quick sketch of a proof
By gzt (Thu Apr 14, 2011 at 10:36:06 AM EST) gzt, trigonometry (all tags)
that the box is cos(π/12) x cos(π/4).

Any rectangle with diagonal 1 can fit into the box and this is the smallest box by area for which this is the case.

We have trivially established that one of the sides needs to be, at minimum, cos(π/4). The question is whether the other side can be of length less than 1.

A rectangle with diagonal 1 will have a height cos(θ) and a width sin(θ) for some θ. I'm not going to explain why, but the critical rectangles for us are the ones with an angle between 0 and π/8. You can check up to π/4 if you want, but they fit trivially.

So there are two ways a rectangle can fit inside the box: sitting flush against the edge and diagonally. So if you imagine that the BOX we're trying to fit our rectangle in is situated with vertices at (0,0), (0,H), (sin(π/4),0), and (sin(π/4),H), we can calculate, for each possible rectangle, what the needed height H will be. It will be the minimum of cos(θ) and Q, the highest point of the rectangle when it is sitting skew. So now we have an ugly trig problem and a minimax problem to solve. By the way, you solve the minimax problem by equating Q and cos(θ). Easy peasy.

Wait, so what's this θ, by the way? Situate the rectangle so the long side is along the y-axis, draw the diagonal, it's the smaller internal angle. Okay, great.

Now, tilt your rectangle over so it fits in your box askew. Now the angle between the diagonal of the rectangle and the y-axis is π/4, so the angle the side of your rectangle makes with the y-axis = β = π/4 - θ (why: you have a diagonal of 1 and a base of sin(π/4)). Now draw a bunch of triangles, labeling the right angles, the θ angles, the β angles, and do a lot of projections with sines and cosines to get something like the required Q = sin(θ)sin(β)+cos(θ)cos(β) or something like that, I don't have my scrap paper with me. Which solves for θ = π/12.

I might make some diagrams later or something, but I have some crap I'm supposed to be doing.

quick sketch of a proof | 5 comments (5 topical, 0 hidden) | Trackback
Pity I can't remember my calculus here.... by wiredog (2.00 / 0) #1 Thu Apr 14, 2011 at 10:50:34 AM EST
As this seems like a limit problem as n->0

Earth First!
(We can strip mine the rest later.)

no, this is actually all geometry. by gzt (4.00 / 1) #3 Thu Apr 14, 2011 at 11:07:27 AM EST
If you do a diagonal of length N instead of 1, well, just multiply by N. So it's easier to just deal with the case of 1.

As for the minimization, set the two equal to each other and it suddenly becomes a trig problem.

[ Parent ]
I follow it by sasquatchan (2.00 / 0) #2 Thu Apr 14, 2011 at 11:04:48 AM EST
but the "trivial" parts, aren't .. (or at least link back to Ambrosen's diary where you established it .. )  (my geometry was 9th grade, so, uh, 1989 ..)

The trivial parts actually are by gzt (2.00 / 0) #4 Thu Apr 14, 2011 at 11:12:35 AM EST
The tricky part is the hand-wavey part with all the trig.

The minimum dimension on one side follows from the fact that you have a square of that size that you need to fit.

The π/8 instead of π/4 can be verified with a diagram (the short answer is that their height makes them fit in the boxes that accomodate the other angles), but, if skeptical, you can just do up to π/4.

So those are the two things I labelled trivial. One can be ignored - everything fits in my box and nothing smaller will fit the ugly rectangle that demands it - and one truly is trivial.

[ Parent ]
So... by ana (2.00 / 0) #5 Fri Apr 15, 2011 at 11:10:51 AM EST
I did a bunch of trig, setting things up the way you did (I think).  I get the width of a rectangle that's just big enough to contain a rectangle with aspect angle theta and tilt angle beta to be W=sin(theta+bet), and the height H=sin(theta+pi/2-beta).

So the deal then is to extremize WH with respect to beta for any given theta. You do the partial derivative of WH with respect to beta (i.e. holding theta fixed), and I get cos(2*beta), oddly independent of theta. Set that to zero and you get beta=pi/4, but it turns out that's a maximum, not a minimum. So you look at the endpoints, and discover the minimum area rectangle that'll circumscribe a given rectangle is at beta=0 or pi/2. I think this means that tilting the rectangle only makes the problem worse.

So then it reduces to finding the area of the largest rectangle required to contain one of diagonal meausrement 1, for any angle theta (arctan of the ratio of height to width). The largest area of the little rectangle itself is for the square (and it's 1/2 in these units), but that's not relevant. The height of the envelope rectangle needs to be at least 1/sqrt(2), and the width 1, so I think that's the answer: the minimum area of an envelope that'll contain any rectangle with diagonal measuring unity is 1/sqrt(2).

"And this ... is a piece of Synergy." --Kellnerin

quick sketch of a proof | 5 comments (5 topical, 0 hidden) | Trackback