| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 542.1 | X marks the spot. | CHOVAX::YOUNG | Chi-Square | Wed Jul 23 1986 17:36 | 25 |
|
+---+---+---+---+---+---+---+---+
| | * | | | * | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | X | | |
+---+---+---+---+---+---+---+---+
| * | | | * | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | * | |
+---+---o---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | * | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | * | |
+---+---+---+---+---+---+---+---+
How this Lynn?
-- Barry
|
| 542.2 | re. .0 on Problem 1 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed Jul 23 1986 17:38 | 28 |
| re. .0
> Problem 1: Find another point in the diagram which has the same properties
> as 'o'.
How's this?
+---+---+---+---+---+---+---+---+
| | * | | | * | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---o---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| * | | | * | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | * | |
+---+---o---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | * | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | * | |
+---+---+---+---+---+---+---+---+
Kostas G.
|
| 542.3 | 1 answer to #2 | CHOVAX::YOUNG | Chi-Square | Wed Jul 23 1986 17:43 | 13 |
| #2 :
The cartesian plane, with only integer vertices distinguished, has
an infinite number of such rotation points.
I guess you where thinking finite though, huh?
Also, did you intend that +90, -90 degrees be the only rotations
allowed?
-- Barry
|
| 542.4 | | CLT::GILBERT | It's a Dusey | Wed Jul 23 1986 20:49 | 1 |
| re .2 Sorry, Barry's got it right.
|
| 542.5 | | CLT::GILBERT | It's a Dusey | Wed Jul 23 1986 21:23 | 12 |
| Problem 2: Does there exist an arrangement of tiles with at least two
distinguished points which has THREE or more centers of rotation?
I doubt it. Here's how I've been proceeding....
Without loss of generality, we may assume one of the centers of rotation
is at (0,0) in the xy-plane, and that the two 'tile' points at greatest
distance from (0,0) are at (1,0) and (0,1). Now consider where another
center of rotation may be, so that (1,0) and (0,1) will rotate somewhere
back into the circle x�+y�<=1. This locus of possible centers is simply
(x-�)�+(y-�)� <= � -- i.e., a circle centered at (�,�) and passing through
(0,0), (0,1) and (1,0).
|
| 542.6 | Look a little deeper... | MODEL::YARBROUGH | | Thu Jul 24 1986 09:36 | 13 |
| re .2: close, but no cigar. See .1 for the correct solution.
Since we are talking about square tiles being rotated, clearly only
multiples of 90 deg. are relevant.
Notice that, for the two centers of rotation in note .1 that
the squares that rotate into one another are completely different.
The fact that there are seven squares involved is relevant. To see
this more clearly, assign letters A-G to the distinguished tiles
and write down the pairs that are separated by 90 deg in each case.
Also note that the radii are different in the two cases.
Lynn
|
| 542.7 | | CLT::GILBERT | It's a Dusey | Thu Jul 24 1986 11:54 | 31 |
| No, I don't think the fact that the diagram has seven points is particularly
significant. Here is the diagram, the centers, and the 'squares' (the As and
the X; the Bs and the X). The centers of the 'squares' are marked with Cs.
+---+---+---+---+---+---+---+---+
| | A | | | A | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+-C-+---+---+---+---+---+
| | | | | | O | | |
+---+---+---+---+---+---+---+---+
| A | | | X | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | B | |
+---+---o---+---+---+---+---+---+
| | | | | C | | |
+---+---+---+---+---+---+---+---+
| | | | B | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | B | |
+---+---+---+---+---+---+---+---+
Now, what is relevant is that the Os and Cs form a square, and that the
'squares' have 180� rotational symmetry around the Cs. With this, you
can create any number of similar problems.
Note that I *haven't* shown that this technique produces all such diagrams
with two centers, although I expect this is true. Even if true, it wouldn't
mean that diagrams with three centers are impossible.
- Gilbert
|
| 542.8 | 8 rotations, 7 points | GALLO::JMUNZER | | Thu Jul 24 1986 12:59 | 35 |
| Seven points seem to fit naturally:
+---+---+---+---+---+---+---+---+
| | C | | | E | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | X | | |
+---+---+---+---+---+---+---+---+
| G | | | A | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | D | |
+---+---o---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | B | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | F | |
+---+---+---+---+---+---+---+---+
If R means rotation of 90 degrees about X, and
if r means rotation of 90 degrees about o, then
r R r R
A -----> B -----> C -----> D -----> A
and
R r R r
A -----> E -----> F -----> G -----> A
Seven points, each connected to another point by an R or an r.
Part of the difficulty of Problem #2 is that a third rotation can't fit
connections together so easily.
John
|
| 542.9 | four, not seven | MODEL::YARBROUGH | | Thu Jul 24 1986 13:27 | 6 |
| As pointed out in .-1, there are two cycles of four tiles each which
satisfy the rotational condition independently. So in fact seven
is not the relevant number, but four is.
My current suspicion is that three centers of rotation are possible,
but not in the 8x8 confines; I think a larger space might work out.
|
| 542.10 | No, seven is irrelevant | CLT::GILBERT | It's a Dusey | Thu Jul 24 1986 14:15 | 31 |
| Find the two centers.
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | X | X | | X | | | |
+---+---+---+---+---+---+---+---+
| | | | | X | | X | |
+---+---+---+---+---+---+---+---+
| X | X | | | | | X | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
Find the two centers.
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | X | X | | | |
+---+---+---+---+---+---+---+---+
| | | | | X | | X | |
+---+---+---+---+---+---+---+---+
| | | X | X | | | X | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
P.S. I wonder whether whether I can prove three centers are impossible
before Lynn finds an example. :-).
|
| 542.11 | Party of the first part... | CHOVAX::YOUNG | Chi-Square | Thu Jul 24 1986 18:48 | 19 |
|
+---+---+---+---+---+---+---+---+
| | | | 1 | | | | |
+---+---+---+---+---+---+---+---+
| | X | X | | X | | | |
+---+---+---+---+---+---+---+---+
| | | | | X | | X | |
+---+---+---+---+---+---+---+---+
| X | X | | | | | X | |
+---+---+---+---+---+---+---+---+
| | | | 2 | | | | |
+---+---+---+---+---+---+---+---+
How's this?
-- Barry
|
| 542.12 | Party of the second part. | CHOVAX::YOUNG | Chi-Square | Thu Jul 24 1986 19:00 | 19 |
| Here we go:
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | X |<X>| | | |
+---+---+---+---+---+---+---+---+
| | | | | X | | X | |
+---+---+---+---+---+---+---+---+
| | | X | X |<2>| | X | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
Note : <1> is on an 'X'.
-- Barry
|
| 542.13 | | CLT::GILBERT | It's a Dusey | Fri Jul 25 1986 10:59 | 69 |
| I've been working on a characterization of diagrams with two centers
of rotation, and found the following, which is not equivalent to an
object with 180� symmetry and its translation and rotation by 90�.
+---+---+---+---+---+---+---+
| | | X | | X | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| X | | | | | | X |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | X | | X | | |
+---+---+---+---+---+---+---+
The way this was discovered was as follows. Consider the complex plane.
Now the rotation of a point X by 90� (clockwise) around point A is
A+(X-A)(-i); and counterclockwise: A+(X-A)i.
Suppose we're interested in following the 'path' of a single point X
as we rotate it around points A and B, alternately. We eventually want
it to return to itself, so we need only consider an even number of rotations.
Let us write 'a' for a clockwise rotation (of 90�) around point A, and
'A' a counterclockwise rotation (of 90�). Similarly for 'b' and 'B'.
Let 'Ab...' denote a counterclockwise rotation around point A, followed
by a clockwise rotation around point B. These are functional operators,
so (if we let point A be the origin, and let B = 1+0i) we have the equations:
X 'ab' = -X + (1+i)
X 'aB' = X + (1-i)
X 'Ab' = X + (1+i)
X 'AB' = -X + (1-i)
The second equation means that if we rotate any point clockwise around A,
then counterclockwise around B, we've simply translated it by (1-i).
In general, we have:
X '...' = � X + m + n i, where m and n are integers, and m = n (mod 2).
Now if we set this equal to X (so that the point returns to where it started),
we have either:
X + m + n i = X or -X + m + n i = X
The first has solutions only if m=n=0; in this case, any point returns
to its start ('abab', which occurs in the original problem, is the simplest
example of this; there are others). The second equation can be solved for X.
The above diagram was generated by X 'abAbaB' = X, which gives the equation
3+i = 2X. I suspect that this fact is of little use in solving the above
problem, whereas there are other techniques (that you've probably discovered)
that work quite well.
The following was the result of solving X 'abAbaBaB' = X.
+---+---+---+---+---+---+---+---+
| | | | X | | | | |
+---+---+---+---+---+---+---+---+
| | | X | | X | | | |
+---+---+---+---+---+---+---+---+
| | X | | | | X | | |
+---+---+---+---+---+---+---+---+
| | | X | | X | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
|
| 542.14 | | GALLO::JMUNZER | | Fri Jul 25 1986 18:37 | 24 |
| Three centers of rotation:
+---+---+---+---+---+---+---+---+---+
| | * | | * | | * | | * | |
+---+---+---+---+---+---+---+---+---+
| * | | * | | * | | * | | * |
+---+---+---+---+---+---+---+---+---+
| | * | | * | | * | | * | |
+---+---+---+---+---+---+---+---+---+
| * | | * | o | * | o | * | | * |
+---+---+---+---+---+---+---+---+---+
| | * | | * | o | * | | * | |
+---+---+---+---+---+---+---+---+---+
| | | | | * | | | | |
+---+---+---+---+---+---+---+---+---+
| | | | * | | * | | | |
+---+---+---+---+---+---+---+---+---+
| | | * | | * | | * | | |
+---+---+---+---+---+---+---+---+---+
| | | | * | | * | | | |
+---+---+---+---+---+---+---+---+---+
John
|
| 542.15 | WOW. | CHOVAX::YOUNG | Chi-Square | Fri Jul 25 1986 21:48 | 9 |
|
You know, I would have bet anything that three centers of rotation
could not exist.
VERY nice.
-- Barry
|
| 542.16 | | CLT::GILBERT | like an eager child | Sat Jul 26 1986 22:46 | 1 |
| Oh, Wow! Amaze us again.
|
| 542.17 | | CLT::GILBERT | It's a Dusey | Mon Jul 28 1986 15:13 | 28 |
| Somewhere this argument is flawed...
Three centers cannot be in a row.
Assume we can have three centers in a row, at points (a,0), (b,0) and (c,0)
(with a < b < c), let x0 and x0 be the x-coordinates of the (marked) tiles
having least and greatest x-coordinate, respectively. Let y0 be the
y-coordinate of the tile having the greatest y-coordinate. By symmetry, if
for each marked tile (x,y), we also mark the tile (x,-y), we will also have
a solution, so we have -y0 as the least y-coordinate.
(x0,y0) +-------------------------------+ (x1,y0)
| |
| |
| |
| A B C |
| |
| |
| |
(x0,-y0) +-------------------------------+ (x1,-y0)
Now if point A is able to rotate a right-most tile somewhere into the
rectangle, we must have 'x1 <= a+y0'. For point B, we have 'y0 <= x1-b'.
But
y0 <= x1-b <= (a+y0)-b implies b <= a.
Note that we haven't even considered point C, so we might reasonably
conclude that *two* points cannot be in a row!
|
| 542.18 | >, not <. | MODEL::YARBROUGH | | Mon Jul 28 1986 17:29 | 2 |
| 'x1 <= a+y0' is incorrect; should read
'x1 >= a+y0'.
|
| 542.19 | | CLT::GILBERT | It's a Dusey | Mon Jul 28 1986 18:29 | 3 |
| No, that's not it. The equation should put an upper bound on x1;
if x1 is *too* large, there's no way point A can rotate it into
the rectangle.
|
| 542.20 | | GALLO::JMUNZER | | Mon Jul 28 1986 18:50 | 12 |
| Re .17:
> Now if point A is able to rotate a right-most tile somewhere into the
> rectangle, we must have 'x1 <= a+y0'.
Okay.
> For point B, we have 'y0 <= x1-b'.
No, it's 'x1 <= b+y0', just like A.
John
|
| 542.21 | More on Lynn's bathroom :-) | CLT::GILBERT | It's a Dusey | Mon Jul 28 1986 20:13 | 48 |
| Ah, now I see it.
> > For point B, we have 'y0 <= x1-b'.
>
> No, it's 'x1 <= b+y0', just like A.
Actually, what I meant was that at least one of the following must be
true: either 'y0 <= x1-b' or 'y0 <= b-x0'. That is, a point on the top
when rotated about B will be in the rectangle. From this point, the
rest of the proof follows....
We have:
x1 <= a+y0 (rightmost point rotated about A must be in rectangle)
x0 >= c-y0 (leftmost point rotated about C)
y0 <= x1-b or y0 <= b-x0 (top point rotated about B)
From 'y0 <= x1-b' we have 'y0 <= x1-b <= a+y0-b', or 'b <= a'; while
from 'y0 <= b-x0' we have 'y0 <= b-x0 <= b-(c-y0)', or 'c <= b'.
In either case, we have two coincident centers. Thus, three distinct
centers cannot be colinear.
Here's an interesting way to find the centers of rotation of a diagram.
Given some x-y axes, draw the smallest aligned rectangle that surrounds the
points. From the corners of this rectangle, draw lines at angles of �45�
from the corners of the rectangles. Now any centers must be on these lines!
(In fact, if the rectangle) (...and the centers must be)
(is not a square, we have...) (on these 'diagonal' lines)
\ / \ /
+-----------------------+ +-----------------------+
| \ \ / / | | \ / |
| \ X / | | X |
| \ / \ / | | |
| X X | | |
| / \ / \ | | |
| / X \ | | X |
| / / \ \ | | / \ |
+-----------------------+ +-----------------------+
/ \ / \
Now, we can rotate the diagram slightly (i.e., choose different alignment)
to get a different rectangle, and different lines. The intersections between
these give the locations of the centers of rotation.
A proof of this amazing property is forthcoming.
|
| 542.22 | | CLT::GILBERT | It's a Dusey | Tue Jul 29 1986 16:28 | 60 |
| Given perpendicular axes, draw the smallest aligned rectangle that surrounds
the points. This we'll call the bounding rectangle. From the corners of the
bounding rectangle, draw lines at angles of �45� with the sides. Now any
centers of rotation must be on these lines.
Rough Proof:
In the following diagrams, a center of rotation cannot be in a shaded area.
In the first diagram, a center cannot be in the shaded area, since if it were,
it would not be able to rotate a righmost point (one on the right edge of the
bounding rectangle) into the rectangle. The other diagrams show the excluded
areas for rotating topmost, leftmost and bottommost points.
ZZZZZZZZZ\
ZZ+-----------------------+ +-----------------------+
ZZ|ZZZZZZZZZ\ / | | \ / |
ZZ|ZZZZZZZZZZZ\ / | | \ / |
ZZ|ZZZZZZZZZZZZZ\ / | | \ / |
ZZ|ZZZZZZZZZZZZZZZX | | \ / |
ZZ|ZZZZZZZZZZZZZ/ \ | | \ / |
ZZ|ZZZZZZZZZZZ/ \ | | X |
ZZ|ZZZZZZZZZ/ \ | | /ZZZ\ |
ZZ+-----------------------+ +-----------------------+
ZZZZZZZZZ/ /ZZZZZZZZZ\
/ZZZZZZZZZ \ZZZZZZZZZ/
+-----------------------+ZZ +-----------------------+
| \ /ZZZZZZZZZ|ZZ | \ZZZ/ |
| \ /ZZZZZZZZZZZ|ZZ | X |
| \ /ZZZZZZZZZZZZZ|ZZ | / \ |
| XZZZZZZZZZZZZZZZ|ZZ | / \ |
| / \ZZZZZZZZZZZZZ|ZZ | / \ |
| / \ZZZZZZZZZZZ|ZZ | / \ |
| / \ZZZZZZZZZ|ZZ | / \ |
+-----------------------+ZZ +-----------------------+
\ZZZZZZZZZ
Thus, we see that the only part of the plane where a center of rotation can
possibly be is on the following rays:
\ /
+-----------------------+
| \ / |
| X |
| |
| |
| |
| X |
| / \ |
+-----------------------+
/ \
Q.E.D
As it happens, the 4 points that are on the bounding rectangle can be used to
easily determine at most 4 places where the centers of rotations may be.
Stay tuned!
- Gilbert
|
| 542.23 | finding possible centers, given the convex hull | CLT::GILBERT | It's a Dusey | Tue Jul 29 1986 18:48 | 116 |
| In note 542.22, we determined that, given an aligned bounding rectangle, any
center of rotation must be on a line through a corner of the rectangle, at an
angle of 45� with the sides of the rectangle. We'll call these diagonal lines.
Suppose that each side of the bounding rectangle touches only one point in
the diagram (this can be accomplished by appropriate choice of the alignment).
Now, we can rotate the alignment slightly to get a different bounding rectangle,
and slightly different diagonal lines. Since the centers of rotation must be
on either set of diagonals, the intersections between the two sets of diagonal
lines give the locations of all places where the centers of rotation may be.
Consider the diagram, with the four points on the bounding rectangle marked:
+-------B---------------+
| |
| C
| |
| |
A |
| |
| |
+------------D----------+
If we change the alignment slightly, the corner between A and B will
still form a right angle with points A and B. Thus, this corner must
remain on the circle having AB as its diameter (remark: the locus of points
forming some fixed angle with two given points is a circle that passes
through the given points, and this circle satisfies the angle requirement).
A line through this corner at angles of 45� with the A and B sides must
pass through a (fixed) point of the same circle: the point that bisects
the (semicircle) chord AB (the above remark applies here, too).
Any possible center of rotation must be at one of these fixed points,
since it must remain on one of the diagonals (actually, it must remain
on two). Thus, we have a simple way to find the centers of rotation
(if any) for a diagram.
Find four points in the diagram that 'support' a bounding rectangle (i.e.,
each side of the rectangle touches no other point in the diagram). Taking
the four pairs of points from connecting sides, for each pair, construct the
square having those points as opposite corners. The inward (to the rectangle)
point that is a third corner of the square gives the possible location of a
center of rotation. Any center of rotation for the diagram must be one of
these four constructed points.
An example should help. In the following, the points supporting a bounding
rectangle are labelled A, B, C, and D (the rectangle is a slight counter-
clockwise rotation of the enclosing grid). Labels on interior points indicate
which support points form a square with this point; these four points are the
only places where centers of rotation may be.
+---+---+---+---+---+---+---+---+---+
| | B | | * | | * | | Y | |
+---+---+---+---+---+---+---+---+---+
| X | | * | | * | | * | | C |
+---+---+---+---+---+---+---+---+---+
| | * |a-b| | | | | * | |
+---+---+---+---+---+---+---+---+---+
| A | | |c-d| |a-d| | | * |
+---+---+---+---+---+---+---+---+---+
| | * | | |b-c| | | * | |
+---+---+---+---+---+---+---+---+---+
| | | | | | | | | |
+---+---+---+---+---+---+---+---+---+
| | | | | | | | | |
+---+---+---+---+---+---+---+---+---+
| | | * | | * | | Z | | |
+---+---+---+---+---+---+---+---+---+
| | | | W | | D | | | |
+---+---+---+---+---+---+---+---+---+
There are other possible sets of support points in the above diagram, for
example, W, X, Y, and Z. In general, considering these may yield other
sets of four possible centers of rotation -- except that the places where
the centers actually are will be common to all.
Here are two examples from note 542.13. The first:
+---+---+---+---+---+---+---+
| | | B | | * | | |
+---+---+---+---+---+---+---+
| | | |d-a| | | |
+---+---+---+---+---+---+---+
| A | |a-b| |c-d| | C |
+---+---+---+---+---+---+---+
| | | |b-c| | | |
+---+---+---+---+---+---+---+
| | | * | | D | | |
+---+---+---+---+---+---+---+
And the second. Here, the point labelled 'a-b' also forms a square
with points B and C.
+---+---+---+---+---+---+---+---+
| | | | B | | | | |
+---+---+---+---+---+---+---+---+
| | | * |d-a| * | | | |
+---+---+---+---+---+---+---+---+
| | A | |a-b|c-d| C | | |
+---+---+---+---+---+---+---+---+
| | | * | | D | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
We've seen that there can be no more than four centers of rotation.
Stay tuned to see that there may really be no more than three.
Here is a related problem for which I have no answer. Any takers?
Given any three points that are not colinear, can a diagram be constructed
having the three points as centers of rotation?
- Gilbert
|
| 542.24 | | CLT::GILBERT | schmaltzy | Wed Jul 30 1986 20:13 | 68 |
| Note 542.22 showed that any centers of rotation must be on the following
'V's that are at 45� angles with the sides of the bounding rectangle.
\ /
+-----------------------+
| \ / |
| X |
| |
| |
| |
| X |
| / \ |
+-----------------------+
/ \
When the alignment for the bounding rectangle is rotated slightly, the 'V's
also rotate and shift slightly. Yet any centers must still be on the 'V's.
No more than two centers of rotation may be on one of the 'V's. If there were
three or more centers, then one leg of the 'V' would pass through two centers,
and a slight rotation of the 'V' would be unable to still pass through both.
Incidentally, we again see that there are no more than four centers.
Let's assume that some diagram has four centers of rotation.
As the diagram has four centers, there must be two per 'V'. Note that (from
the original construction of the 'V's) right angles are formed by the 'V's,
the line passing through the two 'X's below has the same axes of alignment
as the bounding rectangle, and must bisect the 'V's.
\ | /
o | /
\ | o
X
|
|
X
o | \
/ o
/ \
Considering just one 'V', the locus of points where the 'X' may be is the
circle having the two centers of rotation as opposite sides of a diameter.
And the line passing through the two 'X's must also pass through a (fixed)
point on the same circle: the point that bisects the (semicircle) chord
between the two centers of rotation, and is on the 'outer' part of the
diagram and circle.
The same considerations apply to the other 'V', and so we have found
two fixed points that the line connecting the 'X's must pass through.
However, this implies that we cannot change the alignment of the bounding
rectangle, and still have the line connecting the 'X's still pass through
these two points. Thus, the assumption that some diagram has four centers
of rotation must be false.
A diagram may have no more than three centers of rotation. J.Munzer has
demonstrated the existance of diagrams with three centers of rotation.
As we have seen, given three points, it is not always possible to construct
a diagram having these points as centers of rotation (i.e., the points may
not be colinear). Later we will examine the constraints on the centers.
Stay tuned.
- Gilbert
P.S. Woe to us if Lynn's bathrooms had had hexagonal tiles!
|
| 542.25 | Centers cannot form an obtuse triangle | CLT::GILBERT | eager like a child | Tue Aug 05 1986 12:13 | 37 |
| For a diagram having three centers of rotation to exist, the centers
must form an acute (or right) triangle.
To see this, consider the case when they do not. Without loss of
generality, we can assume the centers are positioned as:
o
o o
That is, two centers are on a horizontal line, and the third center
is positioned above and to the right. Now consider what happens when
the bounding rectangle is at a 45� angle with the horizontal line and
the lines at 45� with its corners are horizontal and vertical. We see
that the centers must be on the lines:
|
|
o
|
--o------o-+-----------
|
|
|
|
Note however, that with a slight rotation of the alignment axes, we
cannot produce two 'V's that pass through the three points. Thus, the
three centers cannot form an obtuse triangle.
I believe this is the sole constraint on three centers such that they
are able to form a diagram, having formed some other diagrams (besides
J.Munzer's) where the centers form an acute triangle. A proof that any
three points forming an acute (or right) triangle can be used as centers
is in the works....
- Gilbert
|
| 542.26 | Here's the tile-rotation note: | CHOVAX::YOUNG | Where is our Laptop VAXstation? | Sun Oct 21 1990 15:05 | 1 |
| Mentioned at the math dinner...
|