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Turkey NMO 2007 1st Round - P01 (Geometry)

Source:

October 4, 2012
geometry

Problem Statement

Let ABCABC be a triangle with m(A^)=90∘m(\widehat{A})=90^\circ, ∣AB∣=4|AB|=4, and ∣AC∣=3|AC|=3. Let DD be the foot of perpendicular from AA to [BC][BC]. If PP a point on [BD][BD] such that 5∣AP∣=13∣PD∣5|AP|=13|PD|, what is ∣CP∣|CP|?
<spanclass=′latex−bold′>(A)</span> 9+435<spanclass=′latex−bold′>(B)</span> 5615<spanclass=′latex−bold′>(C)</span> 145<spanclass=′latex−bold′>(D)</span> 3713<spanclass=′latex−bold′>(E)</span> 55+35 <span class='latex-bold'>(A)</span>\ \frac{9 + 4\sqrt 3}{5} \qquad<span class='latex-bold'>(B)</span>\ \frac{56}{15} \qquad<span class='latex-bold'>(C)</span>\ \frac{14}{5} \qquad<span class='latex-bold'>(D)</span>\ \frac{37}{13} \qquad<span class='latex-bold'>(E)</span>\ \frac{5\sqrt 5 + 3}{5}
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