## Task A-1.6.

Given is a triangle $A B C$ with an area of 1. Point $D$ is the midpoint of segment $\overline{B C}$, point $E$ is the midpoint of segment $\overline{A D}$, point $F$ is the midpoint of segment $\overline{B E}$, and point $G$ is the midpoint of segment $\overline{C F}$. Determine the area of triangle $E F G$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9afe9556dd3f84037ce6g-05.jpg?height=923&width=1494&top_left_y=281&top_left_x=201)

Let $P(X Y Z)$ denote the area of an arbitrary triangle $X Y Z$.

Let $v_{b}$ be the length of the altitude from vertex $B$ in triangle $A B D$. This altitude coincides with the altitude from vertex $B$ in triangle $B D E$, so we have

$$
P(B D E)=\frac{1}{2} \cdot|D E| \cdot v_{b}=\frac{1}{2} \cdot \frac{|A D|}{2} \cdot v_{b}=\frac{1}{2} P(A B D) .
$$

Similarly, we conclude that $P(C D E)=\frac{1}{2} P(A D C)$.

Therefore,

$$
P(B E C)=P(B D E)+P(C D E)=\frac{P(A B D)+P(A D C)}{2}=\frac{P(A B C)}{2}=\frac{1}{2} . \quad 2 \text { points }
$$

If $v_{c}$ is the length of the altitude from vertex $C$ in triangle $B C E$ (and triangle $C F E$), by similar reasoning as before, we conclude

$$
P(C F E)=\frac{1}{2} \cdot|E F| \cdot v_{c}=\frac{1}{2} \cdot \frac{|B E|}{2} \cdot v_{c}=\frac{1}{2} P(B C E)=\frac{1}{4}
$$

Finally, if $v_{e}$ is the length of the altitude from vertex $E$ in triangle $E F C$ (and triangle $E F G$), we get

$$
P(E F G)=\frac{1}{2} \cdot|F G| \cdot v_{e}=\frac{1}{2} \cdot \frac{|C F|}{2} \cdot v_{e}=\frac{1}{2} P(E F C)=\frac{1}{8}
$$