8,9 |  |

A circle with its center on the diagonal $A C$ of parallelogram $A B C D$ touches the line $A B$ and passes through points $C$ and $D$. Find the sides of the parallelogram if its area $S=\sqrt{2}$, and $\angle B A C=\arcsin \frac{1}{3}$.
Let $O$ be the center of the circle, $F$ the point of tangency of the circle with the line $AB$. Denote $OC = OD = OF = R$, $AB = CD = x$, $AD = BC = y$, $\angle ACD = \angle BAC = \alpha$. Then

$$
\sin \alpha = \frac{1}{3}, \cos \alpha = \frac{2 \sqrt{2}}{3}
$$

Extend the radius $OF$ to intersect the chord $CD$ at point $M$. Since $OF \perp AB$ and $CD \parallel AB$, then $OM \perp CD$, so $M$ is the midpoint of $CD$. From the right triangle $OMC$, we find that

$$
R = OC = \frac{MC}{\cos \alpha} = \frac{x}{2 \cos \alpha} = \frac{3x}{4 \sqrt{2}}, OM = OC \sin \alpha = \frac{1}{3} R = \frac{x}{4 \sqrt{2}}
$$

Then

$$
FM = OF + OM = R + \frac{1}{3} R = \frac{4}{3} R = \frac{4}{3} \cdot \frac{3x}{4 \sqrt{2}} = \frac{x}{\sqrt{2}},
$$

and since $MF$ is the height of the parallelogram $ABCD$, then $S = CD \cdot MF$, or $\sqrt{2} = x \cdot \sqrt{\sqrt{2}}$, from which $AB = x = \sqrt{2}$. Let $P$ be the projection of vertex $C$ onto the line $AB$. Then $CP = MF = \sqrt{2}$. From the right triangle $APC$, we find that

$$
AC = \frac{CP}{\sin \angle PAC} = \frac{CP}{\sin \alpha} = \frac{\sqrt{2}}{\frac{1}{3}} = \frac{3x}{\sqrt{2}} = 3
$$

By the cosine rule

$$
y^2 = BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos \angle BAC = 2 + 9 - 2 \cdot \sqrt{2} \cdot 3 \cdot \frac{2 \sqrt{2}}{3} = 11 - 8 = 3
$$

Therefore, $BC = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-17.jpg?height=1146&width=1835&top_left_y=170&top_left_x=154)

## Answer

$\sqrt{2}, \sqrt{3}$.