Watch that drop
Four identical spheres, each with diameter 20cm, are placed in a cylinder. The cylinder gives room for three of them to fit tightly at the bottom of the cylinder. The fourth sphere is placed on top of the other three. How much liquid is needed to cover the four spheres?
Problem source: Mattenøtt, NRK.
November 24th, 2009 at 1:03 am
Let the bottom 3 spheres have centres A, B, C, and the top sphere have centre D. Let the centre of triangle ABC be O; let the sphere centre A touch the wall of the cylinder at R. Then the cylinder’s radius is:
OR = AR + OA
= (1 + sec 30°)(10 cm)
= (1 + 2/sqrt(3))(10 cm).
and its cross-sectional area is
pi(1 + 2/sqrt(3))²(100 cm²)
= pi(1 + 4/sqrt(3) + 4/3)(100 cm²)
= pi(7 + 4 sqrt(3))(100 cm²)/3
Now, how high must we fill the cylinder? 10cm to cover the bottom halves of the bottom 3 spheres and reach the level of the OABC plane, then OD, the height of tetrahedron ABCD, then 10cm to cover the top half of the top sphere.
What’s the height of a regular tetrahedron? I didn’t know but I thought of it this way. Set up Cartesian coords so that D is at (0,0,0) and A, B, C are at the cyclic perms of {0, 1, 1}. Then O is (2/3, 2/3, 2/3), so the height OD is (2/3)(sqrt(3)) = 2/sqrt(3). Each side of the tetrahedron has length sqrt(2), and the height is 2/sqrt(3). So if the edge length is 1, the height is (2/sqrt(3))/sqrt(2) = sqrt(2/3).
Thus the height to fill the cylinder is
(1 + sqrt(2/3) + 1)(10 cm)
= (2 + sqrt(2/3))(10 cm)
and the cylinder occupied by spheres and surrounding liquid has volume
= pi(7 + 4 sqrt(3))(2 + sqrt(2/3))(1000 cm^3)/3
= pi(14 + 7 sqrt(2/3) + 8 sqrt(3) + 4 sqrt(2))(1000 cm³)/3
The volume of a sphere of radius r is
(4/3)pi.r³
so the volume of the spheres is
4 pi (1000 cm³)
= 12 pi (1000 cm³)/3
and the volume of the liquid is
= pi(2 + 7 sqrt(2/3) + 8 sqrt(3) + 4 sqrt(2))(1000 cm³)/3
November 24th, 2009 at 1:05 am
Correction. The volume of the spheres is
16 pi (1000 cm³)/3
and the volume of the liquid is
= pi(7 sqrt(2/3) + 8 sqrt(3) + 4 sqrt(2) – 2)(1000 cm³)/3