Cube  [ Properties of parts obtained by cutting ]

The surface of a 3 x 3 x 3 Rubik's cube consists of 54 cells. What is the maximum number of cells that can be marked so that the marked cells do not share any vertices?
In Fig. 1, it is shown how to mark 7 cells on three adjacent faces of a cube. On the three "invisible" faces, seven cells symmetrical to these need to be marked.

Now let's prove that it is impossible to mark more than 14 cells in the required manner. We will do this in two ways.

The first way. Let's count the total number of vertices of the cells of a Rubik's cube. There are 8 vertices of the cube itself, 2 more vertices on each of the 12 edges, and 4 more vertices on each of the 6 faces. In total:

$8 + 12 * 2 + 6 * 4 = 56$ vertices.

Each of these vertices belongs to no more than one marked cell by the condition. If there were more than 14 marked cells, then there would be more than $14 * 4 = 56$ vertices, since each cell has 4 vertices. Therefore, there cannot be more than 14 marked cells.

The second way. Cut the surface of the Rubik's cube into parts. Cut three adjacent faces into 7 parts as shown in Fig. 2. Cut the other three faces into the same 7 parts. It is easy to see that any two cells that fall into one part have a common vertex. Therefore, no more than one marked cell can be in each part. Thus, no more than 14 cells can be marked in total.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-11.jpg?height=434&width=486&top_left_y=2327&top_left_x=60)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-11.jpg?height=434&width=486&top_left_y=2327&top_left_x=617)

Fig. 2

## Answer

14 cells.

\section*{[ Examples and counterexamples. Constructions ]

Problem