7. (10 points) The bisector of angle $A L$ of triangle $A B C$ intersects the circumcircle of the triangle at point $D$. It turns out that $A L: L D=\alpha$. Find the perimeter of triangle $A B C$, if $B C=\beta$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_285474b343983a91d99cg-07.jpg?height=856&width=828&top_left_y=360&top_left_x=571)

Since $AL$ is the angle bisector, then

$$
\frac{BA}{BL}=\frac{CA}{CL}=k \quad \text{and} \quad AL^2 = BA \cdot CA - BL \cdot CL
$$

We get

$$
\alpha^2 x^2 = k^2 bc - bc = (k^2 - 1) bc
$$

We also note that the power of point $L$ is $\alpha x^2 = bc$. Then, we get

$$
\alpha^2 x^2 = (k^2 - 1) bc = (k^2 - 1) \alpha x^2 \Longrightarrow \alpha = k^2 - 1 \Longrightarrow k = \sqrt{\alpha + 1}
$$

Let's calculate the perimeter of triangle $ABC$

$$
P_{ABC} = kc + kb + c + b = (k + 1)(c + b) = \beta(k + 1) = \beta \sqrt{\alpha + 1} + \beta
$$

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 4.2 | 4.4 | 6.9 | 7.2 | 7.5 | 10.4 | 10.8 | 11.2 | 8.7 | 12.4 | 12.8 |