So the question goes like this:
Given a triangle ABC, there is a point M within that triangle such that [AMB]=[AMC]=[BMC]. Prove that M is the centroid of the triangle. ([AMC] denotes the area of triangle AMC)
I can easily prove the converse, that if M is the centroid, then the areas are equal, but I don't know how to prove this. Can anyone help me? I have tried placing the triangle in a coordinate system, among other things, but could not get anywhere.
On
Let $P$ be any point inside $\triangle ABC$ such that $[ABP] = [APC]$, then
$$|AB||AP|\sin(\angle PAB) = 2[ABP] = 2[APC] = |AP||AC|\sin(\angle CAP)\\ \implies \sin(\angle CAP) : \sin(\angle PAB) = |AB| : |AC| $$ Since $\angle CAP + \angle PAB = \angle CAB$ is independent of choice of $P$, the last equality forces $\angle CAP$ and $\angle PAB$ to be independent of choice of $P$. This means the locus of any point $P$ which satifies the condition $[ABP] = [ABC]$ lies on a line passing through $A$. It is easy to see this line passes through the mid-point of $BC$ too. This tells us the locus is the median of $\triangle ABC$ through $A$.
If you don't want to use trigonometric fucntions, you can extend the line $AP$ and let it hit $BC$ at $X$. It is easy to see
$$\begin{align} & [ABX] : [ABP] = |AX| : |AP| = [AXC] : [APC]\\ \implies & |AX| : |XC| = [ABX] : [AXC] = [ABP] : [APC] = 1 : 1 \end{align}$$ From this, we can conclude $X$ is the mid-point of $BC$ and once again, $P$ lies on the median $AX$ of $\triangle ABC$.
Back to original problem. If a point $M$ satisfies $[AMB] = [BMC] = [CMA]$, then $M$ lies on the intersection of the three medians of $\triangle ABC$. By definition, $M$ will be the centroid of the triangle.
On
Since $\triangle MBC$ has one-third the area of $\triangle ABC$, yet has the same "base", $\overline{BC}$, it follows that $M$'s distance to that base is one-third $A$'s distance; therefore: $M$ is on the line, parallel to $\overline{BC}$, through the "lower" trisection point of the altitude from $A$. Likewise, $M$ is on lines parallel to $\overline{CA}$ and $\overline{AB}$ through corresponding trisection points on the altitudes from $B$ and $C$. We know that the three lines have the centroid in common, and that (because they are distinct) they can share no other point. Therefore, any "area-trisecting point" $M$ must coincide with the centroid itself. $\square$
If the comment was not helpful enough: suppose we know that the centroid divides the triangle into 3 triangles of equal area.
Given $M$ and $M_c$, the point in question and the centroid respectively, we have that: $$[AMC]+[AMB]+[BMC]={\text Area}(\Delta ABC)=[AM_cC]+[AM_cB]+[BM_cC]$$ Now, letting the heights of the triangles corresponding to $M$ be $$h_{AC}, h_{AB},h_{BC}$$ and the heights of the triangles corresponding to $M_c$ be $$H_{AC}, H_{AB},H_{BC}$$ we have the following: $$\frac{1}{2}\left(\overline{AC}\cdot h_{AC}+\overline{AB}\cdot h_{AB}+\overline{BC}\cdot h_{BC} \right)=\frac{1}{2}\left(\overline{AC}\cdot H_{AC}+\overline{AB}\cdot H_{AB}+\overline{BC}\cdot H_{BC} \right)$$ From here, can you deduce that the points $M_c$ and $M$ are equal?