Interesting math problem

[dauerbach]

A band is placed around the earth at the equator, touching the surface. You add one meter to the length of the band. What animals can now walk under the band: Flea, mouse, cat, large dog?

[I am Sparticus]

Does it bunch up so that the extra meter is all in one place? Or is the extra meter spread out over the entire circumferance of the earth? In situation one, you could probably get any animal on your list under it. In situation two, you probably couldn't even get the flea under it, and certainly not a mouse or anything larger.

[dauerbach]

The earth is then centered in the middle of the larger band, so yes, it is spread out over the entire circumferance of the earth and not eccentric.

[Desmostylus]

I don't know. What's 15.9 cm tall?

[K364]

D = C/pi is the old Diameter where C=circumference
D' = (C + 1)/pi is the new Diameter

the difference between the new and old is:
(C + 1)/pi - C/pi = 1/pi

or .318 meters. Divide by 2 to get the increase in radius on one side = .16 meters.

so, Cat?

[Achernar]

The tricky part is, it doesn't matter the size of the sphere! You could do the same experiment on a baseball, and the same size animal could still pass under.

[Manduck]

We know that the circumference is pi times the radius, which means that the radium is the circumference divided by pi. We increased the circumference by a meter, so we must have increased the radius by a meter divided by pi, which is about the height of a cat.

[Manduck]

Grrr those other responses weren't the when I started

[Desmostylus]

I think you missed something there, Manduck.

[dauerbach]

I'ld say all but the large dog.

Ok, now put a band around a basketball. Add 1 meter to its length. Which animals can now fit under it?

[mmmiiikkkeee]

Achernar answered that one.

[dauerbach]

Ooops, guess that's already been answered. It is really counter-intuitive (to me at least) that the size of the original sphere doesn't matter.

[MonkeyMensch]

Originally posted by dauerbach

Ok, now put a band around a basketball. Add 1 meter to its length. Which animals can now fit under it?

Of course, none of them. Because the basketball is sitting on the ground and there is no room for them to pass underneath.

FWIW

[Enderw24]

Originally posted by MonkeyMensch
Of course, none of them. Because the basketball is sitting on the ground and there is no room for them to pass underneath

Well, true, but the Earth is resting on the shell of a turtle so all those people in China will have a hard time squeezing through too.

[astro]

Assuming a perfectly spherical earth there is ribbon around the earth touching the surface. The earth is approximately 40,076.5 km or 40,075,500 meters in circumference. You add one (1) little meter to that length and the height of the band from the earth's surface (at all points) is now .159 meters.

That really is non-intutive - I see the geometrical logic but the overall height difference as a result of an infintesimal increase in relative length is really a poser when you think about the problem.

What are some other interesting, non-intuitive math/geometry problems that are understandable by layman.

[The Flying Dutchman]

I see a unwarranted assumption here, probably based on experience with the now popular (add a yard to the string around the earth) brain teaser.

Look at the question carefully. It is entirely appropriate to draw the band taut around the earth, save for 20cm. One could form a bridge with ther band 40 cm high and 20 cm wide for example.

Now 40 cm is 16 inches high. Kind of debatable if that qualifies as a large dog.

[RM Mentock]

Originally posted by astro
That really is non-intutive - I see the geometrical logic but the overall height difference as a result of an infintesimal increase in relative length is really a poser when you think about the problem.

What's boggling my mind is that a meter is infinitesimal to you. You must be huge.

Here's another interesting problem, that I think we've discussed before, but since you asked: Take a sphere and drill a hole in it that is eight cm long from rim to rim. How much material is left of the sphere?

[cdhostage]

Originally posted by RM Mentock

Here's another interesting problem, that I think we've discussed before, but since you asked: Take a sphere and drill a hole in it that is eight cm long from rim to rim. How much material is left of the sphere?

? I can't do it - I dcon't see enough information here. How wide is the diameter of this tunnel? I'm assuming that the tunnel (hole) goes straight through the center and out the diametrically opposite point. What is the diameter of the sphere?

I remember doing the old Earth-string thing back in highschool physics. I still don't believe it, even though the math works.

[RM Mentock]

Yes, the tunnel goes straight through the center of the sphere. Other than that, I don't think you need any more information than that (that's a hint on how to get the answer--the harder part is how to prove that that is actually the answer)

I don't think I've screwed up the problem statement...

[Gomez]

So, what you're saying with the earth/string problem is that if you have a sphere, circumference 40,075,500 and a string of the same length, wrapped around the centre of the sphere, and then added just 1 metre of slack to the rope the rope would be raised 15.9 centimeters all the way around the sphere? I'm no mathematician but if ever you wanted a textbook case of a counter intuitive maths problem, this'd be it.

I think I speak for the majority when I say

[Nightime]

RM Mentock - the material left in the sphere has a volume equal to the volume of an 8 cm diameter sphere.

Method: If there is a solution, the answer cannot depend on the diameter of the hole, because it is not given. Therefore the hole can have a diameter of zero. And if it does, we are left with an 8 cm diameter sphere. This is probably not the right way to solve it, but I like it.

[Ximenean]

A non-intuitive logic puzzle that has been doing the rounds recently is this:
Three men are sitting in a room. Someone randomly places a red or blue hat on each of their heads, so that they can see the other two hats but not their own hat. On the count of three, each of them must either guess the colour of his own hat, or say nothing.
To win the prize, which will be shared among them, at least one of them must correctly identify the colour of his hat, and nobody must call out the wrong colour. How can they give themselves a better than 50% chance of winning the prize? (They can agree a plan beforehand, but are not allowed to communicate in any way during the 'game'.)

Intuitively, you might wonder how seeing the other hats could possibly improve their chances of guessing their own colour, but there is a way.

[astro]

Originally posted by RM Mentock
What's boggling my mind is that a meter is infinitesimal to you. You must be huge.

Here's another interesting problem, that I think we've discussed before, but since you asked: Take a sphere and drill a hole in it that is eight cm long from rim to rim. How much material is left of the sphere?

Ummm... non-math adept WAG - you can only go precisely from true maximum edge ie "rim to rim" with an infinitely small diameter hole so material excised is infinitely small and thus volume unchanged. A hole of any measurable diameter would violate the true "rim to rim" requirement by taking in some portion of curvature from the "rim"?

[Hari Seldon]

Originally posted by Usram
A non-intuitive logic puzzle that has been doing the rounds recently is this:
Three men are sitting in a room. Someone randomly places a red or blue hat on each of their heads, so that they can see the other two hats but not their own hat. On the count of three, each of them must either guess the colour of his own hat, or say nothing.
To win the prize, which will be shared among them, at least one of them must correctly identify the colour of his hat, and nobody must call out the wrong colour. How can they give themselves a better than 50% chance of winning the prize? (They can agree a plan beforehand, but are not allowed to communicate in any way during the 'game'.)

Intuitively, you might wonder how seeing the other hats could possibly improve their chances of guessing their own colour, but there is a way.

It is important to add to this puzzle the fact that the three people have met beforehand and agreed on a common strategy. Given that they can increase their chances to 75%. But I will refrain from giving the solution. It can be found on the web, so no looking.

[Nightime]

astro - the rim refers to the hole itself, not the original sphere.

Usram - Good puzzle. Think of the three men as A, B, and C. Before they enter the room they decide that if A cannot see any blue hats, he will guess blue. If he can see any blue hats, he will be silent. B will look only at C, and if he sees a red hat, he will guess blue. If C has a blue hat B remains silent. C will always guess blue if A and B were silent.

This strategy means they will win every time unless they all have red hats. I'm not sure if I violated the "no communication" clause though. Does seeing someone remain silent count as communication? That is an important factor.

[Nightime]

Given that they can increase their chances to 75%.

But my strategy is better than 75%! Which means I must be wrong...

[astro]

Here's an old physics one.

A car is going 30 MPH on a collision course with another, absolutely identical ) car going at the same speed toward it on a perfectly straight road and crashes into the other car absolutely head on.

A car is going 30 MPH on a perfectly straight road on a collision course with an immovable concrete wall facing the side of a mountain and crashes into it absolutely head on.

In which scenario does the car suffer more damage?

[Achernar]

While I understand where everyone is coming from when they say that the OP's puzzle is counterintuitive, I think it actually makes sense when you think about it the right way. Imagine a cord 100 meters long lying on the ground. Take a second cord placed 16cm directly above the first one, with the ends directly above the ends of the original cord. How long would it have to be? Since the earth is practically flat over 100 meters, the answer is clearly "just a tiny little bit over 100 meters". Extremely small addition. Now, for 200 meters, this extra length is twice as much, but still incredibly small. Now imagine how long the original cord would have to be for the extra length to be one whole meter! Pretty big, huh? As big as... the earth?

[GTPhD1996]

Originally posted by Usram
ng.
To win the prize, which will be shared among them, at least one of them must correctly identify the colour of his hat, and nobody must call out the wrong colour.

If nobody can call out the wrong color, then doesn't that mean they all have to be correct to win? Or is it that they must guess red or blue and not guess something like yellow?

GTPhD1996

[Ximenean]

Nightime, they must make their guesses simultaneously i.e. with knowing what, if anything, the other two will say.

[Ximenean]

and GTPhD1996, they each have the option of remaining silent. They can say "Yellow" if they want to, but that would obviously count as a wrong guess.

[Uncommon Sense]

Originally posted by astro
Here's an old physics one.

A car is going 30 MPH on a collision course with another, absolutely identical ) car going at the same speed toward it on a perfectly straight road and crashes into the other car absolutely head on.

A car is going 30 MPH on a perfectly straight road on a collision course with an immovable concrete wall facing the side of a mountain and crashes into it absolutely head on.

In which scenario does the car suffer more damage?

Both the same.
I must admit I used no math just common sense/

[GTPhD1996]

Ok. I reread the problem and saw that they can remain silent. So if they do guess a color, they must be correct.

Nightime, I think they all need to call out the color at the same time, so your plan wouldn't work.

GTPhD1996

[Ximenean]

Uh, I meant to say "i.e. without knowing what, if anything, the other two will say."

[The_Peyote_Coyote]

Astro: If I remember my physics class correctly, the car striking the mountain will take more damage because that would be an inelastic collision.
Am I right? Do I win a no-prize?

[Cisco]

Ursam - You don't say anything unless the other two are wearing identical hats, in which case you say the opposite color. Of course, this doesn't work in the case that you all three got the same colored hat, but it gives you a 75% chance.

[Achernar]

Unless I'm mistaken, astro's question has been done to death in an old thread, which I'd rather not link to, for fear of its being resurrected. The thread ID is 30041, but please don't post unless you have something really important to add.

[The Flying Dutchman]

For those of you into brain teasers ,logic problems etc. theres a web site called The Grey Labyrinth. The best puzzles are way back in the Archives, and you'll find lots lots of good puzzles presented by that board's members. I was heavily involved there for over a year until I found the SD message board .

[RM Mentock]

What about the banana and camels problem?

[rowrrbazzle]

The OP's puzzle is "counterintuitive" because although we all agree that the length increase is infinitesimal compared to the length of the band, we tend to forget that the height increase will also be infinitesimal only compared to the diameter of the earth.

[Desmostylus]

rowrrbazzle: Yes, that's what I don't get. All of these people have presumable done a least a bit of calculus, and should be familiar with ratios of "infinitesmals". I had trouble understanding why people would think this problem counterintuitive at all.

[Ximenean]

I think what rowrrbazzle was saying is that we are accustomed to measuring heights relative to the earth's surface, not its centre. It's obvious that adding a mere metre to the band doesn't increase its circumference noticeably, so, intuitively, it shouldn't increase the height of the band noticeably either. And it doesn't, if you measure height relative to the centre of the earth, but people generally don't, which is why it is counterintuitive.

[cdhostage]

8 cm diamter-sphere's volume left? Hmm.

Is this another trick of the mind like the band strecthing around the Earth? Nah! There's no numbers given. I'll make up some numbers.

1000 m - diameter sphere.
1 m - dimater tunnel straight through its heart.

Ignoring the fact that there's a slight fudge factor (I'll calcuclate for a plain old cylinder - dunno how to figure out the curved ends), here goes:

The volume of the undistbured sphere is 3.14 (500) 500 (500) =3925000 cubic meters.
The volume of a rod 1000 cm in length and 1 m in diameter is 3.14 (.5) .5 (1000) = 785 cubic meters.

The remaining volume is 3924215cubixc meters.

I'm lazy and am not doing any more math, but that's more than the volume of an 8 cm sphere.

[Achernar]

Originally posted by cdhostage
There's no numbers given. I'll make up some numbers.

Indeed, there is one number given, and you managed to ignore it! The hole has to be 8cm long.

Okay, here's what I got. It seems to give the right answer. The sphere with a hole through it can be characterized by two numbers, the radius of the sphere, which I'll call R, and the radius of the circular hole, which I'll call d. If you draw a diagram, I think you'll see that the length of the hole will be given by 2x, where x² + d² = R². For the problem as stated, x = 4cm. Now, the volume that's left will be the total spherical volume, minus the volume of the cylindrical core, minus the volume of the caps on the end:

V = V_S - V_C - 2V_E

V_S = 4/3 pi R³

V_C = pi d² × 2x = 2pi R²x - 2pi x³

Those were easy enough. I think there's a way to get V_E without calculus, but I couldn't figure it out. I did it by determining the volume of the spherical sector defined by the circle at the end, minus the cone that you'd have to subtract, and I got this:

V_E = 2/3 pi R³ - pi R²x + 1/3 pi x³

Put it together and what've you got?

V = 4/3 pi x³

[Hari Seldon]

Originally posted by Cisco
Ursam - You don't say anything unless the other two are wearing identical hats, in which case you say the opposite color. Of course, this doesn't work in the case that you all three got the same colored hat, but it gives you a 75% chance.

Hat's off to Cisco.

One analysis to show that 75% is the best possible to point out that you are really guessing and 50% of all the guesses will be wrong. But you have arranged it so that when there is one wrong guess, they all guess wrong so there will be three wrong guesses for every three right guesses and you cannot improve on that. With seven people, there is a strategy so that one time in 8, all seven guess wrong and the other seven times there is one correct guess and no wrong ones. There are similar strategies for every number that is one less than a power of 2. For other numbers, such perfect strategies do not exist, but very good ones do.

[RM Mentock]

You are going to carry apples to a applesauce factory 1000 furlongs away, by horse. The horse has to eat one apple per furlong, and can only carry 1000 apples at a time. If you start with 3000 apples, how many apples can make it to the factory?

[jackelope]

Originally posted by RM Mentock
You are going to carry apples to a applesauce factory 1000 furlongs away, by horse. The horse has to eat one apple per furlong, and can only carry 1000 apples at a time. If you start with 3000 apples, how many apples can make it to the factory?

As near as I can figure late at night and half-lit, I think it's 500.

Possibly 750 though; it's clear that the entire enterprise needs to be figured in 250-furlong units.

[Nightime]

I'm barely awake right now, but I get 500 too.

[RM Mentock]

You can eke out a few more, apparently. Not near as many as 750, however.

[Nanook of the North Shore]

How do you come up with 500?