12.139. An equilateral triangle $A_{1} B_{1} C_{1}$ is inscribed in an equilateral triangle $A B C$. Point $A_{1}$ lies on side $B C$, point $B_{1}$ - on side $A C$, and point $C_{1}$ - on side $A B$. The angle $A_{1} B_{1} C$ is equal to $\alpha$. Find the ratio $A B : A_{1} B_{1}$.
Solution.

In $\triangle A_{l} B_{l} C$ (Fig. 12.12) $\angle C=60^{\circ}, \angle B_{l} A_{l} C=120^{\circ}-\alpha$.

\[
\begin{aligned}
& \frac{A B}{A_{1} B_{1}}=\frac{A_{1} C+B_{1} C}{A_{1} B_{1}}=\frac{A_{1} C}{A_{1} B_{1}}+\frac{B_{1} C}{A_{1} B_{1}}=\frac{\sin \angle A_{1} B_{1} C}{\sin \angle C}+\frac{\sin \angle B_{1} A_{1} C}{\sin \angle C}= \\
& =\frac{\sin \alpha+\sin \left(120^{\circ}-\alpha\right)}{\sin 60^{\circ}}=\frac{2 \sin 60^{\circ} \cos \left(60^{\circ}-\alpha\right)}{\sin 60^{\circ}}=2 \sin \left(30^{\circ}+\alpha\right)
\end{aligned}
\]

Answer: $2 \sin \left(30^{\circ}+\alpha\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0900.jpg?height=438&width=507&top_left_y=80&top_left_x=126)

Fig. 12.11

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0900.jpg?height=507&width=574&top_left_y=13&top_left_x=674)

Fig. 12.12