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[Fossaman]

If I've got a right triangle (ABC) where B = 90 degrees, and I'm told the coordinates of B are (10,8), and A's are (2,4). How do I get C's coordinates if I know it sits somewhere on the x axis (i.e. C(x,0) )?

Since I didn't really comprehend what the post above mine was talking about, I figured you might not either. Here's what I would do:

*Find the equation of line AB
*Use this to find the equation of the perpendicular line BC
*Solve the equation of line BC for y=0

Point-slope form is probably going to be your friend here, although I'm sure there are other methods that would work.

[Darvi]

（・Ａ・）

[Christes]

So I'm in the process of studying for quals, and I'm trying to regularly do problems.  Here's an interesting little problem that I can seem to find the right approach for:

Let G be a finitely generated group where every nontrivial element is of order 3.  Prove that G is finite.

I can show it for small numbers of generators by writing out strings of generators and reducing, but I'm not sure about the general case.  I have seen the proof before, and remember that it was slick, but can't seem to reproduce it.

[ZetaX]

I managed to prove it by induction on the number of generators. It suffices to show:
Let G be a group such that x³=e for all x in G, let H be a finite subgroup and let y in G. If G is generated by y and H, then G is finite.
Sketch of proof:
For any h in H we have (yh)³ = e = y³h³, thus hyhy=y²h², giving yhy = h'y²h² = h'y'h' (with ' denoting inverses). This shows that whenever we have something like yhy we can write it as ay'b with some a,b in H. Similiarily, we can always replace y'hy' by ayb for some a,b in H. Now yhy' = (yhy)y = ay'by and similiarily y'hy = ycy'd for some a,b,c,d in H.
Let's conclude: we have "replacement rules" of the following types, where X always stands for an arbitrary element of H, two X's not necessarily meaning the same element:
a) yXy      -> Xy'X
b) y'Xy'    -> XyX
c) XyXy'X -> Xy'XyX
d) Xy'XyX -> XyXy'X

Now every element g of G is by assumption something of type g = XYXYX...XYX, where the X's denote (possibly different) elements of H and each Y is either y or y'. Then by applying the rules a) and b) we can combine neighbouring Y's that are the same, and using c) and d) allows us to interchange y and y' (possibly changing the X's in the process, but that's irrelevant). We can do this until at most one y and at most one y' remains, and we may assume that if both occur, then y is to the left of y'. Thus g is of type X or XyX or Xy'X or XyXy'X, where there are only |H| many options for each X. This shows that |G| <= |H|+|H|²+|H|²+|H|³, which is finite.


Edit: And using that inequality and Cauchy's theorem, one actually gets the inequality |G| <= 3^(3^(n-1)), n denoting the number of generators.

[Christes]

And that was exactly what I was trying to do.  Well done!  I'm sure that's not the proof I saw before, but I'll take it.

[Fossaman]

I'm trying to distribute a number of points on the surface of a sphere so that they're equidistant from their adjacent points, as well as symmetric, so that each eighth of the sphere has the same pattern on it. Anybody have any idea how I would go about this? My best idea at this point is taking a compass to a ball of some sort, and I don't see that working very well.
The sphere has a diameter of 56 units and the points need to be < 11 units apart.

[ed boy]

You're going to be more thorough about what you mean by 'adjacent points'. I'm going to assume for now that by 'adjacent piont' you mean the closest one, which results in each point being some distance d from the next closest one. If you mean something different by adjacent, then the answer will change.

You also need to be more specific about what you mean by symetric. I'm going to assume roational symetry, but again a change would result in a different answer.

Since each octant needs to be symetric, we will work out what the positive octant must be (that is, if the sphere is centred at (0,0,0), the octant in the region of positive x,y and z).

For any one point, we can imagine a circle of diameter d centred at the point. We must find a bunch of circles of radius d/2 such that each circles has at least one tangential circle, and no two circles intersect. We alse require that d is less than or equal to 11.

The simplest solution is to have no points, but you will likely not find that helpful.

For each d, we can find an a such that 28a=d. d is the arc length between points, so a is the angle in radians between the points.

You can make an infinite number of solutions be having each of the points have its nearest neighbour on the same octant. Just make sure that there are at least two points on the octant, and each point has a distance of at least a/2 from the edge of the octant.

Failing that, you can find solutions where there are points that make an angle of a/2 from the edge. As long as each point has at least one point of distance d on the same octant, or has a corresponding point an angle of a/2 from the opposite edge, then it will work.

[Fossaman]

You're going to be more thorough about what you mean by 'adjacent points'.

The closest points, yes. Perhaps 'neighboring' would be a better term to use?

You also need to be more specific about what you mean by symetric. I'm going to assume roational symetry, but again a change would result in a different answer.

Yeah, rotational symmetry. Octahedral, if you want to be really specific.

Using 28a=d, 2pi/a needs to equal some multiple of 4 in order to place points symmetrically, since any given ring of points will only pass through four octants. So a value of d=11 gives me 2pi/a=16 to a working approximation. (This is for lighting the surface of a spherical chamber in Minecraft, so it's not like I need to be precise past one decimal place.) So that's handy.

Here's where I start getting muddled. How do I set up the math to go from the known location of my first point to its neighbors? I've got some sort of hazy conception that I'd need to convert the angle to slope in the equation of a line, but I have no clue where to go from there. And once I have the first ring of points set up, I think I'd need to be able to triangulate from two points to start the next ring.

As a disclaimer, I haven't gotten as far as trigonometry in my math education. Most of this is terra incognita to me.

[ed boy]

You don't need to have the pattern pass through multiple octants. You can have a single pattern local to a single octant that is repeater over the sphere.

[Sowelu]

Octahedral, multiple of 8 points, let's use 8.
Make a cube inscribed in the sphere.  Bam.

[Fossaman]

Octahedral, multiple of 8 points, let's use 8.
Make a cube inscribed in the sphere.  Bam.

That would work, except I have to have a maximum distance of 11 units between each point. That's not enough points to evenly cover the sphere.

You don't need to have the pattern pass through multiple octants. You can have a single pattern local to a single octant that is repeater over the sphere.

True, but if I'm doing this properly it should come out the same way anyway. If a point is d/2 distant from the edge of the octant, it should also be d distant from its symmetric partner point in the neighboring octant.

[RedWarrior0]

Octahedral, multiple of 8 points, let's use 8.
Make a cube inscribed in the sphere.  Bam.

That would work, except I have to have a maximum distance of 11 units between each point. That's not enough points to evenly cover the sphere.

Icosahedron/icosahedra then?

[ed boy]

Here's a simple algorithm that will generate you a bunch of solutions:

-Pick some d. the smaller the better.
-On each octant, place one point in the centre of the octant and one point a distance d away from it
-You now have a solution. You can move these points wherever you want on the octant, as long as you are careful near the edges
-If you want another solution with more points, add a third point a distance s away from the second
-You can keep adding points to make a chain of points. Depending on what you chose d to be, there is an upper limit to the number of points possible.
-If you want more points than the upper limit allows, start over with a smaller d.

[Dsarker]

Does infinity/infinity have a solution? The only possible answers I can see are infinity or one. Am I wrong, or is one of these correct?

[Darvi]

Not really. Same deal as 0/0 (OHSHI-).