Problem 3. In the square $\mathrm{ABCD}$, the point $\mathrm{M} \in(\mathrm{BC})$ is taken such that $\frac{B M}{B C}=\frac{1}{4}$. The line $\mathrm{DM}$ intersects $\mathrm{AC}$ at $\mathrm{Q}$ and the extension of $(\mathrm{AB})$ at $\mathrm{P}$. Find the value of the ratio $\frac{Q M}{D P}$.

Marius Farcas
## Solution and Scoring:

From $\mathrm{BM} \| \mathrm{AD}$, we get $\triangle \mathrm{PBM} \sim \triangle \mathrm{PAD}$, so: $\frac{B P}{P A}=\frac{B M}{D A}=\frac{M P}{D P}=\frac{1}{4}$. It follows that $M P=\frac{1}{4} D P$ $1 \text{ point}$

From $\frac{B P}{A P}=\frac{1}{4}$, using derived proportions, we obtain $A P=\frac{4}{3} A B$

From $\mathrm{DC} \| \mathrm{AP}$, we get $\triangle \mathrm{QDC} \sim \triangle \mathrm{QPA}$, so: $\frac{Q D}{Q P}=\frac{D C}{A P}=\frac{Q C}{Q A}$ and we obtain $\frac{Q D}{Q P}=\frac{A B}{\frac{4}{3} A B}, \mathrm{QD}=\frac{3}{7} \mathrm{DP}$... $2 \text{ points}$
$\mathrm{QM}=\mathrm{DP}-\mathrm{DQ}-\mathrm{MP}=\mathrm{DP}-\frac{1}{4} \mathrm{DP}-\frac{3}{7} \mathrm{DP}=\frac{9}{28} \mathrm{DP}$. Finally, $\frac{Q M}{D P}=\frac{9}{28}$