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.

I'm not sure I understand what you mean by "No answer". If every scalene triangle gives the same fraction (strongly implied by the problem), then by taking a limit (scalene triangles that approximate an equilater one better and better), we would expect an equilateral triangle to give the same fraction as well, so that you've found the answer is 95.
If you're just looking for a different approach, I like barycentric coordinates.


*

*$P_1$ is $\left[\dfrac23,\dfrac13,0\right]$

*$Q_1$ is $\left[\dfrac23,0,\dfrac13\right]$

*So $\overleftrightarrow{BQ_1}$ is $\left[\dfrac{2t}3,1-t,\dfrac{t}3\right]$ and $\overleftrightarrow{CP_1}$ is $\left[\dfrac{2s}3,\dfrac{s}3,1-s\right]$, with intersection $\left[\dfrac12,\dfrac14,\dfrac14\right]$.

*$Q_2$ is $\left[\dfrac13,0,\dfrac23\right]$

*So $\overleftrightarrow{BQ_2}$ is $\left[\dfrac{t}3,1-t,\dfrac{2t}3\right]$ and the intersection with $\overleftrightarrow{CP_1}$ is $\left[\dfrac27,\dfrac17,\dfrac47\right]$.

*$P_2$ is $\left[\dfrac13,\dfrac23,0\right]$

*So the intersection of $\overleftrightarrow{BQ_2}$ with $\overleftrightarrow{CP_2}$ ($\left[\dfrac{s}3,\dfrac{2s}3,1-s\right]$) is $\left[\dfrac15,\dfrac25,\dfrac25\right]$.


Now we can use the barycentric area formula to find the area of the right half of the quadrilateral, which is half of the area of the whole quadrilateral by symmetry. So the whole quadrilateral has area:
$$2*45*\det\begin{bmatrix}\dfrac15&\dfrac25&\dfrac25\\\dfrac27&\dfrac17&\dfrac47\\\dfrac12&\dfrac14&\dfrac14\end{bmatrix}$$
$$=9*\det\begin{bmatrix}1&2&2\\\dfrac27&\dfrac17&\dfrac47\\1&\dfrac12&\dfrac12\end{bmatrix}$$
$$=\dfrac{9}{14}*\det\begin{bmatrix}1&2&2\\2&1&4\\2&1&1\end{bmatrix}$$
$$=\dfrac{9}{14}*\det\begin{bmatrix}1&2&2\\0&0&3\\2&1&1\end{bmatrix}$$
$$=-\dfrac{27}{14}*\det\begin{bmatrix}1&2\\2&1\end{bmatrix}$$
$$=\dfrac{81}{14}$$
