Covering a unit ball with balls half the radius This is a direct (and obvious) generalization of the recent MO question, "Covering disks with smaller disks":

How many balls of radius $\frac{1}{2}$ are needed to cover completely a ball of radius 1?

The answer may be in the paper "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n$,"
by Jean-Louis Verger-Gaugry, which I cannot immediately access (Springer link here).
If anyone knows the answer for $\mathbb{R}^3$, I'd be curious to learn of it.
And it would be especially interesting if there were a proof as satisfying as  Noam Elkies's for seven half-disks covering one in $\mathbb{R}^2$.  Thanks!
Update. [29May2012] A series of contributions by Will Jagy, Gerhard Paseman, Karl Fabian, and zy, have reduced the upper bound from $56$ to $33$ to $22$, with a lower bound of $16$.
Update. [5Aug2012] Ed Wynn settled the question with a careful analysis: $21$ balls are needed, and suffice!
 A: Alrighty, it can be done with $56$ in an evident pattern. Take a four by four by four cube of smaller cubes, each of those of edge $1 / \sqrt 3$. Circumscribe each with a ball of diameter $1$, or radius $1/2$. These then cover the larger cube. Place the ball of diameter $2$, radius $1$, centered at the center of the large cube. It is covered by the smaller balls. Now remove the eight corner cubes.  These are superfluous, the spheres circumscribing these were actually tangent to the large ball. $64 - 8 = 56$.

        [Graphic added by J.O'Rourke:]
        
The same method with truncated octahedra may be a little worse, maybe a little better. 
Either way, it makes the soccer ball thing seem a bit unlikely.  But it would be nice to see some pictures.  


*

*http://en.wikipedia.org/wiki/Bitruncated_cubic_honeycomb 

*http://en.wikipedia.org/wiki/Truncated_octahedron 
The third method coming from the same idea, 


*

*http://en.wikipedia.org/wiki/Rhombic_dodecahedron#Honeycomb
EDIT, Monday morning. Thanks for the figure, @Joseph. I looked up some things in SPLAG by Conway and Sloane, pages 31-34. The best (most efficient) lattice covering is traditionally called the body-centered cubic lattice, proved optimal by Bambah in 1954. The Voronoi cell is the truncated octahedron, as mentioned. Note that this polyhedron can be inscribed in a sphere, in that all vertices are the same distance from the center, true for the cube also but false for the rhombic dodecahedron. Indeed, the lattice for best covering is dual to the lattice for best sphere-packing. If we had very small radii instead of 1/2, the truncated octahedron covering would certainly win. As we have radius 1/2, hard to say. It is also uncertain whether best to put a vertex at the origin, as I did with the cubes, or the center of one of the cells, which is computationally easier, by hand anyway.  
A: Extending the idea of W. Jagy, 
this is a Mathematica code visualizing that 33 spheres with radius 1/2 centered at the origin and the midpoints of the faces of a soccerball with circumradius  3/4 cover the unitsphere.
coord = PolyhedronData["TruncatedIcosahedron", "VertexCoordinates"];
faces = PolyhedronData["TruncatedIcosahedron", "FaceIndices"];
f6 = Select[faces, Length[#] == 6 &];
f5 = Select[faces, Length[#] == 5 &];
len = Norm[coord[[1]]] // Simplify;
Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], 
   Sphere[ Mean[coord[[#]]], len*2/3] & /@ f6, 
   Sphere[ Mean[coord[[#]]], len*2/3] & /@ f5, Opacity[1], 
   Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], 
   PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]

        [Graphic from the above code added by J.O'Rourke:]
        
Plotting only three outer spheres as in
Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], 
   Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f6, 2], 
   Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f5, 1], Opacity[1], 
   Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], 
   PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]

        [Graphic from the above code added by J.O'Rourke:]
        
shows that the two intersection points of three neighboring outer spheres
lie either outside the sphere with radius 1 or inside the sphere of radius 1/2.
This can easily be made rigorous by calculation:
mcl = Chop[ 
   Join[Mean[coord[[#]]] & /@ Take[f6, 2], 
    Mean[coord[[#]]] & /@ Take[f5, 1]]];
erg = Solve[(Norm[{x, y, z} - #] == len*2/3) & /@ mcl, {x, y, z}] // N;
((Norm[{x, y, z}]/len*3/4) /. #) & /@ erg  
that yields the distances 
{0.216794, 1.06042}
A: Here is an idea which should generalize to dimensions 2 and greater.  I will start with dimension 2.
Let us place a circle of radius  1/2 in the center of the radius 1 ball.  We will place most, if not all, of the rest of the balls at a distance such that the center of the small ball is sqrt(3)/2 from the center of the large ball.  This placement is chosen so that the angle of arc cut out of the two concentric circles is the same, which turns out to be 60 degrees.  Now a convexity argument should show that every thing between the 60 degree arc on the small circle and the corresponding arc on the large circle will be covered by the same ball.  The general covering problem is now reduced to a covering of the surface of the smaller (or the larger) sphere by circular caps which extend 60 degrees of arc
For n=2, this is a matter of taking the ratio 360/60.  For n=3, I propose 6 caps around the equator, and for each hemisphere 6 more caps appropriately spaced with centers at latitude 30 degrees, and 6 more at latitude 60 degrees, sharing central longitude lines with the equatorial circles.  Even if I messed up and two polar circles are needed, that gives a total of 33 spheres, but I think 31 balls suffice.
I am not familiar with higher dimensional sphere coverings, so I'll let someone else take over.  I imagine that someone else can come up with a lower bound based on this style of arrangement. (Hey Noam Elkies, care to try out more dimensions?)
If Joseph understands this, maybe we will be graced with a few illustrations of it.
Edit 2012.05.31 I decided not to wait any longer for Noam Elkies.  Here is my idea of a lower bound argument.  It can probably be extended to open balls; I prefer to use compactness and closed balls for simplicity.
Let there be a covering of the closed unit ball by finitely many closed balls of radius 1/2. Any covering ball which contains the center, call it c, of the unit ball contains at most one point, call it p, on the boundary B of the unit ball.  Since B minus p is open with respect to B, p is contained in one of the other covering balls which does not contain c.  So we can assume the boundary B is covered by balls none of which contain c.  The covering now has a finite number of balls which cover B plus at least one more ball covering c, and perhaps others.
Now replace the covering above with a new (perhaps identical) covering: shift each ball toward or away from c so as to maximize its intersection with B.  This places each covering ball center at distance sqrt(3)/2 from c.  B is still covered, and this new covering along with a ball of radius 1/2 placed with its center also at c is another (perhaps the same) covering with the same or fewer number of balls.  Thus the posted problem is (essentially) the same as optimally covering B with caps of spherical radius of 30 degrees.
Elsewhere I noted Neil Sloane had a cover of a 3-d sphere with 20 caps each of radius slightly less than 30 degrees.  I now claim an
upper bound of 21 for the posted problem. Assuming Sloane's expertise with sphere packing, I expect 21 to be an exact bound.
You can ask him for the covering number for dimensions greater than 3.
END Edit  2012.05.31
Gerhard "Ask Me! About System Design" Paseman, 2012.05.27 
A: Yes, the paper does cover that question, and gives bounds. However, programming the author's upper bound in mathematica reveals that in three dimensions you need at most $-50$ balls, which looks a bit suspicious. On the other hand, the author cites earlier results of Rogers, which only works for covering balls of radius at least $3/(2\log 3)\approx 1.65$ by balls of radius $1/2.$ Anyway, you can try to make sense of all this yourself.
