A unit side regular hexagonal lattice point is reflected to the midpoint of the six sides. Calculate the area of the resulting hexagon.

Consider an arbitrary internal point of a regular hexagon with a side length of one unit. This point is reflected to the midpoint of each of the six sides. Calculate the area of the resulting hexagon.
Solution.

Let's create a diagram. The vertices of the hexagon are denoted by $A_{1}, A_{2}, \ldots, A_{6}$, and the midpoints by $F_{1}, F_{2}, \ldots, F_{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7117adb620f2d3ae4716g-1.jpg?height=531&width=596&top_left_y=266&top_left_x=754)

Let's examine, for example, the triangle $P P_{4} P_{5}$, where $P_{4}, P_{5}$ are the reflections of point $P$ over $F_{4}$ and $F_{5}$, so $P F_{4}=F_{4} P_{4}$ and $P F_{5}=F_{5} P_{5}$. Therefore, $F_{4} F_{5}$ is the midline of the triangle $P P_{4} P_{5}$, so $T_{P P_{4} P_{5}}=4 \cdot T_{P F_{4} F_{5}}$. This is true for the other triangles as well,

![](https://cdn.mathpix.com/cropped/2024_05_02_7117adb620f2d3ae4716g-1.jpg?height=49&width=1784&top_left_y=935&top_left_x=165)
applying to the triangle $F_{1} A_{2} O$: $r^{2}+\left(\frac{1}{2}\right)^{2}=1^{2}$, from which $r=\frac{\sqrt{3}}{2}$.

$$
\begin{gathered}
T_{O F_{1} F_{2}}=\frac{r^{2} \cdot \sin 60^{\circ}}{2}=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{16} \\
T_{F_{1} F_{2} F_{3} F_{4} F_{5} F_{6}}=6 \cdot T_{O F_{1} F_{2}}=6 \cdot \frac{3 \sqrt{3}}{16}=\frac{9 \sqrt{3}}{8}
\end{gathered}
$$

Therefore, the area of the hexagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6}$ is: $T=4 \cdot \frac{9 \sqrt{3}}{8}=\frac{9 \sqrt{3}}{2}$.