As the sphere diameter approaches the height of the cylinder the radius of
the cylinder tends towards zero. So the volume tends to that of a sphere
with radius 3. I did it algabreicly by subtracting the volume of the
cylinder and the two pole caps. It works out as 36 and is invariant with
the sphere radius.

On Mon, 7 Mar 2022, 13:18 Raymond West, raywest@raywest.com wrote:

Yes, you misread my explanation.  The REAMINING volume is that of a sphere
of the same diameter as the height of the cylinder (6").

The drilled sphere is of undetermined diameter as it is dependent upon the
diameter of the cylinder (using a lot of trig and geometry...and maybe some
calc).

On Mon, Mar 7, 2022 at 7:18 AM Raymond West raywest@raywest.com wrote:

This is from "Aha! insight" by Martin Gardner. The trick is to see
that IF the problem has a solution, THEN it must be "the volume of the
sphere as the diameter of the cylinder tends to zero" per Nop Head's
argument.

When you're good enough with sin's and cos's you can do the math and
find that indeed, independent of the starting radius of the sphere,
the volume always comes out to the same number.

The other one in the book is: (The story is about a tower and a
painter getting stuck with his ladder).

Two concentric circles have an area between them (so area larger -
area smaller). Find this area. Given: ghe longest straight line in the
sought area is 10 units. (this line begins and ends on the outer
circle and touches the inner one).

The Aha! insight is that given that this has a solution, it MUST be
that the radius of the inner circle doesn't matter, and you can tend
that to zero, leaving an outer cicle with a 10 unit diameter.

Roger. 

On Mon, Mar 07, 2022 at 01:37:10PM +0000, nop head wrote:

--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110 **
**    Delftechpark 11 2628 XJ  Delft, The Netherlands.  KVK: 27239233    **
f equals m times a. When your f is steady, and your m is going down
your a is going up.  -- Chris Hadfield about flying up the space shuttle.

There is a valuable clue in Larry's statement of the problem:

And yes, the question has all the information required.

If that is true then the answer is independent of the sphere diameter.
Trusting his formulation of the puzzle the answer is trivial: the volume of
the sphere with a diameter equal to 6. Not trusting it requires the
additional math nophead mentioned.

Is this what Larry called "a logical way to solve it"?

Em seg., 7 de mar. de 2022 às 13:38, nop head nop.head@gmail.com escreveu:

Yes, by my way of understanding.

The "logical" solution is actually the clue to the setup for the rigorous
solution.  Knowing that the "answer" is revealed by pushing a parameter
toward a "limit" suggests the framework for the calc required.  This is
true also for the concentric circle solution.

On Mon, Mar 7, 2022 at 8:40 AM Ronaldo Persiano rcmpersiano@gmail.com
wrote:

Agreed, I misread your statement. I was part way through calculating it
at the time, which got a bit tedious.

On 07/03/2022 13:40, FF Systems wrote:

On Mon, 2022-03-07 at 14:39 +0000, Ronaldo Persiano wrote:

Indeed it is.

I used to devour Martin Gardner's /Mathematical Games/ articles, as
well as the /Amateur Scientist/ column every month from about the time
I was 12 or 13.

This puzzle was my favourite, and I got it (finally) without using
math.

My favourite /Amateur Scientist/ column was about the Hilsch tube.