Trying to construct a rectangle inside a rectangle

Let me back up one step and say that I am ASSUMING that ugly beast of an
equation can be solved for b

 

Rewriting that equation in terms of b...
f(b) = 4b^4 - 4yb^3 + (x^2 + y^2 - 4w^2)b^2 + 2w^2yb + w^2(w^2 - x^2)

Remember that x,y, & w are known (constant distances), so if
there is a solution, we also know that the following must be true:

0 < x
0 < y
0 < w <= (whichever is greater, x or y) (correct?)

Finally the range of b is:
0 < b <= w

Can we assume that if a solution exists it will be unique? I think it must
be.
If so, then the graph of Y = f(b) over the range (0,w]
must cross (or touch) the X-Axis (0) at some point (our solution)

If it crosses the X-Axis, then the Y value of one range endpoint of the
curve ought to be positive and the other negative.
In this case, a linear slope (m) could be calculated from the curve
endpoints and an estimated solution could then be interpolated from the
result and perhaps refined further...
LINE:
Y = mX + B
B = f(0)
m = (f(w) - f(0))/w
Y = 0 at w = -B/m <= First estimate

Calculate f(-B/m) for new curve segment endpoint.
OR calculate the curve at two endpoints b = -B/2m and b = (w + B/m)/2
IOW, two points each halfway from the original endpoints to our estimate.
Then recalculate slopes, etc to create a second estimate... and so on.

If it only touches the X-Axis, then the slope of the curve at that point
must be 0 (horizontal).
Slope of original equation at any point will be its first derivative:

f'(b) = 16b^3 - 12yb^2 + 2(x^2 + y^2 - 4w^2)b + 2w^2y

Where f'(b) = 0 the slope is 0. I assume you could interpolate things as
above and refine further.

Fun stuff...
(Joe's routine is probably faster.

Best regards,
David Kozina

 

David:
That's very likely the same fourth-order polynomial my son came up with,
mentioned in my Monday posting (I don't have it here to compare, but it had
a comparable number of terms, with all those different powers up to 4 on
your b term). He didn't think it was reducible to b = something only in
terms of w x & y. But if there are other whizzes out there who might be
able to do something with it....

 

David,

I guess it would be fun, if I had the faintest idea what you and Kent are talking
about.

Let's assume, as you suggested, the solution you are thinking about still requires
interpolation. I suspect that would be faster than my method of physically rotating
rays, finding intersection points, and comparing distances a couple thousand times.

You may be faced with what I was looking at. I didn't see a way to short code the
thing in order to predict how it might turn out.

Regards
Joe Burke

 

hi, this problem was solved by Boutora Kamal,
in thread "un rectangle impossible à dessiner ?" 25/07/2004
news://news.planetar.net/autocad.general

;;; Programme pour le calcul d'un rectangle inscrit dans un autre
;;; connaissant la diagonale du 'petit' triangle
;;; Boutora kamal
;;;
;;; lancer la commande icare.

(defun c:icare (/ p1 p2 di ctr r xh yh xd yd h l binf bsup found dcalc oldos
oce dmax)

(setq oce (getvar "cmdecho"))
(setvar "cmdecho" 0)
(setq oldos (getvar "osmode"))
(if (and (setq p1 (getpoint "\nCoin 1 du rectangle"))
(setq p2 (getpoint p1 "\nCoin 2 du rectangle"))
(setq di (getdist "\nDiagonale du rectangle à inscrire"))
)
(progn
(setvar "osmode" 0)
(setq ctr (list (/ (+ (car p1) (car p2)) 2.0) (/ (+ (cadr p1) (cadr
p2)) 2.0))
l (abs (- (car p2) (car p1)))
h (abs (- (cadr p2) (cadr p1)))
)

;;; Conditions de départ pour une recherche dichotomique.
(setq binf (/ (max h l) 2.0)
bsup (distance ctr p1)
)
(setq found nil)
(setq xd (+ (car ctr) (/ l 2.0)))
(setq yh (+ (cadr ctr) (/ h 2.0)))
(setq yd (+ (cadr ctr) (sqrt (- (* binf binf) (* 0.25 l l)))))
(setq xh (+ (car ctr) (sqrt (- (* binf binf) (* 0.25 h h)))))
(setq dmax (distance (list xh yh) (list xd yd)))
(if (<= dmax di)
(progn
(princ (strcat "\nLa largeur maximale du rectangle est de " (rtos dmax
2 4)))
(setq binf bsup)
)
)

(while (and (null found) (> (abs (- bsup binf)) 1E-6))

;;; Calcul du point d'intersection du cercle de centre ctr et de rayon r
;;; Avec l'angle en haut à droite du grand rectangle.
;;;
(setq r (/ (+ bsup binf) 2.0))
(setq xd (+ (car ctr) (/ l 2.0)))
(setq yh (+ (cadr ctr) (/ h 2.0)))
(setq yd (+ (cadr ctr) (sqrt (- (* r r) (* 0.25 l l)))))
(setq xh (+ (car ctr) (sqrt (- (* r r) (* 0.25 h h)))))
(setq dcalc (distance (list xh yh) (list xd yd)))

;;; Modification des bornes pour la recherche trigonométrique.
(if (EQUAL dcalc di 1E-5)
(setq found 't)
(progn
(if (< dcalc di)
(setq bsup r)
(setq binf r)
)
)
)
)
(if found
(progn
(command "_pline"
(list
(+ (car ctr) (/ l 2.0))
(+ (cadr ctr) (sqrt (- (* r r) (* 0.25 l l))))

)
(list

(+ (car ctr) (sqrt (- (* r r) (* 0.25 h h))))
(+ (cadr ctr) (/ h 2.0))

)
(list
(- (car ctr) (/ l 2.0))
(- (cadr ctr) (sqrt (- (* r r) (* 0.25 l l))))
)
(list
(- (car ctr) (sqrt (- (* r r) (* 0.25 h h))))
(- (cadr ctr) (/ h 2.0))

)
"_close"
)
)
(princ "\nRien trouvé désolé pour vous !!!")
)
)
)
(setvar "osmode" oldos)
(setvar "cmdecho" oce)
(princ)
)

 

BTO,
Nice to have a mathematical solution.

Thanks.

Too bad it doesn't work if the width of the inside rectangle is specified to be 1/2 or more of the outside rectangle width.

Bill

 

Having had only one year of French, about 30 years ago, I'm not sure what's
happening everywhere here, without plowing through all the code to figure
out exactly what steps it's taking. But it looks like another
trial-and-error approach, because of:

which looks like repeatedly comparing to a very small distance, rather than
calculating an absolute answer.

[I also wonder whether it's usable for the original post's purposes, because
of:

which looks like it's asking for the diagonal, rather than the width, of the
inner rectangle.]

I'm still [pipe-]dreaming of a purely mathematical solution....

Kent Cooper, AIA


----- Original Message -----
From: "BTO"

 

I also wonder whether it's usable for the original post's purposes, because
of:

which looks like it's asking for the diagonal, rather than the width, of the
inner rectangle.]<<<

Yes, You pick the LL & UR corners and then give a width of the end of inside rectangle desired.
If you put in 1/2 or more of the outside rectangle end width, the message says "can't find, feels bad" or something like that.
I'm 1/2 French but never learned the language.
I'm still [pipe-]dreaming of a purely mathematical solution....<<

Me too.

Bill

 

BTO,
Thanks for finding this routine. It actually produces the results I was
wanted with the interior angles of the inner rectangle at 90 degrees. I'm
working on something else right now, so it will be a week or so before I can
follow the math. Thanks to all who looked into this problem.
Alan

 

Based on the common interest of this, I passed the problem to MathCad
discussion Forum.
There has come out two solutions since now ( 30.1.2005).
Interested persons can join the Collaboratory at www.mathcad.com and check
them.

It is there on the Puzzles & Games section under the name SquareInsideSquare
( wrong name actually ).

Regards Matti