, 2014

In the following, let $[XYZ]$ denote the area of the triangle $XYZ$.


Since we are only dealing with ratios of areas and parallels to the sides of the triangle, we can assume without loss of generality that $ABC$ is equilateral for the following argument. (If it is not, an affine transformation will yield this case without changing the ratios involved.)
Let $CX$ be the median of $ABC$ through $C$ and assume without loss of generality that $P$ lies to the left of $CX$, i.e. in the interior of $AXC$. We then have
$$\overline{QP} < \overline{PR} \quad \text{and} \quad \overline{QE} < \overline{DR}, \quad \text{since} \quad \langle APQ = \langle DPR \rangle \langle BPR = \langle EPQ \rangle$$
With the angle equality $\langle EQP = \langle DRP \rangle$, this implies
$$[PEQ] < [PRD].$$

Equality of the areas of these two triangles therefore implies that $P$ lies on the median through $C$.
The second equality $[PQE] = [PET]$ implies $\overline{QE} = \overline{ET}$. Since triangles $PQT$ and $BAC$ are similar (corresponding sides are parallel), triangle $PQT$ is also equilateral and $PE$ is perpendicular to $QT$. It therefore follows that $BE$ is perpendicular to $AC$ and therefore a median in $ABC$. It follows that there is only one point with the required properties, namely the centroid of $ABC$. $\square$