Auxiliary similar triangles Quadratic equations. Vieta's Theorem

In triangle $A B C$ with sides $A B=18$ and $B C=12$, a parallelogram $B K L M$ is inscribed, with points $K, L$, and $M$ lying on sides $A B, A C$, and $B C$ respectively. It is known that the area of the parallelogram is $4 / 9$ of the area of triangle $A B C$. Find the sides of the parallelogram.
Let $B M=K L=x, B K=L M=y, S_{A B C}=S$. Then $S_{B K L M}=4 S / 9, 2 S / 9=1 / 2 S_{B K L M}=S_{B K M}=x / 12 \cdot y / 18$, so $x y=48$. Triangle $A K L$ is similar to triangle $A B C$, so $K L: B C=A K: A B$, or ${ }^{x} / 12=18-y / 18$, from which $3 x+2 y=36$. From the system $x y=48, 3 x+2 y=36$ we find that $x=8, y=6$ or $x=4, y=12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_eab8c6bbe58aee0b55b3g-05.jpg?height=506&width=1075&top_left_y=-1&top_left_x=503)

## Answer

8,6 or $4,12$.