Trisecting a Triangle Given a (non-degenerate) triangle $PQR$ in the Euclidean plane, does there exists a point $A$ in the interior of the triangle such that, the triangles $APQ$, $AQR$, and $ARP$ have same area? If it exists, is it unique?
(I thought about this question while reading the book "Proof without Words- R. B. Nelsen".)
 A: For every (non-degenerate) triangle, such $A$ exists and it is unique.
In the following, we'll show that only the barycenter $C$ of a triangle is such point. Let $[PQR]$ be the area of $\triangle PQR$. 
First, let us prove that $[CPQ]=[CQR]=[CRP]$. Let $S$ be the intersection point of $CP$ and $QR$. Since $PS:CS=3:1$, we can see that $[CQR]=(1/3)[PQR]$. We can get $[CPQ]=(1/3)[PQR],[CRP]=(1/3)[PQR]$ in the same way as above. So, we have $[CPQ]=[CQR]=[CRP]$.
Second, let us prove that $C$ is the only point such that $[CPQ]=[CQR]=[CRP]$. Draw three lines $CT, CU, CV$ where $CT,CU,CV$ are parallel to $PQ,QR,RP$ respectively. The point $A$ has to exist on each of $CT,CU,CV$, and three lines $CT, CU, CV$ intersect (of course only) at $C$, which leads that $C$ is the only point such that $[CPQ]=[CQR]=[CRP]$. 
Therefore, the proof is completed. Q.E.D.
A: An analytic solution, based on the fact that the area of the triangle with vertices at $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is
$$
\frac{1}{2}\det
\begin{bmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3
\end{bmatrix}
$$
provided the path goes counterclockwise. Without loss of generality, we can assume the vertices are $O(0,0)$, $A(a,0)$ and $B(b,c)$, with $a\ne0$ and $c\ne0$. Let $P(x,y)$ be the point we're looking for. Then we must have
$$\begin{cases}
\det\begin{bmatrix}
1 & 1 & 1 \\
a & b & x \\
0 & c & y
\end{bmatrix}
=
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & a & x \\
0 & 0 & y
\end{bmatrix}
\\[3ex]
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & x & b \\
0 & y & c
\end{bmatrix}
=
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & a & x \\
0 & 0 & y
\end{bmatrix}
\end{cases}
$$
which becomes
$$
\begin{cases}
by-ac-cx-ay=ay\\
cx-by=ay
\end{cases}
$$
or
$$
\begin{cases}
-cx+(b-2a)y=ac\\
cx-(a+b)y=0
\end{cases}
$$
which easily gives $3ay=ac$ or $y=c/3$ and similarly $x=(a+b)/3$. The point
$$
\left(\frac{a+b}{3},\frac{c}{3}\right)
$$
is indeed the barycenter of the given triangle and is the unique solution.
