
Re: Classification of quadrilaterals
Posted:
Dec 20, 1994 7:00 PM


I like Professor Conway's classification of ngons 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 "pseudocyclic" 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 ngons 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 ngons by three properties: 1) parallity of sides, 2) equality of sidelengths, and 3) equality of angles. I'll do this by mapping an ngon to three ntuples 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 ntuples 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_{n1},...,u2), (v1,vn,v_{n1},...,v2), (wn,w_{n1},...,w1)). Notice that rcr=c^{1}.
Finally let C be C' modulo this action of D2n.
Now we define a map M from ngons to C.
Start at an arbitrary edge and label the edges going around (either clockwise or counterclockwise) 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_{l1}) if el is not parallel to any of e1 through e_{l1}), 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_{l1}) if el is not the same length of any e1 through e_{l1}), 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_{l1}) if al is not the same angle as any a1 through a_{l1}), 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 welldefined (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 pseudocyclic quadrilateral??? (uhg) (1234)(1123)(1211) ???Isosceles biangular pseudocyclic quadrilateral??? (ouch) (1234)(1123)(1123) ???Isosceles triangular pseudocyclic quad.??? (woe is me) (1234)(1213)(1234) ???Isosceles quadrilateral??? (1234)(1213)(1231) ???Isosceles triangular quadrilateral??? (1234)(1234)(1112) ???Biangular quadrilateral??? (1234)(1234)(1213) ???Pseudocyclic 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

