# Interesting math problem

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?

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.

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.

I don’t know. What’s 15.9 cm tall?

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?

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.

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.

Grrr those other responses weren’t the when I started :mad:

I think you missed something there, Manduck.

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?

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.

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

FWIW

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.

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.

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.

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?

? 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.

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…

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 