45. Given a triangle $A B C, \overline{A B}=20$. On the sides $A B$ and $A C$ of triangle $A B C$, points $X$ and $Y$ are chosen such that $\overline{A X}=\overline{A Y}=15$ and the area of triangle $A B C$ is twice the area of triangle $A X Y$. Find the ratio of the areas of triangles $B X Y$ and $C X Y$.
Solution. Let's denote the blanks we have

and the angle at vertex $A$ as $\alpha$. For the area of the two triangles,

$$
\begin{aligned}
P_{\triangle A B C} & =\frac{\overline{A B} \cdot \overline{A C} \sin \alpha}{2} \\
& =10 \overline{A C} \sin \alpha=2 P_{\triangle A X Y}
\end{aligned}
$$

$$
P_{\triangle A X Y}=\frac{\overline{A X} \cdot \overline{A Y} \sin \alpha}{2}=\frac{225 \sin \alpha}{2}
$$

Then according to the condition

![](https://cdn.mathpix.com/cropped/2024_06_05_47715ec139c950209997g-11.jpg?height=533&width=642&top_left_y=1327&top_left_x=804)

$10 \overline{A C} \sin \alpha=225 \sin \alpha$
And therefore $\overline{A C}=\frac{45}{2}$. Now $\overline{X B}=5$ and $\overline{A C}=\frac{15}{2}$. Triangles $A X Y$ and $B X Y$ have a common height dropped from vertex $Y$, so $\frac{P_{\triangle A X Y}}{P_{\triangle B X Y}}=\frac{\overline{A X}}{\overline{B X}}=3$. Similarly, triangles $C Y X$ and $A Y X$ have a common height dropped from vertex $X$,
so $\frac{P_{\triangle A X Y}}{P_{\triangle C X Y}}=\frac{\overline{A Y}}{\overline{B Y}}=2$. Then the required ratio of the areas of triangles $B X Y$ and $C X Y$ is $\frac{P_{\triangle} B X Y}{P_{\triangle C X Y}}=\frac{\frac{P_{\triangle A X Y}}{3}}{\frac{P_{\triangle} A X Y}{2}}=\frac{2}{3}$.

## IV year