Note that we can bisect any segment by overlapping its ends. In addition, we can fold the paper along a line passing through a given point, perpendicular to a given line. In other words, we can construct the midpoints of segments and a perpendicular to a given line passing through a given point.

We will use these constructions in solving the problem, along with the fact that $2016=32 \cdot 63$.

Fold the square $A B C D$ along the diagonal $B D$, and then six times along lines perpendicular to it so that this diagonal is divided into 64 equal parts. Let $X_{1}$ and $X_{2}$ lie on the diagonal $B D$, such that $B X_{1}=X_{1} X_{2}=1 / 64 B D$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-46.jpg?height=697&width=686&top_left_y=0&top_left_x=695)

Let the line $A X_{1}$ intersect $B C$ at point $Y$. Then $B Y: B C=B Y: A D=B X_{1}: X_{1} D=1: 63$, so $B Y=32$. Considering the point $Z$ - the projection of $X_{2}$ onto $B C$, we get that $B Z: B C=B X_{2}: B D=1: 32$, so $B Z=63$. Therefore, by folding along the lines $A X_{1}$ and $X_{2} Z$, and then along the line passing through $Y$ and perpendicular to $B C$, we obtain a segment of length $2 B Y - B Z = 1$.

## Answer

It is possible.
Zassovstiy A.A.

Given a square sheet of paper with a side length of 2016. Is it possible, by folding it no more than ten times, to construct a segment of length 1?