Given two planes $\left(A_{1}, A_{2}\right)$, two distances $\left(\varrho_{1}, r_{1}\right.$ and $\left.\varrho_{2}, r_{2}\right)$, and a plane $(T)$. Two spheres are to be constructed:

1) whose radii are $r_{1}$ and $r_{2}$.
2) which cut out circles of radii $\varrho_{1}$ and $\varrho_{2}$ from the planes $A_{1}$ and $A_{2}$.
3) which have a common tangent plane at a common point, parallel to $T$. One of the two spheres should touch the first picture plane.
The center of the first sphere $\left(O_{1}\right)$ can be in the planes $\mathbf{B}_{1}$ and $\mathbf{B}_{2}$, which are at a distance of $\sqrt{r_{1}^{2}-\varrho_{1}^{2}}$ from the plane $\mathbf{A}_{1}$. The center of the second sphere $\left(O_{2}\right)$ can be in the planes $\mathbf{C}_{1}$ and $\mathbf{C}_{2}$, which are at a distance of $\sqrt{r_{2}^{2}-\varrho_{2}^{2}}$ from the plane $\mathbf{A}_{2}$.

If the common tangent plane of the two spheres is parallel to T, then their centers are perpendicular to T and $\overline{O_{1} O_{2}}=r_{1} \pm r_{2}$.

The construction is as follows. From an arbitrary point $(P)$ on $\mathbf{B}_{1}$, we draw a line parallel to the axis and measure the distance $r_{1}+r_{2}$ on it. Thus, we get a point $Q$. By drawing a plane through $Q$ parallel to $\mathbf{B}_{1}$, this plane intersects C in the line $m$. The plane through $m$ parallel to the axis intersects $\mathbf{B}_{1}$ in a line $n$ parallel to $m$. The distance between $m$ and $n$ in the direction of the axis is $r_{1}+r_{2}$. $O_{2}$ is on $m$, and $O_{1}$ is on $n$. The first image plane can touch both spheres on two sides, so there are four solutions here. Since there are two planes, the number of solutions is 8; but there are two $\mathbf{B}$ planes, so the number of solutions is 16; however, from $P$ we can measure $\left(r_{1}+r\right)$ in two directions. 32 solutions; and this is valid for $\left(r_{1}-r_{2}\right)$ as well, so the total number of solutions can be 64.

George Bizám (Bolyai g. VII. r. o. Budapest V.)