...and in another of my infrequent forays into the physics rooms, I have come across a rather interesting individual called Jeffrey Stapleton.
Not only does this geyser (Hey Diamond and Marianne (you work it out!)) proclaim himself a genius, but he also catagorically refuses to admit he is wrong in anything!
He is at www.jeffreystapleton.com and when confronted with my "hole-in-sphere" problem, claimed to have solved it with the answer 94! Check his site out, he is unique!
The problem is this: "A cylindrical hole 6 inches long of maximal radius is drilled straight through the center of a solid sphere. What is the volume remaining in the sphere?"
The answer I will leave to all you real geniuses.
It can be solved three different ways. Algebra (arduous), Calculus (beautiful), or Logic (ingenious).
Sad thing is, the Stapleton (spin -0, look at his site to see my comment) gave the right answer (but didn't know it was correct), and continued to try and explain why his "better" answer was 94. Strange, since the problem has a unique answer, and not 94 (logic gates, sorry Bill, look out!).
All in the physics chatroom!
These guys are so FUN to be with (Sirius (Yes, I still think Mr Black is Alive) Cybernetics Corporation).
Oh well, come the revolution....... all particles of spin -0 will be up against the wall....!
Yippee!
Monday, September 12, 2005
Physics Room Gold Medal Award Winner!
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12 comments:
Yeah, pretty amazing as you have answered the wrong problem. There are no cones in the question, nor in the answer, which is 36 Pi. Viz:
The problem seems insoluble as it lacks sufficient data for a solution. But it would not have been given unless there was a solution, a unique solution. The size of the sphere is not known, nor is the size of the cylindrical hole, as the sphere size determines it. If the problem has a unique solution, therefore, the volume remaining must be a constant, even if the hole were zero radius. This mean the residue is the same as a sphere of diameter 6, hence the residue is 36Pi.
Even more amazing, when you think about it really hard.
Hi nerd. Thanks for the support. I tryto go to Physics(1) as often as possible. Only physics room where there aren't too many arabs. As for cutting and pasting, I hardly ever do it, except for answering questions like asl, etc.
The problem states "maximal" not "maximum", Jeffrey.
The only place I go to are the science rooms in Yahoo messenger. The only place science even rears its head is Physics 1. The rest ar full of arabs chatting in arabic.
Even in Physics 1, you rarely get any sensible physics chat.
If you wanna go, trustworthy IDs are vincentvangogh, ione, lalady, fiftypushups, h_dcase. Don't talk to Buphalo Bill, he is nasty, and steer clear of Seasky, who sometimes pretend to be me using geisha_humble_girl07
i have to think somewhere along the line, a key part of this problem was misplaced or misrearranged.
let R be the radius of the sphere.
let B be the radius of the base of the cylinder.
the height of the cylinder is already given as 6.
let Vs be the volume of the sphere.
let Vc be the volume of the cylinder.
let Vr be the volume remaining, i.e., Vs - Vc.
Vc = 6*pi*B^2
Vs = (4/3)*pi*R^3
R = sqrt(B^2 + 9) (pythagorean)
Vs = (4/3)*pi*(sqrt(B^2 + 9)^3)
Vr = Vs - Vc
= (4/3)*pi*(sqrt(B^2 + 9)^3) - 6*pi*B^2
we now have an equation for our remaining volume, Vr, as a function of B. in the solution by Logic given in a previous post, a calculation is made with the assumption that the radius of the cylinder is zero. letting B=0, we see that we do indeed get the answer 36*pi.
however, letting B=4, we see that we get a different answer, (500/3)*pi - 96*pi = (212/3)*pi. this undermines the assumption, made in the Solution By Logic, that the resulting answer is a constant, regardless of the radius of the cylinder.
basically, the confusion is that there is really is no restriction on the radius of the cylinder, and therefore "maximal" radius would kind of just be infinity or something, which makes no sense for this problem. if you can picture it, as the radius of the cylinder increases with constant height of 6, it is simply an expanding disc of constant thickness. no matter how large the disc gets, there always exists some sized sphere (whose volume is increasing at a greater rate than that of the disc) into which it can be inscribed.
i think the problem should be different. the radius of the SPHERE should be the given, say 6. and then you should say the VOLUME of the cylinder should be maximal. since the cylinder is inscribed within the sphere, its height would be limited to 12 and its radius would be limited to 6. at each of these limits, the volume of the cylinder would be zero, thus the maximal volume would be at some intermediate non-trival value that could actually be worked out.
hmmm... i wonder why arabic people would be more prone to hang out in the physics rooms..
The answer does not involve cones, Jeffrey!
you can certainly have a maximal value without having a minimal value, e.g., a function which is a parabola which opens up downward has a maximum, but no minimums.
when one speaks of maximal, it has to be maximal with reference to something. you, dotcomer, are thinking of it as a maximal radius with reference to the position along the axis of individual cylinders where you should be thinking of it as the maximal radius within a set of multiple differently-sized cylinders. if we were to truly extend your logic of maximal/minimal radius, your shape wouldn't necessarily even have to be a cone. with your assumption, who's to say that the radius must be linearly increasing/decreasing between the two radiuses at either end of the "cynlinder"? it could be shaped like some sort of a vase instead. in fact, it could be shaped as a hemisphere, or the whole sphere, within which it is being described.
take a question such as "what is the maximal radius of a cylinder of height 6 which fits inside a cube whose edge is 6?" you can see that a cylinder of height 6 and with radius 1, or 2, would fit inside the cube. but that would not be maximal. at a radius of 3, the cylinder finally becomes so big that it could not possibly be any bigger and still fit into the cube. thus, you see that you could have many cylinders of radius < 3 fit into the cube, yet only the one where radius = 3 is the one with the maximal radius.
also notice that in every case, we are still considering a cylinder, just as specified in the original question. changing the cylinder to a cone would be changing the problem. then you would be solving a problem that you made up yourself.
Check this chat which I just had with Stapletom/dotcomer:
geisha_humble_girl007: No cones in the question. How did you figure there was a cone there when a cylinder was proposed?
geisha_humble_girl007: Al the solutions involve cylinders
genius_lq: s cone is a cylinder
geisha_humble_girl007: and they all have the same asnwer
geisha_humble_girl007: "s cone"?
genius_lq: a cone*
geisha_humble_girl007: a cone is a cylinder?
genius_lq: yes
geisha_humble_girl007: Run that by me one more time
genius_lq: a cone is a cylinder
He said twice that a cone is a cylinder.
Really!
nerd31416 said:
{
Duca, please abstain doing those calculations. It irritates me to see big people doing elementary math... it is a big NO! It is almost if I saw an adult counting with fingers, so please abstain...
}
sorry to irritate you, but an answer to the problem was stated in the second post. given that this whole blog entry is pretty much about this math problem, and since the stated answer was incorrect, yet nobody seemed to notice, i had to point it out. did you want me to the just say "no, that is wrong." and neglect to prove it?
yes, i agree that it is elementary math, so why didn't YOU do it? yes, it's true that i would have never thought i'd have to post such trivial math up on here, but if you had done the math yourself, you'd know, too, that the answer was wrong. how can you be wrong, and then bash the math that proves you are wrong, simple or not? was it just too "elementary" for you to bother doing? so instead, you've just backed up Geisha in this battle with Mr. Stapleton unwaveringly. as much as Geisha hates Mr. Stapleton (and, i fear, me, now that i've apparently become irritating), she IS a true scientist. as much as my elemenary math may be beneath you, don't you think a true scientist wants to know when he/she is wrong? i know i hate being wrong, so if i am, i appreciate being corrected so that i can reevaluate and proceed to be right.
we all know Geisha is as brilliant as she is beautiful, and i feel that she must have recalled the more-amazing elegant solution to the problem, but then simply mangled the less-memorable statement of the problem itself, and then neglected to redo this mangled form of the problem. because the solution is logic-based, it would not obviously contradict the stated problem since the key flaw is the base assumption that the problem "would not have been given unless there was a solution, a unique solution." all we need now is the unmangled original problem.
Thanks for the praise, Lancelot. The sphere problem I quoted from Martin Gardner. It was I who modified it by one word - "maximal"!!!
Otherwise, it was a perfect quote. Anyone who can read can see it in an old Penguin book.
Which reminds me of the problem of the island with two tribes, the truth-tellers and the liars......
The correspondence between this blog and Mr Jeffrey Stapleton is now CLOSED.
May he rest in his piece.
Mr Healey, this is for you.....
http://www.puzzles.com/PuzzlePlayground/HoleInTheSphere/HoleInTheSphereSol.htm
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