Posted by Denis Borris on June 18, 2002 at 03:37:32:
In Reply to: Better formula posted by Joel on June 17, 2002 at 23:21:45:
: : : The number of bearings in the pyramid stack is N*(N+1)*(2N+2)/6, where N is the number of layers.
: : : The height is 1+(N-1)/sqrt(2).
: : : The number of bearing in the other (tetrahedron) stack is M*(M+1)*(M+2)/6. The height is
: : : 1+(N-1)*sqrt(2/3).
: : : If the pyramid has 34 layers (13,685 bearings), it is 23.83 high. You can make a tetrahedron of 42
: : : layers with these which is 33.98 high and have 441 = 21^2 bearings left over.
: : You're right - my height for the pyramid is wrong; it should be 1 + (n-1)/sqrt(2) as you said, but whereja get that formula for the number of bearings in the pyramid? It gives 1.33, 6, 16, 33.33 as the number of bearings for stacks 1, 2, 3 and 4 high. Should be 1, 5, 14, 30, shouldn't it?
: : Next problem: as the problem was posed, there must be "just enough" bearings to complete the second arrangement, whatever it is, with one ball on top. I think that states pretty explicitly that there shouldn't be any extra bearings left over. Let's work on it some more.
: Bearings in pyramid: n*(2n+1)*(n+1)/6