Austria 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 $$\overline{QP}<\overline{PR}\text{and}\overline{QE}<\overline{DR},\text{since}⟨APQ=⟨DPR⟩⟨BPR=⟨EPQ⟩$$ With the angle equality $⟨EQP=⟨DRP⟩$, 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 $ABC$. $\square$