At most, how many parts can 4 planes that pass through a single point divide the space into?
I. solution. If there is a line through which more than two of the given planes pass, then the number of regions formed is 8 - if the four planes pass through the same line -, or 12, if one of the planes intersects the common line of the other three planes.

If there is no such line, then by translating one of the planes in parallel, a new - bounded - region is created: the interior of a tetrahedron whose four faces are the translated plane and the other three planes.

The faces of this tetrahedron divide the space into 15 regions, since one region is connected to each vertex (4), edge (6), and face (4), and the 15th region is the interior of the tetrahedron.

By translating the translated plane back to its original position, this bounded region disappears, and the unbounded regions remain, so we get that the four planes divide the space into at most 14 regions.

Tóth Szilveszter (Tata, Eötvös J. Gymn., 11th grade)

II. solution. Let's examine how the number of regions changes when adding an $(n+1)$-th plane to $n$ planes. This plane intersects each of the existing $n$ planes in a line; these $n$ lines pass through the common point (points) of the planes, and the $(n+1)$-th plane is divided into at most $2n$ regions. Each of these regions defines exactly one new region, so the number of regions increases by at most $2n$ with the addition of the $(n+1)$-th plane. This is possible if there are no identical lines among the $n$ lines, and at most two of the given planes pass through any line in space.

Based on the above result, we get the following table:

| number of planes $(n)$ | number of regions |
| :---: | :---: |
| 1 | 2 |
| 2 | $2+2 \cdot 1=$ |
| 3 | $4+2 \cdot 2=$ |
| 4 | $8+2 \cdot 3=$ |

According to these, if four planes pass through a point, they divide the space into at most 14 regions.

Remarks. 1. The first solution can also be explained more vividly: consider a tetrahedron, or the 15 regions into which its faces divide the space. If we move "very far" from the tetrahedron, it "shrinks to a point," while the 14 unbounded regions remain, so the answer to the problem is 14.

2. We can also reach the goal with another method. Draw a sphere around the common point of the planes. Each plane intersects the sphere in a great circle, which divides the sphere's surface into regions. Each region corresponds to exactly one such domain on the sphere's surface.

The question, then, is how many regions 4 great circles divide the sphere's surface into. If we project the sphere onto a plane tangent to the sphere at a point $P$ through which no circle passes, it is known that the image of the sphere's surface is the entire plane, and the image of the circles is a circle. Thus, we have reformulated the original question: into how many regions does the plane divide at most by 4 circles. The two problems are therefore equivalent, and the latter form is much better known.

3. In the second solution, we get the recursion $R_{1}=2 ; R_{n+1}=R_{n}+2 n$ for the number $R_{n}$ of regions formed, and the solution to the recursion is: $R_{n}=n^{2}-n+2$.