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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    03 Oct '11 08:05
    For this I have found a numerical answer only, accurate but not exact. Hopefully someone can find a more elegant solution?

    Start with a piece of paper in the shape of a circle. Cut it into sectors, and bend each of those so it becomes a cone. How should those cones be cut, if the idea is to make the difference in their volumes as big as possible?
  2. 08 Oct '11 04:39
    Assuming a sector is cut from the circle (like a piece of pizza), and both remaining pieces are folded into cones, I get the following solution, albeit through trial and error. I estimate the circumference (or radius) of the larger cone is 85.2% of the circumference (or radius) of the original circle. Is that what you get?

    I presume we would need some sort of limit / derivative function to prove this?
  3. Standard member talzamir
    Art, not a Toil
    08 Oct '11 07:37
    Yes, that is what I meant.. points A and B from the circumference of a circle with center O, the area cut into two sectors by the line AOB means the same but I prefer "like a pizza" as it tells exactly what it is about. Our answers are the same too. 85.2% of the angle or circumference on one side, the rest on the other.

    Some form of f'(x) = 0 would sound like the way to go. But getting the exact solution from that has proven difficult.
  4. 08 Oct '11 10:54 / 1 edit
    Say the paper circle has circumference C
    It is cut into two pieces, the smaller piece has circumference xC, and the larger has circumference (1-x)C

    side length of both cones = Radius of paper circle = C/(2*pi())

    Radius of smaller cone = xC/(2*pi)
    Radius of large cone = (1-x)C/(2pi)

    cone_height = sqrt(Side^2 - Radius^2)

    So height of smaller cone = sqrt((1-x^2)C^2/(4*pi^2))
    height of large cone = sqrt((2x - x^2)C^2/(4*pi^2))

    Volume of a cone = 1/3 * pi * r^2 * h

    volume of small cone = 1/3 * pi * ( xC/(2*pi) )^2 * sqrt((1-x^2)C^2/(4*pi^2))
    volume of large cone = 1/3 * pi * ( (1-x)C/(2pi) ) ^ 2 * sqrt((2x - x^2)C^2/(4*pi^2))

    we are trying to maximise large volume over smaller volume

    dividing one by the other and cancelling out common terms:

    large_vol_over_small_vol = ((1-x)^2 * sqrt(2x - x^2)) / (x^2 * sqrt( 1 - x^2))

    hmm, looks simplish, to get rid of the square roots, maybe we need to square it all and maximise the ratio of the squares of the volume, then differentiate a quotient?
  5. 08 Oct '11 19:42 / 4 edits
    Originally posted by iamatiger
    Say the paper circle has circumference C
    It is cut into two pieces, the smaller piece has circumference xC, and the larger has circumference (1-x)C

    side length of both cones = Radius of paper circle = C/(2*pi())

    Radius of smaller cone = xC/(2*pi)
    Radius of large cone = (1-x)C/(2pi)

    cone_height = sqrt(Side^2 - Radius^2)

    So height of smaller c ...[text shortened]... quare it all and maximise the ratio of the squares of the volume, then differentiate a quotient?
    Hmm, that doesn't seem to work, has no maxima. Ah, I see, we need the difference in the volumes, not the ratio.

    volume of small cone = 1/3 * pi * ( xC/(2*pi) )^2 * sqrt((1-x^2)C^2/(4*pi^2))
    volume of large cone = 1/3 * pi * ( (1-x)C/(2pi) ) ^ 2 * sqrt((2x - x^2)C^2/(4*pi^2))

    large - small = K ((1-x)^2*sqrt((2x-x^2) - x^2*sqrt(1-x^2))

    so we need to find the x value, between 0 and 1, that gives the maximum value of
    (1-x)^2*sqrt(2x-x^2) - x^2*sqrt(1-x^2)

    Pasting that into wolfram:
    http://www.wolframalpha.com/input/?i=max%28+%28%281-x%29^2*sqrt%282x-x^2%29+-+x^2*sqrt%281-x^2%29%29%29

    We get x = 0.147977, which agrees with you to, now we just need to differentiate it...

    hmm, typing:
    differentiate (1-x)^2*sqrt(2x-x^2) - x^2*sqrt(1-x^2)

    into wolfram, and clicking on show steps seems to imply this is not a sum I really want to do by hand that much.
  6. Standard member talzamir
    Art, not a Toil
    08 Oct '11 20:11
    A numerical approach gives a very nice s-shaped curve, showing that the min and max points are there for vol1 - vol2 and where they approximately are.. as accurately as we want, but they are still approximations. Finding the exact values for x for which f'(x) = 0 has proven fairly challenging.
  7. 10 Oct '11 00:11
    Wolfram appears to give an exact answer, I can see how it finds the differential, although it is extremely long winded, but I can't work out how it then finds the points where that differential are zero, as it is such a complex function.