Find Rectangle

I need to enclose any given rectangle with another rectangle.

The enclosing rectangle must be of a specific ratio - eg width is twice its height for example. In actual fact the ratio I need for the enclosing rectagle is that the width is 1.62344 times the height.

So given any rectange r with heigh h and width w I need to find rectangle r2 with height h2 and width w2 such that r sits inside r2 and r2 is just big enough to enclose r but no bigger than it needs to be.

Many thanks

Define the given rectangle

Hi bjhopper

The rectangle to be enclosed is not defined. It can be any rectangle - well lets say any rectangle with a width >= 1 and < = 100000 and height >= 1 and <= 100000

Thanks

Dave

Code:

---

  A          1.62344c          B

  E        b          F
                              c
  a

  D                    G      C

---

Rectangle EFGD (height=a, width=b) is inside rectangle ABCD (height=c, width=1.62344c).
a and b are givens, with b>a.
Is this what you mean?

Hello, daveywavey!

Quote:

---

I need to enclose any given rectangle with another rectangle.

The enclosing rectangle must be of a specific ratio - e.g. width is twice its height, for example.
In actual fact, the ratio I need for the enclosing rectagle is that the width is 1.62344 times the height.

So given any rectange with height and width ,
I need to find rectangle with height and width
such that sits inside , and is just big enough to enclose
. . but no bigger than it needs to be.

---

Code:

---

      *-------------------*  -
      |                  |  :
    - *-------------------*  :
    : |                  |1.848
    1 |        R        |  :
    : |                  |  :
    - *-------------------*  -
      : - - - - 3 - - - - :

---

Code:

---

      : - - 3 - - :
      *-----------*---*
      |          |  |
    2 |    R    |  |
      |          |  |
      *-----------*---*
      : - - 3.247 - - :

---

Hi both and thanks for your help.

I needed this for some computer code I was writing and have already got it working with an algorithm which I guess is pretty simillar to Sorobans as follows:

private void MaintainAspectRatio()
{
double canvasWidth;
double canvasHeight;

canvasWidth = _croppedWidth + 4;
canvasHeight = canvasWidth * _aspectRatio;

if (canvasWidth < _croppedWidth || canvasHeight < _croppedHeight)
{
canvasHeight = _croppedHeight + 4;
canvasWidth = canvasHeight / _aspectRatio;
}
etc

Thanks for your solution though, I can make my test more elegant with this. I was also kind of wondering if this could be expressed in one formulae but I don't know if it can?

Quote:

---

Originally Posted by daveywavey

I was also kind of wondering if this could be expressed in one formulae but I don't know if it can?

---

Only in this style: IF [whatever] ELSE [whatever]

Like, impossible to have ONE straightforward formula giving you the cost of apples AND of bananas, right?

Ok thanks Wilmer. I only did O level maths and some A level maths at college so I didn't know what you can do these days - I just remember Russell Crowe in a beautiful mind writing lots of algorithms and was thinking of that sort of thing I think!

Many thanks for youre help :)

Also this is my first question on here - are there protocols for expressing thanks, giving points etc? I'm sure there must be as I see it says 40 Thanks next to your name.

If we let r = ratio (1.62344 in your example),
a = height inner rectangle, b = width inner rectangle,
u = height outer rectangle,v = width outer rectangle,

then you can use (assuming a*r <> b):

IF ar > b then u = a and v = a*r ELSE v = b and u = b/r

Thanks Wilmer thats very succint, thanks!