3.6. In a right-angled triangle, the area $S$ and an acute angle $\alpha$ are given. Find the distance from the point of intersection of the medians of the triangle to the hypotenuse.
3.6. By the condition, $\angle A C B=90^{\circ}, \angle C A B=\alpha$, $S_{\triangle A B C}=S$ (Fig. 3.10). Let $A B=2 c$; then $C D=A D=D B=c$. Further, let $A C=b$, O - the point of intersection of the medians; then $O D=\frac{1}{3} C D=\frac{c}{3}$. Draw $C F \perp A B$ and $O E \perp A B$ (segment $O E$ is the desired one). Since $\triangle O E D \sim$ $\sim \triangle C F D$ and $O D=\frac{1}{3} C D$, then $O E=\frac{1}{3} C F$, where $C F=b \sin \alpha$ (from $\left.\triangle A F C\right), b=$ $=2 c \cos \alpha$ (from $\triangle A B C$ ). Therefore, $C F=2 c \cos \alpha \sin \alpha=c \sin 2 \alpha$, from which $O E=\frac{c \sin 2 \alpha}{3}$. The value of $c$ will be found from the equality $S=\frac{1}{2} b \cdot 2 c \sin \alpha$; we have

$$
S=b c \sin \alpha ; S=2 c^{2} \cos \alpha \sin \alpha=c^{2} \sin 2 \alpha
$$

from which $c=\sqrt{\frac{S}{\sin 2 \alpha}}$. Finally, we get

$$
O E=\frac{c}{3} \sin 2 \alpha=\frac{\sin 2 \alpha}{3} \sqrt{\frac{S}{\sin 2 \alpha}}=\frac{1}{3} \sqrt{S \sin 2 \alpha} .
$$