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Grex > Tutoring > #14: Algebra, Geometry, Calculus, Trig, all that good stuff |  |
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| 25 new of 132 responses total. |
dang
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response 73 of 132:
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Jul 18 19:59 UTC 1997 |
Sure it does. One sign must be true. Hense, two signs must be false. One
of those two signs, then, must be false. If one of them is false, then the
other one is true. Hense, one of those signs must be true, and sign I must
be false.
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aruba
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response 74 of 132:
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Jul 18 22:23 UTC 1997 |
It is in fact the case that if one of the two signs (II and III) is false,
then the other must be true. *That* is what you need to conclude that one of
them must be true. It's not enough to say that "if one is true, then the
other is false", because that doesn't eliminate the possibility that they are
both false. (And it's *not* the case that one sign must be true. All we know
is that *at most one* of the signs is true.)
With that correction, though, dang's (and i's) solution is a very nice one.
I'll enter the next trial when i get home.
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aruba
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response 75 of 132:
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Jul 19 07:58 UTC 1997 |
The Tenth Trial
Again there was only one lady and two tigers. The king explained to
the prisoner that the sign on the door of the room containing the lady was
true, and that at least one of the other two signs was false.
Here are the signs:
I II III
A TIGER IS A TIGER IS A TIGER IS
IN ROOM II IN THIS ROOM IN ROOM I
What should the prisoner do?
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remmers
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response 76 of 132:
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Jul 19 12:41 UTC 1997 |
If the lady were in Room (II), then sign (II) would be false,
contradicting the fact that the sign on the room containing the
lady is true. Therefore, the lady is not in room (II).
If the lady were in room (III), then tigers would be in the
other two rooms, implying that signs (I) and (II) are both true,
contradicting the fact that at least one of the signs on the
tiger rooms is false. Therefore, the lady is not in room (III).
Hence the lady is in room (I). (Note that this implies that
sign (I) is true and sign (III) is false, consistent with the
king's preconditions.)
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aruba
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response 77 of 132:
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Jul 19 16:19 UTC 1997 |
Yes, John has put it very nicely. On to The Eleventh Trial:
First, Second, and Third Choice
In this more whimsical trial, the king explained to the prisoner that
one of the three rooms contained a lady, another a tiger, and the third
room was empty. The sign on the door of the room containing the lady was
true, the sign on the door of the room with the tiger was false, and the
sign on the door of the empty room could be either true or false. Here
are the signs:
I II III
ROOM III THE TIGER IS THIS ROOM
IS EMPTY IN ROOM I IS EMPTY
Now, the prisoner happened to know the lady in question and wished to
marry her. Therefore, although the empty room was preferable to the one
with the tiger, his first choice was the room with the lady.
Which room contains the lady, and which room contains the tiger? If
you can answer these two questions, you should have little difficulty in
also determining which room is empty.
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drew
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response 78 of 132:
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Jul 19 17:22 UTC 1997 |
The lady can't be in room III, because this would make sign(III) false.
If she were in room II, this would put the cat in room I with sign(I) being
true.
So she has to go in room I, making sign(I) true and room III empty, leaving
the cat in room II.
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aruba
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response 79 of 132:
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Jul 20 01:30 UTC 1997 |
Very nice, Drew. Ok, here's the grand finale:
THE FOURTH DAY
"Horrible!" said the king. "It seems I can't make my puzzles hard
enough to trap these fellows! Well, we have only one more trial to go,
but this time I'll *really* give the prisoner a run for his money!"
A Logical Labyrinth
Well, the king was as good as his word. Instead of having three rooms
for the prisoner to choose from, he gave him nine! As he explained, only
one room contained a lady; each of the other eight either contained a
tiger or was empty. And, the king added, the sign on the door of the room
containing the lady is true; the signs on doors of all rooms containing
tigers are false; and the signs on doors of empty rooms can be either true
or false.
Here are the signs:
I II III
THE LADY THIS ROOM EITHER SIGN V
IS IN AN IS EMPTY IS RIGHT
ODD-NUMBERED OR SIGN VII
ROOM IS WRONG
IV V VI
SIGN I EITHER SIGN II SIGN III
IS WRONG OR SIGN IV IS WRONG
IS RIGHT
VII VIII IX
THE LADY THIS ROOM THIS ROOM
IS NOT IN CONTAINS CONTAINS
ROOM I A TIGER A TIGER
AND ROOM IX AND SIGN VI
IS EMPTY IS WRONG
The prisoner studied the situation for a long while.
"The problem is unsolvable!" he exclaimed angrily. "That's not fair!"
"I know," laughed the king.
"Very funny!" replied the prisoner. "Come on, now, at least give me a
decent clue: is Room Eight empty or not?"
The king was decent enough to tell him whether Room VIII was empty or
not, and the prisoner was then able to deduce where the lady was.
Which room contained the lady?
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janc
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response 80 of 132:
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Jul 20 15:56 UTC 1997 |
Ugh. I got as far as figuring that if room VIII is not empty then the lady
is in room one or seven, but then I decided to get a life.
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remmers
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response 81 of 132:
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Jul 20 16:56 UTC 1997 |
(In order to "get a life", the prisoner's course of action would
have to be rather different from yours... :-)
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drew
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response 82 of 132:
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Jul 20 18:24 UTC 1997 |
In order for sign(VIII) to yield any useful information, it must be
ascertained to have some boolian value. Thus room VIII can't be empty.
It can't have the lady either, since sign(VIII) would have to be true, and
it would not be. So there's a kitty in room VIII. This is permitted by the
AND function in sign(VIII) so long as there is something in room IX.
Room IX also has to have a kitty, for similar reasons that room VIII does.
This makes sign(VI) *right* (true).
This of course makes sign(III) false.
Sign(III) is an OR function (it was stated that either-or was *inclusive*-or
in an earlier problem) which requires *both* clauses to be false.
so, sign(VII), meaning that the lady is not in room I.
Also, NOT sign(V), which is another OR function, both of whose switches must
be turned off.
Thus, NOT sign(IV); sign(I) is right, and the lady must be in either room I,
III, V, VI, and IX.
We've already eliminated rooms III, V, and IX, due to their signs being false,
which leaves rooms I and VII; and we had already eliminated room I due to
sign(VII) being true. This leaves the lady in room VII.
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aruba
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response 83 of 132:
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Aug 16 19:04 UTC 1997 |
Sorry it has taken me so long to get back here; Drew is right, of course, and
his solution is very succinct. I'm going away for a while - anyone else want
to enter a puzzle?
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aruba
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response 84 of 132:
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Aug 29 17:47 UTC 1997 |
Ok, so I've started reading Jack Chalker's Well World books. The Well
world is supposed to be a planet which is divided into 1560 environments,
each of which is shaped like a hexagon. "Each hex is identical in size -
each one of the six sides is just a shade under three hundred fifty-five
kilometers, and they're a shade under six hundred fifteen kilometers
across."
So the planet is tiled with hexagons. It's not clear whether 1560 is the
total number of hexagons, though, because Chalker implies later that some
hexes are cut in half by the equator, and I'm not sure whether those count
in the 1560 number (I think not).
So the question is: did Chalker think this out before he wrote it? Is it
possible to tile a sphere with hexagons in a way that fits the
description?
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rcurl
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response 85 of 132:
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Aug 30 02:00 UTC 1997 |
There are only five regular polyhedrons - all sides the same shape and
a sphere touching each side at the center can be inscribed. I don't know,
however, how close one might get to the Well World - say, a mismatch at
just the poles.
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awijaya
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response 86 of 132:
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Aug 30 03:48 UTC 1997 |
Hi, Math always the worst experience in college . I get E
on both Calculus I & II. The only way to study it is
to do 10,000 exercises. (take a long time). The 2nd year,
I got A and B. It is a nightmare! Remembering all those
Integral formula etc. They just prove that Newton is
wrong in Gravitation theory. There is a WEAK force
that cause in-accurate result. Fortunately physic is
an interesting subject (nuclear physic) Regards AW
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awijaya
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response 87 of 132:
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Aug 30 03:59 UTC 1997 |
Hi, about the book upgrade: what about buying CD, instead of books?
IMO it will be more exciting, easy to understand and save
tons of wood from Tropical rainforest. Perhaps they
can use rewritable CD to keep the cost down? What about bookstore
with CD-Writer? IMO it will be cheaper than printed books.
Regards (AW)
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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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