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".)

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.
