mersenneforum.org  

Go Back   mersenneforum.org > Fun Stuff > Puzzles

Reply
 
Thread Tools
Old 2007-09-12, 13:40   #23
Orgasmic Troll
Cranksta Rap Ayatollah
 
Orgasmic Troll's Avatar
 
Jul 2003

641 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
And others criticize me for insulting him.
I gave everybody else on the board a $50 bill first
Orgasmic Troll is offline   Reply With Quote
Old 2007-09-12, 13:55   #24
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

194A16 Posts
Default

Quote:
Originally Posted by mfgoode View Post


Well! Well! Davie old boy!

You state you can’t digest my long posts (attention deficiency?) so I’ll oblige and make it short but not necessarily sweet!

....


Re-reading your subsequent posts there is every evidence that you are not at all aware of this theorem. .
There is every evidence that you have NOT digested my other
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).
davieddy is offline   Reply With Quote
Old 2007-09-12, 16:35   #25
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Quote:
Originally Posted by davieddy View Post
By additionally considering the intersection of
a plane tangential to the cylinder with the sphere (a circle) I can deduce
from Mally's theorem that the largest area of Mally's cross section is
that of this circle (9PI) whatever the diameter of the sphere.
However it is far from obvious that the area of other rings stacked in
the 6 inch height are similarly independent of the sphere's diameter.

Am I missing something?

David
Yes I was missing a trick!
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 2007-09-12 at 16:38
davieddy is offline   Reply With Quote
Old 2007-09-13, 07:37   #26
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default Critique

Quote:
Originally Posted by mfgoode View Post
Here is a more rigorous solution by an analogy in two dimensions. Take a cross section of the 'completed' residue of the sphere. You will get two concentric circles. Let us call it a circular track. Pl. draw a figure for a better view.
Now given the longest straight line that can be drawn on the circular track of ANY dimensions. This is a tangent to the smaller circle.
The area of the track will be equal to the area of a circle having the straight line as a diameter. No assumptions here as it can be proved! Convince your self if you like by proving it. It should be called a theorem!
So far so good.

Quote:

Now reduce the dia. of the central smaller circle. The straight line (Tangent) will become longer.
As you showed later, and Wacky pointed out, you were
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 2007-09-13 at 08:07
davieddy is offline   Reply With Quote
Old 2007-09-13, 09:02   #27
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default 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 2007-09-13 at 09:16
davieddy is offline   Reply With Quote
Old 2007-09-13, 12:52   #28
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

647410 Posts
Default Construction

Quote:
Originally Posted by mfgoode View Post

You have done a grave injustice to the greatest mathematician Archimedes, The Father of Calculus! Yes! long before Newton and co-discoverer Leibniz and Gauss was a Greek called Archimedes.

.
See my post #17. (Your whole rant was completely misdirected).

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 2007-09-13 at 13:11
davieddy is offline   Reply With Quote
Old 2007-09-14, 14:50   #29
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

22·33·19 Posts
Arrow 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
mfgoode is offline   Reply With Quote
Old 2007-09-14, 19:20   #30
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

Quote:
Originally Posted by mfgoode View Post

David, I am having a bout of malaria
Sorry to hear that. Hope you recover soon.
Quote:
so I cannot continue our dialogue.
Some aspects of it are best dropped, but I shall reply to
your points about the puzzle.
Quote:
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 analogy is only there if the length of the tangent is a constant.
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:
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.
No. The volume is 4r/3 times the area of the great circle.
It just happened that r=3 in this case.
Quote:
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.
We are agreed that this is the clumsiest approach.

David

Last fiddled with by davieddy on 2007-09-14 at 20:05
davieddy is offline   Reply With Quote
Old 2007-09-15, 07:41   #31
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default 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 2007-09-15 at 07:59
davieddy is offline   Reply With Quote
Old 2007-09-15, 08:31   #32
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

205210 Posts
Thumbs up 3D analogy.

Quote:
Originally Posted by davieddy View Post
Let's make the analogy more explicit by comparing these two puzzles:

~ ~

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
Thank you David for your in depth analysis. Your 3D analogy is completely original. Its the kind of thing that keeps me going.

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
mfgoode is offline   Reply With Quote
Old 2007-09-15, 12:55   #33
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

Quote:
Originally Posted by mfgoode View Post
Thank you David for your in depth analysis. Your 3D analogy is completely original. Its the kind of thing that keeps me going.

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.
Thankyou Mally.
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
davieddy is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
4D-Volume of a 4D-Sphere davar55 Puzzles 16 2007-07-05 22:16
Triangles on a sphere davieddy Puzzles 15 2007-04-06 20:16
A proof with a hole in it? mfgoode Puzzles 9 2006-09-27 16:37
Volume of a Sphere Patrick123 Puzzles 19 2006-02-20 15:00
Cylinder in sphere Greenbank Puzzles 17 2006-01-26 17:18

All times are UTC. The time now is 16:09.


Tue Jul 27 16:09:49 UTC 2021 up 4 days, 10:38, 0 users, load averages: 3.65, 3.39, 3.24

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.