8,9

In triangle $A B C$, the ratio of side $B C$ to side $A C$ is 3, and $\angle A C B=\alpha$. From vertex $C$, two rays are drawn, dividing angle $A C B$ into three equal parts. Find the ratio of the segments of these rays contained within triangle $A B C$.
If $a$ and $b$ are the sides of a triangle, and $\alpha$ is the angle between them, then the bisector $l$ of this angle can be calculated by the formula

$$
l=\frac{2 a b \cos \frac{\alpha}{2}}{a+b}
$$

## Solution

Let $M$ and $N$ be the points of intersection of the specified rays with side $AB$ (with $M$ between $A$ and $N$). Denote $AC=x$, $CM=y$, $CN=z$. Then $BC=3x$.

$$
S_{\Delta \mathrm{ACN}}=\frac{1}{2} x z \sin \frac{2 \alpha}{3}=S_{\Delta \mathrm{ACM}}+S_{\Delta \mathrm{MCN}}=\frac{1}{2} x y \sin \frac{\alpha}{3}+\frac{1}{2} y z \sin \frac{\alpha}{3}
$$

From this, we find that $y=\frac{2 x z \cos \frac{\alpha}{3}}{x+z}$. From triangle $MCB$, we similarly find that

$$
z=\frac{6 x y \cos \frac{\alpha}{3}}{y+3 x}
$$

Express $x$ from the obtained equations:

$$
x=\frac{y z}{2 z \cos \frac{\alpha}{3}-y}, x=\frac{y z}{6 y \cos \frac{\alpha}{3}-3 z}
$$

By equating the right-hand sides of these expressions, we get the equation

$$
2 z \cos \frac{\alpha}{3}-y=6 y \cos \frac{\alpha}{3}-3 z
$$

Divide both sides of this equation by $z$ and find the required ratio $\frac{y}{z}$:

$$
\frac{y}{z}=\frac{2 \cos \frac{\alpha}{3}+3}{6 \cos \frac{\alpha}{3}+1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-31.jpg?height=440&width=1152&top_left_y=978&top_left_x=453)

## Answer

$\frac{2 \cos \frac{\alpha}{3}+3}{6 \cos \frac{\alpha}{3}+1}$.