12. As shown in Figure 4, with the vertices of the equilateral triangle $\triangle ABC$ with side length 1 as the centers and its side length as the radius, draw three equal circles, obtaining intersection points $D, E, F$. Connect $CF$ to intersect $\odot C$ at point $G$. With point $E$ as the center and $EG$ as the radius, draw an arc intersecting side $AB$ at point $M$. Find the length of $AM$.
12. According to the problem, $\angle A C G=30^{\circ}$. As shown in Figure 7, connect $E A$, $E C$, and $E M$.

Since the radii of $\odot A$ and $\odot C$ are both 1, $\triangle A C E$ is an equilateral triangle, so $\angle A C E=60^{\circ}$.
Thus, $\angle E C F=90^{\circ}$.
In the right triangle $\triangle E C G$, we get $E G=\sqrt{2}$.
Therefore, $E M=\sqrt{2}$.
Draw a perpendicular line from point $E$ to line $A B$, intersecting the extension of $B A$ at point $H$. Then
$\angle E A H=60^{\circ}$.
Thus, in the right triangle $\triangle A E H$,
$A H=\frac{1}{2} A E=\frac{1}{2}$,
$E H=\sqrt{A E^{2}-A H^{2}}=\frac{\sqrt{3}}{2}$.
In the right triangle $\triangle M E H$,
$H M=\sqrt{E M^{2}-E H^{2}}=\frac{\sqrt{5}}{2}$.
Therefore, $A M=H M-A H=\frac{\sqrt{5}-1}{2}$.