Solution of the problem #22 in AMC 12 A 2012

AMC (American Math Competitions) Test Problem #22
This is the simplified formulation of this problem.

A straight line segment connects every two midpoints of the edges within each face of a cube. A collection of distinct planes is such that the intersection of the union of all these planes with the surface of the cube consists of all such segments. What is the difference between the maximum and the minimum numbers of planes that such collection can have?

(A)   8   (B) 12   (C) 20   (D) 23   (E) 24

Solution

This diagram shows 3 of the total 6 faces of the cube. Altogether, there are 6 x 6 = 36 segments described in this problem. Let’s call the segments that are parallel to the edges “edge-parallel” and the remaining segments - “diagonal”. So, there are 2 x 6 = 12 “edge-parallel” segments and 4 x 6 = 24 “diagonal” segments.

Our goal is to identify the collection of the minimum number of distinct planes that satisfy the conditions of this problem (we will call it “the minimal collection of planes”) and the collection of all possible distinct planes, such that intersection of each plane with the surface of the cube consists of several such line segments (we will call it “the maximal collection of planes”).

The diagram shows midpoint A of one of the edges of the cube. Let's identify all the planes containing all possible pairs of segments that emanate from point A and lie in two perpendicular faces of the cube. In this diagram, numbers 1,2,3 are assigned to 3 such segments in each of these two faces of the cube.

We can identify the following types of planes by placing 3 x 3 = 9 pairs of segments in four groups, so that all the symmetrical pairs are in one group. (Note: considering the symmetry of all edges of the cube, it is clear that the midpoint of any other selected edge belongs to the same types of planes as point A.)

Case I:      Segments 2-2
Case II:     Segments 1-1 and 3-3
Case III:    Segments 1-2, 2-1, 3-2, 2-3
Case IV:    Segments 1-3 and 3-1

Now, let's calculate the numbers of planes in each group.

Case I. This type of plane divides the cube into two congruent rectangular parallelepipeds (shown in the diagram below) that intersects with the surface of the cube at 4 “edge-parallel” segments.

It is clear that there are 3 distinct, pairwise perpendicular, planes of this type: two vertical planes and one horizontal plane. These 3 planes cover 3 different sets of 4 “edge-parallel” segments. So, altogether, they cover all 12 “edge-parallel” segments.

Case II. This type of plane cuts off a tetrahedron whose one vertex is also a vertex of the cube (as shown in the diagram below).

It satisfies the conditions of the problem because its intersection with the surface of the cube consists of 3 “diagonal” segments adjacent to one of the vertices of the cube. Since there are 8 vertices of the cube, there are 8 such distinct planes that cover 8 different sets of 3 “diagonal” segments. So, altogether, they cover all 24 “diagonal” segments.

Case III. This type of plane cuts off a triangular prism whose one edge is also one full edge of the cube (as shown in the diagram below).

It satisfies the conditions of the problem because its intersection with the surface of the cube consists of 2 “diagonal” segments and 2 "edge-parallel" segments. Since there are 12 edges of the cube, there are 12 such distinct planes that, altogether, cover all 24 “diagonal” segments and all 12 "edge-parallel" segments (each "edge-parallel" segment is shared by two such planes).

Case IV.  This type of plane also satisfies the conditions of this problem, but it's more difficult to see than for the first three types. This is probably why this is problem #22 of AMC 12.

This type of plane cuts off a solid shape shown in the diagram below.

This solid shape is a heptahedron (meaning that it has 7 faces) that has a regular hexagon as its base and 3 alternating pairs of triangles and pentagons as its side faces. For the aficionados and connoisseurs of geometry, we can notice that it has 15 edges and 10 vertices (even though this information is not relevant to the solution of this problem).

The edges of the hexagon at the base of this heptahedron are 6 “diagonal” segments. Each of these 6 “diagonal” segments belongs to exactly one of 6 faces of the cube. Consequently, considering the symmetry of "diagonal" segments, faces, edges, and everything else in the cube, there are 4 such planes that, altogether, cover 4 different sets of 6 “diagonal” segments. So, they cover all 24 “diagonal” segments.

Strictly speaking, we must prove that the entire set of such 6 segments (in which one “diagonal” segment is properly selected from each face of the cube)  belongs to one plane. It can be done, for example, by calculating the coordinates x,y,z of the ends of each of these segments and by using the formula that verifies that all these points belong to one plane. The coordinates of midpoints of the properly selected 6 edges of the cube are shown in the diagram below.

A plane is a graph of the linear function z = ax + by + c.
The plane that contains the first 3 midpoints, (0,1,0), (1,0,0), (2,0,1), is the graph of the following linear function:
z = x + y -1
(the values of a, b, c can be easily calculated by plugging the coordinates of first 3 midpoints in the places of of x, y, z in the general equation z = ax + by + c).
Now, we can simply check that the coordinates of the remaining 3 midpoints, (2,1,2), (1,2,2), (0,2,1), satisfy this equation. They do.

Here is another proof of this fact. If we draw a straight line from each end of each of the six diagonal segments to vertex A of the cube (as shown on the picture below), two of these line segments belong to the left face of the cube, two other segments - to the front face of the cube, and the last two segments - to the bottom face of the cube. Each of these line segments is the hypotenuse of a right triangle congruent to the triangle marked with green color on the diagram. Thus, the length of each hypotenuse is equal to sqrt(1 + 1/4) = sqrt(5)/2. Hence, they form six congruent isosceles triangles that form a regular hexagonal pyramid. It proves that six diagonal segments belong to one plane since they are the edges of the base of this pyramid.

Note: if we connect the same six points to the opposite vertex of the cube, A1, it will form the other regular hexagonal pyramid, congruent to the first one. Its lateral edges belong to the right, back, and top faces of the cube. This is another proof that all the midpoints 1 - 6 belong to one plane: these six points are equidistant from points A and A1, and we can use the theorem that the locus of all points equidistant from two fixed points in 3-dimensional space is the plane perpendicular to the line segment connecting these two points.

Let's summarize our calculations. One type of planes covers all 12 "edge-parallel" segments:
      I.             3 planes that cut off the rectangular parallelepipeds

and three other types of planes can each cover all 24 “diagonal” segments:

II.                    8 planes that cut off the tetrahedra

III.                  12 planes that cut off  the triangular prisms (these planes also cover all 12 "edge-parallel" segments)

IV.                 4 planes that cut off the heptahedra

Types I and IV are the clear winners to be included in the minimal collection of planes, since the total number of planes in this collection is 3 + 4 = 7. Of course, they must be counted in the maximal collection of planes as well. So, they can be eliminated from the calculations of the difference between the maximum and the minimum numbers of planes.

As a result, the only two types of planes that belong to the maximal collection and do not belong to the minimal collection are types II and III. The total number of distinct planes in these two types is 8 + 12 = 20.

The answer: (C).