Trisected sides of a scalene triangle Scalene triangle $\bigtriangleup ABC$ has area 45. Points $P_1$ and $P_2$ are located on side $AB$ such that $AP_1 = P_1P_2 = BP_2$. Additionally, the points $Q_1$ and $Q_2$ are located on side $AC$ such that $AQ_1 = Q_1Q_2 = CQ_2$. The area of the intersection of triangles $BQ_1Q_2$ and $CP_1P_2$ can be expressed as a common fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A) } 15   \qquad \textbf{(B) } 47   \qquad \textbf{(C) } 79 \qquad \textbf{(D) } 95 \qquad \textbf{(E) } 257 $
So to solve this, I assumed that an equilateral triangle wouldn't change the answer (I'm very lazy), so I fakesolved with an equilateral triangle.

I scaled it down so that the area was $\sqrt{3}$.
Then, the coordinates of the kite in the middle (in counter-clockwise order) are:
$(1,\frac{\sqrt{3}}{5})$
$(\frac{4}{7},\frac{2\sqrt{3}}{7})$
$(1,\frac{\sqrt{3}}{2})$
$(\frac{10}{7},\frac{2\sqrt{3}}{7})$
The product of the two diagonals divided by 2 is: $\frac{3}{7} \cdot \frac{3\sqrt{3}}{10}=\frac{9\sqrt{3}}{70}$. Multiplying by $\frac{45}{\sqrt{3}}=15\sqrt{3}$ gives $\frac{9 \cdot 15 \cdot 3}{70}=\frac{405}{70}=\frac{81}{14} \implies 95$.
No answer, can you guys check this? Thanks.


Let [.] denote areas and I = [ABC] = 45. Observe that the intersection area [DEGF] is equal to
$$[DEGF] = [DBC] - [EBC] - [FBC] + [GBC]\tag 1$$
Evaluate [EBC], one of the four triangle areas on the RHS, as follows. 
$$\frac{[EBC]}{[P_2BC]}=\frac{EC}{P_2C} 
= \frac{[Q_1BC]}{[Q_1P_2BC]} = \frac{[Q_1BC]}{I - [AP_2Q_1]} 
= \frac{\frac23I}{I - \frac13\cdot\frac23I}=\frac67 $$
Then,
$$[EBC] = \frac67[P_2BC] = \frac67\cdot \frac13I = \frac27I$$
Following the same procedure to obtain the areas of the other three triangles,
$$[FBC] = \frac27I, \>\>\>\>\> [DBC] = \frac12I, \>\>\>\>\> [FBC] = \frac15I$$
Substitute the four areas and $I=45$ into (1),
$$[DEGF] = \left(\frac12 - \frac27- \frac27 + \frac15\right)I=\frac9{70}\cdot 45 = \frac{81}{14}$$
