41 new of 132 responses total.
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.
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.)
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.
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.
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....
Those would be about as irregular as it's possible to get.
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.
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!)
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.
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).
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.)
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....)
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?
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.
Re #103: Yeah, but what's the ANSWER? :)
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?
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.
Where did you get that 1/6 from, Drew?
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.
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
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.
(Compared to the actual growth rates, 25% per generation [less early mortality, infertility, etc.] would have been wonderfully low, and India would be in much better shape today.) Steve's "H O W E V E R..." in #111 is the correct and "best" solution. There are simpler "add things up" solutions, but this is one of those problems where the student's approach tells the teacher more about his/her understanding than anything else. (The instructor threw this problem at a class of PhD-track math (not stats) grad students I was in. The initial class consensus was for a wrong answer [and they almost agreed on which one].) On to a more familiar mathematical topic - Sets. Apples and oranges are elements of the set of fruits, Mr. Figston's 3rd grade is a subset of the set of students at Washington Elementary School, the intersection of the set of wet thing with the set of dry things is the empty set, and all that fun. One very popular set is the Universal Set, which contains EVERYTHING - real, abstract, imagined or undiscovered, simple or complex, it's all there. But the idea that such a set can exist suffers from a fatal logical flaw. It is not that the Universal Set can't contain itself as a subset - but that's a good hint on where to start looking.
Did you mean to say "contain itself as a subset" or "contain itself as an element"?
resp:112 Hey, that's cool, Mark. I learned how to solve that silly series using differentiation. Your way is much better.
Would the Universal Set, then try to contain 'null' and 'infinity' at the same time?
The Universal Set (U for short), by "definition", contains null, infinity, and anything else you can think up. Yourself included. Re #114: U must do both. I believe that a contradiction can be derived either way, but the "nontraditional" one requires minimal knowledge of power sets.
Well, every set contains itself as a subset, so I figured you must have meant "contain itself as an element". The most familiar contradiction based on the notion of sets containing themselves as elements is the Russell Paradox, which goes as follows: Let S be the set of all sets that are not elements of themselves. Then if S is an element of itself, then by definition of S, S is not an element of itself. Conversely, if S is not an element of itself, then again by definition of S, S is an element of itself.
Mike shaves all men that do not shave themselves. Does Mike shave himself?
re #119: Your statement of Russell's "Barber" paradox is insufficient to indicate the paradox since it says nothing about whether or not the barber shaves some of those who shave themselves (which would only include the barber..) or whether or not the restriction applies to the barber (e.g. is the barber a woman?) You need a restriction more like "the barber shaves all those and *only* those who don't shave themselves.."
Ok, fine. But if you are going to be particular, remember that the defining relative pronoun is *that*, so it has to be stated as "the barber shaves all those and only those *that* don't shave themselves."
Re: #118 - my meaning was that a contradiction can be obtained by looking either at sets which are elements of U or at sets which are subsets of U (the power sets are used in the latter). I did't want to give too big a hint. This Statement Is False. (Is the above statement a paradox? Is it not a paradox only for "poetic" reasons - the contradiction is too poorly hidden, insufficiently interesting, etc.? What is a paradox?)
Hey, I know Barry & Sally Childs-Helton, they both have Phds, they are a Paradox.
re 122: two places to moor ships. Speaking of Russel's paradox, Greg (my roommate, flem) was just reading his autobiography and mentioned it to me last night. Interesting coincidance.
Isn't a paradox something that appears to be true, even though it is not?
Er, not exactly. I would describe a paradox as an assertion that seems to be contradictory. Almost the opposite of your definition.
I agree with Steve. A paradox appears to be internally contradictory. Paradoxes may have resolutions, or they may not. Something that "appears to be true, even though it is not" is just an error. I think there is an expression for it, such as "plausible but mistaken".
Chalk one up to fuzzy-headedness. *sigh*
paradox is self-contradictory.
Re #129: A paradox is something that *appears* so be self-contradictory.
That's what I said in #127....... ;)
Apparently, he missed it the first time. :-)
You have several choices: