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Diagonal Garden Path

  1. Feb 24, 2008 #1
    I found this problem in Erwin Brecher's "Journey through Puzzleland" and I find it baffling. There is a garden 55 yards by 40 yards. There is a diagonal path through the garden 1 yard wide. How long is the path? Notice that it is not quite a 40-55-11-root-185 diagonal because ALTERNATE "corners" of the path terminate on the corners of the garden: that is, you enter and leave the garden on the "long" sides (the 55's). So the trajectory of the path is just slightly off from the true diagonal. If you draw the garden in "landscape" perspective, the path is just a little steeper. I hope I've described it well enough.

    The problem I'm having is that I can grind out the solution by brute force, but in fact it turns out to be a very simple solution. I just can't find an easy way of getting it. Any ideas?

  2. jcsd
  3. Feb 25, 2008 #2
    Do you mean that the path is on the diagonal of a 54 x 40 yard rectangle?
  4. Feb 25, 2008 #3
    No, that path wouldn't be one yard wide. The width of the path is measured perpendicular to the trajectory of the path.
  5. Feb 25, 2008 #4
    Do you mean that the path is 1 yard wide and is on the diagonal of a 54 x 40 yard rectangle that sits within the 55 x 40 yard garden?
  6. Feb 25, 2008 #5
    No. The path is a long thin parallelogram with its acute corners on the corners of the 55x40 rectangle. The short faces of the parallelogram lie on the long edges of the rectangle. The transverse width of the parallelogram measured normal to its long faces is 1 yard.
  7. Feb 25, 2008 #6
    OK. Now cut away the top half yard and the bottom half yard of the garden. What is left?
  8. Feb 25, 2008 #7
    If I do as you suggest I am left with an elongated hexagon whose greatest dimension is the hypotenuse of a 54-40 right triangle. But this hypotenuse is not the length I am after; the line which it traces is not parallel to the long faces of my original parallelogram.
  9. Feb 25, 2008 #8
    When I drew it, it was parallel. I'm not sure that I understand the problem. Is there any way you could draw a picture?
  10. Feb 25, 2008 #9
    Here is a picture (attempt to attach bmp file:)

    Attached Files:

  11. Feb 25, 2008 #10
    Thanks. Now I see my error.
  12. Feb 26, 2008 #11
    Answer is hidden.

    Remove the path (say of length x) from the garden, we have the area equation
    (55)(40) - x = 2 \sqrt{x^2 - 40^2} (40)
    and that gives x = 48.202 yards.
  13. Feb 26, 2008 #12
    To Doodle:

    Your method is OK but you have an extra factor of 2 somewhere. In any case you can see that the answer must be greater than 55.
  14. Feb 26, 2008 #13
    Oops, it should have been:

    Remove the path (say of length x) from the garden, we have the area equation
    (55)(40) - x = \sqrt{x^2 - 40^2} (40)
    and that gives x = 66.67 yards.
  15. Mar 25, 2008 #14
  16. Mar 25, 2008 #15
    You are close, but Doodle had the exact answer of 66.666 which you can see if you hilite the white text in the previous message. In fact, this number arises from a very simple 3-4-5 geometry which I was unable to notice until after I saw the answer. What baffled me was the idea that you needed to solve such a difficult quadratic equation to arrive at such a simple solution. I have the general idea that if there is a simple answer, there should be an easy way of getting there. This philosophy doesn't seem to work with this problem.

    I now have an idea why the equation appears so complicated when the solution ends up being so simple. If anyone is interested in my thoughts I will elaborate.

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