dissecting a square into similar triangles: Torsten Sillke
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Problem 2139. Triangulated Squares by Robert T. Wainwright, New Rochelle,
New York, Journal of Recreational Mathematics, 26:2 (1994) 147
How many ways can a square be dissected into six similar triangles?
-- Bob Wainwright --
Hess comunicated this problem to Nob.
Nob publishes it in his puzzletopia 101., Aug. 1995
Nob listed 97 dissections.
Only dissections with right angled triangles have been found.
different ratios: 1, 1/2, 2/3, 1/3,
.552, .570, .632, .648, .715 (cubic roots)
He gives no references!
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-- Torsten Sillke -- Jan 1994
The square can be dissected into n>=7 similar triangles, where no two
once are of the same size. There are different solutions with the
(45³, 45³, 90³)-triangle (ratio=1).
I worked on this problem after seeing the Hikimi-Puzzle
of a rectangular dissection of 9 (45³, 45³, 90³)-triangles of
different size at the Martin Gardner Gathering in Atlanta 1993.
Hikimi Shimani (Puzzland Hikimi)
Triangle-Areas: 16, 25, 32, 36, 50, 64, 72, 121, 128.
Rectangular-Area = 544 = 16*34 (ratio=8/17)
-- Torsten Sillke -- Jan 1996
Dissection of equal right angled triangles into different
(by iterated bisection).
Let a and b be the reduction factors (of the size), if one dissects the
triangle into two (a^2 + b^2 = 1).
If one have two triangles of the same size, one can
dissect one to get the three triangles of size 1, a, b.
Three equal triangles can be dissected into 8 parts
of size: 1, a, b, aa, ab, bb, aab, abb.
It is assumed that a^n = b^m iff n=0=m. If for example a = b*b, then
no iterated halving is possible to aviod triangles of the same size.
Four equal triangles can not be dissected by iterated halving into
different parts.
2 -> 1 3 -> 1 4 -> 1 -> 1 -> not resolvable
1 1 1 1 1 1 1 1
1 1 1 2 4 2 2 1 2
1 1 1 1
2 4 2
-- Torsten Sillke -- Jan 1996
Dissection of a (45³, 45³, 90³)-triangles into similar triangles:
minimal dissections are (6 parts):
49 -> 1,2,4,8,16,18 (area messure)
50 -> 1,2,4,9,16,18 (area messure)
minimal rectangular dissections (5 parts)
24 -> 1,2,4,8,9 (area messure) ratio: 3:2 or 4:3
minimal square dissections (7 parts)
98 -> 1,2,4,8,16,18,49 (area messure)
100 -> 1,2,4,9,16,18,50 (area messure)
-- Torsten Sillke -- Mai 1996
The (30, 60, 90)-triangle can be dissected into (30, 30, 120)-triangles
and vice versa. Call pairs of triangles with this property friends.
Are there other friends?
As the (30, 30, 120)-triangle is not on M. Laczkovich's list,
the square cannot be dissected into (30, 60, 90)-triangles.
There are other polygonal friends:
- the square and polyominoes tiling a square.
- the monoamond and the trianmond.
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==== E-Mail from the math-fun mailing-list. ====
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Date: Tue, 24 Oct 1995 17:21:14 -0400
From: Bob Wainwright
To: math-fun@cs.arizona.edu
Subject: Dissection of a square into six similar triangles
How many different ways can a square be dissected into six similar triangles?
Re-arrangements of a given tile set do not count.
Bob Wainwright
Life1ine@aol.com
Date: Wed, 25 Oct 1995 12:22:44 MST
From: "Richard Schroeppel"
To: math-fun@cs.arizona.edu
Subject: square dissections
I was thinking about Bob Wainwright's square dissection problem,
(into 6 similar triangles, how many ways?)
and discerned the following shapes of triangles that work:
90-45-45 can cut up the square many different ways;
90 - atan(1/2) - atan2 (half of a 1x2 rectangle)
90 - atan(1/3) - atan3 (half of a 1x3 rectangle).
(A right triangle can always be split into two similar right triangles
by dropping a perpendicular from the right angle to the hypotenuse.)
So we can take any similar-triangle-dissection of the square (with
right triangles) and cut the triangles as needed to get up to six
pieces. This will create a lot of 90/45/45 dissections.
This raises a related question: Suppose we want to dissect a square
into similar triangles, any finite number of pieces, but all the
same shape: What kinds of triangles might we use? Are there shapes
that require more than one size?
I can achieve 90 - atan A/B - atan B/A easily enough, by cutting
the square into rectangles of size 1/A x 1/B, and halving the
rectangles. Are there other possible shapes? Any "wrong" (left?)
triangles?
Rich rcs@cs.arizona.edu
Date: Thu, 26 Oct 1995 13:46:51 MST
From: "Richard Schroeppel"
To: math-fun@cs.arizona.edu
Subject: dissected squares
Re: Dissecting a square into similar triangles.
The dissection plan below allows some triangles with irrational
tangents, answering one of the questions in my message yesterday.
__________
| / /|
| / / |
| / / |
|/____/___|
A
The square has side = 1. The dissecting triangle has base A.
The central parallelogram can be dissected into a bunch of smaller
copies of the triangle, if the hypotenuse sqrt(1+A^2) is an
integer multiple of the base of the small copy. The small copy has
hypotenuse 1-A and base (1-A) * A/sqrt(1+A^2).
sqrt(1+A^2) = N A (1-A) / sqrt(1+A^2)
This rearranges to
(N+1) A^2 - N A + 1 = 0,
A = N +- sqrt( N^2 - 4N - 4 )
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2 (N+1)
Taking N = 5 makes the radicand a square, and A = 1/2 or 1/3.
The A = 1/2 example is particularly easy to see.
Since the radicand is (N-2)^2 - 8, there are an infinite number
of N which make it a square, in a Fibonacci-ish like series.
In fact, rational N also work for the dissection, and the full
set of rational solutions will have a parametric solution
reminiscent of the pythagorean triangle formula.
If we plug in random N > 2+sqrt8, say N=6, we get triangles with
irrational tangents:
N=6 -> A = (3 +- sqrt2)/7.
Revised open questions: Can the square be dissected into 30-60-90
triangles? Describe the set of allowed triangles. Wrong triangles?
Rich
Date: Thu, 26 Oct 1995 18:12:47 -0400 (EDT)
From: John Conway
Subject: Re: dissected squares
To: math-fun@cs.arizona.edu
" Can the square be dissected into 30-60-90 triangles? "
No! I assert that any convex polygon that's dissected into
such triangles can be redone so that all its vertices
are in points of K = Q(root(-3)). For let its bottom edge be
rescaled so that its two ends are at rational points on the real
axis, and define the lengths a,b,c,... of the edges of the triangles
that lie along it be taken as parameters. Do likewise with all the
other lines along which triangle-edges lie, except that the parameters
for lines at angles of 30 or 90 degrees to the horizontal are the
edge-lengths of the corresponding triangles divided by root3.
Then the consistency conditions for the parameters are certain
linear equations with rational coefficients. Since these have a solution
with an irrational parameter (coming from the height of the square), there
is at least a 1-parameter family of solutions. Follow this away from the
square until the topology first changes. This can only be by some
triangle edge-length passing through zero - by making it zero we
get a rectangle dissection with one fewer triangle. If there was an
n-parameter solution to the original equations, we could kill n-1
triangles like this and still have a rectangular solution of
variable shape.
Well, I thought I had a complete proof, but am now stuck; though
I hope only temporarily so. I don't actually believe you can build
a continuously variable-shape rectangle, but can't quite see how to
disprove it. JHC.
From: Bob Wainwright
Date: Sat, 28 Oct 1995 12:07:52 -0400
To: math-fun@cs.arizona.edu
Subject: Re: Similarity
Mainly to Dan Asimov:
> I forget whether a mirror-image of an arbitrary triangle is considered
> to be similar. Can you remind me?
Good question. With regard to dissection of a square into six similar
triangles, yes, the mirror image IS considered to be similar.
Dan, I'm not really sure how this problem eventually got posted but did you
receive a message indicating the total number of combinations?
Bob Wainwright
Life1ine@aol.com
Date: Mon, 30 Oct 1995 14:32:32 -0800
From: Dean Hickerson
To: math-fun@cs.arizona.edu
Subject: Re: Dissecting a square into similar triangles
The problem of cutting a square into similar triangles is allegedly
discussed in:
Tilings of polygons with similar triangles, by M. Laczkovich,
Combinatorica 10 (1990) 281-306.
I haven't seen the article itself, only a reference to it (in the book
"Algebra and Tiling: Homomorphisms in the Service of Geometry" by
Sherman Stein and Sandor Szabo), so i'm not sure if Laczkovich
completely determines which triangles work. He did show that there
are only 3 triangles that aren't right triangles which work, namely
those with angles (pi/8, pi/4, 5 pi/8), (pi/4, pi/3, 5 pi/12), and
(pi/12, pi/4, 2 pi/3). Also, it is impossible to tile a square with
triangles all of whose angles, when measured in degrees, are even
integers.
Dean Hickerson
dean@ucdmath.ucdavis.edu