areaeuclidean-geometryplane-geometrytriangles
Let $G$ be a point inside triangle $ABC$ such that $[GBC]=[GCA]=[GAB]$, where $[XYZ]$ is the area of a triagle $XYZ$. Show that $G$ is the centroid of the triangle $ABC$.
My attempt: Since that $[GBC]=[GCA]=[GAB]$, so we have $CG$, $AB$ and $GB$, are the $3$ medians, so $G$ is centroid of $ABC$.
I'm not sure about it.
Yes, this triangle exists. (Found using brute force approximation.)
Coordinates: $$A\approx(0.182,0.260)\quad B=(0,0)\quad C=(1,0)\\ D\approx(0.229,0.120)\quad E\approx(0.394,0.087)\quad F\approx(0.182,0.571)$$
Angles: $$a=d\approx107.2957\quad b=e\approx55.0744\quad c=f\approx 17.6299$$
Without loss of generality, suppose that $[XYZ]$ (the area of $\triangle XYZ$) is $1$, and the ratio of similarity between $\triangle DEF$ and $\triangle XYZ$ is $r<1$ (so that $[DEF] = r^2$).
Let $a, b, c$ be the distances between $EF$ and $YZ$, between $ZX$ and $FD$, and between $XY$ and $DE$, respectively.
Then we have $[AEF] = \frac a2 \cdot EF$, $[BFD] = \frac b2 \cdot FD$, and $[CDE] = \frac c2 \cdot DE$ by the formula for triangle area; adding them together, we have $$[ABC] - [DEF] = \frac a2 \cdot EF + \frac b2 \cdot FD + \frac c2 \cdot DE.$$
On the other hand, we have $[AEY] = \frac a2 \cdot AY$, $[AFZ] = \frac a2 \cdot AZ$, $[BFZ] = \frac b2 \cdot BZ$, $[BDX] = \frac b2 \cdot BX$, $[CDX] = \frac c2 \cdot CX$, and $[CEY] = \frac c2 \cdot CY$; adding them together and noting that for example $YZ = AY + AZ$, we have $$[XYZ] - [ABC] = \frac a2 \cdot YZ + \frac b2 \cdot ZX + \frac c2 \cdot XY.$$
Because $r$ is the ratio of similarity between $\triangle DEF$ and $\triangle XYZ$, we have $EF = r \cdot YZ$, $FD = r \cdot ZX$, and $DE = r \cdot XY$, which tells us that
$$
[ABC] - [DEF] = r([XYZ] - [ABC]).
$$
Recall that we assumed $[XYZ] = 1$ and $[DEF] = r^2$, so we now have $[ABC] - r^2 = r(1 - [ABC])$. Solving, we get $[ABC] = r$, so $[ABC] = \sqrt{r^2 \cdot 1} = \sqrt{[DEF] \cdot [XYZ]}$.
Best Answer
Not really, unless the triangle $ABC$ is equilateral.
But this suggests a line of reasoning if you can use affine transformations. We have the following facts:
Under an affine transformation, the ratio between two areas is constant.
If $(ABC)$ and $(A'B'C')$ are two non-degenerate triangles, then there exists an affine transformation that maps one onto the other.
Consequently, to solve the problem in general it is sufficient to solve it for an equilateral triangle. And there you have it.