volleyballs by the plane $A B C$ (Fig. 9). This is three mutually tangent circles with a radius of 18. Therefore, the points $A, B, C$ are located at the vertices of an equilateral triangle with a side length of 36. Let the center of this triangle be denoted by $O$; then $O A=18 / \cos \left(30^{\circ}\right)=12 \sqrt{3}$ (from the right triangle $O A M$, where $M-$ is the midpoint of $A B$).

By symmetry, it is clear that the points $X, D$, and $O$ lie on the same vertical line. Consider the section of the tennis ball and one of the volleyballs by the plane $A D O$ (Fig. 10). The triangle $A D O$ is a right triangle, $A D=18+6=24$, by the Pythagorean theorem we have

$$
O D=\sqrt{A D^{2}-O A^{2}}=12
$$

The point $O$, as well as the entire plane $A B C$, is at the height of the radius of the volleyball, which is 18; the point $D$, as we found, is 12 units higher; and the point $X$ is 6 units higher than point $D$, which is the radius of the tennis ball. In total, we get $18+12+6=36$.