<s>[INST] Find the area of the shaded region. (Apparently this is a Chinese primary school math question!) 
Rectangle $ABCD$ has an area of $1$. $AE = ED$. $3BF = AB$. What is the shaded area?
 [/INST] The simplest way, in my opinion, is to write the equations of lines $BE$, $EC$, $FC$, $FD$.  Because scaling doesn't change the ratios of areas, we can assume that we have a square of unit side length, where
$ B = (0, 0)$
$ E = (\frac{1}{2}, 1)$
$ C = (1, 0) $
$ F = (0, \frac{1}{3} )$
$ D = (1,1) $
It is straight forward to derive the following equations
$BE: y = 2 x $
$EC: y = -2 (x - 1) $
$FC: y = -\frac{1}{3} ( x- 1 ) $
$FD: y = \frac{1}{3} + \frac{2}{3} x $
Now, we proceed to find the intersection points
$H: BE \& FC : (\frac{1}{7}, \frac{2}{7} ) $
$G: FD \& BE :  (\frac{1}{4}, \frac{1}{2} ) $
$I: EC \& FD : (\frac{5}{8} , \frac{3}{4} ) $
Finally, applying the so-called shoe-lace formula, we get
$[CIGH] = \frac{1}{2} \left( C_x I_y - C_y I_x + I_x G_y - I_y G_x + G_x H_y - G_y H_x + H_x C_y - H_y C_x \right) $
And this is
$[CIGH] = \frac{1}{2} \left( (1)(\frac{3}{4}) - 0 + (\frac{5}{8})(\frac{1}{2}) - (\frac{3}{4})(\frac{1}{4}) + (\frac{1}{4})(\frac{2}{7}) - (\frac{1}{2})(\frac{1}{7}) + 0 - (\frac{2}{7})(1) \right)$
which reduces to
$[CIGH] = \frac{1}{2} \left( \frac{3}{4} + \frac{5}{16} - \frac{3}{16} + \frac{1}{14} - \frac{1}{14} - \frac{2}{7} \right) $
And finally this comes to
$[CIGH] = \frac{1}{2} \left( \frac{ 84 + 35 - 21  - 32 }{(16)(7)} \right) = \boxed{\frac{33}{112}} $
This is assuming we have a square of side $1$. For a rectangle of area $A$, we get
$[CIGH] =\boxed{ \frac{33}{112} A } $
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