Results 1 to 13 of 13

Math Help - Impossible Puzzle?

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    6

    Impossible Puzzle?

    Hi. I came up with this question a while ago, and am finally ready to give up on finding the answer myself. I hope you can find one. Good luck.

    _______________________________
    You are standing on a planet which is perfectly spherical with radius r.

    A rope was been wrapped around its equator. Adding x meters to this rope how high will you be able to lift the rope with it still being in the plane of the equator?
    ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,912
    Thanks
    775
    Hello, powotae!

    You are standing on a planet which is perfectly spherical with radius r.

    A rope was been wrapped around its equator. Adding x meters to this rope,
    how high will you be able to lift the rope with it still being in the plane of the equator?

    The planet has a radius of r (the units don't matter).
    At the equator, the circumference (length of the rope) is: . 2\pi r

    Add x meters to the rope.
    The rope is now a circle with circumference  2\pi r + x.

    Let R be the radius of this new (larger) circle.
    Then the circumference of the circle is: 2\pi R

    So we have: . 2\pi R = 2\pi r + x

    Solve for R\!:\;\;R \:=\:r + \frac{x}{2\pi}

    We see that the new radius is \frac{x}{2\pi} meters larger.


    Therefore, the rope can be raised \frac{x}{2\pi} meters (all around the equator).

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    The sphere should have a known radius R.
    Although the additional x length of the rope is "unknown" here, it is considered as a fixed value or a known quantity here.
    So the only known quantities here are R and x.

    Draw the figure on paper. It is a right triangle whose
    --->hypotenuse = (R+y), in meters
    --->angle included by the two legs = theta, in radians.
    --->one leg = R, in meters
    --->the other leg, the short leg opposite theta, = x/2 +R*theta, in meters.

    where, y is the to-find-for height the rope could be lifted. -----***

    Why is the short leg opposite the angle theta is (x/2 +R*theta)?
    Because:
    When the rope is equal to the circumference of the equator, the height of the lift is zero. When x is added to the rope, the (height of the lift) = y meters.
    Half of the lifted rope is the short leg in question. Its length is the length of the circular arc whose central angle is theta, plus half of x. Since arc = R*(central angle, in radians), then
    short leg = R*theta +x/2

    So, in the right triangle as figured,

    tan(theta) = (R*theta +x/2)/R
    tan(theta) = theta +x/(2R) -------------(1)
    The x/(2R) is known once you plug in the values of x and R.
    Eq.(1) is hard to solve for theta. But you can use iteration to find theta.

    cos(theta) = R /(R+y) ------------(2)
    Substitute into (2) the theta you found in (1), and voila, you have y!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2007
    Posts
    6
    Quote Originally Posted by Soroban View Post
    Hello, powotae!


    The planet has a radius of r (the units don't matter).
    At the equator, the circumference (length of the rope) is: . 2\pi r

    Add x meters to the rope.
    The rope is now a circle with circumference  2\pi r + x.

    Let R be the radius of this new (larger) circle.
    Then the circumference of the circle is: 2\pi R

    So we have: . 2\pi R = 2\pi r + x

    Solve for R\!:\;\;R \:=\:r + \frac{x}{2\pi}

    We see that the new radius is \frac{x}{2\pi} meters larger.


    Therefore, the rope can be raised \frac{x}{2\pi} meters (all around the equator).
    Thankyou. But I was thinking more of lifting the rope at one point as high as possible aware from the globe. Sorry if it was not clear.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2007
    Posts
    6
    ticbol I am not sure where you get that right angle triangle from with hypotenuse = (R+y).
    Could you explain please.

    \pi
    Last edited by powotae; August 24th 2007 at 04:23 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by powotae View Post
    ticbol I am not sure where you get that right angle triangle from with hypotenuse = (R+y).
    Could you explain please.
    Sure.

    So you lift the rope at one point as high as you can (using a crane, maybe) until the rope is taut. No sag, no slack.
    So you have now an "inverted angle" or two rays/straight lines from the highest point.
    The highest point is y meters from the equator.
    The ends of the two rays are touching the equator equidistantly from the point on the equator directly, or vertically, below the highest point.
    Any of these two rays is tangent to the equator at the point of contact.

    We concentrate on only one ray.
    ---If we imagine a radius R touching the point below the highest point, the total distance of the highest point from the center of the planet is (R+y).
    ---A radius R to the point of tangency of the ray will place that point R away from the center.
    ---Let's call the acute angle separating these (R+y) and R as theta.
    ---The ray is tangent to the equator, so the ray and the R are perpendicular, Or, the angle separating the ray and the R is a right angle or is 90degrees or is pi/2 radians.
    ----So, the R is one leg of the right triangle, the ray is the other leg, the (R+y) is the hypotenuse.
    Last edited by ticbol; August 20th 2007 at 04:45 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2007
    Posts
    6
    Thanks I can follow it now.
    You have done similer to what i did.

    I was just trying to solve it to be able to get a general equation for theta from equation (1).

    tan(theta) = theta +x/(2R) -------------(1)

    Which is the part I do not think is possible, even though there must be a unique solution for it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by powotae View Post
    Thanks I can follow it now.
    You have done similer to what i did.

    I was just trying to solve it to be able to get a general equation for theta from equation (1).

    tan(theta) = theta +x/(2R) -------------(1)

    Which is the part I do not think is possible, even though there must be a unique solution for it.
    Eh? I've done similar to what you did and you could not follow my right triangle? .

    Why not give values to x and R, then try iteration.
    Come on, give values to x and R.

    You know, if you think only that it is not possible, even without trying first, then it is really not possible. Buit if you try first, maybe you'd change your mind.

    Another way is to graph the two curves,
    y = tan(theta) ----(i)
    y = theta +x/(2R) ----(ii)
    and find their intersection.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Aug 2007
    Posts
    6
    Quote Originally Posted by ticbol View Post
    Eh? I've done similar to what you did and you could not follow my right triangle? .

    and find their intersection.
    Ye. sorry its been a few months since I have looked at it, and did not understand a bit of the terminology you were using.


    Quote Originally Posted by ticbol View Post
    Why not give values to x and R, then try iteration.
    Come on, give values to x and R.

    You know, if you think only that it is not possible, even without trying first, then it is really not possible. Buit if you try first, maybe you'd change your mind.

    Another way is to graph the two curves,
    y = tan(theta) ----(i)
    y = theta +x/(2R) ----(ii)
    I am after a general solution though. Giving X and R values would not provide that.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by powotae View Post
    Ye. sorry its been a few months since I have looked at it, and did not understand a bit of the terminology you were using.




    I am after a general solution though. Giving X and R values would not provide that.
    That is the general solution.

    Of course if all the unknowns are variables, then you can not find any solution.
    You have only two equations with 4 unknowns (theta, x, R, y) so how can you get a solution?

    Again, giving the x and R values will give a solution.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Aug 2007
    Posts
    6
    Still open to other suggestions.

    From the way I see it. For any given r and x there is a unique y.
    So in theory there should be an equation y = y(r,x)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Nov 2007
    Posts
    5

    solution

    I made a picture to clarify things:

    edit: sorry for using the letter "L" so much... bad choice of letters

    we want to find "h"
    we know L0 the original length was 2\pi*R
    L1 is length after the increase of x is 2\pi*R+x
    look at the angle t, and the ark Lt
    the length of Lt is tR
    the complement of the ark has length of (2\pi-t)R
    and the new length of the rope L1 is then the complement plus twice "L"
    L1=2\pi*R+x=(2\pi-t)R+2L
    L=\frac{tR+x}{2}

    from the right angle triangle we get C^2=R^2+L^2
    C=\sqrt{R^2+\frac{(tR+x)^2}{4}}
    and
    h=C-R
    h=\sqrt{R^2+\frac{(tR+x)^2}{4}}-R

    you win!


    Yuwie.com | invite friends. hang out. get paid.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Nov 2007
    Posts
    5
    Quote Originally Posted by Manu View Post
    ....
    L=\frac{tR+x}{2}
    Ooops, the angle "t" is not given is it?
    need to somehow extract it from R and x

    sorry
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. impossible problem =P
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 6th 2009, 06:17 AM
  2. HELP.. HELP .. Another Impossible question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: March 8th 2008, 10:17 AM
  3. HELPPP ...... Impossible... PLZ
    Posted in the Statistics Forum
    Replies: 3
    Last Post: March 8th 2008, 03:34 AM
  4. Replies: 1
    Last Post: March 5th 2008, 11:12 PM
  5. HELP!!!! These are impossible.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: December 10th 2006, 06:56 PM

Search Tags


/mathhelpforum @mathhelpforum