15
$\begingroup$

Given two rectangles $R_1(l_1, b_1)$ and $R_2(l_2, b_2)$ where $l$ and $b$ are their length and breadth respectively, how to check if $R_1$ can fit inside $R_2$ or vice versa.

If $R_1$ and $R_2$ lie in the same plane and there exists an orientation of $R_1$ such that it lies completely inside $R_2$, then $R_1$ fits inside $R_2$.

$\endgroup$
2
  • $\begingroup$ Nice question. The rectangle $R_1$ rotated by $\theta$ radians fits in an axis-aligned box of length $l_1\lvert\cos\theta\rvert+b_1\lvert\sin\theta\rvert$ and width $l_1\lvert\sin\theta\rvert+b_1\lvert\cos\theta\rvert$. So you want to determine whether there is any $\theta$ such that these are less than $l_2$ and $b_2$ respectively. Of course, one can restrict $\theta$ to $[0,\pi/2]$ and drop the absolute value signs. This leaves you with inequality constraints on a couple of trigonometric functions, and I don't feel like working it out after that... $\endgroup$
    – user856
    Sep 3, 2012 at 12:05
  • 2
    $\begingroup$ This reminds me of a story where a train allowed only items of a max length/breadth/depth, so to get his over-long umbrella onto the train the passenger put it into a box. $\endgroup$
    – binn
    Sep 3, 2012 at 12:55

2 Answers 2

6
$\begingroup$

Look at: http://www.jstor.org/stable/2691523

$\endgroup$
1
  • 2
    $\begingroup$ This theorem is wrong. Check for rectangles (a x b) = (8 x 8) and (p x q) = (10 x 10). Both conditions (p > a and b >= fraction) are satisfy (it is necessary and sufficient condition) but square (10x10) doesn't fit into square (8x8). $\endgroup$ Mar 10, 2017 at 14:42
6
$\begingroup$

Joseph Malkevitch's answer is perfect.

For all who can't read the complete article, here's the most elegant solution (necessary and sufficient condition for a rectangle pxq to fit in a rectangle axb, provided p≥q, a≥b and p>a):

Rectangle pxq fit in rectangle axb

=((a+b)/(p+q))^2+((a-b)/(p-q))^2

Two obvious day-to-day applications: rug on a floor, tray in an oven.

$\endgroup$
12
  • 2
    $\begingroup$ p≥q? What happened with formula if p=q? $\endgroup$ Mar 10, 2017 at 14:47
  • $\begingroup$ @PiotrWasilewicz I agree the condition is incorrect: if you cannot fit a square rug in a room, you won't fit it by rotation. But I love the answer. $\endgroup$ Mar 13, 2017 at 20:15
  • $\begingroup$ @WeatherVane But answer is totally incorect! Check for a = 8, b = 1, p = 10 and q = 9. This theorem sais that rectangle 10 x 9 fits in rectangle 8 x 1. $\endgroup$ Mar 14, 2017 at 8:38
  • 1
    $\begingroup$ @PiotrWasilewicz yes I see that now, was just starting to implement it - it seems to work for some values. Perhaps there must be a condition too that a*b > p*q. $\endgroup$ Mar 14, 2017 at 8:57
  • $\begingroup$ @WeatherVane "work for some values" = doesn't work. I think a*b > p*q is not only condition that you need (but I'm not sure yet). Futhermore if you really only need a*b > p*q (and we have also p>a and also a ≥ b) we have theorem which is working not for "some values" it will be not enough for most examples. $\endgroup$ Mar 14, 2017 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.