In an isosceles triangle $ABC (AB = AC)$, the angle bisectors $AA_1$, $BB_1$, and $CC_1$ are drawn. The area of triangle $ABC$ is to the area of triangle $A_1B_1C_1$ as $\frac{9}{2}$. Find the ratio of the perimeter of triangle $A_1B_1C_1$ to the perimeter of triangle $ABC$.
Apply the property of the bisector of a triangle.

## Solution

Let $B C=a, A C=A B=b, S_{\Delta \mathrm{ABC}}=S$. By the property of the bisector of a triangle,

$$
\frac{A B_{1}}{A C}=\frac{b}{a+b}, \frac{C B_{1}}{C A}=\frac{a}{a+b}
$$

$$
\begin{aligned}
& S_{\Delta \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}}=S_{\Delta \mathrm{ABC}}-2 S_{\Delta \mathrm{A}_{1} \mathrm{CB}_{1}}-S_{\Delta \mathrm{B}_{1} \mathrm{AC}_{1}}= \\
& = \\
& =-2 \cdot \frac{1}{2} \cdot \frac{a}{a+b} \cdot S-\frac{b^{2} S}{(a+b)^{2}}=\frac{a b S}{(a+b)^{2}}
\end{aligned}
$$

Thus, we have the equation

$$
\frac{a b}{(a+b)^{2}}=\frac{2}{9}, \text { or } 2 a^{2}-5 a b+2 b^{2}=0
$$

From this, we find: $b=\frac{a}{2}$ (which is impossible) or $b=2 a$. Then the perimeter of triangle $A B C$ is $5 a$,

$$
A A_{1}=\sqrt{A C^{2}-C A^{2}}=\sqrt{4 a^{2}-\frac{a^{2}}{4}}=\frac{a \sqrt{15}}{2}
$$

Let $D$ be the midpoint of $B_{1} C_{1}$. Since

$$
\frac{A D}{A A_{1}}=\frac{A B_{1}}{A C}=\frac{b}{a+b}=\frac{2}{3}
$$

then

$$
A_{1} D=\frac{1}{3} A A_{1}=\frac{a \sqrt{15}}{6}, B_{1} C_{1}=\frac{2}{3} B C=\frac{2 a}{3}, A_{1} B_{1}=\sqrt{A_{1} D^{2}+B_{1} D^{2}}=\frac{a \sqrt{19}}{6}
$$

Therefore, the perimeter of triangle $A_{1} B_{1} C_{1}$ is $a\left(\frac{2}{3}+\frac{\sqrt{19}}{3}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-37.jpg?height=426&width=426&top_left_y=1496&top_left_x=816)

## Answer

$\frac{2+\sqrt{19}}{15}$.