12.006. 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.
Solution.

In the right triangle $ACB$, we have: $\angle ACB=90^{\circ}, \angle CAB=\alpha$, $S_{\triangle ABC}=S, AB_1=B_1C$ and $AC_1=C_1B, BB_1 \cap CC_1=O$ (Fig. 12.9);

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-720.jpg?height=302&width=562&top_left_y=269&top_left_x=111)

Fig. 12.8

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-720.jpg?height=494&width=342&top_left_y=77&top_left_x=781)

Fig. 12.9

We need to find the distance from $O$ to $AB$. Draw $CD \perp AB$ and let $CD=h$. From $\triangle ADC$ and $\triangle CDB$, we find $AC=\frac{h}{\sin \alpha}, CB=\frac{h}{\sin \left(\frac{\pi}{2}-\alpha\right)}=\frac{h}{\cos \alpha}$. Since $S=\frac{1}{2} AC \cdot BC=\frac{h^2}{\sin 2 \alpha}$, we have $h=\sqrt{S \sin 2 \alpha}$. Draw $OF \parallel AB$, $OF \cap AC=F, OF \cap CB=E$ and $OF \cap CD=K$. Since $\triangle COK \sim \triangle CDC_1$, we have $\frac{CK}{CD}=\frac{OC}{CC_1}=\frac{2}{3}$, hence $CK=\frac{2}{3} CD$, and $KD=\frac{1}{3} CD=\frac{\sqrt{S \sin 2 \alpha}}{3}$, which is the distance from $O$ to $AB$, since $O \in FE \parallel AB$.

Answer: $\frac{1}{3} \sqrt{S \sin 2 \alpha}$.