## Task A-3.5. (4 points)

A cylinder of height $10 \mathrm{~cm}$ is given. Points $A$ and $B$ are marked on the edges of its bases such that $\overline{A B}$ is parallel to the axis of the cylinder. If we connect points $A$ and $B$ with the shortest line that goes once around the cylinder (along the lateral surface), its length will be $15 \mathrm{~cm}$. What is the length of the shortest line that goes around the cylinder twice and connects points $A$ and $B$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-18.jpg?height=352&width=758&top_left_y=652&top_left_x=660)

Let's cut the cylinder's lateral surface along the line $\overline{A B}$.
![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-18.jpg?height=354&width=1494&top_left_y=1219&top_left_x=290)

This way, we obtain a rectangle $A A^{\prime} B^{\prime} B$ for which $\left|A B^{\prime}\right|=15 \mathrm{~cm}$. From this, we conclude that the circumference of the base of the cylinder (the length of side $\overline{A A^{\prime}}$) is

$$
\left|A A^{\prime}\right|=\sqrt{15^{2}-10^{2}}=5 \sqrt{5}
$$

If we denote by $d$ the length of the shortest line that wraps around the cylinder twice and connects points $A$ and $B$, and by $C$ the intersection point of this line with segment $\overline{A B}$, then we have

$$
d=\left|A C^{\prime}\right|+\left|C B^{\prime}\right|=2\left|A C^{\prime}\right|=2 \sqrt{\left|A A^{\prime}\right|^{2}+\left|A^{\prime} C^{\prime}\right|^{2}}=2 \sqrt{(5 \sqrt{5})^{2}+5^{2}}=10 \sqrt{6}
$$

Therefore, the length of the shortest line that wraps around the cylinder twice and connects points $A$ and $B$ is $10 \sqrt{6} \mathrm{~cm}$.