3. The bisectors $A A_{1}$ and $B B_{1}$ of triangle $A B C$ intersect at point $O$, and $A O: O A_{1} = 2: 1$. The bisector of the external angle at vertex $C$ of triangle $A B C$ intersects line $A B$ at point $C_{1}$. Find the angle $B_{1} A_{1} C_{1}$ and the length of segment $A_{1} C_{1}$, if $A B=2, A C=4$. (16 points)
# Solution.

$$
\begin{gathered}
a=BC, b=AC=4, c=AB=2 \\
AO: OA_{1}=2: 1
\end{gathered}
$$

1) By the properties of angle bisectors, we have

$$
\begin{gathered}
AB: BA_{1}=AO: OA_{1}=2: 1, BA_{1}=\frac{c}{2} \\
AC: CA_{1}=AO: OA_{1}=2: 1, CA_{1}=\frac{b}{2} \\
BC=\frac{b+c}{2}=3
\end{gathered}
$$

2) We will prove that $\frac{BC_{1}}{AC_{1}}=\frac{a}{b}$. For the areas of triangles $AC_{1}C$ and $BC_{1}C$, we have the following relationships

$$
\begin{gathered}
\frac{S_{AC_{1}C}}{S_{BC_{1}C}}=\frac{BC_{1}}{AC_{1}^{\prime}} \\
S_{AC_{1}C}=b \cdot C_{1}E, S_{BC_{1}C}=a \cdot C_{1}D
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-09.jpg?height=777&width=905&top_left_y=497&top_left_x=1044)

$C_{1}E=C_{1}D$. The segments $C_{1}E$ and $C_{1}D$ are the altitudes of triangles $AC_{1}C$ and $BC_{1}C$. Since $CC_{1}$ is the angle bisector of $\angle DCE$, the right triangles $CC_{1}D$ and $CC_{1}E$ are congruent, and $C_{1}E=C_{1}D$. Therefore, $\frac{BC_{1}}{AC_{1}}=\frac{a}{b}$. By the properties of angle bisectors, $\frac{BA_{1}}{CA_{1}}=\frac{c}{b}$, $\frac{AB_{1}}{CB_{1}}=\frac{c}{a}$. Then $\frac{BA_{1}}{CA_{1}} \cdot \frac{CB_{1}}{AB_{1}} \cdot \frac{AC_{1}}{BC_{1}}=\frac{c}{b} \cdot \frac{a}{c} \cdot \frac{b}{a}=1$. By Menelaus' theorem, points $A_{1}, B_{1}, C_{1}$ lie on the same line. $\angle B_{1}A_{1}C_{1}=180^{\circ}$, $B_{1}C_{1}=B_{1}A_{1}+A_{1}C_{1}$.

3) $\frac{AB_{1}}{CB_{1}}=\frac{c}{a}=\frac{2}{3}, AB_{1}=\frac{8}{5}, CB_{1}=\frac{12}{5} ; CA_{1}=2 ; \quad \frac{BC_{1}}{AC_{1}}=\frac{3}{4}, \frac{BC_{1}}{2+BC_{1}}=\frac{3}{4}, BC_{1}=6, AC_{1}=8$. $\alpha=\angle BAC, \cos \alpha=\frac{4+16-9}{16}=\frac{11}{16}, \gamma=\angle ACB, \cos \gamma=\frac{9+16-4}{24}=\frac{7}{8}$.

4) $C_{1}B_{1}^{2}=AC_{1}^{2}+AB_{1}^{2}-2AC_{1} \cdot AB_{1} \cos \alpha=64+\frac{64}{25}-2 \cdot \frac{64}{5} \cdot \frac{11}{16}=\frac{1224}{25}, C_{1}B_{1}=\frac{6\sqrt{34}}{5}$.

$$
\begin{gathered}
A_{1}B_{1}^{2}=CA_{1}^{2}+CB_{1}^{2}-2CA_{1} \cdot CB_{1} \cos \gamma=4+\frac{144}{25}-2 \cdot \frac{24}{5} \cdot \frac{7}{8}=\frac{34}{25}, A_{1}B_{1}=\frac{\sqrt{34}}{5} \\
A_{1}C_{1}=B_{1}C_{1}-A_{1}B_{1}=\sqrt{34}
\end{gathered}
$$

Answer: $180^{\circ}, \sqrt{34}$.