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Posers and Puzzles

Posers and Puzzles

  1. Standard member uzless
    The So Fist
    15 Jul '08 17:15 / 2 edits
    Five balls of various sizes are placed in a conical funnel. Each ball is in contact with the adjacent ball(s). Also, each ball is in contact all around the funnel wall.


    The smallest ball has a radius of 8mm. The largest ball has a radius of 18mm. What is the radius of the middle ball?
  2. 15 Jul '08 17:31
    I initially suspect the answer to be either 12 cm or 13 cm, but I am pondering the relationship between adjacent balls, whether they be multiplicative or additive.

    I suspect it is multiplied (by 1.5^0.5, answer 12 cm) rather than a constant added (2.5, answer 13 cm).
  3. Subscriber AThousandYoung
    It's about respect
    15 Jul '08 17:39
    Originally posted by uzless
    Five balls of various sizes are placed in a conical funnel. Each ball is in contact with the adjacent ball(s). Also, each ball is in contact all around the funnel wall.


    The smallest ball has a radius of 8mm. The largest ball has a radius of 18mm. What is the radius of the middle ball?
    The top of the smallest ball is 4mm higher than it's center. The next ball's center is it's own radius higher than that.

    Hmm. This might be harder than it looks.
  4. Standard member PBE6
    Bananarama
    15 Jul '08 17:40
    Originally posted by uzless
    Five balls of various sizes are placed in a conical funnel. Each ball is in contact with the adjacent ball(s). Also, each ball is in contact all around the funnel wall.


    The smallest ball has a radius of 8mm. The largest ball has a radius of 18mm. What is the radius of the middle ball?
    12mm.
  5. Standard member uzless
    The So Fist
    15 Jul '08 17:59
    Originally posted by PBE6
    12mm.
    You have to show your work grasshopper....now, PAINT THE FENCE!
  6. Standard member PBE6
    Bananarama
    15 Jul '08 18:19
    Originally posted by uzless
    You have to show your work grasshopper....now, PAINT THE FENCE!
    Allow me to wax-on, poetic...

    Consider any two adjacent spheres in the cone. The small sphere has a radius "r" and the larger sphere has a radius "R". The ratio between the two radii is then simply R/r. Each pair of adjacent spheres exhibits the same ratio between radii since each pair must meet the same stipulations given in the question (touch each other, touch the sides of the cone). Therefore, to find the radius of the next sphere in line, we multiply the radius of the preceeding sphere by R/r. There are 5 spheres, so we have to scale up by this ratio 4 times to get the radius of the largest sphere, giving us the equation:

    18 = 8 * (R/r)^4

    Therefore:

    (R/r)^4 = 18/8
    (R/r) = (18/8)^(1/4) ~= 1.224744871

    The middle sphere is scaled up twice from the original sphere, so its radius is 8 * (R/r)^2 = 12.

    Alternatively, one could argue that since the radii form an ascending geometric sequence, the middle sphere must have a radius equal to the geometric mean of the two outside spheres:

    R(mid) = SQRT(8*18) = SQRT(144) = 12.
  7. Standard member uzless
    The So Fist
    15 Jul '08 18:24
    Originally posted by PBE6
    Allow me to wax-on, poetic...

    Consider any two adjacent spheres in the cone. The small sphere has a radius "r" and the larger sphere has a radius "R". The ratio between the two radii is then simply R/r. Each pair of adjacent spheres exhibits the same ratio between radii since each pair must meet the same stipulations given in the question (touch each othe ...[text shortened]... l to the geometric mean of the two outside spheres:

    R(mid) = SQRT(8*18) = SQRT(144) = 12.
    Congratulations Danielson.
  8. Standard member Palynka
    Upward Spiral
    16 Jul '08 16:21
    Originally posted by PBE6
    Alternatively, one could argue that since the radii form an ascending geometric sequence, the middle sphere must have a radius equal to the geometric mean of the two outside spheres:

    R(mid) = SQRT(8*18) = SQRT(144) = 12.
    Nice and elegant.