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On May 13 2009 01:13 evanthebouncy! wrote:Show nested quote +On May 13 2009 01:10 Khenra wrote:+ Show Spoiler +The answer is: the area of all private points is equal to the area of the surface of one sphere. Example:
If you have 2 spheres, half of both spheres is private (this is easy to imagine).
If you have 3 spheres, and align them in a triangle, 1/3 of each sphere is private. If you however put 1 of the spheres directly inbetween the 2 others, one sphere has no private points but the other 2 spheres each have 1/2.
This rule remains true for n spheres, having an average private area of 1/n * (area of surface of 1 sphere) per sphere, totalling n/n = 1 area of surface of a sphere.
+ Show Spoiler +
you have yet to prove why you can freely re-align them to different shapes? how come such operation do not change the total amount of private areas?
by your definition i have a better proof, say, i have n spheres, line them up in a role, and it's obvious total private area is 1 sphere, but what gives me the right to freely move the spheres? that's something you'd have to prove
I knew it wasn't proof by any means, but as you said in the opening post, at least I took a shot at it. I have never been a star at proving mathematical issues, I'm afraid 
Nice work, to the guys who posted above me. I take it you guys have studied mathematics?
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On May 13 2009 03:38 Muirhead wrote:+ Show Spoiler +Here's a more formal sketch
I defined a "map" F from the real projective plane to pairs of points on the spheres. The map F is well-defined unless the moving plane's final point of contact or initial point of contact with the spheres is shared by more than one sphere (i.e. your A.). One can see that the set of points in the real projective plane which are messed up in this fashion is a finite union of curves (subsets of projections of great circles on the sphere). Thus, in reality F maps an open, dense subset of the real projective plane onto pairs of points on the spheres. If n is the total number of spheres, there is a natural n-fold covering map G from the n spheres onto a single sphere. One sees that F followed by G is an "inverse" for the canonical projection from the sphere to the projective plane. EDIT: Gondolin got there first ^^
which class u learn these cool stuff, i'm taking algebraic geometry nxt semester
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On May 13 2009 06:22 Khenra wrote:Show nested quote +On May 13 2009 01:13 evanthebouncy! wrote:On May 13 2009 01:10 Khenra wrote:+ Show Spoiler +The answer is: the area of all private points is equal to the area of the surface of one sphere. Example:
If you have 2 spheres, half of both spheres is private (this is easy to imagine).
If you have 3 spheres, and align them in a triangle, 1/3 of each sphere is private. If you however put 1 of the spheres directly inbetween the 2 others, one sphere has no private points but the other 2 spheres each have 1/2.
This rule remains true for n spheres, having an average private area of 1/n * (area of surface of 1 sphere) per sphere, totalling n/n = 1 area of surface of a sphere.
+ Show Spoiler +
you have yet to prove why you can freely re-align them to different shapes? how come such operation do not change the total amount of private areas?
by your definition i have a better proof, say, i have n spheres, line them up in a role, and it's obvious total private area is 1 sphere, but what gives me the right to freely move the spheres? that's something you'd have to prove
I knew it wasn't proof by any means, but as you said in the opening post, at least I took a shot at it. I have never been a star at proving mathematical issues, I'm afraid Nice work, to the guys who posted above me. I take it you guys have studied mathematics?
yeah that's the whole ponit <3
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+ Show Spoiler +
Cleanest way I can show this:
Each sphere is a translation of a sphere of the same size from the origin, call that sphere S. For each point on S, there corresponds at most one private point on any of your spheres. I assume the spheres themselves do not intersect.
Proof: assume there are at least two such corresponding private points on the spheres. Draw a line connecting both of them (selecting 2). If the line segment does not go into the interior of the sphere, then it is not a private point. However, if we consider a translation of one sphere to the other sphere (which we can do since we are moving along corresponding points), it is impossible for the line segment to go inside both spheres. Therefore there is at most one private point corresponding to a point on S.
Now we show that there is at least one private point corresponding to a point on S.
Proof:
Draw tangent planes of each of the corresponding points on every sphere. The planes are parallel. If we reorient space to make these planes tangent to the z-axis, there is obviously a plane with the greatest z value. The point corresponding to that highest plane is a private point since the tangent plane will have all of the spheres on one side.
Therefore it is one to one between the private points on all the spheres and a sphere of the same radius, so the area of the private spheres is the area of one of the spheres.
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EPT
