[ Ratio of areas of similar triangles ]

In a right triangle, the sine of the smaller angle is $\frac{1}{3}$. A line perpendicular to the hypotenuse divides the triangle into two equal areas. In what ratio does this line divide the hypotenuse?
Let $M$ be a point on the hypotenuse $AB$ of the right triangle $ABC$, $\sin \angle A = \frac{1}{3}$, and $N$ be a point on the leg $AC$ such that $MN \perp AB$ and $S_{\triangle AMN} = \frac{1}{2} S_{\triangle ABC}$.

Denote $BC = t$. Then,

$$
AB = \frac{BC}{\sin \angle A} = \frac{t}{\frac{1}{3}} = 3t.
$$

Triangle $ANM$ is similar to triangle $ABC$ with a similarity ratio equal to the square root of the ratio of the areas, i.e., $\frac{1}{\sqrt{2}}$, so $MN = \frac{t}{\sqrt{2}}$.

From the right triangle $ANM$, we find that

$$
AN = \frac{MN}{\sin \angle A} = 3MN = \frac{3t}{\sqrt{2}}, \quad AM = \sqrt{AN^2 - MN^2} = \sqrt{\frac{9t^2}{2} - \frac{t^2}{2}} = 2t,
$$

thus $MB = AB - AM = 3t - 2t = t$. Therefore,

$$
\frac{AM}{MB} = \frac{2t}{t} = 2
$$

## Answer

$2: 1$