Two Geometric Problems with Elegant Solutions?

1) Consider a unit sphere with the center in P. Mark the points of its intersections with the rays. Note that the distance between any two marked points is fixed because its the third side of a triangle with two unit side and a fixed angle between them, let's denote it R. Restating the same thing we say that on a sphere with center in any marked point of radius R lie all the other points, so now we are to maximize the number of such spheres. Taking three of these spheres with three corresponding marked points we notice that they have only 2 common points with the distance between them more than R (equal to Sqrt(8/3)*R), so only one of them can be in the construction with no more than 4 totally.

For problem number 1, in planar case, say we have k rays r_1, r_2, .. r_k. Let's call the angle between any two of them phi. Start with r_1, and we put r_2 phi apart from it. The rest of the rays r_3, r_4, ... all are same angle away from r_1 and r_2. But there is only one ray with such property which proves r_3, r_4, ... all overlap in one ray. Similarly in three dimensions, start with r_1 and place r_2 phi away. This forces you to place r_3 phi away from both to form a tripod. Now r_4, r_5, .. r_k should all be phi away from the three. But this only happens when all of them collapse in one ray, because there is only one ray with equal angle to each of the three rays. Therefore maximum is 4.

Problem two will be solved if we can find an example of such octahedron. Because of symmetry of the regular octahedron, we know its inscribed sphere should have its center at the geometrical center of the regular octahedron. Here is the best picture I could draw:

http://img147.imageshack.us/img147/3360/octahedron...

Also because of symmetry, we can show the unit sphere will touch each face on its perpendicular bisector. These are the red lines in the picture. Then OC = 1 because it's a unit sphere. Let's say the sides of the octahedron is a. AB = sqrt(3)/2 a and OA = sqr(2)/2 a. Then 2* area of AOB = OB*AO = a/2 * sqrt(2)/2 a= a² sqrt(2)/4 = OC*AB = 1 * sqrt(3)/2 a. a = sqrt(6). Then Volume of the regular octahedron is given by (1/3) sqrt(2) a^3 = 4sqrt(3) ~ 6.9. I have tried to inscribe the sphere in a "hexagon tower" whatever it's called but it gave me the exact same volume 4 sqrt(3)!

I'm still thinking how to reduce this volume. I'll update my answer. If you find any errors please let me know.

By the way, if you want elegance and neatless see Zo Maar's answer here:http://answers.yahoo.com/question/index;_ylt=AjyDH... I always learn from him.

EDIT: Wow so you found more 4sqrt(3)'s!

EDIT: 3 * Volume = Surface_area * Inscribed_sphere_radius

Am I the only one that doesn't see why?! It is a neat formula/theorem. Does it have a name or proof? Using it things would be a bit simpler. See, the problem with "non-standard" octahedrons is that the computation of their volume gets messy.

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