<s>[INST] In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of
$\triangle ABC$.
I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.
 [/INST] This is not a trivial geometry problem... Let $x=AF/FB, y=BD/DC, z=CE/EA$, then we have $xyz=1$. Furthermore, we can calculate that the area of $AFB$ is $\frac{x}{(1+z+yz)(1+x)}$. Similarly we can calculate two others, so, we have
$$\frac{x}{(1+z+xz)(1+x)}=\frac{y}{(1+x+yx)(1+y)}=\frac{z}{(1+y+zy)(1+z)}$$
Notice that $(1+z+xz)=(xyz+z+xz)=z(xy+1+x)=z(xy+xyz+x)=zx(y+yz+1)$, so we have
$$\frac{x}{(1+x)}=\frac{yz}{(1+y)}=\frac{z^2x}{(1+z)}$$
So we have $x(1+z)=z^2x(1+x)$ and $x(1+y)=yz(1+x)$. Substitute $y=1/xz$, we have $x(1+1/xz)=z/(xz)(1+x)$, or, $x(1+xz)=z(1+x)$. This yields a linear equation of $z$, that is, $z=\frac{x}{1+x-x^2}$. Substitute it in, we have
$$(1+\frac{x}{1+x-x^2})=(\frac{x}{1+x-x^2})^2(1+x)$$
Or,
$$(1+2x-x^2)(1+x-x^2)=x^2(1+x)$$
Expanding, we have
$$x^4-4x^3-x^2+3x+1=0$$
Factorizing, we have
$$(x-1)(x^3-3x^2-4x-1)=0$$
One solution is $x=1$, the other positive solution is located $x>4$, where $z=\frac x{1+x-x^2}<0$ (which means $P$ is outside), contradiction. Therefore, the only valid solution is $x=y=z=1$.
</s>