In the following figure, $A B C$ is an equilateral triangle, $D, E$ and $F$ are its midpoints, and $P$ is its center. What fraction does the shaded area represent of the total area of triangle $A B C$?

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-01.jpg?height=397&width=407&top_left_y=1250&top_left_x=904)

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Solution

The points $I, G$ and $H$ are the intersections of the segments $D E, E F$ and $D F$ with the segments $A P$, $B P$ and $C P$. By the symmetry of the figure, the areas of the quadrilaterals EIPG, DIPH and $F G P H$ measure the same value $x \mathrm{~cm}^{2}$. Moreover, by the same argument, the areas of the triangles $A E I$ and $A D I$ also measure the same value $y \mathrm{~cm}^{2}$. The triangles $A D E, D E F, D B F$ and $C E F$ have the same area $S \mathrm{~cm}^{2}$ and thus $x=S / 3$ and $y=S / 2$. The sought fraction is

$$
\begin{aligned}
\frac{A_{A E G P}}{A_{A B C}} & =\frac{y+x}{4 S} \\
& =\frac{S / 2+S / 3}{4 S} \\
& =\frac{5}{24}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-02.jpg?height=394&width=394&top_left_y=263&top_left_x=712)