In the figure, there is a square $A B C D$, in which four identical isosceles triangles $A B E, B C F, C D G$, and $D A H$ are placed, all shaded gray. The sides of the square $A B C D$ are the bases of these isosceles triangles. We know that the gray areas of the square $A B C D$ together have the same area as its white area. Furthermore, we know that $|H F|=12 \mathrm{~cm}$. Determine the length of the side of the square $A B C D$.

(L. Šimünek)

![](https://cdn.mathpix.com/cropped/2024_04_17_4274193d47a1b7e4f9cbg-2.jpg?height=526&width=594&top_left_y=588&top_left_x=731)

Idea. Appropriately divide the figure.
In the square $ABCD$, we mark both diagonals and the lines connecting the midpoints of opposite sides. The four thus added segments intersect at a single point $S$ and divide the figure completely into eight congruent triangles. One of them is marked as $STC$ in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_4274193d47a1b7e4f9cbg-2.jpg?height=528&width=596&top_left_y=1409&top_left_x=730)

These eight triangles are also congruent in their gray shaded parts, so the given condition about areas can be applied to each of these triangles individually. In the case of triangle $STC$, it follows that its gray and white areas, i.e., triangles $FTC$ and $SFC$, have the same area. Both triangles have the height $TC$. To have the same area, the lengths of the sides perpendicular to this height must also be the same, i.e., $|FT| = |SF|$. The length of segment $SF$ is half of the length given in the problem, so it is $6 \, \text{cm}$. The length of segment $ST$ is then $6 + 6 = 12 \, \text{cm}$, and the length of the side of square $ABCD$ is $2 \cdot 12 = 24 \, \text{cm}$.

Another solution. In all the gray isosceles triangles, we mark the height perpendicular to the base. This divides the original triangles into eight congruent right triangles, which we rearrange inside the square $ABCD$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_4274193d47a1b7e4f9cbg-3.jpg?height=523&width=594&top_left_y=241&top_left_x=731)

In the square $ABCD$, we have obtained two congruent gray rectangles and one white rectangle. The sides of these three rectangles that are vertical in the figure have the same length. The horizontal side of the white rectangle is $12 \, \text{cm}$. For the gray areas and the white area to have the same area, the horizontal sides of the two gray rectangles together must also be $12 \, \text{cm}$. Therefore, the side length of the square $ABCD$ is $24 \, \text{cm}$.