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Distance to the corner of a rectangle

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    This question is taken from 2011 Malaysian Mathematical Olympiad.
    Mary is standing in a rectangular garden. Her distance to the four corners of the garden are 6 m, 7 m, 9 m and d m, respectively, where d is an integer. Find d.


    2. Relevant equations

    Triangle inequality. a + b < c, a + c < b, b + c < a, where a, b, and c are the lengths of the three sides of the triangle.



    3. The attempt at a solution

    I tried to denote the length of the rectangular garden as a and b, respectively, then from the four triangles formed, I formed some inequality and try to see if the value of d is bounded, but it yields nothing. I have also tried to solve for d by using the concept of area. Also, I tried using the law of cosine and the Pythagorean theorem. But still, I can't find the value for d.

    Any other ideas how to approach this problem? Thanks. :)
     
  2. jcsd
  3. Nov 20, 2011 #2

    eumyang

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    Pythagorean Theorem is the way to go. Draw two perpendicular lines "through" Mary. You'll have four right triangles, with 6, 7, 9 and d being the hypotenuses. Use the Pythagorean Theorem four times, and through some manipulation, you'll be able to find d.
     
  4. Nov 20, 2011 #3
    Thanks for the clue. Now I have a clearer direction. I get d = √94, not an integer though, but still, at least I can get the value of d. :)
     
  5. Nov 20, 2011 #4

    eumyang

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    I didn't get that answer. Can you double-check?

    EDIT: I think I know why our answers differ. It depends on how you label the four distances from Mary to the corners. I took "6, 7, 9 and d, respectively" to mean that you label the line segments clockwise in that fashion. It looks like you labeled them as "6, 9, d and 7," going clockwise, or something similar. Are you looking at a diagram?
     
    Last edited: Nov 20, 2011
  6. Nov 20, 2011 #5
    [itex]v_{1}^{2}+h_{1}^{2}=6^2=36[/itex]
    [itex]v_{1}^{2}+h_{2}^{2}=7^2=49[/itex]
    [itex]v_{2}^{2}+h_{1}^{2}=9^2=81[/itex]
    [itex]v_{2}^{2}+h_{2}^{2}=d^2[/itex]

    [itex]d^2=v{2}^{2}+h_{2}^{2}=130-36=94[/itex]
     
  7. Nov 20, 2011 #6
    Yup. I should have used the "6, 7, 9, d" clockwise. I recount, and get √68. :)

    EDIT: No diagram was given for the question.
     
  8. Nov 20, 2011 #7
    Since the question says d is an integer, I try to use other types of 'combination', and when I tried with '6, 9, 7, d' going clockwise, I get d = 2. Yay! Thanks again, eumyang, for the help :)
     
  9. Nov 20, 2011 #8

    eumyang

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    I had forgotten about the "d is an integer" part when I last posted. :redface: I'm glad you got the answer.
     
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