5. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, and $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the sine of the angle between the diagonals of this quadrilateral, if its area takes the smallest possible value under the given conditions.
Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the given condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\frac{1}{2} x_{1} y_{2} \sin \alpha=\frac{1}{2} x_{2} y_{1} \sin \alpha$. Hence, $\frac{x_{1}}{x_{2}}=\frac{y_{1}}{y_{2}}$, and therefore, triangles $BOC$ and $AOD$ are similar by the first similarity criterion: two sides ($x_{1}$ and $y_{1}$) of triangle $BOC$ are proportional to two sides ($x_{2}$ and $y_{2}$) of triangle $AOD$, and the angles formed by these sides ($\angle BOC$ and $\angle AOD$) are equal. Let $k=\frac{x_{1}}{x_{2}}=\frac{y_{1}}{y_{2}}$ be the similarity coefficient of triangles $BOC$ and $AOD$. Denote by $S$ the areas of triangles $ABO$ and $CDO$ (by the given condition $S=\frac{3}{2}$). Then $S_{BOC}=k \cdot S$ and $S_{AOD}=$ $S / k$. In the end, the area of quadrilateral $ABCD$ can be represented as:

Interregional School Olympiad in Mathematics for Students of Vedomstvenny Educational Organizations

$$
S_{ABCD}=S_{AOD}+S_{CDO}+S_{BOC}+S_{ABO}=2 S+S\left(k+\frac{1}{k}\right)
$$

It is known that for $k>0$ the minimum value of the expression $k+\frac{1}{k}$ is achieved at $k=1$. Therefore, $x_{1}=$ $x_{2}$ and $y_{1}=y_{2}$, which means that the diagonals of the quadrilateral are bisected by their point of intersection, so $ABCD$ is a parallelogram. Its area $S_{ABCD}=4 S=6$.

To find the sine of the angle between the diagonals, we use the fact that the area of the quadrilateral is half the product of the diagonals and the sine of the angle between them:

$$
S_{ABCD}=\frac{1}{2} \cdot AC \cdot BD \cdot \sin \alpha \Rightarrow \sin \alpha=\frac{2 S_{ABCD}}{AC \cdot BD}
$$

To find the lengths of the diagonals, we calculate the side $CD$ by writing the formula for the area of the parallelogram

$$
S_{ABCD}=4 S=AD \cdot CD \cdot \sin \angle ADC \Rightarrow CD=\frac{4 S}{AD \cdot \sin \angle ADC}=\frac{4 \cdot \frac{3}{2}}{3 \sqrt{2} \sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^{2}}}=2 \sqrt{5}
$$

Now we find the diagonals $AC$ and $BD$ using the cosine theorem from triangles $ADC$ and $BCD$

$$
\begin{aligned}
& AC=\sqrt{AD^{2}+CD^{2}-2 \cdot AD \cdot CD \cdot \cos \angle ADC}=\sqrt{2} \\
& BD=\sqrt{AD^{2}+CD^{2}+2 \cdot AD \cdot CD \cdot \cos \angle ADC}=\sqrt{74}
\end{aligned}
$$

Substituting the found values into relation (1), we get $\sin \alpha=\frac{6}{\sqrt{37}}$.

Answer: $\frac{6}{\sqrt{37}}$