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is the incentre and point 
is defined such that 
and 
is the perpendicular bisector of 
with 
. Denote the orthocenter of 
as 
, the intersection of 
and 
as 
. Compute the length of 
.
shown below has area 
. Find the area of the rectangle 
.
, to each of the vertices creating 
equilateral triangles. The area of 
is clearly equal to the area of 
. Therefore each of the triangles and 
is equal in area to one of the equilateral triangles. The remaining area, which is the rectangle, is just 
of the equilateral triangles or 
.
. Triangles and 
are both 
isosceles triangles, meaning that they have side lengths of 
with an altitude of 
, and the area of one of the triangles is 
, so the area of two is 
. And the area of the rectangle is 
. Putting these together we have 
, which means 
, and since we are looking for the value of , we just divde both sides by 
getting 
.
. Now connect 
. Note that ![$[FAB]=\frac12 x^2 \sin{120}=\frac{x^2\sqrt3}4$](http://latex.artofproblemsolving.com/a/7/8/a78c973decff9c59da89799fdb9e04bd02268264.png)
. Note that 
is a 
so ![$[FBC]=\frac{\sqrt3x}2$](http://latex.artofproblemsolving.com/a/4/1/a417561fa423eb04bb5b3f510cb2967e2d2f2778.png)
. Hence, ![$\frac{[FBC]}{[FAB]}=\frac{2}{1}$](http://latex.artofproblemsolving.com/4/7/6/4769345a29717e2c05759fcce0481e1b2ce6a3ba.png)
and the answer is 
.
