```From:           fc3a501@math.uni-hamburg.de
To:             sci.math
Subject:        cute geometric quickie
Date:           Thu, 14 Sep 2000 11:49:40 -0700
```

```Cut a square into two rectangles so that the
larger rectangle circumscribes the smaller.

(For our anal retentives: One cut parallel
to a squares side, and each edge of the
smaller rectangle lies on a different
side of the larger rectangle :-)
--
Hauke Reddmann <:-EX8                  BRANDNEW,IMPROVED SIG!
Send all spam to buggeth@thee.off
Send all personal e-mail to fc3a501@math.uni-hamburg.de
Send all e-mail for our chemistry workgroup to fc3a501@uni-hamburg.de
```

```From:           David Eppstein <eppstein@ics.uci.edu>
To:             sci.math
Subject:        Re: cute geometric quickie
Organization:   UC Irvine, Dept. of Information & Computer Science
Date:           Thu, 14 Sep 2000 13:10:31 -0700
```

```In article <8pq97u\$7mn\$2@rzsun03.rrz.uni-hamburg.de>,
fc3a501@math.uni-hamburg.de (Hauke Reddmann) wrote:

> Cut a square into two rectangles so that the larger rectangle
> circumscribes the smaller.

It is indeed a cute problem.

> (For our anal retentives: One cut parallel to a squares side, and each
> edge of the smaller rectangle lies on a different side of the larger
> rectangle :-)

I assume you actually mean: the four vertices of the smaller rectangle are
strictly interior to the sides of the larger rectangle (to rule out
"solutions" like cutting the square in half and making the two rectangles
coincide).

.
.
.
s
p
o
i
l
e
r
.
.
.

Draw the square, divided horizontally into two rectangles S (the smaller
one) and L (the larger one).  Let S1 be a copy of S, inscribed in L, and
look at the two vertices p and q of S1 that are on opposite vertical sides
of the square.  Then p and q differ horizontally by the side of the square,
and have Euclidean distance equal to the diagonal of S, so one can place
another copy S2 of S, with horizontal and vertical sides, sharing the same
two vertices p and q.

Now S2 is placed symmetrically in L, so it divides L into pieces in some
proportion b:a:b, where a is the side length of S.  Thus the horizontal
edges of S and S2 together divide the square in the proportion a:b:a:b.  In
particular, one of p and q is on the midpoint of its square edge.

Consider the four triangles at the corners of L, cut off by the edges of
S1.  They are similar right triangles, and the larger of the two has
hypotenuse equal to the square's side and one side equal to half the
square's side.  So they are 30-60-90 right triangles.

From this one can calculate in turn the lengths of the remaining side of
the large triangle, the short side of the small triangle, and the
hypotenuse of the small triangle, concluding that the width of S is
(2-sqrt(3)) times the width of the square, and that the width of L is
(1+sqrt(3)) times the width of the square.
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```From:           fc3a501@math.uni-hamburg.de
To:             sci.math
Subject:        Re: Reply to "cute geometric quickie"
Date:           Sun, 17 Sep 2000 09:32:05 -0700
```

```I didn't need any surds (in the inbetween computation) at all -
here is how I did it:

First I generalized: Start not with a square but with a
rectangle 1*(a+b) and cut it into 1*a and 1*b (a<b).
(So you can add questions# like "When does the 1*a inscribe
into the 1*b and that in the 1*(a+b)", which is also
interesting). I just handle a<1, but a>1 gives another
diagram.
When you cut out the 1*a from the 1*b, two pairs of similar
triangles 1:x:y=a:1-y:b-x remain. Together with one Pythagoras you
immediately get x=(b-a)/(1-a**2),y=(1-ab)/(1-a**2),
(1-a**2)**2-(1-ab)**2=(b-a)**2. You can divide this by (a-b)
(the "super anal retentive degenerate case" :-) to get
b=a*(3-a**2)/(1+a**2). a+b=1 now gives the 2-V3 solution.

# When is...one of the three rectangles similar to another...
one of them a square...b maximal...and so on :-)
--
Hauke Reddmann <:-EX8                  BRANDNEW,IMPROVED SIG!
Send all spam to buggeth@thee.off
Send all personal e-mail to fc3a501@math.uni-hamburg.de
Send all e-mail for our chemistry workgroup to fc3a501@uni-hamburg.de
```

```From:           lew@ihgp160h200.ih.lucent.com
To:             sci.math
Subject:        Re: cute geometric quickie
Date:           Mon, 18 Sep 2000 16:28:24 -0700
```

```In article <8ps9r0\$lv5\$1@cantuc.canterbury.ac.nz>,
Bill Taylor <mathwft@math.canterbury.ac.nz> wrote:
>fc3a501@math.uni-hamburg.de (Hauke Reddmann) writes:
>
>|> Cut a square into two rectangles so that the
>|> larger rectangle circumscribes the smaller.
>|>
>|> (For our anal retentives:
>
>between  <  and  <= .
>
>Presumably the answer you want is to cut a sliver of a rectangle off from near
>one edge, and lay it diagonally in the remaining near-square.
>
>Seems a bit trivial for sci.math.    Try rec.puzzles for these.   I leave it
>to someone with TOTH to find the limiting width of the sliver rectangle...

This is kind of interesting, actually. I kept bogging down
when I tried to set it up brute force, and I finally hit
on the following which turns out pretty slick:

If you circumscribe a rectangle about another rectangle
with sides touching the corners and making angles theta
and (90-theta) with the original sides, the sides of the
circumscribed rectangle will have length given by:

K' = C*L + S*K
L' = S*L + C*K

... in terms of the given sides K and L
( C and S are cos and sin of theta .)

If we want the rectangles to be the pieces of a square
formed by a cut parallel to a side, we have the requirement
L' = L, K' = L-K . Then we have the equations:

(C-1)*L + (S+1)*K = 0
(S-1)*L +   C  *K = 0

... and we must have C*(C-1) + ( 1 - S^2 ) = 0 = C*(2*C - 1 )
so C= 1/2 and S = sqrt(3)/2 and K = (2-sqrt(3))*L ~= 0.268 * L
( Passes a sanity check with paper and scissors. )

Lew Mammel, Jr
```

```From:           Ayatollah Potassium <no@spam.net>
To:             sci.math
Subject:        Inscriptible Rectangles
Date:           Wed, 20 Sep 2000 13:02:05 -0700
```

```Hauke Reddmann wrote:

> Cut a square into two rectangles so that the
> larger rectangle circumscribes the smaller.

I think a non-square AxB rectangle inscribes
in a CxD rectangle when:

((C+D)/(A+B))^2 + ((C-D)/(A-B))^2 = 2

and  C + D > A+B.

(suppose AxB is placed at an angle Theta
to the coordinate axes, so that its projections
on the axes are of length C and D. the
squared expressions are sin+cos and
sin-cos  of Theta.)
```