Tasks

premature, we did not teach the possibility of “attacking the corner.”
If the cross is on the diagonal of the defended square, then it threatens almost twice as many fields outside the square as in a “frontal attack” on the side of this square, for example, on figure 8 it threatens, as before (Fig. 5), thirteen fields on each side, and a total of 25 fields.
Are the zeroes really not able to withstand this attack? The ability to take the sides of a large square as a defense plazar comes to the aid of the zeros. Let’s take a larger square (much larger than figure 9) and draw a square in it, spaced six squares from the side of the first one. Let’s take the center of the square in the cell where the first cross was placed. As long as the crosses are inside the smaller square, the side of the square is not threatened by any attack, and while the crosses are moving towards the side (or corner) of the inner square, the crosses will be able to strengthen their positions in the corners of the large square, for example, placing 11 zeros in each corner (Fig. 9), thereby the “attack on the corner” will be meaningless, and we have learned to repel an attack on the side if it starts from a distance of six cells from the side. To put 11 zeros in each corner (in fact, you can get by with fewer), you will need an inner square with a side of 87 cells, and therefore an outer square will have a side of 99 cells.
We leave it to the readers to precisely formulate the “winning” strategy of the zeros and explore other variants of this game (for example, to find out at what $p$ and $q$ zeros they cannot “lock” the crosses if $p$ crosses and $q$ zeros are placed in one move).

Exercise

1. On sides $AB, BC, CA$ of triangle $ABC$ points $C_1, A_1, B_1$ are selected respectively. Prove that the circles circumscribed around the triangles $ABC_1, BCA_1, CAB_1$ intersect at one point. Generalize this statement to the case of six points chosen on the sides of the teradir.

2. Given the pyatulnik $ABCDE$. A circle is drawn through the ends of each side and the intersection point of two adjacent sides. Prove that these circles intersect at a fifth point (different from the vertices of the fifth point) lying on the same circle.

3. Through the point $P$ lying inside $BAC$, two equal circles are drawn, one of which touches the line $AB$, and the other touches the line $AC$.

4. Two circles touch (on the same side) line $l$ at point $C$, and line $m$ is parallel to $l$ and intersects both circles at points $S, T, X, Y$ (sequentially). Prove that the point $C$ lies on the bissertrix of one of the two angles formed by pairs of tangents to the circle at points $S, T$ or $X, Y$.

5. A triangle is inscribed in a circle, and the square lengths of its sides are proportional to the lengths of perpendiculars drawn from opposite vertices to a fixed tangent to the circle. Prove that one of the sides lies on the diameter of the circle.

6. Two circles of radii $R$ and $r$ are arranged so that the square of the distance between the centers of the circles is $R^2 + 14Rr + r^2$. Prove that there exists an infinite multitorus of circles tangent in pairs to each other and to both given circles.