```From:           Joshua Bao <xxb1@psu.edu>
Date:           Tue, 26 Aug 1997 17:42:30 +0000
Newsgroups:     sci.math
Subject:        CHALLENGES (Dissection)
```

```I have some dissection challenges.  Please respond by email.  I don't
3) through 5).  Share these with your math friends.

1) Dissect a 3,4,5 triangle into 4 pieces and form a square.
2) Dissect a square into 5 pieces and form 3 squares of different sizes.

*3) Generalize 1) to all right triangles.  Which right triangles can be
dissected
into a finite number of pieces and rearranged to form a square ?
Calculate the minimum
number of pieces needed for an arbitrary right triangle.

*4) Generalize 2) to an arbitrary number of squares.  How many pieces
are needed?

*5) Generalize 3) and 4) to all convex polygons.

- Josh Bao
```

```From:           eppstein@ics.uci.edu
To:             xxb1@psu.edu
Subject:        CHALLENGES (Dissection)
Date:           Thu, 28 Aug 1997 16:09:16 -0700
```

```    1) Dissect a 3,4,5 triangle into 4 pieces and form a square.

You know, by the way, that this is also possible for the equilateral triangle?

2) Dissect a square into 5 pieces and form 3 squares of different sizes.

Any ideas what ratios of the four squares are possible?  Obviously they
must satisfy a^2+b^2+c^2=d^2 where d is the side length of the large
square and a,b,c are the three smaller ones.

I can solve it for (a,b,c,d)=(2,3,6,7), (1,4,8,9), (2,6,9,11),
(3,4,12,13), and (8,9,12,17).  Let me know if you want pictures of
these.  All my solutions have axis-aligned cuts; I don't know how to do
it for irrational ratios of side lengths (for which some diagonal cuts
would be necessary).  I also don't know of a solution for (2,5,14,15)
or (2,10,11,15).  All other integer solutions have d>=19; I haven't
tried any of them.
```

```From:           eppstein@ics.uci.edu
To:             xxb1@psu.edu
Subject:        CHALLENGES (Dissection) ... continued
Date:           Thu, 28 Aug 1997 16:15:18 -0700
```

```	2) Dissect a square into 5 pieces and form 3 squares of
different sizes.

Any ideas what ratios of the four squares are possible?  Obviously they
must satisfy a^2+b^2+c^2=d^2 where d is the side length of the large
square and a,b,c are the three smaller ones.
I can solve it for (a,b,c,d)=(2,3,6,7), (1,4,8,9), (2,6,9,11),
(3,4,12,13), and (8,9,12,17).

It seems that my (2,3,6,7) and (3,4,12,13) solutions generalize
to an infinite family of solutions of the form (a,a+1,a^2+a,a^2+a+1).
The next solutions in the family are (4,5,20,21) and (5,6,30,31).
--
David Eppstein		UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu	http://www.ics.uci.edu/~eppstein/
```

```To:             eppstein@ics.uci.edu, xxb1@psu.edu
Subject:        CHALLENGES (Dissection)
Date:           Fri, 17 Oct 1997 16:44:23 -0500
From:           Greg Frederickson <gnf@cs.purdue.edu>
```

```Hi David and Joshua,

Geometry Junkyard.  On Joshua's second question,
"Dissect a square into 5 pieces and form 3 squares of different sizes.",
part of chapter 8 in my new book "Dissections: Plane & Fancy" is about
dissecting 3 squares to 1 square in 5 pieces.

There are several infinite families with 5-piece solutions.
David has identified one of them, of the form (a,a+1,a^2+a,a^2+a+1).
The problem (2,5,14,15), for which David didn't know a solution,
is in one of the classes; its solution is in my book.
I don't recall offhand whether (2,10,11,15) is in one of
the classes; I would have to check on it.

Part of chapter 9 deals with dissecting 3 equilateral
triangles into one triangle in 5 pieces.

Greg Frederickson
```

```From:           eppstein@ics.uci.edu
To:             gnf@cs.purdue.edu
Subject:        3 squares to 1
Date:           Sun, 26 Oct 1997 22:00:09 -0800
```

```So, is it possible for a five-piece dissection of three different
squares to one larger square to be hinged?  One of my 2-6-9-11
dissections has a hinged pair of pieces for one square but the other
pair is not hinged.
-d
```

```To:             eppstein@ics.uci.edu
Subject:        Re: 3 squares to 1
Date:           Mon, 27 Oct 1997 10:59:45 -0500
From:           Greg Frederickson <gnf@cs.purdue.edu>
```

```I don't know of any five-piece dissection of three different
squares to one larger square that are hinged.  But I wasn't
looking for hinged dissections when I studied three squares
to one.

Your example 2-6-9-11 falls into a class that I call
the square-sum class:
For x^2 +y^2 +z^2 = w^2,
x is a square number,
y is twice a square,
and w = x + y.
(For your example, x=9, y=2, and w=11.)

Five-piece dissections for this class can be derived
from 6-piece dissections of general (i.e., real number)
relationships where
w = x + y, with z < 4 min {x,y}.
The conversion replaces a 3-piece P-slide
with a 2-piece step.

Greg
```

```From:           eppstein@ics.uci.edu
To:             gnf@cs.purdue.edu
Subject:        3 squares to 1
Date:           Wed, 10 Dec 1997 21:34:00 -0800
```

```I was looking at 3-squares-to-1-square dissections again this afternoon,
cleared up one of the cases I didn't know how to handle before,
and found two general families of dissections that were new to me.
So, I was particularly interested to look at chapter eight of my new
copy of your book to see if these or similar dissections were there.
They weren't. Does this mean they're new?

The families are:

(A) x=1, y=2k, z=2k^2, w=2k^2=1 (formed by m=q=k, n=1, p=0)
for which the first few examples are (1,2,2,3) (1,4,8,9) (1,6,18,19).
I've drawn an example of the dissection for (1,8,32,33)
in http://www.ics.uci.edu/~eppstein/junkyard/1-8-32-33.ps.gz
which I hope makes the general construction clear.

(B) x=(k+1)^2, y=2k^2, z=2k(k+1), w=2k^2+(k+1)^2 (formed by n-1=m=q=k, p=0)
for which the first few examples are (2,1,2,3) (9,8,12,17) (8,9,12,17).
I've drawn an example of the dissection for (25,32,40,57)
in http://www.ics.uci.edu/~eppstein/junkyard/25-32-40-57.ps.gz
which I hope makes the general construction clear.

Both of these have w=x+y or w=x+z, like your two square-sum classes.
Maybe all square-sum classes have five piece dissections?  But the
dissection families we know for the different square-sum classes are all
pretty different from each other...

-d
```

```From:           eppstein@ics.uci.edu
To:             gnf@cs.purdue.edu
Subject:        more triple-to-one squares
Date:           Fri, 12 Dec 1997 23:04:25 -0800
```

```Your figure 8.19 is not (as you assert) an isolated occurrence.
It is related to a Pellian equation involving rational approximations to
sqrt(2).

If the two small squares in one solution are x and y
(so the two larger ones are floor and ceiling of x^2+y^2),
then the next solution has x'=x+2y-1, y'=2x+5y-3.

So starting from (1,2,2,3), you get (4,9,48,49) (your figure),
then (21,50,1470,1471), then (120,289,48960,48961), then
(697,1682,1657466,1657467), then (4060,9801,56271600,56271601)...

-d
```

```To:             eppstein@ics.uci.edu
Subject:        Re: more triple-to-one squares
Date:           Mon, 15 Dec 1997 10:15:01 -0500
From:           Greg Frederickson <gnf@cs.purdue.edu>
```

```David,

>  Your figure 8.19 is not (as you assert) an isolated occurrence.
>  It is related to a Pellian equation involving rational approximations to
>  sqrt(2).
>
>  If the two small squares in one solution are x and y
>  (so the two larger ones are floor and ceiling of x^2+y^2),
>  then the next solution has x'=x+2y-1, y'=2x+5y-3.
>
>  So starting from (1,2,2,3), you get (4,9,48,49) (your figure),
>  then (21,50,1470,1471), then (120,289,48960,48961), then
>  (697,1682,1657466,1657467), then (4060,9801,56271600,56271601)...

Nice!  I am unfamiliar with Pellian equations and will have
to look them up.  I played around with the solutions yesterday,
and noticed that the y's are either b^2 = 2 a^2 +1
or b^2 + 1 = 2 a^2.  Whence b/a is a rational approximation of
sqrt(2), which is I guess what you were referring to.

It's interesting that this seems to be a very sparse class
in comparison to the others.

Greg
```

```From:           eppstein@ics.uci.edu
To:             gnf@cs.purdue.edu
Subject:        Re: more triple-to-one squares
Date:           Mon, 15 Dec 1997 11:07:45 -0800
```

```    I am unfamiliar with Pellian equations and will have to look them up.

Should be in any introductory number theory book.

I played around with the solutions yesterday, and noticed that the
y's are either b^2 = 2 a^2 +1 or b^2 + 1 = 2 a^2

Right, these (with 2 substituted for your favorite integer) are exactly
Pell's equations.  If you translate the statements that the two
staircases in the figure work into equations, you get something of the
form 2(y-x)^2 - (2x-1)^2 = 1.

It's interesting that this seems to be a very sparse class
in comparison to the others.

Yes, exponentially sparse.  Probably the members of the family can be
expressed in a closed form involving powers of (I think) 3+-2sqrt(2),
but I didn't work through the generating function stuff to check this.

Note that you can also get an interesting family of Pythagorean triples
from the same sequence of approximations to sqrt(2):
3^2+4^2=5^2, 20^2+21^2=29^2, 119^2+120^2=169^2...
In general these have the form x^2+(x+1)^2=z^2.  I've been trying
unsuccessfully to find a four-piece dissection for 20^2+21^2=29^2, in
the hope that it would extend to the whole family.

I'm also still stuck trying to find a dissection for 2^2 + 10^2 + 11^2 =
15^2, which I think is the only one with w<20 not covered by the families in
your book. This seems to be part of another family (x,y,y+1,w) related
to the same Pellian -- the next members of which seem to be (12,59,60,85)
and (12,348,349,493) -- y, z, and w are formed by halving the numbers
from the Pythagorean triples above but I don't yet understand the x's.

-d
```

```To:             eppstein@ics.uci.edu
Subject:        Re: more triple-to-one squares
Date:           Mon, 15 Dec 1997 16:00:29 -0500
From:           Greg Frederickson <gnf@cs.purdue.edu>
```

```>  Note that you can also get an interesting family of Pythagorean triples
>  from the same sequence of approximations to sqrt(2):
>  3^2+4^2=5^2, 20^2+21^2=29^2, 119^2+120^2=169^2...
>  In general these have the form x^2+(x+1)^2=z^2.  I've been trying
>  unsuccessfully to find a four-piece dissection for 20^2+21^2=29^2, in
>  the hope that it would extend to the whole family.

I tried a number of times without success.  It's interesting
that the only classes (so far) with 4-piece dissections
have one square much smaller than the other two, as one
looks at larger and larger instances.

For the exponentially sparse 3 squares to 1,
there is the corresponding relationship that fits into
chapter 4.  I didn't think about that when Robert sent me
his (then) isolated (4, 9, 48, 49).
The relationship is
x (w - x - y) = z (z - w + x)  for  x < y < z
and the dissection is the 7-piece P-slide analogue
of Robert's dissection.

Greg
```