11.5. Let's call a trapezoid with bases 1 and 3 a "boat," which is obtained by gluing two triangular half-cells to the opposite sides of a unit square. In a $100 \times 100$ square, an invisible boat is placed (it can be rotated, it does not go beyond the boundaries of the square, its middle cell is entirely on one of the cells of the square). With one shot, you can cover any triangular half-cell. If the shot intersects the interior of the boat (i.e., the intersection of the triangular shot with the boat has a non-zero area), the boat is considered sunk. What is the minimum number of shots sufficient to definitely sink the boat?

(S. Berlov, N. Vlasova)
Answer: 4000 shots.

First solution. We will call a boat horizontal or vertical depending on whether its parallel sides are horizontal or vertical.

First, we will show that 4000 shots are enough. We divide the $100 \times 100$ square into 400 squares of size $5 \times 5$, and in each square, we make 10 shots as shown in Fig. 9. It is not hard to see that in each row and each column, a boat cannot be placed between adjacent shots; therefore, one of the shots will definitely sink the boat.

It remains to show that it is impossible to guarantee sinking the boat with fewer than 4000 shots. This time, we divide the board into 2000 horizontal rectangles of size $1 \times 5$ and show that at least two shots must be made in each such rectangle. Indeed, at least one shot must be made in the left three cells of the rectangle, otherwise an unsunk boat could be placed there; the same is true for the right three cells. Thus, if only one shot was made in the rectangle, it could only be in the central cell of the rectangle. Without loss of generality, this shot was made in the lower left triangle of this cell; but then a boat placed as in Fig. 10 would not be sunk.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e804b2a1d2ef00abb53g-7.jpg?height=497&width=581&top_left_y=133&top_left_x=232)

Fig. 9

![](https://cdn.mathpix.com/cropped/2024_05_06_5e804b2a1d2ef00abb53g-7.jpg?height=95&width=410&top_left_y=336&top_left_x=853)

Fig. 10

Second solution. We will provide another proof that at least 4000 shots are required. We divide the square into 100 horizontal strips of size $1 \times 100$ and show that even to guarantee the sinking of a horizontal boat, at least 40 shots are required in each strip.

Let's count the number of different ways to place a horizontal boat in a strip. The central cell of such a boat can be in any cell of the strip except the edge ones. For each of these 98 cells, there are two possible placements of the horizontal boat with this central cell. Therefore, the total number of ways is $98 \times 2 = 196$.

On the other hand, a shot in any triangle in the strip can sink at most five of these possible boats. Indeed, if this triangle, say, is the upper left one in its cell $c$, then it will sink any of the two boats with the central cell $c$, any of the two boats with the central cell immediately to the left of $c$, and one of the two boats with the central cell immediately to the right of $c$. (Note that some of these boats may extend beyond the edges of the board.)

Thus, if fewer than 40 shots are made in the strip, they can sink at most $39 \times 5 = 195$ possible placements of the boat, meaning there will be a placement that will not be sunk. Therefore, at least 40 shots must be made in each strip, totaling at least $40 \times 100 = 4000$ shots.

Remark. A similar argument can be applied to any rectangle of size $1 \times 5$: there are 6 ways to place the boat in it, so at least two shots must be made in it.

Comment. An example is given showing that the boat can be guaranteed to be sunk with 4000 shots - 2 points.

It is only proven that it is impossible to guarantee sinking the boat with fewer than 4000 shots - 5 points.