jmc

Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $A{A}^{\prime }{D}^{\prime }$ and $D^{\prime }{C}^{\prime }E$ are similar. Thus, the sides are proportional: $\frac{A{A}^{\prime }}{A^{\prime }{D}^{\prime }}=\frac{D^{\prime }{C}^{\prime }}{C^{\prime }E}⟹\frac{1-s}{s}=\frac{s}{1-s-CE}$. Simplifying, we get that $s^2=(1-s)(1-s-CE)$. $\angle EAF$ is $60$ degrees, so $\angle BAE=\frac{90-60}{2}=15$. Thus, $\cos 15=\cos (45-30)=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{1}{AE}$, so $AE=\frac{4}{\sqrt{6}+\sqrt{2}}\cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\sqrt{6}-\sqrt{2}$. Since $△AEF$ is equilateral, $EF=AE=\sqrt{6}-\sqrt{2}$. $△CEF$ is a $45-45-90△$, so $CE=\frac{AE}{\sqrt{2}}=\sqrt{3}-1$. Substituting back into the equation from the beginning, we get $s^2=(1-s)(2-\sqrt{3}-s)$, so $(3-\sqrt{3})s=2-\sqrt{3}$. Therefore, $s=\frac{2-\sqrt{3}}{3-\sqrt{3}}\cdot \frac{3+\sqrt{3}}{3+\sqrt{3}}=\frac{3-\sqrt{3}}{6}$, and $a+b+c=3+3+6=\boxed{12}$.