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[Math Puzzle]Day 6

Blogs > evanthebouncy!
Post a Reply
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2009-05-02 17:04:04
May 02 2009 04:22 GMT
#1
<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...
[image loading]

Poll: Did you get yesterday's question?
(Vote): Yeah!
(Vote): No, darnit


*****
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
illu
Profile Blog Joined December 2008
Canada2531 Posts
Last Edited: 2009-05-02 04:31:04
May 02 2009 04:30 GMT
#2
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...
[image loading]

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.
:]
Physician *
Profile Blog Joined January 2004
United States4146 Posts
May 02 2009 04:36 GMT
#3
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?
"I have beheld the births of negative-suns and borne witness to the entropy of entire realities...."
Kentor *
Profile Blog Joined December 2007
United States5784 Posts
Last Edited: 2009-05-02 05:44:37
May 02 2009 05:41 GMT
#4
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.
Everywhere
Profile Blog Joined March 2009
United States97 Posts
Last Edited: 2009-05-02 06:24:38
May 02 2009 06:12 GMT
#5
+ 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....
sigma_x
Profile Joined March 2008
Australia285 Posts
May 02 2009 06:51 GMT
#6
+ 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).

evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
May 02 2009 17:03 GMT
#7
<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.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
Byo
Profile Blog Joined July 2007
Canada207 Posts
May 02 2009 17:59 GMT
#8
+ 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.


Plutonium
Profile Joined November 2007
United States2217 Posts
Last Edited: 2009-05-03 03:57:42
May 03 2009 03:56 GMT
#9
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.
Deleted User 3420
Profile Blog Joined May 2003
24492 Posts
Last Edited: 2009-05-03 04:31:29
May 03 2009 04:29 GMT
#10
im not sure how one would even begin to answer this question lol
at least I think I understand this one lol
Deleted User 3420
Profile Blog Joined May 2003
24492 Posts
May 03 2009 04:38 GMT
#11
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
Ivs
Profile Joined January 2008
Australia139 Posts
May 03 2009 05:38 GMT
#12

+ 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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