3 [ Intersecting lines, angle between them ]

In a regular triangular prism $A B C A 1 B 1 C 1$ ( $A A 1\|B B 1\| C C 1$ ), the angle between the lines $A C 1$ and $A 1 B$ is $\alpha, A A 1$ $=2$. Find $A B$.
Through vertex $A$, draw a line parallel to the diagonal $BA_1$ of the lateral face $ABB_1A_1$. Let this line intersect the extension of edge $A_1B_1$ at point $M$. Denote the side of the base of the prism by $a$.

The quadrilateral $AM A_1 B$ is a parallelogram, so $MA_1 = AB = A_1B_1 = a$. From the isosceles triangle $MA_1C_1$ with legs $A_1M = A_1C = a$ and an angle of $120^\circ$ between them, we find that $MC_1 = a\sqrt{3}$. The line $AM$ is parallel to the line $BA_1$, so the angle between the lines $AC_1$ and $A_1B$ is equal to the angle between the lines $AC_1$ and $AM$, i.e., $\alpha$. From the right triangles $AA_1M$ and $AA_1C_1$, we find that

$$
AM = \sqrt{A_1M^2 + AA_1^2} = \sqrt{a^2 + 4}, \quad AC_1 = \sqrt{A_1C_1^2 + AA_1^2} = \sqrt{a^2 + 4}
$$

Consider the isosceles triangle $AMC_1$. By the cosine rule,

$$
MC_1^2 = AM^2 + AC_1^2 - 2AM \cdot AC_1 \cos \angle MAC_1,
$$

or

$$
3a^2 = 2(a^2 + 4) - 2(a^2 + 4) \cos \alpha
$$

from which

$$
a^2 = \frac{8(1 - \cos \alpha)}{1 + 2 \cos \alpha} = \frac{16 \sin^2 \frac{\alpha}{2}}{1 + 2(1 - 2 \sin^2 \frac{\alpha}{2})} = \frac{16 \sin^2 \frac{\alpha}{2}}{3 - 4 \sin^2 \frac{\alpha}{2}}.
$$

Therefore,

$$
AB = x = \frac{4 \sin \frac{\alpha}{2}}{\sqrt{3 - 4 \sin^2 \frac{\alpha}{2}}}.
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-23.jpg?height=115&width=206&top_left_y=2109&top_left_x=17)