Math puzzle

Clearly mathsman has access to some top secret maths that allow him to determine the answer.

Quote from: mathsman on July 17, 2013, 10:40:00 AM

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

There isn't enough information in that puzzle to solve it.

Yes there is. If you have made your comment after making a fair attempt at the puzzle then all I can ask is that you have another go.
If you made the comment before any attempt then all I can ask is that you have a go before flapping your jaw.
I will post the solution in a couple of days' time if it is not solved by then.

Time for another math puzzle, guys.  I'm thinking of a number.  What is it?

e

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Incorrect.

Is it a trick question? Do you have to "solve" for the answer, or will you just "get it"?

Ok, I see what you mean now. I wasn't sure what you meant by a hole that is "an inch long". Give me a minute to solve it.

Actually, it's a pretty simple thing. It involves similar thought to the OP's puzzle, at least how I solved it.

It's 1/6 pi in. ^3.

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

π³/6

Quote from: mathsman on July 17, 2013, 10:40:00 AM

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

π³/6

No, the radius is cubed, not pi.

Quote from: Junker on July 17, 2013, 01:48:01 PM

Quote from: mathsman on July 17, 2013, 10:40:00 AM

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

π³/6

No, the radius is cubed, not pi.

It is just the napkin ring problem, isn't it?  πh³/6 and in this case, h=1 (in).  Since h=1 it can be removed from the final product, leaving:

π³/6 (in.)

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

4π/3 1/8.
4π/24
π/6.

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

4π/3 1/8.
4π/24
π/6.

I know the napkin ring problem does not depend on the sphere's radius.  I'd have to go back and check though, it has been forever since I did it.

If the Earth was a sphere (I know, lol), and I mean a perfect sphere, and you wrapped a rope around the great circle of the sphere, how much length would you need to add to the rope in order for it to be able to be 1 meter off the ground?

Thankfully,  Zetetic maths has granted me the simple answer. The answer is 1 meter, for you want it to be 1 meter off the ground. Believe, and you will see this to be true.

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

4π/3 1/8.
4π/24
π/6.

I see now what the intended interpretation of the problem is. I maintain that there was insufficient information in the original expression to infer this unambiguously.

Quote from: spoon on July 17, 2013, 02:02:24 PM

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

4π/3 1/8.
4π/24
π/6.

I see now what the intended interpretation of the problem is. I maintain that there was insufficient information in the original expression to infer this unambiguously.

I thought it was ambiguous too at first, but now it doesn't make sense any other way. Perhaps we were hasty to assume there was insufficient information.

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

English is not my primary language, but I thought that the bolded part already fixed the diameter of the sphere, since a hole drilled from one side to the other must necessarily have 2 points of entry and could not be "inside" the sphere. Maybe I misunterstood the problem.

Quote from: spoon on July 17, 2013, 02:02:24 PM

I thought of it as a limit. As the diameter of the sphere approaches 1, the width of the "drillbit"approaches 0. In which case, solve V=4π/3 r^3 where r=1/2.

4π/3 1/8.
4π/24
π/6.

I see now what the intended interpretation of the problem is. I maintain that there was insufficient information in the original expression to infer this unambiguously.

Still flapping the jaw I see, and not bothering to attempt the puzzle.

Congratulations to Spoon for getting the right answer although he does lose some marks for not showing his working.

Here's my solution:

I like my way better.

It is just the napkin ring problem, isn't it?  πh³/6 and in this case, h=1 (in).  Since h=1 it can be removed from the final product, leaving:

π³/6 (in.)

I feel like this is a troll, but hey, I'll humour you.

πh³/6
Let h=1
π*1³/6
π*1*1*1/6
π/6

Also, mathsman, this would be a cool puzzle if you worded it unambiguously. While the mathematics of it is interesting, your English butchered it.

Quote from: Junker on July 17, 2013, 01:56:11 PM

It is just the napkin ring problem, isn't it?  πh³/6 and in this case, h=1 (in).  Since h=1 it can be removed from the final product, leaving:

π³/6 (in.)

I feel like this is a troll, but hey, I'll humour you.

πh³/6
Let h=1
π*1³/6
π*1*1*1/6
π/6

Also, mathsman, this would be a cool puzzle if you worded it unambiguously. While the mathematics of it is interesting, your English butchered it.

But, order of operations, and the weather patterns on the moon.

Also, mathsman, this would be a cool puzzle if you worded it unambiguously. While the mathematics of it is interesting, your English butchered it.

I don't see how my wording was ambiguous but I'm prepared to be enlightened. How would you have written it?

Still flapping the jaw I see, and not bothering to attempt the puzzle.

It is not possible to attempt a mathematics puzzle which is ambiguous without invoking assumptions, in which case you have presented a riddle rather than a puzzle.

Quote from: mathsman on July 18, 2013, 04:04:49 AM

Still flapping the jaw I see, and not bothering to attempt the puzzle.

It is not possible to attempt a mathematics puzzle which is ambiguous without invoking assumptions, in which case you have presented a riddle rather than a puzzle.

First, did you attempt the puzzle?
Second, what neccessary assumptions were missing? I still maintain it contained enough information for a unique solution and my solution is proof of that.

How would you have written it?

I would have gone with a graphical explanation. It's the easiest way to present geometry problems. If you really need to describe it with words, use appropriate language. Useful key words include: "diameter", "cylinder", "depth".

Second, what neccessary assumptions were missing?

Here are a few I've noticed:

• Spheres don't have sides. You have to assume a definition of a "side".
• The sphere's diameter must be at least 1 inch.
• The drilling is straight.
• The drilling is cylindrical in shape^{[nb]This assumption is problematic, to say the least. Many drills are not cylindrical: http://hamsterking.com/temp/stepping-drill.jpg, http://www.blog.hazaveh.net/wp-content/uploads/corkscrew.jpg[/nb]}.
• The drilling goes through the entirety of the sphere's diameter.
• You have to guess which dimension is the "length" of the drilling.

This list is not attempting to be exhaustive, so it likely isn't.

I still maintain it contained enough information for a unique solution and my solution is proof of that.

The solution is only "unique" if you make the mistake of not exploring all possibilities for each ambiguity. Once you stop making that mistake, the amount of unique solutions becomes virtually infinite.

Quote from: mathsman on July 19, 2013, 12:38:53 AM

How would you have written it?

I would have gone with a graphical explanation. It's the easiest way to present geometry problems. If you really need to describe it with words, use appropriate language. Useful key words include: "diameter", "cylinder", "depth".

Quote from: mathsman on July 19, 2013, 04:43:48 AM

Second, what neccessary assumptions were missing?

Here are a few I've noticed:

• Spheres don't have sides. You have to assume a definition of a "side".
• The sphere's diameter must be at least 1 inch.
• The drilling is straight.
• The drilling is cylindrical in shape^{[nb]This assumption is problematic, to say the least. Many drills are not cylindrical: http://hamsterking.com/temp/stepping-drill.jpg, http://www.blog.hazaveh.net/wp-content/uploads/corkscrew.jpg[/nb]}.
• The drilling goes through the entirety of the sphere's diameter.
• You have to guess which dimension is the "length" of the drilling.

This list is not attempting to be exhaustive, so it likely isn't.

Quote from: mathsman on July 19, 2013, 04:43:48 AM

I still maintain it contained enough information for a unique solution and my solution is proof of that.

The solution is only "unique" if you make the mistake of not exploring all possibilities for each ambiguity. Once you stop making that mistake, the amount of unique solutions becomes virtually infinite.

Pointless pedantry to cover the fact that you couldn't solve it. You're just the sort of buffoon who would claim that the reason he can't beat Roger Federer at tennis is not lack of ability but the wrong string tension in the racquet.

Pointless pedantry to cover the fact that you couldn't solve it.

Actually, I only saw the problem after it had been solved. It's difficult to determine whether or not I would be able to solve it, since the answer was known to me in advance. Clearly, you have a greater understanding of my thought processes than I do; or maybe you've just made more assumptions to comfort yourself.

Also, if you consider high-school-level competency in English and mathematics to be "pointless pedantry", I recommend you change your name to "liberalartsman"... no, wait, that wouldn't work either.

Its a frigging math puzzle and Pizza Planet and Parsifal are treating it like a grand jury deposition.  A shining example of pedantry.

First, did you attempt the puzzle?

I just said it was not possible to attempt the puzzle. It seems that you have as much trouble with reading as you do with writing.