Let $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be nine points in space such that $ABCDE$, $ABFGH$, and $GFCDI$ are each regular pentagons with side length $1$. Determine the lengths of the sides of triangle $EHI$.
1. **Understanding the Problem:**
   We are given three regular pentagons \(ABCDE\), \(ABFGH\), and \(GFCDI\) with side length 1. We need to determine the lengths of the sides of triangle \(EHI\).

2. **Analyzing the Geometry:**
   - Since \(ABCDE\) is a regular pentagon, all its sides are equal to 1.
   - Similarly, \(ABFGH\) and \(GFCDI\) are regular pentagons with side length 1.
   - We need to find the lengths of the sides of triangle \(EHI\).

3. **Positioning Points in Space:**
   - Let's place point \(F\) at the origin: \(F(0,0,0)\).
   - Place point \(C\) at \(C(0,1,0)\).
   - Place point \(B\) at \(B\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right)\).

4. **Finding Coordinates of \(G\):**
   - Since \(\angle GFC = 108^\circ\), we can use trigonometry to find the coordinates of \(G\).
   - The \(y\)-coordinate of \(G\) is \(y_G = \cos 108^\circ\).
   - The \(x\)-coordinate of \(G\) lies on the bisector of \(\angle BFC\), which has the equation \(y = -\sqrt{3}x\).
   - Therefore, \(x_G = -\frac{\cos 108^\circ}{\sqrt{3}}\).

5. **Calculating \(z_G\):**
   - Using the distance formula, we have:
     \[
     x_G^2 + y_G^2 + z_G^2 = 1
     \]
   - Substituting \(x_G\) and \(y_G\):
     \[
     \left(-\frac{\cos 108^\circ}{\sqrt{3}}\right)^2 + (\cos 108^\circ)^2 + z_G^2 = 1
     \]
   - Simplifying:
     \[
     \frac{\cos^2 108^\circ}{3} + \cos^2 108^\circ + z_G^2 = 1
     \]
     \[
     \frac{4\cos^2 108^\circ}{3} + z_G^2 = 1
     \]
     \[
     z_G^2 = 1 - \frac{4\cos^2 108^\circ}{3}
     \]
     \[
     z_G = \sqrt{1 - \frac{4\cos^2 108^\circ}{3}}
     \]

6. **Equation of the Plane through \(F\), \(C\), and \(G\):**
   - The normal vector to the plane \(FCG\) is \(\left(\sqrt{1 - \frac{4\cos^2 108^\circ}{3}}, 0, \frac{\cos 108^\circ}{\sqrt{3}}\right)\).
   - The equation of the plane is:
     \[
     x \cdot \sqrt{1 - \frac{4\cos^2 108^\circ}{3}} + z \cdot \frac{\cos 108^\circ}{\sqrt{3}} = 0
     \]

7. **Angle Between Planes:**
   - The normal vector to the \(xy\)-plane is \((0,0,1)\).
   - The angle \(\theta\) between the planes is given by:
     \[
     \cos \theta = \left| \frac{\frac{\cos 108^\circ}{\sqrt{3}}}{\sqrt{1 - \frac{4\cos^2 108^\circ}{3} + \frac{\cos^2 108^\circ}{3}}} \right| = \left| \frac{1}{\sqrt{3} \tan 108^\circ} \right| = \frac{\tan 18^\circ}{\sqrt{3}}
     \]

8. **Midpoint \(M\) of \([FC]\):**
   - The midpoint \(M\) of \([FC]\) is:
     \[
     MI = \frac{1}{2} \cot 18^\circ
     \]
   - Since \(II'\) is perpendicular to the \(xy\)-plane and \(I'\) lies on the \(xy\)-plane, we have:
     \[
     MI' = MI \cdot \cos \theta = \frac{\sqrt{3}}{6}
     \]

9. **Circumcircle of \(\triangle HIE\):**
   - The center \(O\) of the circumcircle of \(\triangle FCB\) is:
     \[
     MO = \frac{\sqrt{3}}{6}
     \]
   - The radius \(R\) of the circumcircle of \(\triangle HIE\) is:
     \[
     R = OI' = \frac{\sqrt{3}}{3}
     \]
   - Therefore, \(\triangle HIE\) is equilateral with side lengths:
     \[
     HI = IE = EH = 1
     \]

The final answer is \(\boxed{1}\)