Trisecting a Triangle

Question

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".)

Answer 1

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.

Answer 2

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.