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Weekly Problem 29 - 2010

Stage: 3 Short Challenge Level: Challenge Level:1

Triangles $PRS$ and $QPR$ are similar because $< PSR = < QRP$ (since $PR =PS$) and $< PRS = < QPR$ (since $QP =QR$).
Hence $\frac{SR}{RP} = \frac{RP}{PQ}$, that is $\frac{SR}{6} = \frac{6}{9}$, that is $SR = 4$.

This problem is taken from the UKMT Mathematical Challenges.

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