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<bump>
Guys I don't care if you can do it or not. I'm not posting these just to get the right answers since I already solved them all...
If I'm posting it and ONLY getting couple people posting the right answer(that's only the answer and didn't show any approaches, i'd be just reading the answer book all over again) and no one TRYING to solve it, I'll be super depressed. Just start typing some approach you did, I don't care if it is right, and what problem you ran into... please.
Last day's puzzle was first solved by Rhaegar99, Spiritofthetuna should be mentioned because he said the answer with a "?" so I'm not sure if he committed :p
+ Show Spoiler [solution] +
He lost 97.
Neighbor gave 100 and got 100 back. No gain/loss there.
Kid gave nothing(fake bill) and gained 79 worth of change, and 18 worth of toy(raw cost). $97 gain.
So the vendor would've lost 97 dollars.
Today's puzzle would be:
You have a sphere with 5 points randomly distributed on its surface. Show that there is a hemisphere[boundary included] on the sphere that will contain 4 points in it.
Again, put answer in spoilers, and we'll see!
-gl hf!
Clarifications:
did u mean what is the probability that 4 out of the 5 randomly chosen points on a sphere will all lie in a single hemisphere?
-no, absolutely
extra:
I should put a poll regarding last day's question...
Poll: Did you get yesterday's question?
(Vote): Yeah!
(Vote): No, darnit 
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On May 02 2009 13:22 evanthebouncy! wrote:Last day's puzzle was first solved by Rhaegar99, Spiritofthetuna should be mentioned because he said the answer with a "?" so I'm not sure if he committed :p + Show Spoiler [solution] +
He lost 97.
Neighbor gave 100 and got 100 back. No gain/loss there.
Kid gave nothing(fake bill) and gained 79 worth of change, and 18 worth of toy(raw cost). $97 gain.
So the vendor would've lost 97 dollars.
Today's puzzle would be: You have a sphere with 5 points randomly distributed on its surface. Show that there is a hemisphere[boundary included] on the sphere that will contain 4 points in it. Again, put answer in spoilers, and we'll see! -gl hf! extra: I should put a poll regarding last day's question... Poll: Did you get yesterday's question?( Vote): Yeah! ( Vote): No, darnit
+ Show Spoiler +Easy.
Choose two points and draw a line across these two points on the sphere. This cuts the sphere into two halves - call them A and B where A and B excludes the boundary. We will have the following cases:
1. No other points fall on the line. Then we have three points leftover, distributed in A or B. Therefore, A:B is 3:0, 2:1, 1:2, or 0:3, where, for example, 3:0 means A has three points and B has none, etc. Then we choose A or B that has at least two points, in addition to the two points on the boundary by construction, we get a solution.
2. There are three points on the line. Then A:B is 2:0, 1:1, or 0:2. Choose A or B that has at least one point.
3. The case where there are more than three points on the boundary is completely trivial.
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Physician
United States4146 Posts
did u mean what is the probability that 4 out of the 5 randomly chosen points on a sphere will all lie in a single hemisphere?
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Kentor
United States5784 Posts
no i think he wants you to prove that given 5 random points, at least 4 of the points will be on a hemisphere.
well... that's pretty intuitive.
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+ Show Spoiler +5 dots
pick any 2, and draw a line connecting them.
trace this line across the entire sphere. this defines your hemispheres.
the other 3 will be on the line = 5 in one hemisphere
OR 2 on the line, 1 on a side = 5 in one hemisphere
OR 1 on the line, 2 on a side (2-0, 1-1, 0-2 <---whichever side doesn't matter) = 4 or 5 dots in one hemisphere
OR 0 on the line, 3 on a side (1-2, 2-1, 3-0, 0-3 <---whichever side doesn't matter) = 4 or 5 dots in one hemisphere
**sorry my explanation is kinda crappy -_-. im tired as hell for some reason and am going to bed now....
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+ Show Spoiler +pick 2 points and draw a great circle. If a third point lies on the great circle, then you're done. So assume a third point does not lie on the great circle. Then either the other three will lie on one side (in which case you are done) or two will lie on one side (in which case you are also done).
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<bump>
Guys I don't care if you can do it or not. I'm not posting these just to get the right answers since I already solved them all...
If I'm posting it and ONLY getting couple people posting the right answer(that's only the answer and didn't show any approaches, i'd be just reading the answer book all over again) and no one TRYING to solve it, I'll be super depressed. Just start typing some approach you did, I don't care if it is right, and what problem you ran into... please.
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+ Show Spoiler +
Here's mine......
The opposite of a hemisphere (boundary included) containing 4 points is a hemisphere (boundary excluded) containing only 1 point.
Given any 2 points, the only case where there does not exist a hemisphere boundary excluded that contains both points is if they are polar opposites. And there can only be 1 polar opposite thus there always exist a hemisphere boundary included that contains 4 points.
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I think I can answer this pretty clearly. The other people probably got it right, but maybe you get points for clarity.
+ Show Spoiler +Any two points on a sphere can be covered by a single hemisphere, because the maximum distance on the surface they can be from each other is opposite each other on the sphere. Any hemisphere with one point on the border will border the other.
Because those two points at maximum distance will form a line through the center, we can rotate the hemisphere along that axis and still contain the first two points. If you throw in a third point, we can rotate the hemisphere to contain it.
Throw in a fourth point, and we can still rotate the hemisphere along the same axis to contain both it and the third, as the maximum distance it will be away from the third point will be directly across it on the sphere, which means the hemisphere would border all four points.
Whether or not the fifth point is contained is irrelevant.
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im not sure how one would even begin to answer this question lol
at least I think I understand this one lol
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wow, im high as a basket full of heck and this question is making my brain feel weird to ponder
if there are infinite points on a sphere, are there infinite hemispheres in a sphere?
and if there are infinite hemispheres, how can u cut the sphere in half in the first place
ugh
its a crazy world we live in
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+ Show Spoiler +Take any two points, draw the circle through it with the same centre as the centre of the sphere. The circle divides the sphere into two hemispheres. There are 3 leftover points, by the pigeon hole principle two must lie on the same hemisphere. Including the original 2, there are 4 on that hemisphere.
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EPT
