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Scale the triangle by $\frac{2}{5}$ so that $AB=4$ and $BC=6$.
Clearly $\Delta PQC\sim \Delta ABC$
If $PQ=x$, then we have
$$x+\frac{3}{2}x=4$$$$x=\frac{8}{5}$$
Given the isosceles triangle $∆ABC$ with $AB = AC = 10$ and $ BC = 15$. Let points $P$ in $BC$ and $Q$ in $AC$ chosen such that $ AQ = QP = P C $. Calculate the ratio of areas of the triangles $[∆PQA]: [∆ABC]$.
Scale the triangle by $\frac{2}{5}$ so that $AB=4$ and $BC=6$.
Clearly $\Delta PQC\sim \Delta ABC$
If $PQ=x$, then we have
$$x+\frac{3}{2}x=4$$$$x=\frac{8}{5}$$