One fold

$\tan\beta=DE=1-x$

$\alpha=2*\beta$

$\cos\alpha = \frac{1-\tan^{2}\beta}{1+\tan^{2}\beta} = \frac{2x-x^{2}}{2-2x+x^{2}}$

$\sin\alpha = \frac{2-2x}{2-2x+x^{2}}$

$\triangle FDE \sim \triangle ECI \sim \triangle GHI$

$EI=\frac{x}{\cos\alpha}$ , $GH=GI\cos\alpha$ , $HI=GI\sin\alpha$

$EI+IH=EH$ , so $\frac{x}{\cos\alpha}+GI\sin\alpha=1$ and solving for $GI$ gives $GI=\frac{\cos\alpha-x}{\cos\alpha*\sin\alpha}$

The area of $\triangle GHI=\frac{1}{2}GH*HI=\frac{1}{2}GI^{2}\cos\alpha\sin\alpha=\frac{1}{2}(\frac{\cos\alpha-x}{\cos\alpha\sin\alpha})^{2}*\cos\alpha\sin\alpha$

after subbing in values for $\cos\alpha$ and $\sin\alpha$ and some simplifying we arive at

Area of $\triangle GHI = A(x) = \frac{x^{4}-x^{3}}{4x-8}$

A quick graph shows us on the right track. Max is a bit under $x=0.8$

Now for the simple calculus. The derivative, after simplifying, is $A'(x)=\frac{4x^{2}(3x^{2}-10x+6)}{8-4x^{2}}$

$A'(x)=0$ implies $x=0$ or $3x^{2}-10x+6=0$ . The solution of interest is $x=\frac{5-\sqrt7}{3}\approx .7847$

$A(\frac{5-\sqrt7}{3})$ after much simplifying is $\frac{316-119\sqrt7}{54}$

So $A=316$ , $B=-119$ , and $C=54$ which means $A+B+C=316-119+54=\boxed{251}$