2. The area of triangle $A B C$ is 1. On the rays $A B, B C$, $C A$, points $B^{\prime}, C^{\prime}, A^{\prime}$ are laid out respectively, such that

$$
B B^{\prime}=A B, \quad C C^{\prime}=2 B C, \quad A A^{\prime}=3 C A
$$

Calculate the area of triangle $A^{\prime} B^{\prime} C^{\prime}$.
Answer: 18.

Solution. We will solve the problem in a general form, assuming that

$$
B B^{\prime}=q A B, \quad C C^{\prime}=r B C, \quad A A^{\prime}=p C A .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_fbb46158359318ba2f7fg-07.jpg?height=616&width=1088&top_left_y=1925&top_left_x=495)

We will calculate the area of triangle \( A^{\prime} B^{\prime} A \):

$$
S_{A^{\prime} B^{\prime} A}=\frac{1}{2} A^{\prime} A \cdot A B^{\prime} \cdot \sin \angle A^{\prime} A B^{\prime}=\frac{1}{2} p b(c+q c) \sin \angle B A C=
$$

$$
=p(1+q) \cdot \frac{1}{2} b c \sin \angle B A C=p(1+q) \cdot S_{A B C}=p(1+q)
$$

(We used the equality of the sines of adjacent angles). Similarly, we find that \( S_{B^{\prime} B C^{\prime}}=q(1+r) \), \( S_{C C^{\prime} A^{\prime}}=r(1+p) \). Adding the areas of the specified three triangles and the original triangle \( A B C \), we find the desired area:

$$
S_{A^{\prime} B^{\prime} C^{\prime}}=1+p+q+r+p q+q r+r p
$$

Evaluation. 13 points for a correct solution.