4. An unconventional piece, which we will call the "sick lady," threatens any square in the row and column where it stands, while on the diagonal, it only threatens the adjacent squares. How many "sick ladies" do we need to place on an $8 \times 8$ chessboard so that they threaten all unoccupied squares?

![](https://cdn.mathpix.com/cropped/2024_04_17_fa1f8aa7943598e9e4d8g-6.jpg?height=400&width=411&top_left_y=374&top_left_x=868)

(Tomáš Bárta, Josef Tkadlec)
Solution. Fields threatened by a sick queen (we will not consider other queens further) are called directly threatened if they lie in the column or row where the queen stands; threatened fields that are not directly threatened are called indirectly threatened. Each queen thus threatens 15 fields directly (including the field on which she stands) and at most 4 fields indirectly.

First, we will show that for any placement of up to four queens on the chessboard, there is at least one field not threatened by any queen, in other words, that four queens do not threaten the entire chessboard. It is clear from this that fewer queens also do not threaten all fields of the chessboard.

Assume that the required placement of four queens exists. We can certainly find 4 rows and 4 columns of the chessboard where none of the four queens stand. These rows and columns intersect at 16 fields, which must all be indirectly threatened. Since each queen indirectly threatens at most 4 fields, each of the four given queens must indirectly threaten exactly 4 fields. Moreover, no two of them stand in the same row or column, because otherwise the number of fields indirectly threatened by our four queens would have to be even greater than 16 (there would be a fifth row or column where no queen stands).

No queen that indirectly threatens 4 fields can apparently stand in the edge row or edge column of the chessboard. The four corner fields of the entire chessboard are thus indirectly threatened in our situation, each by a different one of the four queens, which must necessarily stand on fields diagonally adjacent to the corner fields. They thus stand in pairs in the same rows (and in pairs in the same columns), which contradicts the conclusion of the first paragraph. This proves that up to four queens are not enough to threaten all fields of the chessboard.

As we can see from the following images, 5 queens can be placed on an $8 \times 8$ chessboard so that every field of the chessboard is threatened by at least one queen.

The smallest number of (sick) queens needed is five.
![](https://cdn.mathpix.com/cropped/2024_04_17_fa1f8aa7943598e9e4d8g-6.jpg?height=394&width=916&top_left_y=2119&top_left_x=573)

Fig. 2

Note. From the professional literature, it is known that the smallest number of "normal queens" that can be placed on an $8 \times 8$ chessboard so that they threaten all its fields (the so-called domination number for the queens' graph $Q(8)$) is 5. Thus, the number of "sick queens" must be at least 5, because these queens threaten fewer fields than "normal queens." Furthermore, we can determine by computer simulation that there are only 8 suitable placements of 5 sick queens, all of which arise from the two placements mentioned above by rotation (or symmetry).

For a complete solution, award 6 points. Of this, award 3 points for proving the inequality $n>4$ (including the possibility that a knowledgeable solver declares this inequality for normal queens as known) (tolerate the absence of a mention of values $n<4$). Award 3 points for an example of a correct placement of 5 queens. Award 1 point for just an example of a correct placement of 6 queens.