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Topic: Classification of quadrilaterals
Replies: 17   Last Post: Jan 11, 1995 6:43 AM

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Chenteh Kenneth Fan

Posts: 14
Registered: 12/6/04
Re: Classification of quadrilaterals
Posted: Dec 20, 1994 7:00 PM
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I like Professor Conway's classification of n-gons according to
symmetries very much.

I wish it captured the generic trapezoid in its classification, however.

I want to offer another classification scheme which is much less elegant
than Professor Conway's, but yet restores the generic trapezoid to a happy
place in the classification scheme.

As a consequence of this classification, we also get the kite and
almost get the cyclic quadrilaterals. The cyclic quadrilaterals will appear
as a subset of "pseudo-cyclic" quadrilaterals (one pair of opposing angles
will be equal).

I guess a classification scheme consists of 2 parts:

1) A map from the objects to be classified to a finite or discrete set C,
the classifying space.

2) A way of relating two elements of C.

In Professor Conway's scheme, these two conditions are given by

1) A map fron n-gons to a conjugacy class of subgroups of D2n,
the dihedral group of order 2n.

2) Relating a conjugacy class of subgroups to another by checking
for possible inclusion.

In the scheme I'm about to suggest, the classifying space C is not so easy
to specify. Basically, I'm trying to classify n-gons by three properties:
1) parallity of sides, 2) equality of side-lengths, and 3) equality of angles.
I'll do this by mapping an n-gon to three n-tuples of vectors whose entries
are integers between 1 and n.

It may be easier to skip over this part to the EXAMPLES and deduce the map
and space C from them. It's like coloring parallel sides with the same color,
coloring edges of the same length with the same color, and coloring angles of
the same measure with the same color and looking at the resulting color
patterns. The colors are given numeric names.

Let S' be the set of n-tuples of vectors with entries integers between 1 and n.
The symmetric group acts on S' by sending (v1,...,vn) -> (p(v1),...,p(vn)),
where p is a permutation. Let S be S' modulo this action. For instance, when
n=2, the set S' has 2^2=4 elements, but the set S only has 2, which can be
represented by the orbits of (1,1) and (1,2).

Now let C' = S X S X S, the direct product of three copies of S.
There is an action of the dihedral group D2n on C' as follows. If c generates
the rotations and r is a reflection in D2n, then given (u,v,w) in C', let
c(u,v,w) = ((u2,...,un,u1), (v2,...,vn,v1), (w2,...,2n,w1)) and let
r(u,v,w) = ((u1,un,u_{n-1},...,u2), (v1,vn,v_{n-1},...,v2), (wn,w_{n-1},...,w1)).
Notice that rcr=c^{-1}.

Finally let C be C' modulo this action of D2n.

Now we define a map M from n-gons to C.

Start at an arbitrary edge and label the edges going around (either
clockwise or counter-clockwise)
e1 through en. Let ak be the angle between ek and e_{k+1) (using modular
arithemetic as necessary, so ak is defined for k=1, ...,n).

To construct the three vectors u,v,w, we go as follows. Let u1=1. If uk has
been defined for all k<l, let ul=(a number between 1 and n different from
{u1,...,u_{l-1}) if el is not parallel to any of e1 through e_{l-1}), or let it
equal uk if el is parallel to ek (k<l).

Let v1=1. If vk has
been defined for all k<l, let vl=(a number between 1 and n different from
{v1,...,v_{l-1}) if el is not the same length of any e1 through e_{l-1}),
or let it equal vk if el has the same length as ek (k<l).

Let w1=1. If wk has
been defined for all k<l, let wl=(a number between 1 and n different from
{w1,...,w_{l-1}) if al is not the same angle as any a1 through a_{l-1}),
or let it equal wk if al is equal to ak (k<l).

The reason for all those group actions used to define C is precisely to
make the above map well-defined (independent of all those choices).

For (2), we use closure relations. (It needs to be shown that if c is in C,
the closure of M^{-1}(c) is a union of fibers of M...I haven't checked this,
but I'm hopeful it is true.)

EXAMPLES

If I put ???, it means I have no idea what to call it.

n=3:

(123)(111)(111) Equilateral Triangle.
(123)(112)(122) Isosceles Triangle.
(123)(123)(123) Generic Triangle.

Everything else doesn't exist.

n=4:

(1212)(1111)(1111) Square.
(1212)(1212)(1111) Rectangle.
(1212)(1111)(1212) Diamond.
(1212)(1212)(1212) Parallelogram.

(1232)(1112)(1122) ???Bilateral Trapezoid???
(1232)(1233)(1231) ???Right Trilateral trapezoid???
(1232)(1233)(1234) ???Trilateral trapezoid???
(1232)(1234)(1231) Right trapezoid.
(1232)(1213)(1122) Isosceles trapezoid.
(1232)(1234)(1234) Trapezoid. (as promised, given a happy place in the
classification!)

(1234)(1112)(1231) ???Bilateral Triangular quadrilateral??? (terrible name!)
(1234)(1112)(1234) ???Bilateral quadrilateral??? (ditto)
(1234)(1122)(1232) Kite.
(1234)(1123)(1213) ???Isosceles pseudo-cyclic quadrilateral??? (uhg)
(1234)(1123)(1211) ???Isosceles biangular pseudo-cyclic quadrilateral??? (ouch)
(1234)(1123)(1123) ???Isosceles triangular pseudo-cyclic quad.??? (woe is me)
(1234)(1213)(1234) ???Isosceles quadrilateral???
(1234)(1213)(1231) ???Isosceles triangular quadrilateral???
(1234)(1234)(1112) ???Biangular quadrilateral???
(1234)(1234)(1213) ???Pseudo-cyclic quadrilateral???
(1234)(1234)(1123) ???Triangular quadrilateral???
(1234)(1234)(1234) Generic quadrilateral.

I'm fairly certain that everything else doesn't exist.

Not very elegant, I know, but if the assertions I made are good, then there
it is...an objective classification scheme which includes the trapezoid and
kite!

(Sorry I didn't indicate the closure relations, but the graph provided
by Tad Watanabe and found in "A Problem Solving Approach to
Mathematics for Elementary School Teachers" by Billstein et al.
is a subgraph of the analogous graph for n=4.)

Ken







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