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Shoutbox - 17 comments
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unbread says:
Here's a kind of math question. I'm sure you've heard the problem before.
How can you place 8 queens on a standard chess board so that no queen can attack another queen. This means there is one queen for each row and column and no two queens are in the same diagonal.
I know of one with translational symmetry
I know of one with rotational symmetry
Is it even possible for reflectional symmetry (if so the line of reflection can't be vert, horz or at 45deg)
Can you think of one with no symmetry?
posted Jan 8
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posted Jul 18
Comment replies (1)
lorem-ipsum says:
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posted Jul 30
unbread says:
Okay, so here's a problem I thought up while sitting eating a burger. I've done a little work on it here and there, but I know my results are bad because my results don't allow radii bigger than 1. Back to the drawing board.
You have a cylinder with height, H, and radius, R. You are to make one straight cut through the cylinder such that the sum of the Surface Areas of the resultant objects is maximized. Where is the cut made?
Good Luck.
posted Apr 16
Comment replies (8)
Robert is Where? says:
Seems to me that the answer would be to slice it from one corner to the opposite corner, creating the greatest length for the surface exposed. Though admittedly this would wind up giving an oval surface exposed. Perhaps a rectangle sliced through a diameter would be better then? What do you think?
posted Apr 17
Joe says:
well it obviously depends on the radius and the height, but if my calculations are correct,
you would cut it corner to corner for
H less than [piR plus sqrt(17)*(piR)^2]/8
and down the center for
h greater than [piR plus sqrt(17)*(piR)^2]/8
posted Apr 17
unbread says:
I have reason to believe that the cut is made at an angle somewhere between "corner to corner" and "down the middle", but never at either. The great H/R is, the closer it is to "down the middle", and the smaller H/R is, the closer it is to "corner to corner".
posted Apr 18
Joe says:
calc 3
SA = double integral [base] of sqrt(1 (F[sub]x)^2 (F[sub]y)^2)
and you can understand that it wont be the same for different radius and heights.
ie. lim height->0 cutting it down the middle will make the center area approach 0 and cutting it corner to corner will approach the area of the base.
posted Apr 18
Joe says:
ah, the angle it is cut at is still a function of the ratio of hight:width.
with height->0 or width->infinity maximization approaching corner to corner cut and
height->infintiy or width->0 maximization approaching a cut down the middle.
you should be able to figure the relation from concepts in calculus 3
basically the answer to your problem is not an angle, but an equation of 2 variables. height and width.
posted Jun 7
Hank says:
Would be nice to have some LaTeX rendering in here ;)
int^13_10{2xdx}
I think that's it...
posted Apr 6
Pages: 1 (17 total comments)
unbread says:
Oh, I just thought of another problem/interesting thing.
My geologist friend what over and looking at some publishings on minerals. I asked here what all the triangles with dots in them represented. See said they showed the percentage(placement) of each of 3 minerals for a rock(dot). I'm sure I have the geology part of it wrong but the math part I understood. Each vertex represented a mineral. The closer the dot to the vertex the more the percentage of that mineral, and the closer to the opposite line the less of that mineral (an therefore more of the others). I was curious if the middle (33.3% of each) really was at a height 1/3rd of the altitude. With some simple geometry you can see that it is. I asked if they ever have 4 minerals. She said yes, but then it has to be a pyramid shape. I asked if she meant a tetrahedron, and she said yes (after I made one out of string with my fingers holding the vertices). I was curious then if the center of the polyhedral was really 1/4th the height of the altitude. After even more geometry i found it was. Where I get stuck is when I try to see if it works for a 4-dimensional tetrahedron (is it a hyper tetrahedron?), i.e. the center is 1/5th the altitude.
posted Jan 8