Guest Posted November 8, 2009 Report Share Posted November 8, 2009 (edited) An infinite amount of spheres are placed in a cone with a half angle x (0<x<90) as in the diagram below. Find the ratio: (total volume of spheres) / (volume of cone) I hope this one last more than 30 minutes this time. Edited November 8, 2009 by psychic_mind Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 9, 2009 Report Share Posted November 9, 2009 Well since I made this puzzle up my self I had to work out the solution myself beforehand. That does mean that my answer may be incorrect but I have done a different method to everyone else and my answer agrees with Tuckleton. This method uses geometric series. Let Y = cosec(x) RY = L+r+R and rY = L RY = rY + r + R R(Y-1) = r(Y+1) Ratio between sphere radii = r/R = (Y-1)/(Y+1) The height, h, of the cone is the infinite sum of diameters h = 2R/(1-(Y-1)/(Y+1) ) h = R(Y+1) The ratio between the volume of each sphere is (Y-1)3/(Y+1)3 The total volume taken by the spheres, Vs, is again an infinite sum Vs = (4/3*pi*R3)/(1-(Y-1)3/(Y+1)3) Vs = (2/3*pi*R3)(Y+1)3/(3Y2 + 1) The radius of the cube is a. a = h*tan(x) The volume of the cone, Vc, is a2*h*pi*1/3 Vc = h3*tan2(x)*pi*1/3 Vc = R3(Y+1)3*tan2(x)*pi*1/3 Vs/Vc = [ (2/3*pi*R3)(Y+1)3/(3Y2 + 1) ] / [ R3(Y+1)3*tan2(x)pi*1/3 ] Vs/Vc = 2 / tan2(x)(3Y2 + 1) Vs/Vc = 2 / tan2(x)(3csc2(x) + 1) Vs/Vc = 2 / (3sec2(x) + tan2(x)) Vs/Vc = 2 / (4sec2x - 1) I have checked Tuckleton, your answer does simplify to this. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 9, 2009 Report Share Posted November 9, 2009 My solution is Vs/Vc = Pi/4(Cos(x)+Sin(x)Tan(x)) This may be the same as other solutions. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 9, 2009 Report Share Posted November 9, 2009 I do not understand why it would not be as following [x[piR2 / KX ]/2[0.5bh] Where X is the infinite number of circles. R/KX is the radius divided by the varying constant. BH is just base times. hight area of triangle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 11, 2009 Report Share Posted November 11, 2009 Well since I made this puzzle up my self I had to work out the solution myself beforehand. That does mean that my answer may be incorrect but I have done a different method to everyone else and my answer agrees with Tuckleton. This method uses geometric series. Let Y = cosec(x) RY = L+r+R and rY = L RY = rY + r + R R(Y-1) = r(Y+1) Ratio between sphere radii = r/R = (Y-1)/(Y+1) The height, h, of the cone is the infinite sum of diameters h = 2R/(1-(Y-1)/(Y+1) ) h = R(Y+1) The ratio between the volume of each sphere is (Y-1)3/(Y+1)3 The total volume taken by the spheres, Vs, is again an infinite sum Vs = (4/3*pi*R3)/(1-(Y-1)3/(Y+1)3) Vs = (2/3*pi*R3)(Y+1)3/(3Y2 + 1) The radius of the cube is a. a = h*tan(x) The volume of the cone, Vc, is a2*h*pi*1/3 Vc = h3*tan2(x)*pi*1/3 Vc = R3(Y+1)3*tan2(x)*pi*1/3 Vs/Vc = [ (2/3*pi*R3)(Y+1)3/(3Y2 + 1) ] / [ R3(Y+1)3*tan2(x)pi*1/3 ] Vs/Vc = 2 / tan2(x)(3Y2 + 1) Vs/Vc = 2 / tan2(x)(3csc2(x) + 1) Vs/Vc = 2 / (3sec2(x) + tan2(x)) Vs/Vc = 2 / (4sec2x - 1) I have checked Tuckleton, your answer does simplify to this. This doesn't make sense, because if you approach x=90 (the cone becomes a cylinder), your answer approaches 0 which geometrically does not make sense. As I said before, the solution should approach 2/3. I stand by my answer above. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 11, 2009 Report Share Posted November 11, 2009 This doesn't make sense, because if you approach x=90 (the cone becomes a cylinder), your answer approaches 0 which geometrically does not make sense. As I said before, the solution should approach 2/3. I stand by my answer above. When x -> 90 the cylinder resembles a flat plane. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 11, 2009 Report Share Posted November 11, 2009 (edited) The angle is the angle at the apex. Sorry if I didn't make that clear. If you used the other angle you just need to substitute it for 90-x. Edited November 11, 2009 by psychic_mind Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 11, 2009 Report Share Posted November 11, 2009 (edited) Forcedhand's "Method Only" in #4 is indubitably correct. I had already spotted this simplification independently anyway. A valid solution involves only two tangents, that of x and that of (45ยบ - x/2), and no other trigonometric quantities. To complete the calculation one needs to know the volume of a sphere as well as that of a truncated cone. The latter formula is 1/3 * pi * height * (r12 + r1*r2 + r22). Here r1 is the radius of the lower circle, while r2 is the radius of the upper circle. When the ancient Egyptians were building their pyramids, they seem to have known a formula for the volume of a truncated pyramid, but nowhere do they seem to have spelt out the volume of a completed one. They must have known it, but sadly there is no evidence (not even in the Rhind papyrus.) Edited November 11, 2009 by jerbil Quote Link to comment Share on other sites More sharing options...
Question
Guest
An infinite amount of spheres are placed in a cone with a half angle x (0<x<90) as in the diagram below. Find the ratio:
(total volume of spheres) / (volume of cone)
I hope this one last more than 30 minutes this time.
Edited by psychic_mindLink to comment
Share on other sites
32 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.