From: David G Radcliffe
Subject: Cutting equilateral triangles
Date: 4 Apr 1999 00:36:57 GMT
Newsgroups: sci.math
Keywords: No dissections into incongruent equilateral triangles
In this note I will prove the following theorem.
Theorem 1. If an equilateral triangle is dissected into a finite
number of smaller equilateral triangles, then two of the smaller
triangles are congruent.
This is a consequence of a more general theorem:
Theorem 2. If a parallelogram is dissected into finitely many
equilateral triangles, then two of the triangles are congruent.
If an equilateral triangle is dissected into equilateral triangles,
then we can obtain a dissection of a parallelogram into equilateral
triangles by attaching another (undivided) equilateral triangle along
one of the sides of the original triangle. According to theorem 2, two
of the triangles in this dissection are congruent. But the triangle
that we attached is larger than all of the constituent triangles of
the original dissection. Therefore, two of the triangles of the
original dissection must be congruent.
This shows that theorem 2 implies theorem 1. We now turn to the proof
of theorem 2. Suppose we have are given a dissection of a parallelogram
into equilateral triangles. The parallelogram is drawn so that two sides
are horizontal.
There is a graph associated to this dissection. Each vertex of the
graph corresponds to a maximal horizontal segment, formed from the
sides of the triangles of the dissected parallelogram. The vertices
corresponding to the top and bottom sides of the parallelogram are
called "polar vertices". Each triangle lies between two maximal
horizontal segments, and thus corresponds to an edge of the graph.
There may be multiple edges joining two vertices.
_
__________ O \
\ /\ | \
\ / \ | |
\ /----\ O |
\ / \ / \ /|\ |
\/___\/___\ \|/ /
O_/
dissected associated
parallelogram graph
Now it is "obvious" that this graph is planar. Therefore, Euler's
polyhedron formula V-E+F=2 holds, where V is the number of vertices,
E is the number of edges, and F is the number of faces (including
the unbounded face).
If the graph contains two edges with the same endpoints, then the
corresponding triangles are congruent, and we are done.
Suppose to the contrary that the graph has no multiple edges. Then
each face must have at least 3 sides, and so 3F <= 2E. Combining
this with V-E+F=2, we conclude that 6 = 3V-3E+3F <= 3V-3E+2E =
3V-E, or E <= 3V-6.
The degree of each non-polar vertex is even and is at least 4. If
a non-polar vertex has degree 4, then the dissected parallelogram
must contain two congruent triangles. If this does not hold, then
each non-polar vertex has degree at least 6. Since each polar
vertex has degree at least 2, it follows that the sum of the
degrees is at least 4+6(V-2) = 6V-8. But the sum of the degrees is
equal to twice the number of edges. Therefore E >= 3V-4, and
this contradicts E <= 3V-6. Thus in any case we must have two
congruent triangles, and the theorem is proved.
The foregoing argument is an elaboration of the proof given in
the article "Dissections Into Equilateral Triangles" by W.T. Tutte.
That article appears in "Mathematical Recreations: A Collection in
Honor of Martin Gardner", edited by David A. Klarner.
Any errors are my own.
--
David Radcliffe
radcliff@uwm.edu