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A regular hexagon has been cut to form a shape as shown in the image. The marked points on the perimeter and inside the hexagon divide the respective segments into quarters.

What is the ratio of the areas of the original hexagon and the cut-out shape?

(A. Bohiniková)

![](https://cdn.mathpix.com/cropped/2024_04_17_cc5184b1f392581ce2bcg-6.jpg?height=399&width=456&top_left_y=500&top_left_x=800)

Hint. Divide the shape into suitable smaller parts.
A hexagon is composed of six congruent equilateral triangles. The gray shapes cut out from these triangles are of two types:
![](https://cdn.mathpix.com/cropped/2024_04_17_cc5184b1f392581ce2bcg-6.jpg?height=390&width=918&top_left_y=1192&top_left_x=570)

In the first case, $K$ is the midpoint of segment $AB$ and $L$ is the midpoint of segment $SA$. Triangles $SKL$ and $SKA$ have the same height from their common vertex $K$, and the corresponding side $SL$ is half the length of side $SA$. Therefore, triangle $SKL$ has half the area of triangle $SKA$. For similar reasons, triangle $SKA$ has half the area of triangle $SBA$. In total, the gray part has a quarter of the area of triangle $SBA$.

In the second case, $N$ is the midpoint of segment $BC$, $M$ is the midpoint of segment $BN$, $O$ is the midpoint of segment $NC$, and $P$ is the midpoint of segment $NS$. Similarly to the previous case, we can deduce that triangle $SMP$ has half the area of triangle $SMN$, which has half the area of triangle $SBN$, which in turn has half the area of triangle $SBC$. In total, triangle $SMP$ has an eighth of the area of triangle $SBC$. Triangle $SOP$ is congruent to triangle $SMP$. In total, the gray part has a quarter of the area of triangle $SBC$.

For each of the six auxiliary triangles, the ratio of its area to the area of the gray shape is $4:1$. Therefore, the ratio of the areas of the original hexagon and the cut-out shape is the same.