## Problem 1.

For four non-coplanar points, an equalizing plane is a plane such that the respective distances from each of the points to that plane are all equal. Given a set of four non-coplanar points, how many equalizing planes are there?
## Solution:

Let A, B, C, and $D$ be four non-coplanar points. Let $p$ be an equidistant plane of these points and consider the possibilities:

i) If $A, B, C$ and $D$ are on the same half-space into which $p$ divides the space, it is clear that one of the two planes parallel to $p$, at the same distance as the four points from $p$, contains all these points, contradicting the hypothesis of non-coplanarity. Therefore, the points cannot all be on the same side of the equidistant plane.

ii) Let $A$ be on one side of the plane and $B, C$, and $D$ on the other. It is obvious that $p$ is parallel to the plane $q$ determined by the points $B$, $C$, and $D$ and intersects the segment that projects $\mathrm{A}$ onto the plane $\mathrm{q}$ at its midpoint. Therefore, under these conditions, the equidistant plane exists and is unique. Now, by taking each of the other points as the "isolated point" on one side of the plane, we obtain as many equidistant planes. Consequently, with one point on one side and the other three on the other, there are exactly four equidistant planes.

iii) Now let $A$ and $B$ be on one side of the plane $p$ and $C$ and $D$ on the other. Consider the plane $r$ that contains $A$ and $B$ and is parallel to the segment $C D$ and the plane $s$ that contains $C$ and $D$ and is parallel to the segment $A B$.

Then, the planes $r$ and $s$ are parallel and the equidistant plane $p$ is the plane equidistant from the planes $r$ and $s$, which is also uniquely determined. Given the point $A$, the choice of its "pair" on one of the sides of $p$ determines as many equidistant planes, three in total.

Thus, for four given non-coplanar points, there are exactly seven equidistant planes.