Progetto Olimpiadi della Matematica - GARA di FEBBRAIO

Solution:

The answer is (B). The segment $A'B''$ is the reflection of the segment $AB$ with respect to the point $C$. Therefore $A'B''$ and $AB$ are parallel and congruent. Moreover $AB' = AB$ because $B'$ is the reflection of $B$ with respect to $A$. Therefore the quadrilateral $AB'B''A'$ is a parallelogram because it has a pair of opposite sides that are parallel and congruent.

Let $CH$ be an altitude of $ABC$ and $A'H'$ an altitude of $AB'B''A'$. The triangles $CAH$ and $A'AH'$ are right triangles and have the angle $\widehat{CAH}$ in common, so they are similar. Then, since $AA' = 2AC$, we have $A'H' = 2CH$. Therefore $S(AB'B''A') = AB' \cdot A'H' = AB \cdot 2CH = 4\left(\frac{1}{2} AB \cdot CH\right) = 4S(ABC)$.

Moreover $S(AB'A') = \frac{1}{2} S(AB'B''A') = 2S(ABC)$. Similarly, $S(CA'A''C') = 4S(ABC)$, $S(CA'C') = 2S(ABC)$, $S(BC' C'' B') = 4S(ABC)$ and $S(BC'B') = 2S(ABC)$.

Then $S(A'B'C') = S(AB'A') + S(CA'C') + S(BC'B') + S(ABC) = 7S(ABC)$,
while $S(A'A''C'C''B'B'') = S(AB'B''A') + S(CA'A''C') + S(BC'C''B') + S(ABC) = 13S(ABC)$.

Therefore $S(A'B'C') / S(A'A''C'C''B'B'') = 7 / 13$.