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Coordinate geometry

  1. Jun 27, 2004 #1
    the vertices of a triangle are 0, and the points A (a, 0) and B (0,a) where a>0.
    the point P moves in the plane OAB such that OP square + AP square + BP square = k square, where k is a positive constant. show that k mst be bigger than (2/3)a X square root of 3


    i tried. but cant even know how to start. any hints?
     
  2. jcsd
  3. Jun 27, 2004 #2

    HallsofIvy

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    Let P be (x,y). The OP square= x2+ y2, AP square= (x-a)2+ y2, and BP square= x2+ (y-a)2.
    Now, what is OP square + AP square+ BP square. Keep in mind that x2 and y2 must be positive.
     
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