Let $P'$, $Q'$, $R'$, $S'$ be the midpoints of $AB$, $BC$, $CD$, $DA$, respectively. To prove that $P'Q'R'S'$ is a parallelogram, draw $AC$ and $BD$ and note that $P'S'\parallel BD\parallel Q'R'$ and $P'Q'\parallel AC\parallel S'R'$. $\frac{EP}{EP'}=\frac{EQ}{EQ'}=\frac{ER}{ER'}=\frac{ES}{ES'}=\frac{2}{3}\implies$ $PQ\parallel P'Q', QR\parallel Q'R', RS\parallel R'S', SP\parallel S'P'\implies$ $PQRS$ is a parallelogram We know $[P'Q'R'S']=\frac{1}{2}[ABCD]$, and since $\frac{PQ}{P'Q'}=\frac{QR}{Q'R'}=\frac{RS}{R'S'}=\frac{SP}{S'P'}=\frac{2}{3}$, $\frac{[PQRS]}{[P'Q'R'S]}=\frac{4}{9}\implies [PQRS]=\frac{2}{9}[ABCD]$.

Jutaro wrote: Let $ABCDE$ be a convex pentagon. Denote the centroids $P$, $Q$, $R$, $S$ of the triangles $ABE$, $BCE$, $CDE$, $DAE$ respectively. Show that the quadrilateral $PQRS$ is a parallelogram and $9\cdot [PQRS]=2\cdot [ABCD]$. PROOF. Denote the midpoints $X,Y,Z,U,V$ of the sides $[AB]$, $[BC]$, $[CD]$, $[DE]$, $[EA]$ respectively. Prove easily that $\left\{\begin{array}{c}PQ\parallel XY\parallel AC\ ,\ PQ=\frac{2}{3}\cdot XY=\frac{1}{3}\cdot AC\\\\ QR\parallel YZ\parallel BD\ ,\ QR=\frac{2}{3}\cdot YZ=\frac{1}{3}\cdot BD\\\\ RS\parallel AC\ ,\ RS=\frac{1}{3}\cdot AC\\\\ SP\parallel BD\ ,\ SP=\frac{1}{3}\cdot BD\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}PQ\parallel RS\parallel AC\ ,\ PQ=RS=\frac{1}{3}\cdot AC\\\\ QR\parallel SP\parallel BD\ ,\ QR=SP=\frac{1}{3}\cdot BD\end{array}\right\|$ $\implies$ the quadrilateral $PQRS$ is parallelogram and $m\left(\widehat{AC,BD}\right)=m(\widehat{PQR})=\phi$ $\implies$ $[PQRS]=QP\cdot QR\cdot\sin \phi=$ $\left(\frac{1}{3}\cdot AC\right)\cdot\left(\frac{1}{3}\cdot BD\right)\cdot\sin\phi =$ $\frac{2}{9}\cdot\left(\frac{1}{2}\cdot AC\cdot BD\cdot\sin\phi\right)=$ $\frac{2}{9}\cdot [ABCD]$ $\implies$ $9\cdot [PQRS]=2\cdot [ABCD]$. Remark I. I"ll use the complex numbers. Denote $X(x)$- the point $X$ with the affix $x\in \mathcal C$ Therefore, $\left\{\begin{array}{c}\left\{\begin{array}{c}3p=a+b+e\\\ 3r=c+d+e\end{array}\right\|\implies 3(p+r)=a+b+c+d+2e\\\\ \left\{\begin{array}{c}3q=b+c+e\\\ 3s=a+d+e\end{array}\right\|\implies 3(q+s)=a+b+c+d+2e\end{array}\right\|$ $\implies$ $p+r=q+s$ $\implies$ $PQRS$ is parallelogram. Remark II. Consider the hexagon $ABCDEE$, i.e. suppose that we have identical six balls such that only one ball in each from $A,B,C,D$ and two balls in $E$. $P$ is centroid of $\triangle ABE$ and $R$ is centroid of $\triangle CDE$ $\implies$ the centroid of the hexagon $ABCDEE$ is the midpoint of $[PR]$. $Q$ is centroid of $\triangle BCE$ and $S$ is centroid of $\triangle ADE$ $\implies$ the centroid of the hexagon $ABCDEE$ is the midpoint of $[QS]$. Thus, the centroid of the hexagon $ABCDEE$ is both the midpoint of the segment $[PR]$ and the midpoint of the segment $[QS]$ ! Therefore, the quadrilateral $PQRS$ is parallelogram.

Interesting. It is a nice problem, but I don't think that this problem is appropriately placed on an olympiad...It takes a total of 'no thinking' to coordinate bruteforce with the shoelace formula. The parallelogram thing can be shown by vectors.

Altheman wrote: Interesting. It is a nice problem, but I don't think that this problem is appropriately placed on an olympiad...It takes a total of 'no thinking' to coordinate bruteforce with the shoelace formula. The parallelogram thing can be shown by vectors. Sure, Altheman! But you should take into account that the problem belongs to one of the first Olympiads ever organized in our region (by the way, my country hosted this one, in 2000 ). Also, the competition is aimed to high school students under 16, so the problems are not supposed to be really difficult.