[Cutting into parts with special properties]

What is the minimum number of equal-area triangles that an $8 * 8$ square with a corner cell cut out can be divided into?

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Consider the triangles adjacent to the cut-out cell and evaluate the area of one of them.

## Solution

Let the square $8 * 8$ with a corner cell cut out (its area is 63) be cut into n triangles, each of which has an area of 63/n. Let A, B, C be the vertices of the cut-out cell. Consider the triangles containing point B. It is easy to see that there are at least two such triangles - one of them has one of its sides (call this side a) touching segment AB, and the other has one of its sides (call this side c) touching segment BC. Note that side a cannot extend beyond segment AB and side c cannot extend beyond segment BC at the same time, otherwise the triangles would intersect. Suppose, for definiteness, that side a does not extend beyond segment AB. Then, in the triangle with side a, the length of side a does not exceed 1, and the length of the height dropped to side a does not exceed 7 (otherwise the triangle would extend beyond the square). Thus, the area of this triangle does not exceed 7/2. Therefore, 63/n does not exceed 7/2, from which n is not less than 18. An example of cutting into 18 equal (and even identical) triangles is shown in the picture.

## Answer

18.00

On a $10 * 10$ square field, nine $1 * 1$ cells are overgrown with weeds. After this, the weeds can spread to a cell if at least two of its neighboring cells are already overgrown with weeds. Prove that nevertheless, the weeds will not be able to spread to all cells.

## Hint

How does the perimeter of the area overgrown with weeds change?

## Solution

Consider the boundary of the area overgrown with weeds (i.e., all segments of length 1 between nodes, on one side of which there are weeds, and on the other side there are none). Initially, the length of the boundary was no more than $9 * 4=36$, since the weeds grew only in nine cells. It is easy to notice that during the spread of the weeds, the length of the boundary cannot increase. But if the entire $10 * 10$ field were to be overgrown with weeds at some point, the length of the boundary would become $10 * 4=40$, which contradicts the considerations mentioned above.

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