The Flat Earth Society
Other Discussion Boards => Technology, Science & Alt Science => Topic started by: Son of Orospu on July 14, 2013, 02:33:53 AM
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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?
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Good job. I am going to edit your post so that others have a chance.
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Good job. I am going to edit your post so that others have a chance.
Oh no, oppressive mods :(
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I will abuse my powerz every chance I get.
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When are the next mod elections anyway? :-\
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When are the next mod elections anyway? :-\
You quit, so it does not matter to you.
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I can't hear you.
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7.283r = Circumference? ??? I suck at math.
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2pi(R+1) - 2piR
R = Radius of the Earth Sphere. (not sure what number you wanna use)
I think...
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2pi(R+1) - 2piR
R = Radius of the Earth Sphere. (not sure what number you wanna use)
I think...
You'll know when you get it right, because he'll delete your comment. >:(
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0.006283 km or 6.283 m or roughly 2 pi m
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The size of the sphere doesn't matter.
It's 2 pi meters.
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2pi
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0.006283 km or 6.283 m or roughly 2 pi m
Incorrect. Your first two answers are roughly correct and your last is precise.
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0.006283 km or 6.283 m or roughly 2 pi m
Incorrect. Your first two answers are roughly correct and your last is precise.
6.283 m is roughly equal to 2 pi.
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http://mathforum.org/mathimages/index.php/Rope_around_the_Earth (http://mathforum.org/mathimages/index.php/Rope_around_the_Earth)
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0.006283 km or 6.283 m or roughly 2 pi m
Incorrect. Your first two answers are roughly correct and your last is precise.
6.283 m is roughly equal to 2 pi.
Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.
To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.
2pi(R+1) - 2piR
R = Radius of the Earth Sphere. (not sure what number you wanna use)
I think...
Uh... yes, but:
2pi(R+1) - 2piR =
2piR + 2pi*1 - 2piR =
2pi*1
You don't need to mention R at all in your final answer.
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0.006283 km or 6.283 m or roughly 2 pi m
Incorrect. Your first two answers are roughly correct and your last is precise.
6.283 m is roughly equal to 2 pi.
Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.
To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.
I wasn't aware that there was a 2 pi exact answer. I just did it two different ways and got the same result. Then I noticed it was equal to 2 pi.
So as I said, I was in the right to say roughly.
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0.006283 km or 6.283 m or roughly 2 pi m
Incorrect. Your first two answers are roughly correct and your last is precise.
6.283 m is roughly equal to 2 pi.
Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.
To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.
2pi(R+1) - 2piR
R = Radius of the Earth Sphere. (not sure what number you wanna use)
I think...
Uh... yes, but:
2pi(R+1) - 2piR =
2piR + 2pi*1 - 2piR =
2pi*1
You don't need to mention R at all in your final answer.
Yeah. I really should learn to simplify my equations.
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Why would you simply the one part but not remove the *1? I'm so confused.
Anyway, tau.
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I wasn't aware that there was a 2 pi exact answer. I just did it two different ways and got the same result. Then I noticed it was equal to 2 pi.
Out of curiosity, how did you solve it? The only way that comes to my mind arrives at 2pi with no approximation involved.
Why would you simply the one part but not remove the *1? I'm so confused.
I could have added one more step to remove the "*1", but I felt like that would be wasting space. What I was trying to demonstrate was clear at that point.
Alternatively, I could have dropped it arbitrarily between steps, but I wanted to keep things readable and visually consistent. In this case it might have been overkill, but it often helps a lot.
Anyway, tau.
Maybe in 20 years. Then again, probably not.
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Out of curiosity, how did you solve it? The only way that comes to my mind arrives at 2pi with no approximation involved.
First I simply followed the formula already posted. You can't get 2 pi exactly because of sig figs.(not that I followed them).
Second, to check because I thought I was missing something, I used ratios to get the same result.
2 pi( r+1)*(1-(r/r+1))
I didn't check but it must simplify to 2 pi. I'm sure you will.
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2 pi( r+1)*(1-(r/r+1))
That would work if you didn't mess up the brackets. What you want is 2π(r+1)*(1-r/(r+1))
I'm still surprised you somehow only reached an approximate result, but I suppose it's fair enough if you didn't follow standard procedures and filled in the variables too early.
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2 pi( r+1)*(1-(r/r+1))
That would work if you didn't mess up the brackets. What you want is 2π(r+1)*(1-r/(r+1))
I'm still surprised you somehow only reached an approximate result, but I suppose it's fair enough if you didn't follow standard procedures and filled in the variables too early.
Yes, there is one missing. What I used in google calculator was 2 pi( r+1)*(1-(r/(r+1)). Which is equal to what you posted. That is how I got the same answer as the other formula.
If you have a problem with significant figures, I suggest you contact the proper controlling parties. You cannot get 2 pi if you enter valves for r.
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ITT: Chemists consider algebra to be beneath them.
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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?
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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.
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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?
Depends on what you mean? Is it the volume of wood left in the holey sphere, or the volume of wood total (which would be the same, since drilling only removes wood--just some of the volume would be sawdust).
If you mean the volume of the sphere, what's the radius of the sphere and the radius of the newly-formed hole?
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Clearly mathsman has access to some top secret maths that allow him to determine the answer.
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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.
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Time for another math puzzle, guys. I'm thinking of a number. What is it?
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e
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e
Incorrect.
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Is it a trick question? Do you have to "solve" for the answer, or will you just "get it"?
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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.
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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.
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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
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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.
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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.)
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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Congratulations to Spoon for getting the right answer although he does lose some marks for not showing his working.
Here's my solution:
(http://i.imgur.com/cgZ47lF.jpg)
(http://i.imgur.com/Wn6qmjJ.jpg)
(http://i.imgur.com/uh2Ni5r.jpg)
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I like my way better.
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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.
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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.
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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?
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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.
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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.
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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://hamsterking.com/temp/stepping-drill.jpg) http://www.blog.hazaveh.net/wp-content/uploads/corkscrew.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.
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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://hamsterking.com/temp/stepping-drill.jpg) http://www.blog.hazaveh.net/wp-content/uploads/corkscrew.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.
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.
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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.
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Its a frigging math puzzle and Pizza Planet and Parsifal are treating it like a grand jury deposition. A shining example of pedantry.
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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.
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Mathsman, I gotta side with PizzaPlanet. There were a lot of things that weren't clear. After seeing the solution, I realized that my original thought on how to solve the problem would work, but I didn't attempt it as it was rather lenghty.
A picture would have helped. Or even said "the diameter of the drill bit is 1 inch".
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A picture would have helped. Or even said "the diameter of the drill bit is 1 inch".
That still wouldn't have made sense.
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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.
And yet I and others managed to solve it. The point of the puzzle is that at first sight there isn't enough information. However, if a person picks up pen and paper and works through it they find that the radius of the sphere is immaterial. Mathematics isn't a spectator sport, you have to get in there and get your hands dirty.
Mathsman, I gotta side with PizzaPlanet. There were a lot of things that weren't clear. After seeing the solution, I realized that my original thought on how to solve the problem would work, but I didn't attempt it as it was rather lenghty.
A picture would have helped. Or even said "the diameter of the drill bit is 1 inch".
Certainly a picture would have helped but the picture is half of the solution.
The diameter of the drill bit is not one inch. The length of the hole is one inch. As the radius of the sphere increases so the radius of the hole increases. If a person approaches this problem in an honest and reasonable manner there is no ambiguity.
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Certainly a picture would have helped but the picture is half of the solution.
If your "puzzle's" difficulty is designed to come from ambiguity and misunderstanding, it's a riddle, not a puzzle, and certainly not a maths one. A picture is necessary until you decide to learn the right terminology.
The length of the hole is one inch.
Depth. How many times do I need to repeat this? Using the standard definition of length makes your solution quite simply incorrect. Use appropriate language.
If a person approaches this problem in an honest and reasonable manner there is no ambiguity.
Unless they happen to know anything about mathematics and approach the problem with knowledge, in addition to honesty and reason.
Since you're very set on "flapping the jaw" and not actually approaching your cock-up with honesty, I'll guide your hand. Here are some definitions of length:
"In geometric measurements, length is the longest dimension of an object."[nb]https://en.wikipedia.org/wiki/Length (https://en.wikipedia.org/wiki/Length)[/nb]
"the measurement of something from end to end or along its longest side"[nb]http://dictionary.cambridge.org/dictionary/british/length_1 (http://dictionary.cambridge.org/dictionary/british/length_1)[/nb]
"the linear extent in space from one end to the other; the longest dimension of something that is fixed in place"[nb]http://www.wolframalpha.com/input/?i=length (http://www.wolframalpha.com/input/?i=length)[/nb]
Under your own specification, all drillings 1 inch in diameter whose depth is less than 1 inch need to be considered in the correct solution. Therefore, you either messed up the wording of the puzzle (likely factual), or you screwed up the solution. Your choice.
Think about it this way, mathsman: if I design a "puzzle" that requires you to arbitrarily assume that all rectangles are in fact squares, is it unambiguous? Is it even a puzzle?
And yet I and others managed to solve it.
Yes, and many modern people have managed to establish that geocentrism is true. Just because a certain group of people manages to miss certain important facts and arrive at a conclusion they're comfortable with does not imply that the conclusion is correct.
You accuse others of buffoonery, and yet you're so arrogant that you can't even accept that you misspoke. Silly, silly mathsman.
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And yet I and others managed to solve it.
Incorrect.
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Mathsman, I gotta side with PizzaPlanet. There were a lot of things that weren't clear. After seeing the solution, I realized that my original thought on how to solve the problem would work, but I didn't attempt it as it was rather lenghty.
A picture would have helped. Or even said "the diameter of the drill bit is 1 inch".
Certainly a picture would have helped but the picture is half of the solution.
The diameter of the drill bit is not one inch. The length of the hole is one inch. As the radius of the sphere increases so the radius of the hole increases. If a person approaches this problem in an honest and reasonable manner there is no ambiguity.
Wait what?
So you're saying that when I read your problem I got the whole thing wrong? Yeah, the wording sucked. Just admit your failed use of proper terms and move on.
I just want to add that rereading the original problem with your actual meaning leads me to believe that the sphere has a very specific diameter.
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There's too much pendantry in this thread for my liking. Can we stick to the maths puzzles?
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There's too much pendantry in this thread for my liking. Can we stick to the maths puzzles?
Seriously. I have never seen so much crying over not getting the answer to a puzzle. Time for a diaper change everyone!
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Mathsman, is your deletion of your last post here a way of taking it back?
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Mathsman, is your deletion of your last post here a way of taking it back?
I apologise for the tone of my comments. Life is too short to be disrespectful to people.
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The puzzle was worded in a confusing manner however, once you started to work through it became apparent what the puzzle was about.
I think some of you are being overly pedantic.
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The puzzle was worded in a confusing manner however, once you started to work through it became apparent what the puzzle was about.
I think some of you are being overly pedantic.
Overly pedantic? Here!?!
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The puzzle was worded in a confusing manner however, once you started to work through it became apparent what the puzzle was about.
I think some of you are being overly pedantic.
Overly pedantic? Here!?!
He's new here.