27 January 2014

What is the distance between these poles?

Two poles stand vertically on level ground. One is 10 feet tall, the other 15 feet tall. If a line is drawn from the top of each pole to the bottom of the other, the two lines intersect at a point 6 feet above the ground. What’s the distance between the poles?
You can take a wild guess (and be correct), or view the answer at Futility Closet.

Addendum: If you're like me and need a bit of help in understanding the answer, see the comment by the fourth Anonymous (8:28 PM), which refers to this annotated version of the same diagram:

12 comments:

  1. Can't tell. No matter what the spacing between the poles, the intersection of the top-to-bottom lines will always be 6 feet.

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    1. Can you explain why? True, 15x10/15+10 = 6, but how does that apply?

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  2. Stared at this for 10 minutes. Stared some more - math, obviously, wasn't my subject. Finally gave up and clicked the link,and what do I get? "502 Bad Gateway"

    You're a cruel man, Stan.

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    Replies
    1. The Futility Closet home page is also inaccessible tonight. Someone must have locked the door.

      Delete
  3. This comment has been removed by the author.

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    1. Are you figuring the tangents of equivalent angles? If so, I don't think that the angles you're equating are equal. I'd say:
      4/x1 = 6/x2 = 10/x
      9/x2 = 6/x1 = 15/x
      which is also a way of showing the slopes of the two diagonals.

      I think JPS above is right; the intersection of the lines, regardless of x1 + x2, will always be at 6 high, just as x1:x2 will always be 2:3. Try figuring x1 and x2 for total x of 10 and 100 separately and see what that gives you.

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  4. Those lines will always intersect at 6' up unless far enough apart for the curvature of the surface of the world (or simple topography) to matter

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    1. Why? Because... math, sure. But can you ELI5?

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  5. I can try, using (https://www.dropbox.com/s/womlsye6da89t02/pole_position.jpg) as a reference.

    For starters, the slope of a line is figured as rise / run. So the slope of line L is 10/C, and the slope of line R is 15/C.

    To get to X, we can see that the rise of line L over distance B = rise of line R over distance A.

    The run of a line over a certain distance is the rise / slope, so for line L over distance B, it's X / (10/C). Since (X/1) / (10/C) can be said (X/1) * (C/10), we get to the run for line L over distance B, or simply distance B = XC/10.

    Going through the same exercise with the distance A and slope of line R (15/C), we get that A = XC/15.

    A + B = C, so by swapping out terms, we get XC/15 + XC/10 = C. To further simplify, let's make those denominators common. The least common multiple of 10 and 15 is 30, so by multiplying the first term by 2/2 and the second by 3/3, we get 2XC/30 + 3XC/30 = C. Add 'em up to get 5XC/30 = C.

    Factor the left side (divide top and bottom by 5) to arrive at XC/6 = C. Looks like we can divide both sides by C to get X/6 = 1. We want to see the X stand alone, so multiply both sides by 6, finally getting X = 6.

    Hope that helps!

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    1. Thank you! Now I can understand.

      I've incorporated your dropbox diagram into the body of the post.

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    2. Also, if instead of using 30 as the denominator you use 150, you get

      10XC/150 + 15XC/150 = C or (10 + 15)XC / (10*15)= C

      Divide both sides by XC to get

      (10+15) / (10*15) = 1/X

      Now flip numerator and denominator to get the Futilityn Closet's equation:

      (10*15)/(10+15) = 6

      Delete
  6. What others have said. Using similar triangles, it's easy enough to figure out the 3:2 ratio between B:A and 15:10 and the parts of L and R. Couldn't progress much from there, however, until I tried using Cartesian equations. Any line is definable by using y = mx + b, where m is the slope and b is the y-intercept. Using that, and setting the origin at the bottom left edge, we get

    f(x) = 15/(A+B) + 0 and

    g(x) = -10/(A+B) + 10

    We also know that f(A) = 6 and g(A) = 6

    From f(A) we get 6A + 6B = 15A and
    from g(A) we get 4A + 4B = 10A

    which gives us no new information. The implication is that we don't have enough information
    in the problem. So I tried A = 2 and B = 3 (keeping the 3:2 ratio), and f(2) = g(2) = 6.
    Then I tried A = 4 and B = 6, and still f(4) = g(4) = 6. The intersection point will always be 6 high.

    Weird.

    Lurker111

    ReplyDelete

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