4. Given a triangle $A B C$. On the rays $B A, C B$ and $A C$ are defined points $A_{1}, B_{1}$ and $C_{1}$ respectively, such that

$$
\overline{B A_{1}}=(1+n) \overline{A B}, \overline{C B_{1}}=(1+n) \overline{C B}, \overline{A C_{1}}=(1+n) \overline{A C}
$$

Determine the ratio of the areas of triangles $A B C$ and $A_{1} B_{1} C_{1}$.
Solution. Let $\overline{B A}=c, \overline{C B}=a$ and $\overline{A C}=b$, and we get

$$
\overline{A A_{1}}=n c, \quad \overline{B B_{1}}=n a, \quad \overline{C C_{1}}=n b
$$

Triangles $C_{1} B_{1} B$ and $C_{1} B C$ have equal heights dropped to the bases $C B$ and $B B_{1}$, so

$$
\frac{P_{C_{1} B_{1} B}}{P_{C_{1} B C}}=\frac{n a}{a}=n
$$

Similarly,

$$
\frac{P_{C_{1} B C}}{P_{A B C}}=\frac{n b}{b}=n
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6b1323df06a0e4d4e36cg-5.jpg?height=361&width=451&top_left_y=889&top_left_x=1029)

From (1) and (2) we get $P_{C_{1} B_{1} B}=n^{2} P_{A B C}$. Therefore, $P_{B_{1} C_{1} C}=\left(n^{2}+n\right) P_{A B C}$. Similarly, $P_{A_{1} B_{1} B}=\left(n^{2}+n\right) P_{A B C}$ and $P_{C_{1} A_{1} A}=\left(n^{2}+n\right) P_{A B C}$. Finally,

$$
P_{A_{1} B_{1} C_{1}}=P_{A B C}+P_{A_{1} B_{1} B}+P_{B_{1} C_{1} C}+P_{C_{1} A_{1} A}=\left(3 n^{2}+3 n+1\right) P_{A B C}
$$

i.e. $\frac{P_{A_{1} B_{1} C_{1}}}{P_{A B C}}=3 n^{2}+3 n+1$.

## III year