Super Geometry Problem

Mark the letters $M$ and $N$ , we find $S_{\triangle ABN}+S_{\triangle CDN}=\frac{1}{2}S_{ABCD}$ (You can see a beautiful proof about it at the bottom).

So $S_{\triangle CDN}=1\times\sqrt{3}-\frac{9\sqrt{3}}{22}=\frac{\sqrt{3}}{11}$ .

And because of $EF\perp AD$ , $\angle EDF=30^\circ$ and $S_{\triangle DEF}=\frac{\sqrt{3}}{8}$ , so we can know $EF=\frac{1}{2},FD=\frac{\sqrt{3}}{2}$ .

In addition, we can find $\triangle EFM\sim\triangle CDM$ , so we assume $DM=x$ , so:

$\frac{1}{2}:(\frac{\sqrt{3}}{2}-y)=1:y$

$y=\frac{\sqrt{3}}{3}$

So $S_{\triangle CDM}=\frac{\sqrt{3}}{3}\times1\times\frac{1}{2}=\frac{\sqrt{3}}{6}$ .

and so

$S_{\triangle DMN}=\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{11}=\frac{5\sqrt{3}}{66}$

So $a=3,b=66$ , the answer is:

$3+66=\boxed{69}$

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Beautiful Proof:

Let $EF = a$ . Then $DF = a \cot 30^\circ = \sqrt 3 a$ . Since the area of $\triangle DEF$ is $\dfrac {a \cdot \sqrt 3 a}2\dfrac {\sqrt 3}8 \implies a = \dfrac 12 = EF$ and $DE = \dfrac {\sqrt 3}2$ .

Let the intersection point of $AD$ and $CE$ be $H$ and $DH = x$ . Then $FH = \dfrac {\sqrt 3}2 - x$ . We note that $\triangle EFH$ and $\triangle CDH$ are similar. Then $\dfrac {FH}{DH} = \dfrac {EF}{CD} \implies \dfrac {\frac {\sqrt 3}2-x}x = \dfrac 12 \implies x = \dfrac 1{\sqrt 3} = \dfrac {\sqrt 3}3$ .

Draw $IJ$ parallel to $BC$ . Since the area of $\triangle ABG$ is $\dfrac {IG \cdot 1}2 = \dfrac {9\sqrt 3}{22} \implies IG = \dfrac {9\sqrt 3}{11}$ . Then $GJ = IJ - IG = \sqrt 3 - \dfrac {9\sqrt 3}{11} = \dfrac {2\sqrt 3}{11}$ .

Now area of the green triangle is

$\begin{aligned} [DGH] & = [CDH] - [CDG] \\ & = \frac {DH \cdot CD}2 - \frac {GJ \cdot CD}2 \\ & = \frac {\sqrt 3}6 - \frac {\sqrt 3}{11} \\ & = \frac {5\sqrt 3}{66} \end{aligned}$

Therefore $a+b = 3+66 = \boxed{69}$ .