In an isosceles triangle $A B C(A B=B C)$, the altitudes $A A_{1}, B B_{1}$ and $C C_{1}$ are drawn.

Find the ratio of the area of triangle $A_{1} B_{1} C_{1}$ to the area of triangle $A B C$, if ${ }^{A B} / A_{1} B_{1}=\sqrt{3}$.
$\cos \angle C=\frac{1}{\sqrt{3}}$.

## Solution

Let $\angle C=\angle A=\alpha, S_{A B C}=S$. Triangle $A_{1} C B_{1}$ is similar to triangle $A B C$ with a similarity coefficient of $\cos \alpha=$ $A_{1} B_{1 /} A B=\frac{1}{\sqrt{3}}$. Therefore,

$S_{A_{1} C B_{1}}=S_{C_{1} A B_{1}}=S_{/ 3}$.

Triangle $A_{1} B C_{1}$ is similar to triangle $A B C$ with a similarity coefficient of ${ }^{B A_{1 /}}{ }_{B A}=\cos \left(180^{\circ}-2 \alpha\right)=-\cos 2 \alpha=1-2 \cos ^{2} \alpha$ $=1 / 3$.

Therefore, $S_{A_{1} B C_{1}}=S / 9$. Consequently, $S_{A_{1} B_{1} C_{1}}=S_{A B C}-2 S_{A_{1} C B_{1}}-S_{A_{1} B C_{1}}=S-2 S / 3-S / 9=2 S / 9$.

## Answer

$2: 9$.