Isosceles triangles $A B C(A B=B C)$ and $A_{1} B_{1} C_{1}\left(A_{1} B_{1}=B_{1} C_{1}\right)$ are similar and $A B: A_{1} B_{1}=2: 1$.

Vertices $A_{1}, B_{1}$ and $C_{1}$ are located on sides $C A, A B$ and $B C$ respectively, and $A_{1} B_{1} \perp A C$. Find the angle

B.
Prove that $A_{1} C_{1} \perp B C$ and formulate a trigonometric equation relative to the angle at the base of the isosceles triangle $A B C$.

## Solution

Let $A_{1} B_{1}=B_{1} C_{1}=1, \angle A=\alpha$. Then $\angle A_{1} C_{1} C=\left(90^{\circ}-\alpha\right)+\alpha=90^{\circ}, A A_{1}=\operatorname{ctg} \alpha, A_{1} C_{1}=2 \cos \alpha, A C=2 A_{1} C_{1}$ $=4 \cos \alpha, A_{1} C=A_{1} C_{1 / \sin \alpha}=2 \operatorname{ctg} \alpha$.

Since $A C=A A_{1}+A_{1} C$, then $3 \operatorname{ctg} \alpha=4 \cos \alpha$, that is $\sin \alpha=3 / 4$, from which $\cos \angle B=\cos \left(180^{\circ}-2 \alpha\right)=-\cos 2 \alpha=$ $2 \sin ^{2} \alpha-1=1 / 8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-19.jpg?height=463&width=552&top_left_y=1486&top_left_x=753)

## Answer

$\arccos \frac{1}{8}$.