Here is how to make a 1-holed torus out of 14 triangles. Each triangle
is given below in the following form. First, the vertices it spans
(among the set {1,2,...,7}) are listed. Then the length of the longest
edge is given, followed by the coordinates of the apex point (distance
from the left corner, then distance up from long edge). As a check we
also give the measure of the largest angle in the triangle.
The triangles thus constructed have their outward face facing you, if
assembled so that the vertices given are labelled on the triangle in the
order given, counterclockwise from the lower left.
Notice that the faces are congruent in pairs; indeed the surface is
symmetric with resepct to the z-axis. Faces 1, 2, 13, and 14 are isoceles.
Aside from mirror images, the only edges of the same lengths are
(12) and (34) (each is its own mirror image), (13) and (14) and
their mirror images (24) and (13), and (36) and (57) and their mirror
images (45) and (67). Except for this last short set (length 3), all other
edges have lengths between 4.4 and 8.5.
If you measure lengths in inches you get a reasonable-sized figure (5" tall),
so in keeping with tradition, lengths and coordinates are given in 32nds of
an inch. Apologies for readers in less-backward societies.
Face 1: vertices A1, A2, A3 edge: 8+16/32, coords: 4+8/32, 4+11/32 angle: 88
Face 2: vertices A2, A1, A4 edge: 8+16/32, coords: 4+8/32, 4+11/32 angle: 88
Face 3: vertices A3, A5, A1 edge: 6+23/32, coords: 4+6/32, 4+14/32 angle: 73
Face 4: vertices A4, A6, A2 edge: 6+23/32, coords: 4+6/32, 4+14/32 angle: 73
Face 5: vertices A6, A1, A5 edge: 6+5/32, coords: 2+19/32, 3+21/32 angle: 80
Face 6: vertices A5, A2, A6 edge: 6+5/32, coords: 2+19/32, 3+21/32 angle: 80
Face 7: vertices A3, A4, A5 edge: 8+16/32, coords: 6+12/32, 2+4/32 angle: 117
Face 8: vertices A4, A3, A6 edge: 8+16/32, coords: 6+12/32, 2+4/32 angle: 117
Face 9: vertices A7, A1, A6 edge: 6+18/32, coords: 1+2/32, 2+26/32 angle: 84
Face 10: vertices A7, A2, A5 edge: 6+18/32, coords: 1+2/32, 2+26/32 angle: 84
Face 11: vertices A1, A7, A4 edge: 6+18/32, coords: 3+16/32, 4+31/32 angle: 67
Face 12: vertices A2, A7, A3 edge: 6+18/32, coords: 3+16/32, 4+31/32 angle: 67
Face 13: vertices A3, A7, A6 edge: 5+27/32, coords: 2+29/32, 0+23/32 angle: 153
Face 14: vertices A4, A7, A5 edge: 5+27/32, coords: 2+29/32, 0+23/32 angle: 153
The vertices are, in order from bottom to top,
A1=(-3,3,0)
A2=(3,-3,0)
A3=(-3,-3,1)
A4=(3,3,1)
A5=(1,2,3)
A6=(-1,-2,3)
A7=(0,0,5)
To make a 10-vertex 2-holed torus, make two copies of everything except
faces 1 and 2, and glue vertices A1 and A2 in one figure to A3 and A4 in
the other.
If you have trouble getting your model to hold together, I suggest you not
cut out all the faces separately, but rather cut a large piece of cardboard
with all the faces forming a connected subset, then fold. I don't know the
"best" way to do this.
It is also useful to cut the faces somewhat larger, leaving tabs sticking out
on each side which can then be pasted together. An alternative works well for
sufficiently heavy cardboard: have the tabs stick out perpendicularly from
the faces (more or less) and run a rubber band around the tabs which you would
otherwise paste.
The data are lightly modified from
\'Akos Cs\'asz\'ar, "A polyhedron without diagonals"
Acta Univ Szegendiensis, Acta Scient. Math
v 13 (1949) pp 140-2 [QA1.A258 in my library]
This arrangement of coordinates may differ slightly from that posted in
sci.math on 2/28/95.
dave rusin@math.niu.edu