Hole and sphere.

[davieddy]

The similar result which is most worth remembering
is that the surface area of a sphere radius R between two parallel
planes a distance h apart is 2*PI*R*h.

[Orgasmic Troll]

Originally Posted by mfgoode

Thank you Wacky! I stand corrected and an astute observation on your part.
I welcome this type of correction instead of posters taking digs at presentation or mis spelling or such trivia which breeds unpleasantness.

Mally

we welcome well-formed, intelligently stated puzzles.

[Orgasmic Troll]

Originally Posted by mfgoode

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!

I'm astounded at how much condescension you can cram into your sentences. Perhaps the equally astounding amount of hyperbole infuses your statements with a hyperbolic metric, allowing that much more condescension to fit.

This doesn't even qualify as a lemma. It's patently obvious, and your showboating and windbaggery border on offensive.

[wblipp]

Originally Posted by Orgasmic Troll

It's patently obvious,

We are talking about two concentric circles, and a chord of the larger circle that is tangent to the smaller circle. Is it really "patently obvious" that the area of the annulus is the same as the area of a circle with the chord as diameter?

It's a simple consequence of the Pythagorean Theorem, but I didn't see it until I added the triangle to the sketch. Is there some even simpler method that I missed, or does this qualify as patently obvious in your lexicon?

[davieddy]

Obvious or not, I stand by my original post in which I
pointed out that this lemma didn't really help solve the
problem:

Originally Posted by davieddy

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

[davieddy]

Originally Posted by mfgoode

Hint: Dont go into calculus and the rest. There is also a purely logical solution too! Without that, one must at least solve it by pure geometry though you must arrive at the volume of the caps which requires a formula not too well known.

Anyone reading this might be forgiven for thinking
that Newton/Leibnitz invented calculus to make finding
volumes difficult

David

PS I don't even count quoting (4/3)piR^3 as "avoiding calculus"
(Unless you have a "purely logical" way of deriving it.)
Mind you I suppose Archimedes knew this formula.

[Orgasmic Troll]

Originally Posted by wblipp

We are talking about two concentric circles, and a chord of the larger circle that is tangent to the smaller circle. Is it really "patently obvious" that the area of the annulus is the same as the area of a circle with the chord as diameter?

It's a simple consequence of the Pythagorean Theorem, but I didn't see it until I added the triangle to the sketch. Is there some even simpler method that I missed, or does this qualify as patently obvious in your lexicon?

Dang, I must be sipping Mally's coffee. Yeah, disregard what I said, I'm reading things way too quickly these days.

Granted, the windbaggery and showboating still holds, since that's a result that I've known about since I was like, 9 or 10.

[davieddy]

Originally Posted by akruppa

We had this puzzle some time ago: http://www.mersenneforum.org/showthread.php?t=5460

Alex

And amusingly it seemed to catch Drew completely off guard

[davieddy]

Originally Posted by mfgoode

... one must at least solve it by pure geometry though you must arrive at the volume of the caps which requires a formula not too well known.

One could argue that the formula pi*H³/6 for the volume of
the remainder from a hole of height H is more memorable, and use this
to deduce the volume of the caps.

[mfgoode]

Originally Posted by davieddy

The similar result which is most worth remembering
is that the surface area of a sphere radius R between two parallel
planes a distance h apart is 2*PI*R*h.

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!

Your ‘brilliant’ remark confuses and mystifies one more than it enlightens.

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.

I had to study mathematical history that far back as some of our Indian mathematicians predate even Pythagoras. I don’t want to go into priority of knowledge, and who taught who, as this is a very controversial subject. Recent Excavations reveal that Budhayan used the Pythag. Theorem as far back as 1000 B.C. Evidently he knew it too!

You have isolated a result stemming from two of the most profound theorems enunciated by Archimedes which can be represented by one diagram.

Archimedes himself considered it his greatest achievement, and wished that a replica be put on his tomb stone! Unfortunately he was unceremoniously hacked to death by a Roman soldier and his grave lies in obscurity.

Without even a passing mention of a sphere inscribed in a cylinder and the relationship of area and volume between the two your ‘brilliant’ observation is open to question and proof.

I can imagine the newbie’s scrambling for their calculus text books when there’s no need too. Archimedes demonstrated this very clearly and without ambiguity 300 years before Christ. Finding the area of a spherical cap is as easy as pie if you know Archie’s theorems. And I will stress that one does not need Calculus at all!

Re-reading your subsequent posts there is every evidence that you are not at all aware of this theorem. I sincerely hope that I am wrong. I can face ridicule, but I doubt if your ego will take it. What a great pity! Now this is when the great Silverman should
Step in and call you all the names ‘wit’ ‘twit’ and my own to him ‘twat’

Bernard Shaw made a very wise observation when he quipped “Those who can, DO, Those who can’t, TEACH.” How very true!

In my experience of flying British school children specials on their holidays to and from London, most of them can only go as far back in mathematical History as Newton.
The simple reason I suppose is that the ‘over qualified’ teachers don’t study
Mathematical history themselves! Need I say more ?

Mally
.

[R.D. Silverman]

Originally Posted by Orgasmic Troll

I'm astounded at how much condescension you can cram into your sentences. Perhaps the equally astounding amount of hyperbole infuses your statements with a hyperbolic metric, allowing that much more condescension to fit.

This doesn't even qualify as a lemma. It's patently obvious, and your showboating and windbaggery border on offensive.

And others criticize me for insulting him.