20070912, 13:40  #23 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 

20070912, 13:55  #24  
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
Quote:
posts referring to the puzzle of this thread. I not only unravelled your waffle, but offered a critique and filled in the gaps. This is where you should direct your response (preferably "considered" although this is usually hoping for too much). 

20070912, 16:35  #25  
"Lucan"
Dec 2006
England
194A_{16} Posts 
Quote:
Chords of this circle (diameter 6 inches) are also tangents to the cylinder and hence govern the area of the rings by Mally's "theorem". This is all independent of R. However, this method a) is more convoluted than the direct application of Pythagoras I outlined. b) differs spectacularly from Mally's "explanation". David Last fiddled with by davieddy on 20070912 at 16:38 

20070913, 07:37  #26  
"Lucan"
Dec 2006
England
14512_{8} Posts 
Critique
Quote:
Quote:
erroneously assuming that the diameter of the outer circle remains constant. What we need to show is that the "Tangent" remains of constant length, and hence the area between the circles remains constant. That is what I showed by considering the circular intersection of a plane tangential to the cylinder with the sphere. This circle has a constant diameter of 6 inches and its chords perpendicular to the cylinder are your "Tangents". David Last fiddled with by davieddy on 20070913 at 08:07 

20070913, 09:02  #27 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Completing the picture
A circle with my chord (yourTangent) as diameter has the area of your
"track". Constructing the whole stack of such circles results in a sphere of diameter 6 inches whose volume is that of the remainder. I hope I have retrieved the merit in using your "theorem" as an approach to the problem. David Last fiddled with by davieddy on 20070913 at 09:16 
20070913, 12:52  #28  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Construction
Quote:
Here is a construction which I hope you will acknowledge as a neat way of validating your approach to the problem, which indeed was not mentioned in Akruppa's previous thread. Consider the circle (diameter 6 inches) formed by the intersection of a plane tangential to the cylinder with the spherical surface of the remainder. Construct the sphere S for which this circle is a great circle. Now take a cross section perpendicular to the axis of the cylinder. We have 3 circles. Two are concentric, the outer (O) and inner (I) boundaries of the remainder. The other (C) is from sphere S. Observe that the line joining the intersections of circles O and C is a diameter of C and tangential to circle I. It follows (by Pythagoras or your "theorem") that the area between O and I (the cross section of the remainder) is equal to the area of C (the cross section of sphere S). Since this applies to all cross sections, the volume of the remainder is equal to that of sphere S namely 36pi in^2. Q.E.D. David Last fiddled with by davieddy on 20070913 at 13:11 

20070914, 14:50  #29 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
2 dimensional analogue.
David, I am having a bout of malaria so I cannot continue our dialogue. How ever I can say that the Area problem is only a two dimensional analogue to the Volume problem and it just gives one an idea. It is not meant to solve the Residue problem and they are in no way connected apart from both giving constants and so are independent of the size of the sphere and the radius of the cylinder. The only connection I see is that the volume of the residue will be 4 times the the area of the annulus in the limit. (36Pi = 4 x 9Pi) always for the same circle/sphere. I would have given Martin Gardner's solution but I find some glaring mistakes in it. For instance he gives the volume of sphere as 4pi R^3  3 and similarly for the volume of the cap. I have worked out the volume of the cap by calculus and get pi(h^2)(3R h)/3 where R is the radius of the sphere and h is the height of the cap. Knowing this its a matter of substituting and you will find that all the terms cancel out and you are left with the constant 36Pi. Mally 
20070914, 19:20  #30  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Sorry to hear that. Hope you recover soon.
Quote:
your points about the puzzle. Quote:
Using your two dimensional "theorem", I have shown that areas of cross section of the residue and my sphere are always the same as each other, so they have equal volumes. Quote:
It just happened that r=3 in this case. Quote:
David Last fiddled with by davieddy on 20070914 at 20:05 

20070915, 07:41  #31 
"Lucan"
Dec 2006
England
6474_{10} Posts 
Analogue
Let's make the analogy more explicit by comparing these two puzzles:
2D: A chord of length L is drawn on a circular disc. A concentric circular hole tangential to this chord is drilled. Show that the remaining area is that of a circle diameter L. 3D: A plane intersects a spherical ball in a disc of diameter D. A concentric cylindrical hole tangential to this plane is drilled. Show that the remaining volume is that of a sphere diameter D. Now you can see that the 2D result can be used in the derivation of the 3D result (as Mally attempted to do and I succeeded in doing) David PS It shows the importance of asking the right questions Last fiddled with by davieddy on 20070915 at 07:59 
20070915, 08:31  #32  
Bronze Medalist
Jan 2004
Mumbai,India
804_{16} Posts 
3D analogy.
Quote:
With my nose running and alternate bouts of fever and chills and a terrible, constant head ache I have an escape on my p.c. When I mentioned about Archimedes I am greatly impressed by his two observations. 1) The area of a sphere is equal to the area of the circumscribing right cylinder. Thus if two planes at a distance h between them cut both the cylinder and sphere at right angles to the axis of the cylinder the areas of each 'strip' are equal. This applies to the spherical caps of height h also 2) The volume of a sphere inscribed in a cylinder is two thirds of the volume of The cylinder. These Greeks were great! As Euclid told Ptolemy's son when he asked if there is an easier to geometry, besides learning all the propositions Euclid replied "There is no royal road to Geometry" and sent the prince back to his books. Mally 

20070915, 12:55  #33  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
I've only just read this because your projector problem overwrote it. I wish to repeat my condolences on your indisposition, and apolgize for my riposte on behalf of Retina in the "projector" thread (re malaria). David 

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