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.
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.
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.
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 x[sup]2[/sup] + d[sup]2[/sup] = R[sup]2[/sup]. 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[sub]S[/sub] - V[sub]C[/sub] - 2V[sub]E[/sub]
V[sub]S[/sub] = 4/3 pi R[sup]3[/sup]
V[sub]C[/sub] = pi d[sup]2[/sup] × 2x = 2pi R[sup]2[/sup]x - 2pi x[sup]3[/sup]
Those were easy enough. I think there’s a way to get V[sub]E[/sub] 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[sub]E[/sub] = 2/3 pi R[sup]3[/sup] - pi R[sup]2[/sup]x + 1/3 pi x[sup]3[/sup]
Put it together and what’ve you got?
V = 4/3 pi x[sup]3[/sup]
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.
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.
I’m barely awake right now, but I get 500 too.
You can eke out a few more, apparently. Not near as many as 750, however.
How do you come up with 500?
Carry 1000 apples to 250 furlongs, leave 500, use rest of apples to go back. Repeat. Repeat. We’re now at 250 furlongs with 1750 apples. Repeat twice. We’re then at 500 furlongs with 1000 apples. Head for the factory, arriving with 500 apples.
I can get a few more apples to market…
Start by taking 1000 apples.
1- Go 200 furlongs, leave 600 apples there, and go back. The horse thus eats 400 apples.
2- Get another 1000 apples, go 200 furlongs, leave 600 apples there, and go back. Another 400 apples eaten.
3- Take the last 1000 apples, and go 200 furlongs. The horse has eaten 200 more, so now you have 2000 left and 800 furlongs to go.
4- Load the horse up at the 200 furlong mark so you have 1000 on the horse and leave 1000 behind. Go 333 1/3 furlongs, leave 333 1/3 apples behind, and go back 333 1/3. The horse will have consumed 666 2/3 apples.
5- Pick up the 1000 apples left at the 200 furlong mark and go 333 1/3 furlongs. The horse will have eaten 333 1/3 apples, leaving you room to pick up what you left there earlier.
6- Now you have exactly 466 2/3 furlongs to go and 1000 apples. You will arrive with 533 1/3 apples.
There ya go. Until you get down to just two loads, each furlong is going to require five trips (five apples per furlong), so the first 1000 apples can get you 200 furlongs. Etc.
Very nice problem, RM. I enjoyed it.
Quoting RM Mentock
I read every post and I am not sure if anyone had the confirmed correct answer.
This may sound stupid but the material left of the sphere will be tha same amount as the material thats to the right of the sphere.