Let's denote the midpoints of segments $B E$ and $G D$ as $H$ and $I$, respectively. Then the rectangle $H E G I$ forms half of the rectangle $B E G D$, and point $C$ lies on its side $H I$. We will further divide this rectangle by a perpendicular line dropped from point $C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_a19b727c65e3ebaca7ddg-3.jpg?height=657&width=1071&top_left_y=234&top_left_x=524)

Now it is clear that the ratio of the white and gray areas in rectangle $H E G I$ is $1: 1$. Triangles $C G E$ and $E F G$ are congruent, so the areas of the white and gray parts in pentagon $H E F G I$ are in the ratio $2: 1$. The entire figure is symmetric with respect to the axis $H I$, so the ratio of the areas of the white and gray parts of hexagon $A B E F G D$ is the same.

Note. It can also be solved by appropriately translating triangle $D G C$ and then dividing the resulting shape into six congruent triangles, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_a19b727c65e3ebaca7ddg-3.jpg?height=791&width=1060&top_left_y=1278&top_left_x=538)

Evaluation. 5 points for a correct and justified method; 1 point for the result.

Alternative solution. Since triangle $B E C$ is equilateral, all its internal angles are $60^{\circ}$. From this, it follows that in triangle $C D B$, the internal angles measure $30^{\circ}, 90^{\circ}$, and $60^{\circ}$, so this triangle is half of an equilateral triangle with a side length of $2 \cdot|C D|=2 \cdot|A B|=10 \mathrm{~cm}$. Therefore, $|B D|=10 \mathrm{~cm}$, and using the Pythagorean theorem, we can calculate the length of side $B C$ in triangle $C D B$:

$$
|B C|=\sqrt{10^{2}-5^{2}}=\sqrt{75}=5 \sqrt{3}(\mathrm{~cm}) .
$$

The area of triangle $C D B$ is thus

$$
S_{C D B}=\frac{1}{2}|B C| \cdot|C D|=\frac{1}{2} \cdot 5 \sqrt{3} \cdot 5=\frac{25}{2} \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
$$

The same area is also had by triangles $A B D, C G E$, and $F E G$, as they are congruent to triangle $C D B$. Since triangle $B E C$ is equilateral, $|B E|=|B C|$, and we can calculate the area of rectangle $B E G D$:

$$
S_{B E G D}=|B E| \cdot|B D|=5 \sqrt{3} \cdot 10=50 \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
$$

Then the area of the white part of hexagon $A B E F G D$ is

$$
\begin{aligned}
S_{\text {white }} & =S_{A B D}+\left(S_{B E G D}-S_{C D B}-S_{C G E}\right)+S_{F E G}= \\
& =S_{C D B}+\left(S_{B E G D}-S_{C D B}-S_{C D B}\right)+S_{C D B}= \\
& =S_{B E G D}=50 \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
\end{aligned}
$$

The area of the gray part of hexagon $A B E F G D$ is

$$
S_{\text {gray }}=S_{C D B}+S_{C G E}=2 \cdot S_{C D B}=25 \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
$$

Therefore, the ratio of the areas of the white and gray parts of the hexagon is

$$
S_{\text {white }}: S_{\text {gray }}=50 \sqrt{3}: 25 \sqrt{3}=2: 1 \text {. }
$$

Evaluation. 1 point for calculating the length of segment $B C$; 1 point each for calculating the areas of triangle $C D B$ and rectangle $B E G D$; 1 point each for determining the areas of the gray and white parts; 1 point for calculating the ratio of the areas of the white and gray parts (each calculation must be justified).

Note. The approximate values of the previous quantities expressed using tables without a calculator are: $|B C| \doteq 8.66 \mathrm{~cm}, S_{C D B} \doteq 21.65 \mathrm{~cm}^{2}, S_{B E G D}=S_{\text {white }} \doteq 86.6 \mathrm{~cm}^{2}, S_{\text {gray }} \doteq$ $\doteq 43.3 \mathrm{~cm}^{2}$ and the ratio $S_{\text {white }}: S_{\text {gray }} \doteq 2: 1$. If the solver calculates with approximate values and does not realize at the end that the calculated ratio $2: 1$ is only approximate, award them a total of 5 points.
In the following figure, there is a hexagon $A B E F G D$. Quadrilaterals $A B C D$ and $E F G C$ are congruent rectangles, and quadrilateral $B E G D$ is also a rectangle. Determine the ratio of the areas of the white and gray parts of the hexagon, given that $|A B|=5 \mathrm{~cm}$ and triangle $B E C$ is equilateral.

![](https://cdn.mathpix.com/cropped/2024_04_17_a19b727c65e3ebaca7ddg-2.jpg?height=648&width=1060&top_left_y=1686&top_left_x=538)

(K. Pazourek)