4. Consider the arbitrary triangle $A B C$. Let the lines $d_{1} \| A B, d_{2} \| A C, d_{3} \| B C$ and denote $d_{1} \cap (A C) = \{I\}, d_{1} \cap (B C) = \{F\}, d_{2} \cap (A B) = \{D\}, d_{2} \cap (B C) = \{G\}, d_{3} \cap (A B) = \{E\}$, $d_{3} \cap (A C) = \{H\}, d_{1} \cap d_{2} = \left\{A^{\prime}\right\}, d_{1} \cap d_{3} = \left\{B^{\prime}\right\}, d_{2} \cap d_{3} = \left\{C^{\prime}\right\}$. Assume that

$\operatorname{Area}(A E H) = \operatorname{Area}(C I F) = \operatorname{Area}(B D G) = \operatorname{Area}(A D G C) = \operatorname{Area}(B E H C) = \operatorname{Area}(A I F B)$.

$$
\text { Find } \frac{\operatorname{Area}\left(A^{\prime} B^{\prime} C^{\prime}\right)}{\operatorname{Area}(A B C)}
$$

Student Paul Crestez, Brăila
Solution. Since

$$
\begin{aligned}
& E H \| B C \Rightarrow \triangle A E H \sim \triangle A B C \Rightarrow\left(\frac{A E}{A B}\right)^{2}=\frac{\text{Area}(\triangle A E H)}{\text{Area}(\triangle A B C)}=\frac{\text{Area}(\triangle A E H)}{\text{Area}(\triangle A E H)+\text{Area}(\triangle E B C H)}= \\
& =\frac{\text{Area}(\triangle A E H)}{2 \text{Area}(\triangle A E H)}=\frac{1}{2} \Rightarrow \frac{A E}{A B}=\frac{1}{\sqrt{2}} \Rightarrow A E=\frac{A B}{\sqrt{2}} \Rightarrow E B=A B-A E=A B-\frac{A B}{\sqrt{2}}=A B \frac{\sqrt{2}-1}{\sqrt{2}}
\end{aligned}
$$

$E H \| B C$ and $A B \| I F \Rightarrow E B' F B$ parallelogram $\Rightarrow B' F=E B=A B \frac{\sqrt{2}-1}{\sqrt{2}},(1)$. Since $D G \| A C \Rightarrow$ $\Rightarrow \triangle B D G \sim \triangle B A C \Rightarrow A D=A B \frac{\sqrt{2}-1}{\sqrt{2}}$.

$I F \| A B$ and $D G \| A C \Rightarrow A I A' D$ parallelogram $\Rightarrow A' I=A D=A B \frac{\sqrt{2}-1}{\sqrt{2}},(2)$

$$
\begin{aligned}
& I F \| A B \Rightarrow \triangle C I F \sim \triangle C A B \Rightarrow\left(\frac{I F}{A B}\right)^{2}=\frac{\text{Area}(\triangle C I F)}{\text{Area}(\triangle C A B)}=\frac{\text{Area}(\triangle C I F)}{\text{Area}(\triangle C I F)+\text{Area}(\triangle I F B A)}=\frac{\text{Area}(\triangle C I F)}{2 \text{Area}(\triangle C I F)}= \\
& =\frac{1}{2} \Rightarrow \frac{I F}{A B}=\frac{1}{\sqrt{2}} \Rightarrow I F=\frac{A B}{\sqrt{2}},(3) . \text{ From }(1),(2)(3) \text{ we have } A' B'=I F-I A'-I B'=\frac{A B}{\sqrt{2}}-A B \frac{\sqrt{2}-1}{\sqrt{2}}- \\
& -A B \frac{\sqrt{2}-1}{\sqrt{2}}=A B \frac{3-2 \sqrt{2}}{\sqrt{2}} . \\
& A' B' \| A B \\
& \left.A' C' \| A C\right\} \Rightarrow \triangle A B C \sim \triangle A' B' C' \Rightarrow \frac{\text{Area}\left(\triangle A' B' C'\right)}{\text{Area}(\triangle A B C)}=\left(\frac{A' B'}{A B}\right)^{2} \Rightarrow \frac{\text{Area}\left(\triangle A' B' C'\right)}{\text{Area}(\triangle A B C)}=\left(\frac{3-2 \sqrt{2}}{\sqrt{2}}\right)^{2} \Rightarrow \\
& \left.B' C' \| B C\right\} \\
& \Rightarrow \frac{\text{Area}\left(\triangle A' B' C'\right)}{\text{Area}(\triangle A B C)}=\frac{17-12 \sqrt{2}}{2} .
\end{aligned}
$$