The figure below, composed of four regular pentagons with a side length of $1$, was glued in space as follows. First, it was folded along the broken sections, by combining the bold sections, and then formed in such a way that colored sections formed a square. Find the length of the segment $AB$ created in this way.
[img]https://cdn.artofproblemsolving.com/attachments/0/7/bddad6449f74cbc7ea2623957ef05b3b0d2f19.png[/img]
1. **Define the problem and setup the coordinate system:**
   We are given a figure composed of four regular pentagons with a side length of 1. The figure is folded such that the colored sections form a square. We need to find the length of the segment \( AB \).

2. **Define the vertices and vectors:**
   Denote the vertices of the colored square base as \( A, B, C, D \) and the pentagonal vertices closest to points \( A, B, C, D \) as \( E, F, G, H \) respectively. Place the figure on a coordinate plane such that the vectors \( \overrightarrow{AB} = \langle 1, 0, 0 \rangle \) and \( \overrightarrow{AD} = \langle 0, 1, 0 \rangle \).

3. **Find the coordinates of point \( E \):**
   We want to find \( E \) such that:
   \[
   \overrightarrow{AB} \cdot \overrightarrow{AE} = \overrightarrow{AD} \cdot \overrightarrow{AE} = \cos 108^\circ
   \]
   Assume \( \overrightarrow{AE} = \langle -a, -a, b \rangle \) with \( a, b > 0 \) and \( 2a^2 + b^2 = 1 \). Then:
   \[
   \langle 1, 0, 0 \rangle \cdot \langle -a, -a, b \rangle = -a = \cos 108^\circ
   \]
   Since \( \cos 108^\circ = -\cos 72^\circ \), we have:
   \[
   a = \cos 72^\circ
   \]
   Therefore:
   \[
   b^2 = 1 - 2\cos^2 72^\circ
   \]
   Using the identity \( \cos 72^\circ = \sin 18^\circ \), we get:
   \[
   b^2 = 1 - 2\sin^2 18^\circ = \cos 36^\circ
   \]
   Thus:
   \[
   b = \sqrt{\cos 36^\circ}
   \]

4. **Find the coordinates of point \( F \):**
   Similarly, \( \overrightarrow{BF} = \langle \cos 72^\circ, -\cos 72^\circ, \sqrt{\cos 36^\circ} \rangle \).

5. **Calculate the midpoint \( I \) and the topmost point \( J \):**
   Let \( I \) be the midpoint of \( AB \) and \( J \) be the topmost point of the pentagon with side \( AB \). Then:
   \[
   |\overrightarrow{IJ}| = \frac{1}{2} \tan 72^\circ
   \]
   We have:
   \[
   \overrightarrow{IJ} \propto \frac{1}{2} (\overrightarrow{AE} + \overrightarrow{BF})
   \]
   Simplifying, we get:
   \[
   \overrightarrow{IJ} = k \langle 0, -\cos 72^\circ, \sqrt{\cos 36^\circ} \rangle
   \]

6. **Compare magnitudes and solve for \( k \):**
   Comparing the magnitudes of these two expressions, we have:
   \[
   k^2 = \frac{\tan^2 72^\circ}{4(\cos^2 72^\circ + \cos 36^\circ)}
   \]
   Simplifying further:
   \[
   k^2 = \frac{\tan^2 72^\circ}{4(\cos^2 72^\circ + 1 - 2\cos^2 72^\circ)}
   \]
   \[
   k^2 = \frac{\tan^2 72^\circ}{4(1 - \cos^2 72^\circ)}
   \]
   \[
   k^2 = \frac{\tan^2 72^\circ}{4(\sin^2 72^\circ)}
   \]
   \[
   k^2 = \frac{1}{4\cos^2 72^\circ}
   \]
   \[
   k = \frac{1}{2\cos 72^\circ}
   \]

7. **Calculate the desired length:**
   The desired length is the sum of the length of the square (1) and twice the magnitude of the horizontal projection of \( \overrightarrow{IJ} \):
   \[
   1 + 2 \left| -\frac{1}{2} \right| = 2
   \]

The final answer is \( \boxed{2} \).