15th Czech-Polish-Slovak Mathematics Competition

Denote by $P$, $Q$, and $R$ the centroids of the triangles $ABD$, $BCD$, and $ADE$, respectively. Let $K$ and $L$ be the midpoints of the segments $BD$ and $AD$, respectively. Since $P$ and $Q$ are centroids, they divide in the same ratio the medians $AK$ and $CK$, respectively. That is, $AP:PK=CQ:QK=2:1$, and we have $PQ\parallel AC$. Similarly $PR\parallel BE$. Hence the angle $QPR$ is of the same measure as the angle $CXE$ determined by the lines $AC$ and $BE$ (here, $X$ is the intersection point of $AC$ and $BE$, see Fig. 1).



Fig. 1

Denote by $\phi$ the measure of the inscribed angle determined by the chord $AB$ of the given circle. Since $CD=DE=AB$, we have $\angle CAE=2\phi$, and therefore from the triangle $AXE$ we conclude $$\angle CXE=180^{\circ }-\angle AXE=\phi +2\phi =3\phi .$$ Since $AB>r$, we have $\phi >30^{\circ }$, and so $\angle QPR=3\phi >90^{\circ }$.