| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 735.1 | Vectorize it! | ANGORA::JANZEN | Tom LMO2/O23 296-5421 | Fri Jul 17 1987 20:16 | 11 |
| four points A,B,C,D
__ __ __
vectors BA,BC,AD
__ __ __ __
|BA X BC|= |BA| * |BC|
__ __ __
and AD = BA + BC
unfortunately, you have to choose the point orders correctly, or try all
orders until it passes.
Tom
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| 735.2 | proof for vectorization? | RDVAX::PERRONE | | Mon Jul 20 1987 13:30 | 4 |
| It seems to me that one would need three equations to verify that
four points comprise the vertices of a square.
Can someone give a proof for the answer in note 735.1?
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| 735.3 | 25 proof anyway | ANGORA::JANZEN | Tom LMO2/O23 296-5421 | Mon Jul 20 1987 16:03 | 27 |
| >four points A,B,C,D
A__________ B
| |
| |
| |
| |
D----------C
> __ __ __
>vectors BA,BC,AD
> __ __ __ __
>|BA X BC|= |BA| * |BC|
> __ __
>|BA X BC|= magnitude of the cross product
= area of parallelogram bounded by BA, BC, and two
other equal & parallel line segments.
__ __
|BA| * |BC| = scalar product of two adjacent sides. In order for the
product of two sides to equal the area of the parallelogram, it must be a
square. So points A,B, and C are on a square.
But we don't know yet if point D on this square.
> __ __ __
>and AD = BA + BC
If it's a square, and the point D is found by adding two of the right
sides, then all four points on on the square.
Tom not a proof but more explication
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| 735.4 | Add another condition | ANGORA::JANZEN | Tom LMO2/O23 296-5421 | Mon Jul 20 1987 16:04 | 4 |
| oh I'm wrong. It could be a rectangle.
add
|BA|=|BC|
Tom
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| 735.5 | three conditions | RDVAX::PERRONE | | Mon Jul 20 1987 20:39 | 13 |
| I don't quite believe the condition that
__ __ __
|AD| = |AB| + |BC|
however what about these three conditions:
__ __ __ __
|BAxBC| = |BA|*|BC|
__ __ __ __
|ABxAD| = |AB|*|AD|
__ __
|AB| = |AD|
Can anyone prove that these describe a square?
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| 735.6 | corrected corrected correction | ANGORA::JANZEN | Tom LMO2/O23 296-5421 | Mon Jul 20 1987 20:59 | 11 |
| -< |I| |goofed| |again| >-
> -< three conditions >-
>
> I don't quite believe the condition that
> __ __ __
> |AD| = |AB| + |BC|
__ __ __
oh. well,uh, how about BD = BA + BC
which is what I meant but got carried away with bars and made a mistake too.
Tom
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| 735.7 | | CLT::GILBERT | Builder | Mon Jul 20 1987 22:41 | 15 |
| How about the following?
A+--------+B
| |
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| |
| |
D+--------+C
__ __
AB = DC (opposite sides parallel and of equal length)
__ __
AD = BC (opposite sides parallel and of equal length)
__ __
AB . BC = 0 (dot product; adjacent sides are perpendicular)
|
| 735.8 | Proof not presented | SEMI::NG | | Tue Jul 21 1987 05:09 | 78 |
| Re .7:
A rectangle will satisfy those conditions too.
Add |AB| = |AD|.
I think that should be enough.
Re .0:
Could you tell us something that motivates this problem?
Re (general):
Let's talk about the definition of a square (probably in Euclid's
Elements. Just kidding, I haven't read that book. I don't know where
in my head I got the definition.)
I think a 4-sided planar object is a square iff it has 4 equal sides
and 2 of the sides are perpendicular. (or equivalently, 3 equal sides
and the 2 angles formed are 90 degrees.) I am sure there are other
equivalent definitions, and some definitions may be better than
others in this case.
A+-------+B
| |
| |
| |
D+-------+C
One set of conditions may be:
|AB| = |BC|
|BC| = |CD| 4 sides equal
|CD| = |DA|
and AB . BC = 0 1 right angle
or
|AB| = |BC|
|BC| = |CD| 3 equal sides
and AB . BC = BC . CD = 0 2 right angles
There are other ways to prove equal sides or right angles like
vector equality or Pa..s Theorem (Sorry, I forgot how to spell it;
that one says z[squared] = x[squared] + y[squared]. I'm ashamed.)
Now, how do we prove that 4 points are planar?
One condition may be vector AB is in the same direction as DC, or
similiarly, BC in the same direction of AD.
Therefore, the least constraints for 4 points ABCD to form a square
in 3-D:
AB = DC (vector equality)
|BC| = |AD|
|AB| = |BC|
and AB . BC = 0
This may not be the most efficient way to compute it, by the way.
For example, changing the second constraint to vector equality still
works, and it is more efficient since it is easier to test vector
equality than scalar equality if the points coordinates are specified.
(Calculating distance involves at least square if not square root.)
I think changing the second constraint to vector equality still
makes the 2 cases equivalent.
Re .1:
I think using cross-product works too as that is used to show
orthogality. (Can't seem to spell words right.) However, that operation
will require more additions and mults.
Re .0, .2, .4:
This is not a proof. (at least not in any formal sense. Hope this
is good enough. I don't know how to rigorize it. Not a mathematician.)
To everyone:
If Micheal still want a rigorous proof, please help me to do so.
I don't think that is within my ability.
David Ng
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| 735.9 | One solution. | CHOVAX::YOUNG | Back from the Shadows Again, | Tue Jul 21 1987 12:56 | 9 |
| Four points in space form the corners of a square IFF:
Any 2 of the 6 line segments satisfy the following conditions:
1) They bisect each other
2) They are of equal length
3) They are perpendicular
-- Barry
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| 735.10 | Only three equations? | RDVAX::PERRONE | | Tue Jul 21 1987 14:29 | 20 |
| re .9
These conditions seem to be the most intuitive/elegant; however
computationally bisection is not trivia to verify.
I can think of two ways of doing it:
__ __
First we could find the midpoints of AC (call it E) and BD (F)
and equate E and F. This seems like one equation but its actually
three (one for each component).
__ __
Alternatively, we could find the intersection point of AC and BD
(call it G) and the equate line segments:
__ __ __ __
|AG| = |GC| |BG| = |GD|
Either way this gives more than three equations (i.e. there is some
redundancy somewhere).
So, can it be done with only three equations?
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