Sliced square inspired

Rotate the square $90^{\circ}$ counterclockwise and place $A$ at the origin as shown below.

$AC$ lies along $y = x$ and $EI$ lies along $y=6x - 3$ ; their intersection point $L$ has coordinates $\left( \dfrac{3}{5}, \dfrac{3}{5} \right)$ .

Then, using $AE$ and $CI$ as their respective bases, $\triangle AEI$ has area $\dfrac{1}{2} \left( \dfrac{1}{2} \right) \left( \dfrac{3}{5} \right) = \dfrac{3}{20}$ and $\triangle CIL$ has area $\dfrac{1}{2} \left( \dfrac{1}{3} \right) \left( \dfrac{2}{5} \right) = \dfrac{2}{30}$ .

Thus the combined area is $\dfrac{3}{20} + \dfrac{2}{30} = \dfrac{13}{60}$ , so our answer is $13 + 60 = \boxed{73}$

Consdering that AE=BE, $AE=\frac{1}{2}$ and considering that DI is double of CI, $DI=2CI$ so:

$DI+CI=1$

$(2+1)CI=1$

$3CI=1$

$CI=\dfrac{1}{3}$

Now let the perpendicular distance from L to AB be $h$ . Then the perpendicular distance from L to CD is $(1-h)$ . Note that AEL and CLI are similar triangles which means that:

$\dfrac{AE}{CI}=\dfrac{h}{1-h}$

$\dfrac{\frac{1}{2}}{\frac{1}{3}}=\dfrac{h}{1-h}$

$\dfrac{3}{2}=\dfrac{h}{1-h}$

$3-3h=2h$

$(2+3)h=3$

$5h=3$

$h=\dfrac{3}{5}$

Now the total area is $\dfrac{1}{2} \times (\dfrac{1}{2} \dfrac{3}{5}+\dfrac{1}{3} \dfrac{2}{5})=\dfrac{1}{2} \times (\dfrac{3}{10}+\dfrac{2}{15})=\dfrac{1}{2} \times \dfrac{9+4}{30}=\dfrac{1}{2} \times \dfrac{13}{30}=\dfrac{13}{60}$ .

With $a=13$ and $b=60$ , $a+b=\boxed{73}$ .