, 2019

$2\sqrt{6} - 2$

Let $I$ be the point of intersection of the line segments $CE$ and $FH$. Let also $AG = a$. Five vertices of the regular pentagon $ABCDE$ lie on the circumference of the same circle, and these points divide the circumference of this circle into five equal parts. Consequently, we have
$$\angle BAE = \frac{1}{2} \cdot \frac{3}{5} \cdot 360^\circ = 108^\circ, \quad \angle AEC = \frac{1}{2} \cdot \frac{2}{5} \cdot 360^\circ = 72^\circ,$$
from which it follows that we have $AB \parallel EC$. Similarly, we get $CD \parallel BE$, $DE \parallel CA$. By the equality of alternate angles, we then get $\angle GBF = \angle IEF$, $\angle BGF = \angle EIF$, from which we conclude that the triangles $GBF$ and $IEF$ are similar. Arguing the same way, we obtain from $\angle GAF = \angle ICF$, $\angle AGF = \angle CIF$ the fact that the triangles $GAF$ and $ICF$ are similar. Consequently, we get
$$AG : CI = FG : FI = BG : EI \quad \text{from which it follows that} \quad \frac{CI}{EI} = \frac{AG}{BG} = \frac{a}{4}$$
holds.
By using the same argument as above, we get from $CD \parallel BE$ the fact that the triangles $EFI$ and $CHI$ are similar. Furthermore, from $CD \parallel BE$, $DE \parallel CA$ we see that the quadrilateral $CDEF$ is a parallelogram, from which it follows that $EF = DC = AB = 4 + a$. Therefore, we get
$$\frac{CI}{EI} = \frac{CH}{EF} = \frac{5}{4+a}$$
Putting together the facts we obtained above, we conclude that
$$\frac{a}{4} = \frac{CI}{EI} = \frac{5}{4+a}$$
must hold, and therefore, we get $(4 + a)a - 20 = 0$, from which it follows, since $a > 0$, that $a = 2\sqrt{6} - 2$ is the desired answer.