4B. Given triangle $ABC$ such that $\overline{AB}=20$ and $\overline{AC}=\frac{45}{2}$. Points $X$ and $Y$ are chosen on sides $AB$ and $AC$ respectively such that $\overline{AX}=\overline{AY}$ and the area of $\triangle ABC$ is twice the area of $\triangle AXY$. Find the ratio of the areas of triangles $BXY$ and $CXY$.
Solution. Let's denote the angle at vertex $A$ as $\alpha$. We have $P_{A B C}=\frac{\overline{A B} \cdot \overline{A C} \sin \alpha}{2}=225 \sin \alpha$.

Then, $225 \sin \alpha=2 P_{A X Y}=\overline{A X}^{2} \sin \alpha$. From here we get $\overline{A X}=\overline{A Y}=15$. Now it follows that $\overline{X B}=5$ and $\overline{Y C}=\frac{15}{2}$. Since triangles $A X Y$ and $B X Y$ have a common height from vertex $Y$, we have $\frac{P_{A X Y}}{P_{B X Y}}=\frac{\overline{A X}}{B X}=3$. Similarly for triangles $C Y X$ and $A Y X$ we get $\frac{P_{A X X}}{P_{C Y X}}=\frac{\overline{A Y}}{C Y}=2$. From this we have $\frac{P_{R X Y}}{P_{C X Y}}=\frac{2}{3}$.

## 4th year