# 5 balls

uzless
Posers and Puzzles 15 Jul '08 17:15
1. uzless
The So Fist
15 Jul '08 17:152 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. AThousandYoung
All My Soldiers...
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. 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. 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. 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. 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. 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.