Kosukhin O.N.

After lunch, dark spots of total area $S$ remained on a transparent square tablecloth. It turned out that if the tablecloth is folded in half along any of the two lines connecting the midpoints of its opposite sides, or along one of its two diagonals, the total visible area of the spots will be $S_{1}$. If, however, the tablecloth is folded in half along the other of its diagonals, the total visible area of the spots remains equal to $S$. What is the smallest value that the ratio $S_{1}: S$ can take?
Let's call points covered by stains dirty. For each point $P$ on the tablecloth, denote by $f(P), g(P)$, and $h(P)$ the points symmetric to $P$ with respect to the two midlines and the first diagonal of the tablecloth, respectively, and by $k(P)$ the point symmetric to $P$ with respect to the second diagonal. Note that for any point $P$, the points $k(P)$ and

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-35.jpg?height=61&width=394&top_left_y=2725&top_left_x=0)

Suppose that $S_{1} = 2S / 3$. Similarly, the area of the set of all dirty points $Q$ for which $f(Q)$ is dirty is greater than $2S / 3$. Note that if two figures have areas $s_{1}$ and $s_{2}$, and their union has an area not greater than $S$, then the area of their intersection is at least

$s_{1} + s_{2} - S$.

Therefore, the area of the set of points $Q$ for which both conditions are satisfied is greater than $2S / 3 + 2S / 3 - S = S / 3$. Hence, the area of the set of all dirty points $P$ for which $f(g(P))$ is dirty is greater than $S / 3$. Similarly, the area of the set of all dirty points $R$ for which $g(f(R))$ and $h(R)$ are dirty is greater than $S / 3 + 2S / 3 - S = 0$. Therefore, the area of the set of dirty points $P$ for which $h(f(g(P)))$ is dirty is greater than zero.

On the other hand, $h(f(g(P))) = k(P)$, and from the condition, it follows that the area of the set of dirty points $P$ for which $k(P)$ is dirty is zero. This is a contradiction.

An example of the placement of stains on the tablecloth, where $S_{1} : S = 2 : 3$, is shown in the figure. The sides of the "unit" square tablecloth are divided into six equal parts by the marked points. Here, $S = 1 / 2$, $S_{1} = 1 / 3$;

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-36.jpg?height=526&width=546&top_left_y=697&top_left_x=756)

## Answer

$2 / 3$.