166 replies to this topic

[coolfflame]

You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.

[bonanova]

You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.

Hello coolfflame, and welcome to the Den.

There is enough information to solve this.

A word of advice.
You will earn respect from members when you respect them.
That would mean first assuming a puzzle does have an answer and trying to find it.
If in the process you come to believe that it can't be solved, show that in the form of a proof.

Finally, please read the READ BEFORE POSTING link at the top of the page.
Name calling, even in jest, won't earn you points with the site administrator.

[upthebuzzard]

Hi,

I've waded through 12 pages of comments about this puzzle, and I am happy that it is an interesting geometrical puzzle, and that the calculus-based solutions demonstrate that the volume of the remaining 6 inch high donut/bracelet/sliver/whatever shape has a fixed volume of 36pi for all spheres with a diameter greater than 6inches.

However, way back at the start there was the promise of 'logical' solution than did not need calculus. I've not seen it in any of the 12 pages.

Forgive me if I've missed a subtelty somewhere, but the 'logical' solution being offered appears to be based on the premise that the original question does not specify a diameter, therefore the solution does not depend on the diameter, therefore we just have to establish the remaining post-drilled volume for any one particular diameter of sphere, oh look for diameter=6 the answer is 36pi, and voila we have the answer for all diameters.

That's not a complete solution. That's a solution based on a very big, unproven, assumption.

What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place that the solution does not depend on the diameter?

[bonanova]

What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place
that the solution does not depend on the diameter?

Spoiler for Here you go

Modus ponens

If the puzzle is well posed, then it contains all necessary information.
Following the first rule of puzzle solving, we take the puzzle to be well posed.
Therefore we take that the puzzle contains all necessary information.

Modus tollens

If the diameter is necessary information, then it is contained in the puzzle.
It is not contained in the puzzle.
Therefore the diameter is not necessary information.

If the solution depends on the diameter, then the diameter is necessary information.
The diameter is not necessary information.
Therefore, the solution does not depend on the diameter.

[bonanova]

The question does not state the diameter or length of the hole (a cyclinder with a curved top and bottom) in relation to diameter of the sphere.
This information is essential.
We only know that the diameter of the sphere must be greater than 6 inches.
You can use the same size apple corer for different size apples.

A hole does not have a curved top or bottom.
A curved portion [spherical caps] on either end of the hole is removed when the hole is made.
But inspecting the inner surface of the hole we find a circular cylinder with a well defined length. That length is given as 6".

[bonanova]

xucam raises several issues. Here are the responses, belated.

The problem as originally stated is not solvable, as it could be reasonably interpreted as a 6-inch diameter hole all the way through an object.

Quite right. That has been fixed in the OP.

I would argue that you cannot drill a 6 inch hole through a sphere with radius = 3 any more than you can through a sphere of r = 2. A hole with no volume is not a hole. It may be true that the r=3 case is a boundry which a solution approaches, and obviously the volume of such an intact sphere is 4/3 * 3 * 3 * 3 * pi cu. in.= 36 pi cubic inches.

Having thus obtained the solution, there's no need to drill the hole.

There was also discussion of the 'caps' that would fit on top of the inscribed cylinder in the problem. Clearly, for the example of boring through the earth with a bit whose radius is almost as big as the radius of the earth, these caps would be massive (basically, you've just cut away 6-inch slice of the earth through the center). The volume of these caps is very obviously dependent on the radius of the sphere, and are thus not included in the 'answer'.

Since the volume of the end caps was not asked, it does not belong in the answer.

the stated 'solution' of 36pi isn't a volume at all, whereas 36 pi cubic inches is...

The stated answer is in fact 36pi cu in.

[jsai3334]

So. Um. How would you find the volume of a six-inch drill hole? Because you could just use the Volume of the Sphere-Volume of the drill hole to get the answer... Unless of course the idea of if the sphere is too small is added.

[Scott Strosahl]

Okay, so I don't think any calculus is required here. It's just simple geometry. Feel free to let me know if I made any mistakes in my calculations:

R = radius of sphere
r = radius of cylinder
L = height of cylinder (6)
h = height of cap

V_sphere = 4/3 pi R³
V_cylinder = L pi r² = 6 pi r²
V_cap = 1/3 pi h² (3 R – h)

r² + L²/4 = R² (based on triangle from center of sphere to edge of cylinder up to top of cylinder)
r² = R² – 9 (substituting 6 for L as given in the problem)

h = R – L/2 = R – 3 (this one should be self explanatory)

Therefore, putting everything into terms of R:
V_cylinder = 6 pi (R² – 9) = 6 pi R² – 54 pi
V_cap = 1/3 pi h² (3 R – h) = 1/3 pi (R – 3)² (3R – (R – 3)) = 1/3 pi (R² – 6R + 9) (2R + 3) = 1/3 pi (2R³ – 9 R² +27) = 2/3 pi R³ – 3 pi R² + 9 pi
V_cylinder + 2 V_cap = 6 pi R² – 54 pi + 4/3 pi R³ – 6 pi R² + 18 pi = 4/3 pi R³ – 36 pi

Remaining Volume = V_sphere – (V_cylinder + 2 V_cap) = 4/3 pi R³ – (4/3 pi R³ – 36 pi) = 36 pi

[yakul85]

i was confused for a bit on what the value 6 was corresponding to; once i visualized it, the question was cake

simple solution: just imagine the diameter of the sphere is 6, then the cylinder would have 0 radius and the remaining volume is the volume of the sphere: 4/3pi*R^3 = 36pi

logically, any larger sphere would have the same remaining volume

[Dej Mar]

Spoiler for I like this puzzle!

The geometric problem has been stated in the form of a poem
(which I must attribute to Anonymous as I could not find the author):

Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.

Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!


The answer to the problem is 36*pi cubic inches, the volume of a 6 inch sphere.

Edited by Dej Mar, 17 July 2010 - 11:08 AM.