9.1. In the triangular pyramid $S A B C$, the edges $S B, A B$ are perpendicular and $\angle A B C=120^{\circ}$. Point $D$ on edge $A C$ is such that segment $S D$ is perpendicular to at least two medians of triangle $A B C$ and $C D=A B=44 \sqrt[3]{4}$. Find $A D$ (if the answer is not an integer, round it to the nearest hundredth).
Solution. Since segment $SD$ is perpendicular to two medians of triangle $ABC$, it is perpendicular to the plane $(ABC)$ (see Fig. 3). By the theorem of three perpendiculars, it follows that $DB \perp AB$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a8c31eb257d724803ce6g-10.jpg?height=471&width=765&top_left_y=1158&top_left_x=700)

Fig. 3:

![](https://cdn.mathpix.com/cropped/2024_05_06_a8c31eb257d724803ce6g-10.jpg?height=405&width=808&top_left_y=1765&top_left_x=678)

Fig. 4:

Let $\alpha = \angle BAC, x = AD$. Then, applying the Law of Sines in triangle $ABC$ (see Fig. 4), we have:

$$
\begin{aligned}
\frac{x \cos \alpha}{\sin (\pi / 3 - \alpha)} & = \frac{x + x \cos \alpha}{\sin (2 \pi / 3)} \Longleftrightarrow \sin \alpha = \frac{\sqrt{3} \cos^2 \alpha}{1 + \cos \alpha} \Longleftrightarrow (1 - \cos^2 \alpha)(1 + \cos \alpha)^2 = 3 \cos^4 \alpha \Longleftrightarrow \\
\Longleftrightarrow 4 \cos^4 \alpha + 2 \cos^3 \alpha - 2 \cos \alpha - 1 = 0 & \Longleftrightarrow (\cos 2\alpha + 1)(2 \cos^3 \alpha - 1) = 0
\end{aligned}
$$

Since $\alpha < \pi / 2$, from the last equation: $\cos \alpha = 1 / \sqrt[3]{2}$. Therefore, $AD = AB / \cos \alpha = 88$.

## Answer: 88.