Shapovalov A.v.

Sasha had four colored cubes. Arranging them differently, he sequentially photographed three figures (left image). Then Sasha assembled them into a parallelepiped of size $2 \times 2 \times 1$ and took a black-and-white photo of it (right image). All the visible faces of the cubes in this photo are of the same color. What color is it?
![](https://cdn.mathpix.com/cropped/2024_05_06_76207e7feb35677daaaeg-06.jpg?height=144&width=966&top_left_y=1894&top_left_x=542)
In the first color photo, there are 8 red faces, in the second - 8 blue, and in the third - 8 yellow. Since the four cubes together have \(6 \cdot 4 = 24\) faces and \(8 \cdot 3 = 24\), the first photo shows all the red faces, the second - all the blue faces, and the third - all the yellow faces.

In the photo of the parallelepiped, all four cubes are visible, with only one of them having one visible face, while each of the other three has at least two visible faces. In the first color photo, there are two cubes with only one red face, so the sought color cannot be red. In the third color photo, there is a cube with no yellow faces, so the sought color cannot be yellow. Therefore, all visible faces of the cubes can only be blue. This is possible because in the second color photo: 1) there is a cube with three blue faces sharing a common vertex; 2) there are two cubes, each with two adjacent blue faces; 3) there is another cube with one blue face.

## Answer

Blue.

A point \( M \) is marked inside a square. Can this square be cut into no more than three rectangles and reassembled into a square such that point \( M \) becomes its center? (The cuts should not pass through point \( M \).)

## Solution

Let point \( M \) be located inside the square \( ABCD \) such that its distances to sides \( AB \) and \( AD \) are \( x \) and \( y \) respectively. Without loss of generality, we can assume that the side of the square is 2, and \( x, y \leq 1 \).

If the inequalities are strict, then draw two cuts parallel to sides \( CD \) and \( BC \): at a distance of \( 1-x \) from \( CD \) and at a distance of \( 1-y \) from \( BC \) (see the figure). By moving the two rectangles as shown in the figure, we obtain a square \( EFGH \). In this case, the distance from point \( M \) to each of its sides is 1, meaning \( M \) is the center of this square.

![](https://cdn.mathpix.com/cropped/2024_05_06_76207e7feb35677daaaeg-07.jpg?height=521&width=554&top_left_y=660&top_left_x=755)

If one of the inequalities is an equality, then it is sufficient to cut and move one rectangle. If both cases are equalities, then \( M \) is already the center of the square \( ABCD \), so no cutting is required.

## Answer

Yes.