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Grex > Tutoring > #14: Algebra, Geometry, Calculus, Trig, all that good stuff |  |
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srw
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response 88 of 132:
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Aug 30 06:15 UTC 1997 |
I don't think you can tile a sphere with any number of regular hexagons.
You can tile a plane with them, though.
You need a few pentagons in order to get it to curl up into a sphere.
If you look at a geodesic dome, a la Buckminster Fuller, you will see
that it is made up of triangles, with 5 or 6 coming together at each
point. The "dual" of this structure is made by using the center of each
triangle as a node and connecting each new node to the adjacent ones.
The dual of a geodesic dome is made up of hexagons and pentagons that
tile a sphere. You can have many more hexagons if you like, but you need
to have exactly 12 pentagons to complete the sphere. If you leave the
hexagons out altogether, you have a dodecahedron.
The hexagons don't contribute any curling, which is why you can tile a
plane with them. if they are regular, each one forms an inside angle of
120 degrees at each vertex. Since there must be no more nor less than 3
coming together at each point, they always sum to 360 degrees. Hence, no
curling.
Curiously, although you can't tile a sphere with hexagons, you can tile
a torus with them. To see this, section the torus and unwind it to form
a cylinder, then cut the cylinder the long way and unwind it to form a
rectangle. Tile the rectange as you would a plane, then wind it and
stitch it back up the way you cut it.
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rcurl
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response 89 of 132:
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Aug 30 06:26 UTC 1997 |
That implies that I could roll a cylinder out of paper, and then wind it
into a torus. I can't. Or, please explain further.
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i
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response 90 of 132:
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Aug 30 13:21 UTC 1997 |
You're stretching & compressing the surface of the torus in your unwrapping
& wrapping operations - yes, they're all hexagons, but goodbye regularity.
I'm not familiar with all the official terminology used, but if you're
dividing up the surface of a sphere (curved pieces) instead of building
polyhedral solids (flat pieces), you CAN do it with hexagons. Only 2
are needed, and those unfamiliar with non-Euclidean geometry may have
problems with calling one of them a hexagon.... But they're regular!
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rcurl
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response 91 of 132:
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Aug 30 17:46 UTC 1997 |
The original proposition (#84) was that the hexagons were "all identical in
size" - that would imply all identical in shape, too, but they were not
stated to be regular (spherical) hexagons, thought the sides are nearly
all the same size, and all the hexagons are about the same 'diameter'....
So..what are the shapes of the two irregular hexagons (ala #90) with which
to (spherically) tile a sphere?
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aruba
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response 92 of 132:
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Aug 30 18:56 UTC 1997 |
Re #91: I think what Walter is thinking of in #91 is: take a sphere, draw a
small hexagon on it. Then the one you drew and its complement are two
hexagons which tile the sphere.
Are you sure, Rane, that "same size" (which I take to mean same area) implies
"same shape"? Can you prove that?
Steve W., can you prove that you can't tile a sphere with hexagons?
I am assuming here that the sides of each hexagon should be "lines", which
on a sphere means segments of a great circle which are less than half a
circumference in length.
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srw
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response 93 of 132:
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Aug 30 20:21 UTC 1997 |
No, I can't. The logic I ran off in resp:88 did assume that exactly 3
and no more or fewer hexagons came together at any one vertex. In order
to drop that restruction, you would have to distort the shape of the
hexagons.
Since your hinting above allows that the hexes may be distorted, it
seems that my assumption may have been too great. Perhaps these must be
the rules.
(1) Every shape is bounded by 6 straight sides (straight is "geodesic"
or great circle on a sphere). All sides are the same length.
(2) Every shape has the same size. That is, the same total area.
(3) The entire sphere is covered, so that every point on the sphere is
either in exactly one of these shapes, on the boundary of exactly two,
or at an intersection of three or more.
Under those conditions, I believe I can prove that it cannot be done,
but I am not posting such a proof here yet. The proof will assume that
at least three hexagons come together at each vertex. Perhaps this is
not a requirement. Let me rewrite rule (3) above as follows...
(3) The entire sphere is covered, so that every point on the sphere is
either in exactly one of these shapes, on the boundary of exactly two,
or at an intersection of two or more.
This allows for two hexagons to share two adjacent pairs of edges and
the vertex between them. Such a vertex I might choose to label "bogus",
because only two edges meet. Topologically, these hexes are now a lot
like pentagons, as they may have five (or fewer) neighboring hexagons.
If "bogus" vertices are allowed, then it very likely can be tiled.
In fact, I can modify the example Mark presumed Walter was proposing to
the construct the following tiling of a sphere with 2 hexagons:
Take a sphere of radius r and draw 6 straight line segments of radius
pi*r/3 along the equator so that they are laid out end-to-end and touch
at 6 vertices, equally spaced along the equator. This tiles the sphere
into two equally sized hexagons of area 2*pi*r^2
(OK, it is a bit degenerate, because the line segments are colinear, but
it makes the point.)
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aruba
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response 94 of 132:
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Aug 30 23:06 UTC 1997 |
Yeah, I think it's OK to assume that every vertex is at the corner of at least
3 edges. And each edge is a segment of a great circle.
I don't know if you can call what I was doing "hinting", because I don't
know the answer to this one.
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i
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response 95 of 132:
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Aug 31 04:35 UTC 1997 |
Re: #92, 91 - aruba has the idea. Regularity presents no challenge whatever.
You can even make the two hexagons the same size if you don't mind degenerate
n-gons on a sphere...as steve notes. If what steve calls bogus vertices
are allowed, then 3, 4, and 5-gons can be called hexagons, and the regular
polyhedra provide plenty of regular & symmetrical tilings. Disallowing
180-degree vertices (let's call the results "true" n-gons) messes up the
regularity and symmetry, but the tilings persist.
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rcurl
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response 96 of 132:
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Aug 31 05:02 UTC 1997 |
I don't think it has yet been pointed out that all the interior angles of
a regular spherical hexagon smaller than half the sphere, are greater
than 120 degrees. Hence no three regular spherical hexagons, regardless
of their size, can have a common vertex. My mind boggles, however, at
irregular hexagons of less than 6 edges....
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remmers
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response 97 of 132:
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Aug 31 12:37 UTC 1997 |
Those would be about as irregular as it's possible to get.
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janc
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response 98 of 132:
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Aug 31 14:06 UTC 1997 |
I think Chalker's plan is pretty hopeless. Tiling a sphere is hexagons is
really pretty much the same problem as putting a grid of squares on it (you
can convert from a grid of squares to a grid of hexagons by shifting the odd
numbered rows half a square left and turning the horizonatal boundaries into
zig-zags). But we know how successfull geographers have been at putting a
grid on the earths surface - The squares near the poles are awfully long and
skinny for squares. I think the same problem arrises with hexagons. The
polar hexes get very long and skinny - not even vaguely regular.
I seem to recall some of Chalker's later books admitted that the polar hexes
aren't as regular, and I think there are large polar regions that aren't
tiled.
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i
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response 99 of 132:
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Aug 31 14:14 UTC 1997 |
The interior angles of a regular spherical hexagon *larger* than half the
sphere are still greater than 120 - the size clause in #96 is unneeded.
Convexity (all interior angles <180) seems to be a popular assumption here.
If you don't get hung up with that, then degeneracy (an 180 interior angle,
thus 2 co-linear edges) is a very un-boggling point discontinuity in an
angle vs. # of sides graph.
For greater irregularity, just allow 2-dimensional projections of 3-d
n-gons. You get X-intersections of lines at non-vertex points, double
vertices, 0 length sides, and all sorts of fun. (But this is boring from
the tiling point of view, because ANY tiling of the sphere can be
represented as such a projection of a 3-d n-gon!)
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drew
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response 100 of 132:
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Aug 31 16:06 UTC 1997 |
Someone recently came out with a 100-sided die. I'm not sure whether hexagons
could be circumscribed about the faces, but if a 100 sided die is possible,
then maybe something like hexagon-world could be done if you don't mind some
irregularity and mismatch.
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i
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response 101 of 132:
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Aug 31 22:26 UTC 1997 |
If you're just interested in generating random numbers from 1 to n, then a
fair n-sided die can easily be made. Besides the common 6-siders, 4-, 8-
12-, and 20-sided dice (based on the regular polyhedra) are quite common.
(Dungeons & Dragons uses all these, as do a number of other games.) The
results are often (mathematically and visually) uglier for other values
of n. My guess is that commercial 100-sided dice would either split the
faces of a 20-sided die into 5 pieces (each) or follow an apple-peeler
pattern. But I wouldn't be surprised by something really odd like a little
10-sided die inside a large, clear 10-sided die (read as 1's and 10's).
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remmers
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response 102 of 132:
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Aug 31 23:41 UTC 1997 |
Okay, change-of-pace time! Here's a reasonably straightforward
geometry problem:
Three billiard balls of diameter 1 are resting on a flat
tabletop. Each ball is touching the other two. A fourth ball of
diameter 1 rests on top of the first three. How high above the
surface of the table is the apex of the fourth ball?
(Supply reasoning to justify your answer.)
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rcurl
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response 103 of 132:
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Sep 1 01:55 UTC 1997 |
It is the diameter of a ball plus the height of a tetrahedron with sides
equal to the diameter of a ball. (Reduced to previously solved problem....)
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toking
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response 104 of 132:
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Sep 1 03:20 UTC 1997 |
re 100: 100 sided die have been out for a while...I picked one
up in like 5th grade to use with D&D..
I recently finished "Flatland" by Edwin Abbott Abbott, is the sequel
"spehere Land" and good?
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rcurl
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response 105 of 132:
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Sep 1 06:12 UTC 1997 |
I have a "die" that was made by a statistician which is an aluminum
ten-sided prism with the sides numbered (on the end of the prism) from
0 to 9. It is for generating random decimal numbers.
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remmers
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response 106 of 132:
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Sep 1 12:15 UTC 1997 |
Re #103: Yeah, but what's the ANSWER? :)
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i
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response 107 of 132:
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Sep 1 15:20 UTC 1997 |
Okay, let's roll out the old HS geometry and take a closer look at that
tetrahedron. It's regular with edges of length 1, so all 4 faces are
equilateral triangles with sides of length 1. Bisect the 3 angles in the
base triangle and run the lines through the opposite sides. The result
just oozes symmetries, similarities, and 30-60-90 triangles. Note that
by easy symmetry arguments, the 3 bisectors meet at the center directly
under the vertex. Using the old 1|1/2|sqtr(3)/2 sides rule for those
30-60-90 trianges gives us a distance of 1/sqrt(3) from a corner of the
base to the center. This distance and the (unknown) height are legs of
a right triangle with an edge as hypotenuse (1). Pythagoras says the
height is sqrt(2/3), so (per #103), the answer is 1+sqrt(2/3).
On to a probability problem. It's not hard, but I've seen grad. students
flub it and I'm told that the government of India once made a major policy
decision based on a wrong answer to it.
Assume that the gender ratio at birth is 50-50 and that all births are
independent events. (In other words, having babies is like flipping a
fair coin - heads it's a boy, tails it's a girl.) The following policy
is proposed: Each couple may have two children. If both are girls, then
they may have a third, if that's a girl, a 4th is permitted, etc. (Two
very different ways you could interpret this sexism! The intent was to
guarantee everyone a son who wasn't an only child.) Assume that couples
will have as many children as permitted. Ignore divorce, birth out of
wedlock, fertility running out before a son is born, etc.
What will the gender ratio of children born under this policy be?
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drew
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response 108 of 132:
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Sep 1 21:53 UTC 1997 |
Okay, I'll take a crack at it.
If you stop at two kids, then the ratio is 50/50. However, 25% of these
families will be two girls, and thus can have another kid. Of these, 50% will
have yet another girl, and of these 50% yet another... Thus we have:
M M 1/4
M F 1/4
F M 1/4
F F M 1/8
F F F M 1/16
F F F F M 1/32
.........
(n) F M 2^(-n-1)
In the first two cases, the ratio is one female to three males, multiplied
by 1/2. The rest can be generalized as n females to one male, multiplied by
(1/2)^(n+1). Expressing it as a sum:
R = 1/6 + sum(i=1 to inf) { i / 2^(i+1) }.
The series works out as 1/6 + 1/4 + 1/4 + 3/16 + 1/8 + 5/64 + 3/64 + ...
By experiment, the series seems to converge at 1.166666667 or 7/6. Shouldn't
be too much of a disaster in itself.
But the real problem is the population *growth*, the increase per generation.
This is best state, per couple, as
2/4 + 2/4 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + sum(i=1 to inf) { i / 2^i }.
This series converges, by experiment, at 2.5 children per male-female pair,
or 1.25 multiplication of the population every generation.
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aruba
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response 109 of 132:
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Sep 2 00:59 UTC 1997 |
Where did you get that 1/6 from, Drew?
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drew
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response 110 of 132:
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Sep 3 00:04 UTC 1997 |
The first two cases are taken as the first term in the series - actually not
really a part of the series itself, but a special case. The female to male
ratio among these first two cases is 1:3, and they take up half the outcomes,
for a product of 1/6.
I suspect my answer may not be right. I'll have to think this one over.
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srw
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response 111 of 132:
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Sep 4 05:21 UTC 1997 |
There is a simple proof that the series sum(i=1 to inf) (i / 2^i) = 2
based on the method of differentiating the series...to wit
Start with y= sum[i=0,inf](x^i) = 1/(1-x) in the region 0<x<1
differentiate w.r.t. x - (you can differentiate each term of the series
separately). so it yields
dy/dx = sum[i=0,inf] (i * x^(i-1) ) = (1 - x)^(-2)
so if we substitute x=2^(-1) (another name for 0.5) we have this
sum[i=0,inf] (i / 2^(i-1)) = 4
now divide each side by 2 and note that the i=0 term is 0, so drop it
sum[i=1,inf] (i / 2^i ) = 2 QED
H O W E V E R .....
If you wish to determine the gender ratio you do not need to evaluate a
series. It is 1:1 because each birth is independent. I don't see any
need to argue further.
For skeptics, if you really must do it by the series approach, then
evaluate the expected number of males per family and the expected number
of females. Both evaluate to 1.25 so the ratio really is 1:1
case probability E(M) E(F)
MM 1/4 2/4 0/4
MF 1/4 1/4 1/4
FM 1/4 1/4 1/4
FFM 1/8 1/8 2/8
FFFM 1/16 1/16 3/16
FFFFM 1/32 1/32 4/32
The E(M) series is 1/4 + sum[i=1,inf](1/2^i) = 1/4 + 1
The E(F) series is 1/4 + sum[i=1,inf](i/2^(i+1)) = 1/4 + 1
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aruba
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response 112 of 132:
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Sep 4 17:42 UTC 1997 |
Another way to find the expected number of girls G (which doesn't require
any calculus) is:
G = 0/4 + 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
= 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
==> 2G = 2/4 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
==> 2G-G = 1/4 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
==> G = 1/4 + (1/2)/(1-1/2)
= 1/4 + 1
= 5/4
Drew's interesting observation about the increase in population is quite
right, each generation will be 25% bigger than the last if everyone has
children. That would be kind of a problem.
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