3.88. A line segment connecting a point on the circumference of the upper base of a cylinder with a point on the circumference of the lower base is equal to $l$ and forms an angle $\alpha$ with the base plane. Find the distance from this line to the axis of the cylinder, if the axial section of the cylinder is a square. What are the possible values of $\alpha$?
3.88. Given that $A K L M$ is a cylinder, $O O_{1}$ is its axis, $A$ belongs to the circumference of the upper base, $B$ belongs to the circumference of the lower base, $A B=l, \angle A B K=\alpha$, and the axial section $A M L K$ is a square (Fig. 3.88); we need to find the distance from $O O_{1}$ to $A B$. Draw $B S \parallel A K$ and obtain the section $B S A K$ of the cylinder, which contains $A B$; through $O O_{1}$, draw $(E F T) \parallel (S B K)$. The distance between $O O_{1}$ and $A B$ is equal to the distance between $(E F T)$ and $(S B K)$. Draw $O P \perp B K$; then $O P \perp (S B K)$, since $O P \perp S B$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-556.jpg?height=400&width=314&top_left_y=798&top_left_x=839)

Fig. 3.88 and $S B \cap B K$, i.e., $O P$ is the desired distance. From $\triangle A K B$, we find $B K=l \cos \alpha, A K=l \sin \alpha$; hence $P B=\frac{1}{2} l \cos \alpha, \quad O B=\frac{1}{2} E N=\frac{1}{2} T N=\frac{1}{2} A K=\frac{1}{2} l \sin \alpha$. Finally, from $\triangle O P B$ we get

$$
O P=\sqrt{\frac{1}{4} l^{2} \sin \alpha-\frac{1}{4} l^{2} \cos ^{2} \alpha}=\frac{1}{2} l \sqrt{-\cos 2 \alpha}
$$

where $\cos 2 \alpha<0$, hence $\frac{\pi}{2}<2 \alpha<\frac{3 \pi}{2}$, i.e., $\frac{\pi}{4}<\alpha<\frac{3 \pi}{4}$.