Klepitsyn V.A.

Cells of a $5 \times 5$ board are painted in a checkerboard pattern (corner cells are black). A figure, a mini-bishop, moves along the black cells of this board, leaving a trail on each cell it visits and not returning to that cell again. The mini-bishop can move either to free (diagonally) adjacent cells or jump (also diagonally) over one cell with a trail to a free cell beyond it. What is the maximum number of cells the mini-bishop can visit?
Let's first provide an example of the mini-elephant's route that ensures it visits twelve cells (see the figure, numbers from 1 to 12 show the order of cell visits).

| 1 |  | 4 |  | 6 |
| :--- | :--- | :--- | :--- | :--- |
|  | 2 |  | 5 |  |
| 3 |  | 7 |  | 9 |
|  | 11 |  | 8 |  |
| 12 |  | 10 |  |  |

We will prove that the mini-elephant cannot visit all black cells. Consider the four corner cells. The mini-elephant can leave such a cell either to an adjacent cell or to the central cell. It can go to an adjacent cell from a corner cell only on the first move. Indeed, to enter a corner cell, the mini-elephant can either come from an adjacent cell or from the central cell, jumping over an already visited adjacent cell. In both cases, it cannot go to an adjacent cell again. The mini-elephant can go to the central cell from a corner cell no more than once. Thus, the mini-elephant will leave the corner cells no more than twice, meaning it will visit no more than three corner cells.

## Answer

12 cells.

Author: Berdnikov A.

An equilateral triangle has been cut out of the plane.

Can the remaining part of the plane be tiled with triangles, any two of which are similar but not homothetic?

## Solution

We will provide two examples of such tiling.

Example 1. On the extensions of the sides of the cut-out triangle, mark points \(A_1, B_1, C_1\) such that \(AA_1 = BB_1 = CC_1 = xAB\) (see the left figure). Triangles \(A_1AB_1, B_1BC_1\), and \(C_1CA_1\) are equal, and triangle \(A_1B_1C_1\) is equilateral. Now extend the sides of triangle \(A_1B_1C_1\) and mark points \(A_2, B_2, C_2\) such that \(A_1A_2 = B_1B_2 = C_1C_2 = xA_1B_1\). Triangles \(A_2A_1B_2, B_2B_1C_2\), and \(C_2C_1A_2\) are equal to each other and similar to triangle \(A_1A_2B_1\). Similarly, construct points \(A_3, B_3, C_3\) and so on. The sizes of triangles \(A_kB_kC_k\) grow in a geometric progression, so the entire plane will be covered. Thus, we have obtained a tiling of the plane into similar triangles of the form \(A_iB_{i-1}B_i, B_jB_{j-1}C_j\), and \(C_kC_{k-1}A_k\).

Consider the equilateral triangles \(A_kB_kC_k\). All of them have a common center, and the directions of the sides of adjacent triangles differ by a rotation of angle \(B_1C_1B\). To ensure that the triangles in the tiling are not homothetic, choose the value of \(x\) such that the angle \(B_1C_1B\) is irrational. Then, for any pair of triangles in the tiling, their long sides will not be parallel to each other, and thus the triangles will not be homothetic. Therefore, this tiling satisfies the condition.
![](https://cdn.mathpix.com/cropped/2024_05_06_5dbee4d95f83bceb3891g-10.jpg?height=874&width=1606&top_left_y=1264&top_left_x=228)

Example 2. See the right figure. All triangles are equal. Equal triangles can be homothetic only with a coefficient of -1. But there are no centrally symmetric triangles in the figure.

## Answer

Yes.

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