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2008 AMC 10B Problem 14

Problem 14 of 25IntermediateGeometry

Triangle OABOAB has O=(0,0),O=(0,0), B=(5,0),B=(5,0), and AA in the first quadrant. In addition, ∠ABO=90∘\angle ABO=90^\circ and ∠AOB=30∘.\angle AOB=30^\circ. Suppose that OA‾\overline{OA} is rotated 90∘90^\circ counterclockwise about O.O. What are the coordinates of the image of A?A?

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Solution

Because ∠ABO=90∘,\angle ABO=90^\circ, segment ABAB is vertical, so A=(5, 5tan⁡30∘)=(5, 533).A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right). A 90∘90^\circ counterclockwise rotation about the origin sends (x,y)(x,y) to (−y,x),(-y,x), so the image of AA is (−533, 5).\left(-\tfrac{5\sqrt3}{3},\,5\right). Thus, the correct answer is B.

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Concepts: coordinate geometry · transformation · special right triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.