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Math Help - Geometry Probability

  1. #1
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    Geometry Probability

    A point is chosen at random from the interior of square ABCD. What is the probability that the point is closer to either A or C than it is to the midpoint of the diagonal connecting A and C.

    My attempt at solving the problem follows.

    Let x be the length of a side of the square. Then, there are x^2 points inside of the square. Furthermore, the midpoint of the diagonal connecting A and C is at the point (\dfrac{x}{2},\dfrac{x}{2}).

    For my square, point A is at point (0,x) and point C is at point (x,0). I found the distances between points A and C, point A and the midpoint, and point C and the midpoint. However, I don't think I am approaching the solution.
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  2. #2
    Senior Member Sambit's Avatar
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    Notice the diagram. If a point lies in the YELLOW region, it will be closer to the midpoint of the diagonal than A or C; and if the point lies in the GREEN region, it will be closer to A or C than the midpoint of the diagonal. Let x be the length of each side of the square. So the required probability is total area of the GREEN region / total area of the square = [{(1/2)*(x/2)*(x/2)}*2] / x*x = 1/8
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  3. #3
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    I appreciate your help, particularly your diagram. However, I do not understand how you knew points in the yellow region are closer to the midpoint than to either point A or point B.
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  4. #4
    Senior Member Sambit's Avatar
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    The line EF divides AO in two equal parts, that is, AI=AO. So any point lying in the part AI will be closer to A than to O. Similarly, any point lying in the part IO will be closer to O than to A. In consequence, any point in the triangle AEF will be closer to A and any point in the triangle OEF will be closer to O.
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