[Math Puzzle] Day10

Going to guess that the area of private points equals the area of one of the spheres. Can't prove it though.

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.

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

the total area of the private points is equal to the area of one sphere. Not sure how to prove it though

if you have 2 spheres, then you have half of each sphere as a set of private points

if you have three spheres, and you arrange them into a regular shape in space, it's fairly obvious that the total area of the private points is equal to the area of one sphere. You can see that this holds as well as you move around the three spheres, as it holds for everything from a regular triangle to all 3 balls in a row.

if you have four spheres, it holds for a square, as well as any other configuration that you put the four spheres into.

Not sure how you prove this more rigorously, but by inspection, you get the area of the private points = area of one sphere.

edit: basically the same answer as Khenra -_-. Not sure if it's right or wrong or anything though, as we both used these hand-waving proofs that basically say: since it worked for this configuration, by thinking about it, this should work for all possible configurations of n spheres. Probably not going to cut it.

Without finishing the problem, I think you could start like this:
Have two non-overlapping spheres. The private points are the surface area of one sphere. If you can show a 3rd sphere, placed anywhere, will have the same private area as it makes public in other spheres, then you've got something to work with.

simple example: place the third sphere colinear with the first two. It will make one hemisphere of one of the first two public, but one hemisphere of itself will be private.

My best guess is that all the private points are equall to the surface of one sphere. You can rearange all n sphere in a row on a horizonal plane and notice half of the shpere that begins the row together with half of the sphere that ends it are private points.

I don't really know a way to prove this though. Although I guess the area of all the private points is constant even if n varies. And if that's the case then the area of private points for n spheres is equal to that of 2 spheres.

I'll just wait for someone really smart to solve this.

Take a giant shape around all of the spheres and shrink it so, that in the end you have some spherical areas at the corners of the shape and the rest are straight lines between the spheres. The giant shape can only bend around the corners so in the end it won't be hollow from any spot. Now if you start moving on this shape from one point in such a way that if u move in a straight line, in the end you will be back in one spot, then you will have moved along a curved area of exactly 1 circle, since you can only turn while moving along a spherical area. The area of the spherical areas on this shape is together the area of the private points, since no straight line from the other spheres can reach it and every point inside the shape can be seen by at least 1 sphere. If you just remove all of the straight areas from the shape and put the remaining shapes together, you get a sphere, because if you had moved from any of the points on shape so that u get back to where you were before you would have turned along exactly 1 circle and if u remove all of the straight lines, the turning points are all that remain. Also note that in the starting shape, from one of the private areas, it was possible to move straight in any direction in such a way that you would be back to the starting point. Only a sphere has such attributes.

Draw any plane in space somewhere far away from the spheres and move the plane in the direction of the spheres so that it remains parallel to its previous positions. The plane will first touch the spheres at some point A on some sphere, and last touch the spheres at some point B on some sphere. Then A and B are the only private points whose tangent planes are parallel to the original plane. Consider all possible directions for starting planes and we are done.

What Ninjafetus suggested is essentially a proof by induction, once you have shown that the private area of 2 sphere is equal to the surface area of one sphere, you go on and assume it to be true for n spheres and then prove that it's true for n+1 spheres. I don't think you can actually do that with words, since the orientations of the spheres have infinite variations, so instead I guess the easiest way would be to show it in a graph, then proving it geometrically.

answer=surface area of one sphere
proof- use induction
base case (n=1) area of private points = surface area of 1 sphere since there's no other sphere to 'see' the first sphere

induction: suppose the answer holds true for k spheres. color private points red in this k-sphere world. now add (k+1)th sphere.

since all the spheres are of the same size, given any 2 spheres, the areas of private points for each sphere are exactly the same. (in the same way, areas of non-private points in each sphere are the same too) ---------- (A)

by introducing (k+1)th sphere to the k-sphere world, some of the private points in the k-sphere world might become non-private points. color such points blue, then color the new corresponding private points on the (k+1)th sphere yellow.

by (A), sum of blue = sum of yellow

on the (k+1)th sphere, non-yellow points are private points since we colored all non-private points yellow.

so area of private points in k+1 world
=area of private points in k world - (private points in k but not in (k+1) world) + private points in k+1
=area of one sphere - blue + yellow
=area of one sphere

the way the question is posed, it leads me to believe that there is only 1 possible surface area of all private points. working under this assumption, I can consider the most simple example and just run with it :p

so assume all the spheres are of radius r and exist in the same cylindrical space of radius r somewhere in space. in this scenario, only the 2 end spheres would have any private points, and they would both have exactly half their surface area covered in private points. thus, in this scenario, there would be 4 * r * pi ^ 2 (surface area of a sphere) are worth of private points.

of course, proving that this will be the case for all scenarios is more difficult, but again, the problem's wording really lets us use this as a jumping off point for our thought process. now, of course with that most simple solution, we could have any number of spheres and there would be the same area of private points, which leads to consider some form of induction for proving this. Forgive the informality of this proof:

trying to prove: for any number, n, of spheres of radius, r, in space, there will be exactly 4 * r * pi ^ 2 area of "private" points.

base case: 1 sphere. all points are private as there are no other spheres, so there is 4 * r * pi ^ 2 area of private points

inductive step: if i spheres have 4 * r * pi ^ 2 area of private points, i + 1 spheres also have 4 * r * pi ^ 2 area of private points.

proof: for each area that was private in the i case and not in the i+1 case, there is an equal area on the backside of the i+1th sphere that must now be private. proving this is not entirely trivial, but I don't really have the time to right now (at work :\). it's also necessary to prove the not-quite trivial fact that the rest of the points on the i+1th sphere are not private... I suppose I'll give the proofs of those 2 things some thought, but I know them intuitively to be true...

What he did is to construct a map from the grassmanian Gr(2,3) to the set of private points.
Identify Gr(2,3) with Gr(1,3) = P(R^3)=S^2/+-1.

Now except for a finite number of points, this application is etale and has for differential the homotethy x*Id. Hence the area of the private points is x*Surface(S^2) (since the application is doubly valued).

[Edit: Oops, i meant "except for a *finite union of 1-dimensional subvarieties*, the application is etale blabla"... Thanks Muirhead for the correction, this will teach me to only make drawings in dimension 2...

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.