We place four spheres of radius $r$ on a horizontal plane so that their centers form a square and they touch each other. We then place a fifth sphere of radius $r$ on top of the four spheres, touching all four of them. How far is the highest point of the fifth sphere from the plane? (H)
I. solution. For any of the four spheres on the horizontal plane, we can say that it touches two "adjacent" spheres, while being "opposite" to the fourth one. Let the centers of the two opposite spheres on the horizontal plane be $O_{1}$, $O_{3}$, and the center of the fifth sphere be $O_{5}$, as shown in the top and side views of the two figures. As the diagonal of a square with side length $2r$, $O_{1} O_{3}=2 r \sqrt{2}$. The triangle $O_{1} O_{3} O_{5}$ is isosceles, with base $O_{1} O_{3}$ and legs of length $2r$. Therefore, the height to the base is $r \sqrt{2}$. Thus, the highest point of the fifth sphere is at a distance of $r \sqrt{2}+2 r$ from the horizontal plane.

II. solution. Let's set up a coordinate system in space with $x, y, z$ axes such that the horizontal plane is at $z=0$. If $r=1$, then the centers of the four spheres can be, for example, $(1 ; 1 ; 1)$, $(-1 ; 1 ; 1)$, $(-1 ;-1 ; 1)$, and $(1 ;-1 ; 1)$, with the center of the fifth sphere on the $z$-axis. Let this center be $\left(0 ; 0 ; z_{0}\right)$. Since the fifth sphere touches the sphere centered at $(1 ; 1 ; 1)$, by the distance formula: $1^{2}+1^{2}+\left(z_{0}-1\right)^{2}=2^{2}$, from which $z_{0}=1+\sqrt{2}$.

For any $r$, $z_{0}=(1+\sqrt{2}) \cdot r$. The highest point of the fifth sphere is thus at a distance of $z_{0}+r=$ $r \sqrt{2}+2 r$ from the horizontal plane.

Zsbán Ambrus (Fazekas M. Főv. Gyak. Gimn., 8.o.t.)
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