National Olympiad of Argentina

If $\text{area}(ABC)=S$ then $\text{area}(APC)=\frac{AP}{AB}S=\frac{1}{5}S$. As $\text{area}(PQC)=\frac{4}{25}S$, we have $\text{area}(PQB)=S-\frac{1}{5}S-\frac{4}{25}S=\frac{16}{25}S$. On the other hand $$\text{area}(PBQ)=\frac{BQ}{BC}\text{area}(PBC)=\frac{BQ}{BC}\cdot \frac{BP}{BA}\text{area}(ABC)=\frac{BQ}{BC}\cdot \frac{4}{5}S.$$ Hence $\frac{16}{25}S=\frac{BQ}{BC}\cdot \frac{4}{5}S$ which implies $\frac{BQ}{BC}=\frac{4}{5}$. Because $\frac{BP}{BA}=\frac{4}{5}$, it follows that $PQ\parallel AC$, so that triangles $ABC$ and $PBQ$ are similar. But $PQ=BQ$ as $Q$ belongs to the perpendicular bisector of $PB$. Therefore $BC=AC=7$.