## Task 5 - 310735

If $A B C$ is any triangle, let $S$ be the intersection of the medians $A D$ and $B E$, and let $F_{1}$ denote the area of triangle $A B C$ and $F_{2}$ the area of the (non-convex) pentagon $A B D S E$.

Determine the ratio $F_{1}: F_{2}$ of these two areas for any triangle $A B C$!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0534.jpg?height=486&width=537&top_left_y=645&top_left_x=291)
In every triangle $ABC$, the three medians $AD, BE$, and $CF$ have a common intersection point $S$. The areas of the triangles $AFS, FBS, BDS, DCS, CES$, and $EAS$ are $J_{1}, J_{2}, J_{3}, J_{4}, J_{5}$, and $J_{6}$, respectively.

Because $AF=FB$ and the common vertex $S$ of $AFS$ and $FBS$, it follows that $J_{1}=J_{2}$. Similarly, it follows that $J_{3}=J_{4}$ and $J_{5}=J_{6}$.

Because $AF=FB$ and the common vertex $C$ of $AFC$ and $FBC$, it follows that

$$
J_{1}+J_{6}+J_{5}=J_{2}+J_{3}+J_{4}
$$

From this and the preceding equations, it follows that

$$
J_{1}+2 \cdot J_{6}=J_{1}+2 \cdot J_{3}
$$

and from this, $J_{6}=J_{3}$. Similarly, it follows (e.g.) that $J_{2}=J_{5}$. Therefore, it holds that $J_{1}=J_{2}=J_{3}=J_{4}=J_{5}=J_{6}$, i.e.,

$$
F_{1}: F_{2}=\left(6 J_{1}\right):\left(4 J_{1}\right)=3: 2
$$