6. Given a triangle $A B C$, let $A^{\prime}$ be the symmetric point of $A$ with respect to $C$, $A^{\prime \prime}$ the symmetric point of $A$ with respect to $B$, $B^{\prime}$ the symmetric point of $B$ with respect to $A$, $B^{\prime \prime}$ the symmetric point of $B$ with respect to $C$, $C^{\prime}$ the symmetric point of $C$ with respect to $B$, and $C^{\prime \prime}$ the symmetric point of $C$ with respect to $A$. Determine the ratio between the area of $A^{\prime} B^{\prime} C^{\prime}$ and that of the hexagon $A^{\prime} A^{\prime \prime} C^{\prime} C^{\prime \prime} B^{\prime} B^{\prime \prime}$.
(A) $\frac{6}{13}$
(B) $\frac{7}{13}$
(C) $\frac{3}{7}$
(D) $\frac{1}{2}$
(E) It depends on the initial triangle.
6. The answer is (B). The segment $A^{\prime} B^{\prime \prime}$ is the symmetric of the segment $A B$ with respect to the point $C$. Therefore, $A^{\prime} B^{\prime \prime}$ and $A B$ are parallel and congruent. Moreover, $A B^{\prime}=A B$ because $B^{\prime}$ is the symmetric of $B$ with respect to $A$. Thus, the quadrilateral $A B^{\prime} B^{\prime \prime} A^{\prime}$ is a parallelogram because it has one pair of opposite sides that are parallel and congruent. Let $C H$ be an altitude of $A B C$ and $A^{\prime} H^{\prime}$ be an altitude of $A B^{\prime} B^{\prime \prime} A^{\prime}$. The triangles $C A H$ and $A^{\prime} A H^{\prime}$ are right triangles and share the angle $\widehat{C A H}$, so they are similar. Then, since $A A^{\prime}=2 A C$, we have $A^{\prime} H^{\prime}=2 C H$. Therefore, $S\left(A B^{\prime} B^{\prime \prime} A^{\prime}\right)=A B^{\prime} \cdot A^{\prime} H^{\prime}=$ $A B \cdot 2 C H=4\left(\frac{1}{2} A B \cdot C H\right)=4 S(A B C)$. Furthermore, $S\left(A B^{\prime} A^{\prime}\right)=\frac{1}{2} S\left(A B^{\prime} B^{\prime \prime} A^{\prime}\right)=2 S(A B C)$. Similarly, $S\left(C A^{\prime} A^{\prime \prime} C^{\prime}\right)=4 S(A B C), S\left(C A^{\prime} C^{\prime}\right)=2 S(A B C), S\left(B C^{\prime} C^{\prime \prime} B^{\prime}\right)=4 S(A B C)$ and $S\left(B C^{\prime} B^{\prime}\right)=2 S(A B C)$. Therefore, $S\left(A^{\prime} B^{\prime} C^{\prime}\right)=S\left(A B^{\prime} A^{\prime}\right)+S\left(C A^{\prime} C^{\prime}\right)+S\left(B C^{\prime} B^{\prime}\right)+S(A B C)=$ $7 S(A B C)$, while $S\left(A^{\prime} A^{\prime \prime} C^{\prime} C^{\prime \prime} B^{\prime} B^{\prime \prime}\right)=S\left(A B^{\prime} B^{\prime \prime} A^{\prime}\right)+S\left(C A^{\prime} A^{\prime \prime} C^{\prime}\right)+S\left(B C^{\prime} C^{\prime \prime} B^{\prime}\right)+S(A B C)=$ $13 S(A B C)$. Thus, $S\left(A^{\prime} B^{\prime} C^{\prime}\right) / S\left(A^{\prime} A^{\prime \prime} C^{\prime} C^{\prime \prime} B^{\prime} B^{\prime \prime}\right)=7 / 13$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8e3f264af5d5452f36c1g-09.jpg?height=569&width=1408&top_left_y=1149&top_left_x=381)