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Q: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle ( No Answer,   12 Comments )
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Subject: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
Category: Science > Math
Asked by: joshuac-ga
List Price: $20.00
Posted: 07 Oct 2006 04:42 PDT
Expires: 06 Nov 2006 03:42 PST
Question ID: 771496
Hi, let's say I have a rectangle, 10 by 20.  I wish to be able to fit
a smaller rectangle inside it, at an angle, with the 4 corners of the
smaller rectangle touching the 4 sides of the larger rectangle.

I have the following information:

-Length of the larger rectangle
-Width of the larger rectangle
-Width of the smaller rectangle

From this information I would like to find:

-Length of the smaller rectangle
-"Coordinates" locating the corners of the smaller rectangle on the
sides of the larger rectangle

Sorry my description is odd and using simplistic language, it's 4am
and not my area of expertise at all :)

A simplified real world example:

Large Rectangle Length and Width == 1 (a square)
Small Rectangle Length == 0.707106781

Defining the corners of the larger rectangle as a,b,c,d I would know
the the small rectangle corners would intersect the large rectangle in
this example at .5 from a,b,c, or d (since the bigger rectangle is a
aquare in my example and my smaller rectangle is rotated 45 degrees in
my example the distance of a,b,c,d from a,b,c,d of the larger
rectangle are all the same).

With my limited knowledge of geometry I thought perhaps a lead was to
break the smaller rectangle into two right triangles and calculate it
that way; I didn't get very far.

Again, sorry my language in describing this is so bad.  Feel free to
ask any questions for clarification.
Answer  
There is no answer at this time.

Comments  
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: harrysnet-ga on 07 Oct 2006 11:31 PDT
 
There are infinitely many solutions to the problem, as you describe it. The 
problem is that you don't know how to 'fit' your given width to the two rectangle 
sides. To use your example you don't have to place the smaller side that you 
give exactly in the middle of the two points it can be (say) at position 0.4
in line ab, closer to a. Then it would touch (ad) at point .5831 from a to d 
(that is it is closer to d). Unless you specify where it touches you can't 
have a specific answer.
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: myoarin-ga on 08 Oct 2006 06:56 PDT
 
I disagree with the above.  The definition of the dimensions of the
outer rectangle and the width of the inscribed rectangle give a unique
solution.

Let the length and width of the outer rectangle be A and B, and the
width and length of the inscribed rectangle be W and L.
The corners of the inscribed rectangle will touch the outer one at
points that that make a smaller triangle, the hypotenuse of which has
the length W.  Let's call the other sides "a" and "b", being portions
of A and B.  These two triangles will have the same proportions as the
the larger triangles with sides L, A-a, B-b.  From Pythagoras:

1:  A-a/B-b = b/a    2:  W^2 = a^2 + b^2     3:   L^2 = (A-a)^2 + (B-b)^2

And the sum of the area 
4:  AB = (A-a x B-b) + ab + LW

Shouldn't it be possible to solve for a, b and L with these equations?
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: harrysnet-ga on 08 Oct 2006 14:29 PDT
 
Well, myoarin you are right. The relations you derive make perfect sense, and 
should all be true. 

In order to avoid confusing joshuac I should say that the construction I gave, 
placing the W line arbitrarily as a first step, produces a parallelogram, and
not necessarilly a rectangle (a parallelogram is a sort of tilted rectangle,
with pairs of sides being parallel, but the angles are not necessarily right).

With the relations that myoarin gave it is possible to solve the problem easily. 
First relations 1 and 2 can be used to solve for a and b, and then relation 3 
can solve for L (there are probably some conditions on A, B and W, so 
that a, b and L come out as positive integers).

Relation 4 must also be true, so it must somehow follow from the other three 
using algebra, but I cannot see how, as the relations get complicated really
fast sincve they are not linear. Maybe someone else can see how to do that. 

In any case sorry joshuac, that my comment pointed in the wrong direction. 
The answer seems to exist, in the direction myoarin indicated.
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: myoarin-ga on 08 Oct 2006 14:55 PDT
 
Thanks, Harry.  I'm fine on simple geometry, but copped out on the algebra.  ;-)
Myoarin
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: berkeleychocolate-ga on 10 Oct 2006 20:01 PDT
 
Continuing the development of myoarin above, let theta be the angle
between the hypotenuse of length w and the side of length a in the
triangle he refers to. Then theta appears in several other places in
the figure.

It is not hard to see that [A - w*cos(theta)]/sin(theta) and [B -
w*sin(theta)]/cos(theta) both represent the length L we want to find.
Setting these two expressions equal we get an equation for theta (and
so for L).

Let x = sin(theta) and so sqrt(1-x^2)=cos(theta). Eliminating all the
radicals, we get a 4th degree equation in x. There are formulas for
the general equation of degree 4, but they are messy. The equation I
got was:

4w^2*x^4 - 4 A w x^3 + (A^2 + b^2 - 4w^2) x^2 + 2a w x + w^2 - b^2 = 0.
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: berkeleychocolate-ga on 10 Oct 2006 21:40 PDT
 
Correction to the formula I gave above:

4w^2*x^4 - 4*A*w*x^3 + (A^2 + B^2 - 4w^2)*x^2 + 2*A*w*x + w^2 - B^2 = 0.
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: myoarin-ga on 11 Oct 2006 04:22 PDT
 
Berkeley,
Theta is only known after we have the length of L  - and also a or b -
 so that we can place the inscribed rectangle correctly.

Cheers, Myoarin
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: berkeleychocolate-ga on 11 Oct 2006 11:08 PDT
 
Myoarin,

The formula I presented gives theta as a function of the three known
lengths: A, B and w. The unknown L is a simple function of theta, also
presented above. (Actually each is a function of the other.) Find
theta according to the formula and then find L.  So I don't agree that
"theta is only known after we have L".
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: myoarin-ga on 11 Oct 2006 13:35 PDT
 
Sorry, Berkeley, I stand corrected.  Congratulations!
Myoarin
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: mathtalk-ga on 13 Oct 2006 06:08 PDT
 
A few geometric observations about what dimensions of the smaller
rectangle are feasible, given dimensions L x W of the larger
rectangle.

Let the "larger" rectangle be centered at the origin, sides parallel
to the coordinate axes, say with length L in the x-direction greater
than (or equal to) width W in the y-direction.  The four corners of
this rectangle are then:

(L/2,W/2)

and they lie on a circle of radius R = SQRT(L+W)/2 centered at the origin.

The "smaller" rectangle will also be centered at the origin, though
without sides parallel to the coordinate axes.  Its radius r must be
less than or equal to R, but at least L/2 in order for the corners of
the smaller rectangle to touch each of the four sides of the larger
rectangle.

When L/2 < r < R the circle of radius r at the origin intersects the
larger rectangle in eight points.  Any opposing pair of points
(symmetric wrt the origin) forms a diameter of the circle, and any two
of these diameters can be taken as diagonals of a smaller rectangle. 
Provided that the two diagonals intersect different pairs of opposing
sides of the larger rectangle, we get a smaller rectangle "on an
angle" to the larger one, as desired.

Thus for radius r strictly between L/2 and R, we get four different
smaller rectangles, which correspond by symmetry to two different
possible smaller widths (if by width we mean the lesser of the smaller
rectangle's dimensions).

The eight points are these:

(u,W/2) and (L/2,v)

where u = SQRT(r - W/4) and v = SQRT(r - L/4).

The two possible smaller widths w are then:

w = dist( (L/2,v),(u,W/2) ) OR dist( (L/2,v),(u,-W/2) )

There is the limiting case when r = L/2 when the circle is tangent to
one pair of sides of the larger rectangle.  Then w takes only one
value as v = 0, and the two distances above are the same:

u = SQRT(L - W)/2

w = SQRT( (L/2 - u) + (W/2) )

But this doesn't give a lower bound for the width of the smaller
rectangle.  Instead the widths for cases L/2 < r < R exist as one
value above this and one value below it.  Thus any "smaller" width w
is feasible between 0 and W.

regards, mathtalk-ga
Subject: Re: Formula for fitting a smaller, rotated rectangle inside a bigger rectangle
From: myoarin-ga on 13 Oct 2006 07:26 PDT
 
Yes, well, that seems almost too obvious, just a bit complicated in "mathspeak".

Can't someone solve the equations for L?
Subject: Ferrari's method: solving the quartic equation
From: mathtalk-ga on 04 Nov 2006 17:43 PST
 
Ferrari's method: solving the quartic equation
==============================================

Wikipedia has a good article here:

[Quartic equation -- Wikipedia]
http://en.wikipedia.org/wiki/Quartic_equation

We will present a tailored treatment of the present
problem with an overriding aim of keeping algebraic
expressions as simple as possible.

As we shall see, the gist of Ferrari's method is
almost as elementary as completing the square is
in the context of solving the general quadratic.
However the method requires the solution of a cubic
equation as an intermediate step, much as Cardano's
solution to the cubic equation requires solving an
auxiliary quadratic equation.

Formulation of the quartic equation
===================================

Let the larger rectangle be centered at the origin
with sides parallel to the axes.  Vertices will be
of the form (a,b).  Without loss of generality,
we may scale the rectangle so that b = 1 and choose
an orientation so that a > 1, provided the width w
which we seek to impose on the smaller rectangle is
also scaled (divide all linear dimensions by b).

Width w is bounded above by the diagonal length of
the larger rectangle, 2 sqrt(a^2 + 1).

The smaller rectangle is here understood to have a
vertex incident with each side of the larger, so let
these vertices be (x,1), (a,y), (-x,-1), (-a,-y),
where:

  -a < x < a and -1 < y < 1                [0]

The symmetry of the smaller rectangle's vertices
with respect to the origin may be deduced from the
larger rectangle's vertices and the equal length
of parallel sides of the smaller rectangle.

The diagonals of any rectangle are equal in length
and mutually bisecting.  As applied to the smaller
rectangle this means:

   x^2 + 1  =  y^2 + a^2                   [1]

since its diagonals intersect at the origin.

We may assume that the side from (x,1) to (a,y) has
length w.  For if instead (-x,-1) to (a,y) were of
the prescribed length, then reflecting the smaller
rectangle in the x-axis would redress this.  Thus:

  (x - a)^2 + (y - 1)^2 = w^2              [2]

Together the inequalities [0], the hyperbola [1],
and the circle [2] determine the existence of a
solution (x,y) generating the smaller rectangle
for specified parameters a and w.

While [1] and [2] may be used directly to eliminate
one variable and give a resulting quartic equation
for the other, the form of the expressions will be
significantly simpler if we restate the problem in
terms of two new variables:

  u = x + y  and  v = x - y                [3]

In particular we can revise [1] as follows:

  uv = x^2 - y^2 = a^2 - 1                 [4]

Furthermore [3] can always be inverted to provide
the corresponding values of x,y from u,v:

  x = (u + v)/2 and y = (u - v)/2          [5]

so that [2] can be rewritten and simplified to:

  / u+v     \^2    / u-v     \^2
 (  --- - a  )  + (  --- - 1  )   = w^2
  \  2      /      \  2      /

  (u - (a+1))^2 + (v - (a-1))^2  = 2w^2    [6]

Multiplying both sides by u^2 and applying [4], we
get the following quartic equation for u:

  (u^2 - (a+1)u)^2 + (uv - (a-1)u)^2 = 2w^2*u^2

  u^4 - 2(a+1)u^3 + 2(a^2 + 1 - w^2)u^2
    - 2(a-1)(a^2 - 1)u + (a^2 - 1)^2 = 0   [7]

We now come to Ferrari's idea of expressing the
quartic equation P(u) = 0 as the difference of
two perfect squares, P(u) = Q(u)^2 - L(u)^2, with
Q(u) second-order & L(u) first-order polynomials.
This factors P(u) = (Q(u) + L(u))(Q(u) - L(u))
and consequently reduces the problem to solving
a pair of quadratic equations.

It is the constant term of Q(u) which gives rise
to the "nested" cubic equation we want to solve:

  Q(u) = u^2 - (a+1)u + d                  [8]

as the leading terms of Q(u) are dictated by [7].
Now from:

  Q(u)^2 = u^4 - 2(a+1)u^3 + ((a+1)^2 + 2d)u^2
             - 2d(a+1)u + d^2

subtract zero as expressed by LHS of [7]:

  Q(u)^2 = (2d - (a-1)^2 + w^2)u^2
            - 2(a+1)(d - (a-1)^2)u 
              + d^2 - (a^2 - 1)^2

For the trinomial on the right hand side to be a
perfect square L(u)^2 requires that the product of
the leading coefficient and the constant term be
the square of half the middle coefficient.  Thus:

  (2d - (a-1)^2 + w^2)(d^2 - (a^2 - 1)^2)
       = (a+1)^2(d - (a-1)^2)^2

Collecting terms gives us a cubic equation in d:

  2d^3 + (w^2 - 2(a^2+1))d^2 - (a^2 - 1)^2 w^2 = 0

                                           [9]

in which helpfully the d term already has a zero
coefficient.  Note that whatever the sign of the
d^2 coefficient, by Descartes' Rule of Signs there
will be exactly one positive root d of eqn. [9].

In my next post I will apply Cardano's method for
solving such a cubic.

regards, mathtalk-ga

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