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"Mike the Elder" <zer### [at] wyan org> wrote:
> I've taken the liberty of posting this problem (I dare not call it a puzzle lest
> I face the wrath of the Puzzle Definition Police) out of "Off-Topic" because
> POV-Ray was used to produce the illustration. The problem itself is presented
> in the image. If anyone wants to refesh his or he memory with respect to basic
> geometry formulas, it's OK to look here:
>
> http://www.math10.com/en/geometry/volume.html
>
> I would not consider this "cheating" as this is not intended as a memory quiz.
>
>
> The solution to the problem as posed is here:
> LOOK HERE ONLY IF YOU WISH TO SEE THE SOLUTION:
>
http://a833.ac-images.myspacecdn.com/images01/116/l_c46789151c59aa7c2d8fa08b9022ecc0.jpg
>
>
> If anyone wants to see the solution using calculus referred to in the problem,
> it can be seen here: (a handy site that I have no affiliation with)
>
> http://www.ltcconline.net/greenl/courses/106/areavolume/spherecore.htm
>
>
> Happy Solving,
> Mike C.
the volume of a sphere = pi*(6/2)^3*4/3
but what is the purpose of the 2nd black paragraph in the image?
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"Mike the Elder" <zer### [at] wyanorg> wrote
Isn't adding "this problem is solvable" more or less like writing code like
"if x = true then ..."
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I wish i hadn't scrolled down as I glimpsed the solution in the first post which
spoiled my fun of working it out the non-calculus way. However I had in my mind
gone through how to do it using calculus. I haven't looked at the solution but
I believe it will run something as follows.
Let us start by imagining we cut the cored sphere into infinitely thin slices,
then we will have little discs. Then if we add up the area of all the discs we
will get the volume.
If we represent the inner sphere as having a radius of say R and the sphere as
having a radius T then. Now we can write that the height from the axis (say x)
is given by the formula height^2 = T^2 - R^2
The discs have a radius whose square changes with proportion to the axis of
rotation (say Z), this is because the formula for a circle is x^2+y^2=r^2. Thus
the square of the outer radius of the disc is given by T^2-z^2. Notably at the
origin then we get the radius of the sphere out as z=0. This only works for
values of z between -height and height.
We thus get areas for the disc that are given by: Area = pi*((outside
radius)^2-(inner radius)^2) = pi*(T^2-z^2-R^2) = pi*(height^2-z^2)
Thus to find the volume we must intergrate from -height to height the above area
formula. With a little working we get a mess which equals 2*pi*(2/3*height^3)
or (4*pi*height^3)/3
This I presume is correct given the solution to the non-calculus version is
correct. However potentially this is the proof that the problem is solvable.
Hope noone was too bored by my long mathematical explantion and notably the
calculus involed is not too strenuous! Also thanks Mike as it was a nice little
brain teaser, even to solve the calculus way.
Regards Malcolm
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I realise my last post had one error I used height to mean something different
to on the diagram. My usage of height actually represents half of the height on
the diagram. I.e where i say height i mean 3m not 6m. Sorry for that but didn't
have both pages open at once so was kind of doing it from memory off the
diagram. thus my final solution should have read: Volume =
(4*pi*(0.5*height)^3)/3.
Apologies, Malcolm
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"somebody" <x### [at] ycom> wrote:
> "Mike the Elder" <zer### [at] wyanorg> wrote
>
> Isn't adding "this problem is solvable" more or less like writing code like
> "if x = true then ..."
It's not obvious (at least to most people) that knowing the height of a "cored
sphere" object is sufficient to calculate it's volume. It's an inference that
can be drawn form the fact that the information IS sufficient to find the
please use the links in the original post.)
Best regards,
Mike C.
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somebody wrote:
> Isn't adding "this problem is solvable" more or less like writing code like
> "if x = true then ..."
No. See my rationale in my post.
"This problem is solvable" actually means that you can assume
something about the problem, as I described in that post. Without
knowing that it is indeed solvable it would be quite hard to prove that
it indeed is. Would require quite complicated math.
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"Warp" <war### [at] tagpovrayorg> wrote
> "This problem is solvable" actually means that you can assume
> something about the problem, as I described in that post. Without
> knowing that it is indeed solvable it would be quite hard to prove that
> it indeed is. Would require quite complicated math.
A problem is either solvable, not solvable, or a fields medal candidate. We
can safely ignore the last possibility here I think. If you ask me a
problem/puzzle to find a solution to, and if I trust that you are not
leading me to a wild goose chase, it means it's given that it *is* solvable.
If I don't trust you, no amount of reiterating that it's solvable will
convince me ( "if ((((x=true)=true)=true..." syndrome ).
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somebody wrote:
> If I don't trust you, no amount of reiterating that it's solvable will
> convince me ( "if ((((x=true)=true)=true..." syndrome ).
There are also puzzles of the sort "because I have some of the same
information you do, and because you *can't* solve the problem, I *can*
solve the problem."
http://puzzles.nigelcoldwell.co.uk/twelve.htm
Fun puzzles.
--
Darren New / San Diego, CA, USA (PST)
It's not feature creep if you put it
at the end and adjust the release date.
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somebody wrote:
> A problem is either solvable, not solvable, or a fields medal candidate. We
> can safely ignore the last possibility here I think. If you ask me a
> problem/puzzle to find a solution to, and if I trust that you are not
> leading me to a wild goose chase, it means it's given that it *is* solvable.
> If I don't trust you, no amount of reiterating that it's solvable will
> convince me ( "if ((((x=true)=true)=true..." syndrome ).
Well, it should suffice to say that I couldn't have solved the problem
without the hint given by the "the problem is solvable". While what it
stated is more or less obvious, it was still a rather helpful *hint*. It
made me think about it in a different way than I would have it if hadn't
been there.
In other words, it made the problem easier, at least if you get the
hint and deduce what is it trying to insinuate.
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Warp <war### [at] tagpovrayorg> wrote:
> I haven't looked at the solution. This is what I'm thinking:
>
> ...Thus the problem is reduced to calculating the trivial volume of a
> sphere of diameter of 6m, which should be something like pi*r^3*4/3 =
> pi*3^3*4/3 = pi*36.
>
Nice one, Warp. A good demonstration (to my mind) of problem-solving using
"reductio ad absurdum," if that's the properly-spelled phrase. In this case,
reducing the cylinder down to a radius of zero and thus eliminating it
altogether.
Ken W.
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