Answer to the Friday Puzzle…

63

On Friday I posted this puzzle……There are 2 flagpoles that are each 100 foot high.  A rope that is 150 feet long is strung between the tops of the flagpoles.  At its lowest point the rope sags 25 feet about the ground (see schematic diagram below).  How far apart are the flagpoles?

If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.Answer: The flagpoles are right next to one another.  Did you solve it?  Any other answers?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

 

63 comments on “Answer to the Friday Puzzle…

  1. Gerard Engelage says:

    Only possible answer, under the assumtion that the poles are vertical.

  2. Eh? What sort of “explanation” is that? I need a little more to go on

    • M says:

      Rope = 150 ft
      Half Of Rope (HOR) = 75 ft.
      Pole = 100 ft.

      Distance Above Ground (DAG) = 25 ft.

      Pole – DAG = 100 – 25 = 75 ft = HOR.

      There is nothing left to make the curve, so the distance between the poles = 0.

  3. Michael Sternberg says:

    I started placing coordinates and variables in a sketch, just as my high school physics teacher impressed upon me, and which served me well ever since. But I plugged in the numbers half-way through and realized that the sag (vertical distance between top of the pole and the rope vertex) is equal to half the rope length — doh!

    Of course now the question comes up: What if the rope is longer? I worked out the solution for the general case. I found, somewhat surprisingly, that the pole distance can be written in closed form as function of the sag and the rope length. This is surprising because there are hyperbolic functions involved with an unknown scale parameter both inside and outside. This normally leads to a transcendental equation that can only be solved iteratively. In fact, I’m pretty sure that for the inverse (but doubtless more practical) problem “Given a length of rope and the pole separation, what’s the sag?” that’s still the case.

    Just lik last week, a fun occasion to work through the math, detractors notwithstanding.

    • Steve Jones says:

      I too started down the line of the general solution (which is the route a maths education sends you down) and started researching the catenary curve. It also looked like it wasn’t amenable to a simple analytical approach, so I figured I’d missed something, as this is not what you expect from one of these puzzles. It didn’t take long to spot what I’d missed. I played around a bit with the iterative approach using Excel but that was about the limit if my exploration.

      Anyway, congrats on the analytical solution. I’m not sure that I’d have managed it.

  4. Roland says:

    That is assuming that the ground is flat. If there’s a hill in between, the flagpoles can be anything betwee 0 and 150 meter apart.

  5. Roland says:

    That is assuming that the ground is flat. If there’s a hill in between, the flagpoles can be anything between 0 and 150 meter apart.

  6. Edgar says:

    Well, first I tried with geometry, but I didn’t know how to do it with a parable (it’s not exactly a triangle.

    The I realised the rope is 150 feet long, which is 75+75. 100 feet minus 25 is 75, so the rope had to be straight down to the floor and up again. That meant the poles had to be right next to the other.

    That’s my explanation. No need for geometry.

  7. lilabyrd says:

    Dreamed the puzzle and answer two months ago…nice. one though…..hmmm Friday puzzle….yes very familaior…hmmm…enjoyed!

  8. Seeing as in the real world two objects cannot occupy the same space at the same time I deem this puzzle unsolvable.

    • M says:

      They are right next to each other. They do not occupy the same space

    • Steve Jones says:

      Well, mathematically speaking the tops of the poles would have to be in the same place as the distance between them has to be zero. However, mathematicians live in worlds populated in infinitely thin flagpoles and ropes so they don’t have trouble with such things. Just don’t ask them to put up a washing line.

    • Gus Snarp says:

      It’s OK, the whole thing is measured in feet. The poles can be 2 inches apart and still be 0 feet. It’s all about rounding.

    • AMWhy says:

      The thickness of the rope caters for the difference between the sticks. Mathematically, the rope and sticks are infinity thin. Realistically they both have width which allows force little room to manoeuvre.

  9. Jukka says:

    Fun puzzle, solved it. For extra giggles, and to brush up on calculus of variations, I re-derived the catenary curve as the shape of a free hanging rope.
    Interestingly however, that does not give a correct answer — in fact, the given parameters, ie. length, minimum height, and maximum height of the rope cannot be satisfied by a catenary.

    Cat. curve y(x) = hmin cosh(x / hmin), where hmin min. height.
    Length l = 2 hmin sinh(d / 2 hmin), where d pole separation.
    cosh^2 x – sinh^2 x = 1 => h^2 – (l/2)^2 = hmin^2, where h max. height (flagpole). This last eq. is not satisfied.

    I’ve already “wasted” too much of my time with this, so I’m asking others. Did I make a mistake, and if not why doesn’t the catenary work here, what physical factor besides grav. potential should we include to make minimization of the energy work? Prob. should include energy associated with bending the rope? OR perhaps there’s simply an implicit division by zero somewhere…?

    • Steve Jones says:

      I think there’s an implicit 0/0 when the drop length equals the arc length from my investigations of the catenary as the scale coefficient goes to 0.

    • Jukka says:

      Hi Steve. You may be right, that might be the case. However, I would’ve expected that the cat-curve should work when the pole separation is small but finite. It doesn’t, however.
      But, my mistake may be that I forgot that I can freely move the curve up and down, in other words, there’s one more free parameter that I have not considered.

    • Steve Jones says:

      I though you’d been talking about the actual puzzle values.

      Yes, the vertex of the catenary doesn’t go through the origin (apart from the case at the limit) unless you add another term to make it do so. That, in turn, depends on the scale coefficient of course.

    • Jukka says:

      OK, it works when the curve is shifted up. Do I feel stupid. I really, really should have realized to add the extra constant there, I somehow expected it to arise naturally from solving the variational problem. *sigh*. For reference, the scale constant a is given by

      a = (hmin – h)/2 + l^2 / [8 (h – hmin)],

      which becomes zero when you plug in Richards parameters.

    • Michael Sternberg says:

      Jukka,

      Congratulations on starting from first principles. I actually find it, on first thought, mildly surprising that there is only one parameter to the curve. Then again, the rope is assumed to have no bending penalty, so the only parameter is linear mass density, which combines with g to one scale factor.

      Don’t beat yourself up on the missing shift😉 Your ansatz can be considered correct, except that hmin is not the quantity from the sketch. I made the same mistake at first, and got non-sensical results.

    • Steve Jones says:

      I should add it must surely be possible to solve this problem as a minimum energy one. That is the rope deploys itself to the lowest potential energy distribution

  10. Mad Kev says:

    Unfortunately it is impossible for the rope to sag if the poles are next to each other. There’s misdirection, but also at the expense of being incorrect. A shame.

  11. Noel says:

    I built a scale model with my daughter. The answer was instant. This was one of the only puzzles I got right.

  12. Driffles says:

    It looks like a giant insects bottom.

  13. krar says:

    free hanging rope -> f(x)=cosh(x).
    minimum height of the rope at x=0 -> f(x)=cosh(x)+24
    (+24 because f(0)=cosh(0)=1)

    100 = cosh(x) + 24
    76 = cosh(x)
    x = arcosh(76) = ln (76 + sqrt(76²-1)) = 5,024.

    d = 2 * 5,024 = 10,05 ft.

    q.e.d.

    • Anonymous says:

      An insight into a lonely world.

    • Steve Jones says:

      I’m not sure I recognise that formula for a free hanging rope.

      The formula for the curve of a hanging rope is y=a*cosh(x/a) where a is a scale coefficient. Unfortunately, in this case a = 0 so the lowest point has an indeterminant divide by 0 in it.

      You’d also need the formula for the arc length to solve the general problem as that describes the rope length (or, rather, half the rope length).

      In any case, 10.05 ft is not possible as a horizontal distance, as there’s only enough rope to drop 75 feet and return to the same spot.

  14. Lazy T says:

    Good puzzle, solved when I stopped wondering how to do it and just considered the ‘facts’,

  15. Mickey D says:

    Was too easy! And that made me doubt myself…

  16. The other Matt says:

    I thought that the rope has to be elastic…

  17. astro says:

    ultimately, the puzzle is flawed because a real rope would never hang with a perfect 180 degree angle at the lowest point – it would be a teardrop shape.

    of course, one could always say that ends of the rope are being held at the tops of the towers by spherical cows…

  18. Anonymous says:

    Re. “Only possible answer, under the assumtion that the poles are vertical.” Whether they are vertical or not wouldn’t matter, the question said they were 100 foot high, not 100 foot long and stuck in the ground.

  19. Goliath says:

    How about I tilt the poles towards each other? They can be quite far away on the bottom, while still leaning towards each other on the top. One of the maths heads above can calculate the max distance. I can’t.

    • Steve Jones says:

      Whether the poles are tilted or not shouldn’t matter as they are reported as being 100 feet high, rather than long. You could have poles arbitrarily long if the tips just have to the 100 feet above the ground – that way the distance between the bases could be as lage as you like.

      However, if we take the example where the poles are 100 foot long, it’s easy enough to work out the maximum distance apart with poles tilted towards one another. That would be when the tops are 25 foot above the ground and the rope horizontal (we’ll miss out the little issue that the tension would have to be infinite to do this by imagining we have zero gravity). If we now imagine each pole makes a right angle triangle with the ground with the hypotenuse 100 foot long and one side 25 foot then then horizontal distance of the base (by Pythagoras) would be

      square root((100 x 100) – (25 x 25)) or about 96.8 feet. That would place the base of the poles at about 150 + (2 x 96.8) = 343.6 feet apart.

    • Goliath says:

      Thank you, kind Sir, I am immensely grateful😉

      Seriously, it didn’t occur to me that “long” and “high” have different meanings. Ah well, the fallacy of a foreign language…

  20. Ok, so am I the only one not seeing the official answer?

    What is the correct solution?

  21. Steve says:

    I too started to imagine a curve, but to simplify it, I realized the problem was identical to two poles 75ft high where the rope touches the ground. Once I realized again that the rope was only 150 ft. long, it was clear that the poles had to be touching.

  22. David D says:

    These are typical puzzles from IQ tests. If you didn’t get the answer in less than a minute then imagine the IQ score you will get in a real test! Harsh but true.

  23. Antoine Doumit says:

    the curve is half an ellipse .. the parameter of an ellipse = 2*pi Sqrt (0.5*(a2+b2)) which is equal in our case to 150 x 2 = 300 where a = (height of the tower to ground – the 25 feets sag) . i.e “a” = 100-25 = 75 feets so the ellipse perimeter equation is now left with one unknown which b and b is distance between the two towers the calculation is easy to find b
    2*pi Sqrt(0.5*(75^2+b^2) = 300 implies b =

  24. Marion says:

    Got this one after a minutes thought, I just started by looking at the numbers given and didn’t have to worry about different formulas- just as well really as I don’t know any formulas

  25. I hate to be “that guy”, but it strikes me that accuracy in these things is paramount or the whole thing falls apart. How does a rope sag “about” the ground? Read it. That’s what it says. So is there a cryptic clue in “about”, or is that a red herring designed to catch you out, like the diagram?

    Or it could be frustration at not taking the typo as a typo and spotting the obvious answer.

    And who am I kidding? I love to be “that guy”.

  26. Yeshanew says:

    It is 37.5 fts b/c if the rope is straight it is=150.And it will be 100 fts high from the ground.when the pole come closer to half distance,that’s 75 fts the height from the ground is also halved and would be 50fts.
    like that when 75fts is halved it is=37.5fts with a height of 25fts above the ground.

  27. Waheed says:

    i think it is 37.5ft apart

  28. Eleonore says:

    What a fantastic idea, I cannot wait around to receive my first letter!

  29. hisham says:

    5ln7.89

  30. hisham says:

    25ln7.89

  31. KASHIF says:

    Where is the OFFICIAL ANSWER????????????????/

  32. Yasser says:

    62 ft apart

  33. Somesh Thakur says:

    Another solution using geometry might be if we assume an imaginary circle with an arc of length 150ft that subtends an angle of x radians at the center.

    Clearly, the distance between the two flagpoles is given by:
    150/x* sine(x/2)

    Also, the height of either flagpole can be derived to be:
    (150/x) [1-cosine(x/2)] + 25 feet
    but height is 100ft, so equating the above to 100 yields:
    1-cosine(x/2)=x/2
    to which, x/2=0 is the only solution..

    Thus, x=0 means that the angle subtended by the chord joining the tops of the two towers is 0 radians and hence they are right next to each other.

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