When does the triangle have the smallest area? The following triangle has an area $S$, and the sides $AO$ and $BO$ have the length $a$ and $b$, respectively. There is a fixed point $X$ at $(x,y)$. A point $C$ is put on the line segment $OA$, and the point $D$ is put on the intersection between the line segment $OB$ and the line $CX$. When does the area of the triangle $DCO$ have the smallest value? I think it is either when $DX=XC$, when $D$ is at $B$, or when $C$ is at $A$. Yet, if I try to prove this, calculation becomes so complicated. 

 A: Be warned that there is an instant "flash of insight" solution from sitting and pondering the problem long enough, and you are not far from it, so you might lose some enjoyment reading the answer. 
Hint:

 remove A and B from the picture

Solution.  With increasing detail (move mouse/cursor over the hidden texts to reveal),

  Parallelogram  

which is

  centered at X

and

  made by rotating the figure 180 degrees

A: Drop a perpendicular from D to OA and call that point P.  If 0 $\le$ P $\le$ C the triangle ODC is split into 2 right triangles.  The base of one is P and the base of the other is C - P.  If D has height h, the area of POD = (1/2)hp and the area of the other is (1/2)h(C-P).  Adding them up we get Area = (1/2)hp + (1/2)h(C -p) = C/2.  
So we want C to be as small as possible. To fill the condition that P $\le$ C, the area is minimized when the line CXD is perpendicular to OA.
However, there is a second case, when C < P .  Now drop the perpendicular from C to OD, call it Q.  Using the same logic as above, the area of triangle ODC will be D/2.  How low can D go? Well, until C gets larger than P.  In other words, until DXC is perpendicular to OA.
Same answer as above.
