From: Kurt Foster
Subject: Re: Fibonacci Geometric Fallacy
Date: Tue, 21 Dec 1999 16:45:17 GMT
Newsgroups: sci.math
Keywords: using Fibonacci numbers to make a geometric puzzler
In , Missy said:
[snip]
. If a rectangle with side length of any Fibonacci number is cut, and
. the pieces are rearranged into a rectangle, an interesting problem
. occurs. The area of the sqaure is 169 square units. The area of the
. rectangle, made from the same pieces, is 168 square units. Where did
. the other unit go?
It didn't "go" anywhere. Obviously, the pieces don't quite fit. One
rendition of this [an 8x8 dissected and reassembled into a 5x13] is shown
in Martin Gardner's second SCIENTIFIC AMERICAN Book of Mathematical
Puzzles and Diversions. Here's the general version of that rendition:
Note that, if F_k is the k-th Fibonacci number, then
F_(n+1) * F_(n-1) - [F_n)]^2 = (-1)^n
An F_n x F_n square can be dissected as follows, if n >= 3. First, divide
it into two rectangles, F_n x F_(n-2) and F_n x F_(n-1).
Now, bisect the F_n x F_(n-2) along a diagonal, giving two right
triangles with legs F_n and F_(n-2).
Call the other rectangle ABCD, where
AB = CD = F_n and AC = BD = F_(n-1).
Take the points P on AB and Q on CD so
AP = F_(n-1), PB = F_(n-2); CQ = F_(n-2), QD = F_(n-1).
Divide ABCD along the segment PQ. This gives two trapezoids.
Now, the sides PB and CQ of the trapezoids can be fit to the shorter
legs of the triangles so as to (almost) give right triangles with legs
F_(n-1), and F_(n-1) + F_n = F_(n+1). These can then be (almost) fit
together to form a rectangle with sides F_(n-1) and F_(n+1).
The fallacy is easily seen with these almost-triangles. Bisecting the
F_n by F_(n-2) rectangle gave right triangles with legs F_(n-1) and
F_(n-2). Tacking on the trapezoids supposedly gives larger right
triangles with legs F_(n+1) and F_(n-1). But these larger "right
triangles" *should* be similar to the smaller ones, whereas
F_(n-1)/F_(n+1) does not equal F_(n-2)/F_n.
In other words, the triangle-plus-trapezoid pieces aren't actually
triangles - the "hypotenuses" are bent and have a corner at the join.
[This is most obvious when n = 3; the "trapezoidal" pieces are then
squares, since F_1 = F_2 = 1.]
So the discrepancy in the area of the square and the rectangle is a
parallelogram with a diagonal along the diagonal of a F_(n-1) x F_(n+1)
rectangle. When n is even, this parallelogram appears as a gap between
the pieces of the square when they are reassembled into the oblong
rectangle; when n is odd, it appears as the region where the pieces
of the square overlap when reassembled into the rectangle. In either
case, the area of the parallelogram is of course 1 square unit.
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From: Allan Adler
Subject: Re: Fibonacci Geometric Fallacy
Date: 20 Dec 1999 18:12:01 -0500
Newsgroups: sci.math
Allan Adler writes:
> this must be one of a class of dissection problems based on the identity
> a(n)^2 = a(n-1)*a(n+1) + (-1)^(n+1) for Fibonacci numbers, where a(n)
> is the n-th Fibonacci number and a(1)=a(2)=1. When n is even, the problem
> must be to cut up a square which is a(n) on a side and reassemble the pieces
> so that it looks like one is getting a rectangle which is a(n-1) by a(n+1),
> but the pieces don't really fit exactly and one loses one unit of area in
> a skinny little quadrilateral in the center of the rectangle. Something
> like that. When n is odd, the problem must be to cut up a rectangle
> which is a(n-1) by a(n+1) and reassemble the pieces to nearly get a
> square which is a(n) on a side.
So far, so good.
> I must admit that I never worked through the solution of this one in the
> case of cutting up a chessboard to nearly get a 5 x 13 square. Missy's
> would be the case of cutting up a square 13 on a side and reassembling
> it to nearly get a 8 x 21 rectangle. These both correspond to n even,
> so I don't actually know that the problem is solvable when n is odd.
Bzzzt!!! The case 8x8 is a case of n even, the case of 13 x 13 is a case
of n odd.
> Anyway, I don't know how to do either offhand and would be glad to know
> the answer. I can't work on it myself since I'm bogged down with other
> projects right now.
Another poster gave a web site at Berkeley which I looked at quickly
and realized how to adapt it to solve Missy's problem. Here is the solution.
I hope someone redoes it in graphics somewhere, since the pictures are more
pleasant to look at. This exposition will be prose.
For definiteness and clarity, let me adopt the convention that when I
say a rectangle is X by Y, I mean that X is its height and Y is its width.
First take n even and let S be a square whose side is a(n).
(1) Make a vertical cut to cut S into two rectangles A,B, where A is a(n)
by a(n-2) and B is a(n) by a(n-1).
(2) Draw a diagonal of A, cutting it into two right triangles A1,A2. As for
rectangle B, make a horizontal cut to cut it into two rectangles C,D,
where C is a(n-3) by a(n-1) and D is 2a(n-2) by a(n-1).
(3) Make a diagonal cut in rectangle C, cutting it into two right triangles
C1, C2. Make a horizontal cut in rectangle D, bisecting it into two
rectangles D1,D2, each a(n-2) by a(n-1).
This gives all the pieces, i.e. A1,A2,C1,C2,D1,D2. Now, here is how to
assemble them to get what looks very nearly like a a(n-1) by a(n+1)
rectangle:
(a) Start with a rectangle R which is a(n-1) by a(n+1).
(b) Place the rectangles D1 in the lower right corner of R, with the side
of length a(n-1) of D1 horizontal (as we found it).
(c) Place triangle C1 on top of rectangle D1, with the side of length a(n-1)
of C1 lying on the side of length a(n-1) of D1 and with the right angle
of C1 flush to the right, i.e. the side of length a(n-2) of C1 should
lie along the side of rectangle R.
(d) Place triangle A1 to the left of rectangle D1, with with the side of
length a(n-2) of A1 flush against the side of length a(n-2) of D1
and with the side of length a(n) of A1 lying on the side of the
rectangle R. The right angle of A1 will be to the right, its vertex
on the common edge of A1 and D1.
(e) Now rotate rectangle R 180 degrees and repeat steps (b),(c),(d)
with D2,C2,A2 instead of D1,C1,A1.
This will leave a skinny parallogram of area 1 empty in the interior
of R.
I did the construction this way because I was modifying the pieces found
at the Berkeley web site. However, whereas I use 6 pieces, one can actually
do it with 4. In hindsight, one can see that instead of cutting up rectangle
B into two rectangles and two triangles, one could cut it up into two
trapezoids, each congruent to the trapezoid formed by resting C1 on D1
as in step (c) above. I think it is easier to describe the dissection into
6 pieces and easier to motivate in view of the availability of the Berkeley
web site, but that is a matter of taste.
Ok, now let n be odd and let us start from scratch with our geometrical
notation. Let R be a rectangle which is a(n-1) by a(n+1). Make a
vertical cut to cut it into two rectangles R1,R2, where R1 is actually
a square a(n-1) on a side and R2 is a(n-1) by a(n). Since n-1 is even,
we can use the dissection described above to cut R1 up into pieces
that can be reassambled into a rectangle R3 which is a(n-2) by a(n),
with a skinny parallogram of area 1 missing from the interior of R3.
Now rotate R3 and place R3 and R2 together along sides of length a(n)
to make a square a(n) on a side.
I hope I have written this clearly enough and haven't made too many
typographical errors.
OK, now this class of dissection problems can be viewed as being associated
to the real quadratic field Q(sqrt(5)), since the Fibonacci numbers are
related to the fundamental unit of that field. Can someone come up with
a similar class of dissection problems associated to any real quadratic field?
Allan Adler
ara@altdorf.ai.mit.edu
****************************************************************************
* *
* Disclaimer: I am a guest and *not* a member of the MIT Artificial *
* Intelligence Lab. My actions and comments do not reflect *
* in any way on MIT. Morever, I am nowhere near the Boston *
* metropolitan area. *
* *
****************************************************************************
==============================================================================
From: Nicolas Bray
Subject: Re: Fibonacci Geometric Fallacy
Date: Mon, 20 Dec 1999 10:33:52 -0800
Newsgroups: sci.math
To: Missy
Similarly:
http://www.math.berkeley.edu/~hausel/csalas.gif
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