1. In space, there are 4 points not lying in the same plane. How many planes can be drawn equidistant from these points?
1. Since the four points do not lie in the same plane, the plane equidistant from these points cannot be located on the same side of all of them. Therefore, there are only two possible cases: 1) three points lie on one side of the considered plane, and the fourth point lies on the other side, and 2) two points lie on each side of the plane.

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-090.jpg?height=377&width=490&top_left_y=936&top_left_x=357)

Fig. 27.

Consider the first case. Let points \(A, B\), and \(C\) lie on one side of the plane \(\Pi\), equidistant from points \(A, B, C\), and \(D\), while point \(D\) lies on the other side (Fig. 27). Points \(A, B\), and \(C\) cannot lie on the same line—otherwise, all four points would lie in the same plane. Since the plane \(\Pi\) is equidistant from points \(A, B\), and \(C\), which are located on the same side of it, this plane must be parallel to the plane \(ABC\). For the distance from point \(D\) to this plane to be equal to the distance from it to the three points \(A, B\), and \(C\), the plane \(\Pi\) must pass through the midpoint of the perpendicular \(DP\) dropped from point \(D\) to the plane \(ABC\) (see Fig. 27). Thus, the plane \(\Pi\), equidistant from points \(A, B, C\), and \(D\), with points \(A, B\), and \(C\) on one side and point \(D\) on the other, is uniquely determined.

Similarly, the plane equidistant from the given four points, with only point \(C\) (or point \(B\), or point \(A\)) on one side and the other three points on the other side, can be found. In total, there are four planes equidistant from the given four points such that only one point is on one side of them, and the other three are on the other side.

Consider the second case. Let points \(A\) and \(B\) lie on one side of the plane \(\Pi\), equidistant from points \(A, B, C\), and \(D\), while points \(C\) and \(D\) lie on the other side (Fig. 28). Since the plane \(\Pi\) is equidistant from points \(A\) and \(B\), which lie on the same side of \(\Pi\), it must be parallel to the line \(AB\). Similarly, it can be proven that this plane must be parallel to the line \(CD\). Since points \(A, B, C\), and \(D\) do not lie in the same plane, the lines \(AB\) and \(CD\) must be skew. Draw two parallel planes through the skew lines \(AB\) and \(CD\); it is clear that the plane \(\Pi\) must be parallel to these planes and equidistant from them, i.e., it must pass exactly midway between these planes (see Fig. 28). Thus, there is a unique plane equidistant from points \(A, B, C\), and \(D\) such that points \(A\) and \(B\) lie on one side of it, and points \(C\) and \(D\) lie on the other.

Similarly, there is a unique plane equidistant from our four points such that the point \(A\) and only the point \(C\) (or point \(D\)) lie on the same side, while the other two points (\(B\) and \(D\) or \(B\) and \(C\)) lie on the other side. Therefore, there are three planes equidistant from the given four points such that two of the four points lie on one side of each of them, and the other two lie on the other side.

Thus, the total number of planes equidistant from the given four points in space is \(4 + 3 = 7\). If we consider a triangular pyramid (tetrahedron) with vertices \(A, B, C\), and \(D\),

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-092.jpg?height=455&width=414&top_left_y=407&top_left_x=147)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-092.jpg?height=456&width=415&top_left_y=408&top_left_x=588)
b)

Fig. 29.

then four of these seven planes will pass parallel to the faces of the pyramid through the midpoints of the corresponding heights (Fig. 29, a), and the remaining three will pass parallel to pairs of opposite edges at equal distances from them (Fig. 29, b).