5. Inside a regular hexagon ABCDEF with an area of $30 \mathrm{~cm}^{2}$, a point $M$ is chosen. The areas of triangles $ABM$ and $BCM$ are $3 \mathrm{~cm}^{2}$ and $2 \mathrm{~cm}^{2}$, respectively. Determine the areas of triangles CDM, DEM, EFM, and FAM.
SOLUTION. The problem deals with the areas of six triangles into which a given regular hexagon is divided by connecting its vertices with point $M$ (Fig.3). The entire hexagon, with a given area denoted as $S$, can also be divided into six equilateral triangles, each with an area of $S / 6$ (Fig. 4). If we denote $r$ as the side length of these triangles, $v$ as the distance between the parallel lines $A B$ and $D E$, and $v_{1}$ as the distance from point $M$ to the line $A B$, we get

$$
S_{A B M}+S_{E D M}=\frac{1}{2} r v_{1}+\frac{1}{2} r\left(v-v_{1}\right)=\frac{1}{2} r v=\frac{S}{3},
$$

since $S / 3$ is the sum of the areas of two shaded equilateral triangles. Due to symmetry, the sums $S_{B C M}+S_{E F M}$ and $S_{C D M}+S_{F A M}$ also have the same value of $S / 3$. From this, we can determine the first two unknown areas: $S_{D E M}=S / 3-S_{A B M}=7 \mathrm{~cm}^{2}$ and $S_{E F M}=S / 3- S_{B C M}=8 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-8.jpg?height=516&width=577&top_left_y=2152&top_left_x=385)

Fig. 3

![](https://cdn.mathpix.com/cropped/2024_04_17_beef4017b9d326766e3bg-8.jpg?height=517&width=577&top_left_y=2149&top_left_x=1096)

Fig. 4

How can we determine the remaining two areas $S_{C D M}$ and $S_{F A M}$, given that we only know their sum $S / 3$ so far? Notice that the sum of the given areas of triangles $A B M$ and $B C M$ has a significant value of $S / 6$, which is also the area of triangle $A B C$ (as seen in Fig.4). This equality of areas means that point $M$ lies on the diagonal $A C$. Triangles $A B M$ and $B C M$ thus have the same heights from the common vertex $B$, and the same applies to the heights of triangles $C D M$ and $F A M$ from vertices $F$ and $D$ (points that are equidistant from the line $A C$). For the ratios of the areas of these pairs of triangles, we get

$$
\frac{S_{C D M}}{S_{F A M}}=\frac{|C M|}{|A M|}=\frac{S_{B C M}}{S_{A B M}}=\frac{2}{3}
$$

In the sum $S_{C D M}+S_{F A M}$, which has a value of $S / 3$, the addends are in the ratio $2: 3$. Therefore, $S_{C D M}=4 \mathrm{~cm}^{2}$ and $S_{F A M}=6 \mathrm{~cm}^{2}$.

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