Konyagin S.V.

Given a square with side length 1. From it, four corners are cut off - four triangles, each of which has two sides along the sides of the square and these sides are $1 / 3$ of their lengths. With the resulting 8-sided polygon, the same operation is performed: from each vertex, a triangle is cut off, two sides of which are $1 / 3$ of the corresponding sides of the 8-sided polygon, and so on. A sequence of polygons is obtained (each contained in the previous one). Find the area of the figure that is the intersection of all these polygons (i.e., formed by the points belonging to all polygons).
Let $M_{0}$ be the original square, and $M_{1}, M_{2}, M_{3}, \ldots$ be the polygons obtained from $M_{0}$ by sequentially cutting off corners. It is convenient to also consider the polygon $N_{k}$, whose vertices are the midpoints of the sides of $M_{k} (k=0,1,2, \ldots)$.

Let $A$ be an arbitrary vertex of the polygon $M_{k}$, and $B$ and $C$ be the adjacent vertices. To obtain the polygon $M_{k+1}$ from the polygon $M_{k}$, we need to cut off a triangle $B_{2} A C_{2}$ (see Fig. 8) such that points $B_{2}$ and $C_{2}$ divide, respectively, the segments $[A B]$ and $[A C]$ in the ratio $1:2$. Let $B_{1}$ and $C_{1}$ be the midpoints of the segments $[A B]$ and $[A C]$, i.e., the adjacent vertices of the polygon $N_{k}$; then

$$
\begin{aligned}
& \left|A C_{2}\right|=\frac{2}{3}\left|A C_{1}\right|, \\
& \left|A B_{2}\right|=\frac{2}{3}\left|A B_{1}\right| .
\end{aligned}
$$

When transitioning from the polygon $N_{k}$ to the polygon $N_{k+1}$, the points $B_{1}$ and $C_{1}$ remain its vertices, but they are no longer adjacent: a new vertex $A_{1}$ appears between them, which is the midpoint of the segment $\left[B_{2} C_{2}\right]$. Thus, the polygon $N_{k+1}$ is obtained from the polygon $N_{k}$ by adding a triangle of type $B_{1} A_{1} C_{1}$ to each of its sides $\left[B_{1} C_{1}\right]$. From (eq:234.1) it follows that the areas of the cut-off and added triangles are related by the following ratios:

$$
\begin{aligned}
& S\left(B_{2} A C_{2}\right)=\frac{4}{9} S\left(B_{1} A C_{1}\right) \\
& S\left(B_{1} A_{1} C_{1}\right)=\frac{1}{3} S\left(B_{1} A C_{1}\right)
\end{aligned}
$$

Let $x_{k}$ be the area of the polygon $N_{k}$, and $y_{k}$ be the area of $M_{k}$. Summing (eq:234.2) and (eq:234.3) over all vertices $A$ of the polygon $M'$, we get:

$$
\begin{aligned}
& y_{k}-y_{k+1}=\frac{4}{9}\left(y_{k}-x_{k}\right), \\
& x_{k+1}-x_{k}=\frac{1}{3}\left(y_{k}-x_{k}\right) ;
\end{aligned}
$$

$$
\begin{gathered}
\left(y_{k+1}-x_{k+1}\right)=\frac{2}{9}\left(y_{k}-x_{k}\right) \\
\frac{y_{k}-y_{k+1}}{x_{k+1}-x_{k}}=\frac{4}{3}
\end{gathered}
$$

Mark the points $x_{0}, x_{1}, x_{2}, \ldots$ and $y_{0}, y_{1}, y_{2}, \ldots$ on the number line (see Figs. 9 and 10). From the relation (eq:234.6) it follows that the length of the segment $\left[x_{k+1}, y_{k+1}\right]$ is $4 \frac{1}{2}$ times smaller than the length of the segment $\left[x_{k}, y_{k}\right]$. Consider the point $a$ that divides the segment $\left[x_{k}, y_{k}\right]$ in the ratio $3:4$; then

$$
\frac{y_{k}-a}{a-x_{k}}=\frac{4}{3}
$$

Considering (eq:234.7) and (eq:234.8), we get that

(1)

$$
\frac{y_{k+1}-a}{a-x_{k+1}}=\frac{4}{3}
$$

which means that the point $a$ divides the segment

$$
\left[x_{k+1}, y_{k+1}\right]
$$

in the same ratio $3:4$. Taking the first segment of our sequence: $\left[x_{0}, y_{0}\right]=\left[\frac{1}{2}, 1\right]$; the point $a$ that divides it in the ratio $3:4$ is given by $\frac{1-a}{\frac{1}{4}-\frac{1}{2}}=\frac{4}{3}$, i.e.,

$$
a=\frac{5}{7}
$$

By the above, this point divides all segments $\left[x_{k}, y_{k}\right](k=0,1,2, \ldots)$ in the ratio $3:4$; hence, each subsequent segment of the sequence $\left[x_{k}, y_{k}\right]$ is obtained from the previous one by compressing it towards the point $a$ by a factor of $4 \frac{1}{2}$. Thus, the point $a$ is the only common point of all these segments. (Readers familiar with the concept of the limit of a sequence will certainly notice that $a$ is the common limit of the sequences $x_{k}$ and $y_{k}$; in essence, we have proven this.) From this, it follows that the area of the desired figure $M$ (which is the intersection of all polygons $M_{k}$) is equal to $a$; indeed, $M$ is contained in any $M_{k}$ and contains any $N_{k}$ (prove this (This is a purely set-theoretic fact: if $M_{k} \supset M_{k+1} \supset N_{k+1} \supset N_{k}$ for each $k=0,1,2, \ldots$, and $M=\bigcap_{k} M_{k}$, then $M \supset N_{k}$.)); therefore, the area of $M$ must be between $x_{k}$ and $y_{k}$ (for all $k$), and the only number that satisfies this condition is $a=\frac{5}{7}$.

We could have solved this problem without introducing the polygons $N_{k}$, using only the relation defining the sequence $y_{0}=1, y_{1}=\frac{5}{9}, y_{2}, y_{3}, \ldots$:

$$
y_{k}-y_{k+1}=\frac{4}{9}\left(y_{k}-y_{k-1}\right)
$$

In this case, the area of $M$ is found as the limit of the areas of $M_{k}$ (the sequence $\left(y_{k}-y_{k+1}\right)$ represents an infinitely decreasing geometric progression).

A. Zelenyak (Shakhty) generalized problem M234 to the case where segments are cut off from each side, constituting $\frac{1}{\bar{\rho}}$ of the side, where $p$ is any number greater than 2. The area of the resulting figure is $\frac{p^{2}-2 p+2}{p^{2}-2 p+4}$; for $p=3$, we get the answer $\frac{5}{7}$.