Connect the midpoints of the sides of a rectangle with the opposite vertices. What fraction of the area of the original rectangle is the area of the octagon determined by these eight lines?

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I. solution. Let the area of the given rectangle $ABCD$ be denoted by $T$, and let the midpoints of the sides be $A_1, B_1, C_1, D_1$. Let points $K$, $K_1$, and $K_2$ denote the intersection points of segment $AC_1$ with segments $BD_1$, $DA_1$, and $CD_1$, respectively. Similarly, form points $L, L_1, L_2$, as well as points $M, M_1, M_2$, and points $N, N_1, N_2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_51b8bd2e1756de085d88g-1.jpg?height=539&width=1030&top_left_y=305&top_left_x=537)

Due to the parallelism of opposite sides, the quadrilateral $AA_1CC_1$ is a parallelogram. The height of this parallelogram is equal to the height of the rectangle $ABCD$ corresponding to side $AB$, and its base is half the length of the base $AB$ of the given rectangle; therefore, the area of the parallelogram is $\frac{T}{2}$.

Similarly, the quadrilateral $KLMN$ is also a parallelogram. The height of this parallelogram is equal to the height between the sides $A_1C$ and $AC_1$ of the parallelogram $AA_1CC_1$. Let's calculate the side $KN$ of the parallelogram: segments $D_1K$, $MB_1$, and $C_1L$ are the midlines of triangles $AND$, $CLB$, and $DMC$, respectively. This implies that $AK = KN = LM = MC = 2NC_1$. We also know that $\triangle NAD \sim \triangle NC_1L_1$, and thus

$$
\frac{C_1N}{AN} = \frac{C_1L_1}{AD} = \frac{1}{4}, \quad \text{so} \quad KN = \frac{2}{5} AC_1
$$

Therefore, the area of the parallelogram $KLMN$ is $\frac{T}{5}$.

Point $K_1$ is the intersection of the diagonals of the rectangle $AA_1C_1D$, so $AK_1 = K_1C_1$.

Point $K_2$ is the intersection of the medians $AC_1$ and $CD_1$ of triangle $ACD$, so it is the centroid of triangle $ACD$, and thus $AK_2 = 2K_2C_1$.

From these, we get that

$$
KK_1 = AK_1 - AK = \frac{1}{2} AC_1 - KN = \frac{5}{4} KN - KN = \frac{1}{4} KN
$$

and

$$
K_2N = AN - AK_2 = 2KN - \frac{2}{3} AC_1 = 2KN - \frac{2}{3} \cdot \frac{5}{2} KN = \frac{1}{3} KN
$$

Similarly, we get that $KL_2 = \frac{KL}{3}$. Therefore:

$$
\begin{aligned}
T_{KK_1L_2} & = \frac{1}{2} KK_1 \cdot KL_2 \cdot \sin \angle K_1KL_2 = \frac{1}{2} \cdot \frac{1}{4} KN \cdot \frac{1}{3} \cdot KL \cdot \sin \angle NKL = \\
& = \frac{1}{12} T_{NKL} = \frac{1}{24} T_{KLMN} = \frac{1}{120} T.
\end{aligned}
$$

Similarly, we get that $T_{LL_1M_2} = T_{MM_1N_2} = T_{NN_1K_2} = \frac{T}{120}$.

Thus, finally:

$$
T_{K_1L_2L_1M_2M_1N_2N_1K_2} = \frac{1}{5} T - 4 \cdot \frac{1}{120} T = \frac{6-1}{30} T = \frac{1}{6} T
$$

The area of the resulting octagon is therefore $\frac{1}{6}$ of the area of the original rectangle.

II. solution. The two perpendicular bisectors of the sides of the rectangle (whose intersection is $I$) divide the rectangle into four congruent parts. It is sufficient to examine one of these parts.

Clearly, $BG$ bisects $IF$, and $HC$ bisects $IG$.
![](https://cdn.mathpix.com/cropped/2024_05_02_51b8bd2e1756de085d88g-2.jpg?height=410&width=1200&top_left_y=125&top_left_x=453)

The medians of triangle $IFG$ trisect each other and bisect the area of the triangle. Let the area of triangle $OMN$ be $x$. Since $2NO = OF$, we have $2t_{NOM} = t_{FMO} = 2x$. Since $MN$ is the median of triangle $INF$, we have $t_{INM} = t_{OMN} + t_{OFM} = 3x$. Therefore, $t_{NGF} = t_{INF} = 6x$.

The desired ratio is:

$$
\frac{3x + x}{2 \cdot (3x + x + 2x + 6x)} = \frac{4x}{24x} = \frac{1}{6}
$$

The area of the resulting octagon is one-sixth of the area of the rectangle.