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        riddles >> medium >> Rectangle inside rectangle
        
(Message started by: Barukh on Apr 28^th, 2008, 11:28am)

Title: Rectangle inside rectangle
Post by Barukh on Apr 28^th, 2008, 11:28am

Find necessary and sufficient conditions for a rectangle with sides a, b to be confined inside rectangle with sides A, B.

For clarity, let's suppose a >= b, A >= B.

Title: Re: Rectangle inside rectangle
Post by Aryabhatta on Apr 28^th, 2008, 11:55pm

Interesting problem Barukh!

Seems like the following condition is necessary and sufficient, though I haven't proved it, nor have simplified it and is likely not what you are looking for.

[hide]

There is some x in [0,pi/2] such that

A >= acos(x) + bsin(x)
and
B >= asin(x) + bcos(x)

[/hide]

Title: Re: Rectangle inside rectangle
Post by Hippo on Apr 29^th, 2008, 2:01am

Not solved yet, but big simplification is to consider only positions where the centers of gravity coincide (otherwise such shift does not make situation worse).
The only thing to be tested for such case is ... does both endpoints of an edge of smaller rectangle belong to the bigger rectangle.

It leads to search for appropriate angle ... (in [0, arctg B/A])

Title: Re: Rectangle inside rectangle
Post by Barukh on Apr 29^th, 2008, 3:48am

on 04/28/08 at 23:55:36, Aryabhatta wrote:

Interesting problem Barukh!

Thanks.

Quote:

Seems like the following condition is necessary and sufficient, though I haven't proved it, nor have simplified it and is likely not what you are looking for.

You are right in both cases. ;)

Title: Re: Rectangle inside rectangle
Post by Aryabhatta on Apr 30^th, 2008, 12:49pm

OK. I think we can give a "simpler" condition:

[hide]

Let K = sqrt(a^2 + b^2)

and A' = A/K and B' = B/K

Then we have that:

if A' > 1 then B >= b.

If A' < 1

then
B' >= sin(a + 2y)

where sin(a) = A'
and tan y = b/a. (or maybe a/b, i forget which!)

[/hide]

Title: Re: Rectangle inside rectangle
Post by Grimbal on Apr 30^th, 2008, 1:30pm

Well, I seem to arrive at the following result:
[hide]
If A>=a, you only need to have B>=b.
else you need to have
((A+B)/(a+b))^2 + ((A-B)/(a-b))^2 >= 2
[/hide]

Title: Re: Rectangle inside rectangle
Post by Barukh on May 1^st, 2008, 12:19am

Excellent, Grimbal! :D

Title: Re: Rectangle inside rectangle
Post by Aryabhatta on May 1^st, 2008, 10:51am

Yes, that is a good looking formula!

btw, I think my working is all wrong, so please ignore it.

Title: Re: Rectangle inside rectangle
Post by Hippo on May 3^rd, 2008, 7:12am

on 04/30/08 at 13:30:31, Grimbal wrote:

Well, I seem to arrive at the following result:
[hide]
If A>=a, you only need to have B>=b.
else you need to have
((A+B)/(a+b))^2 + ((A-B)/(a-b))^2 >= 2
[/hide]

This belongs to hard for me :(.
Can you show the reasoning?
I have checked some special cases and it seems to be OK, but I have absolutely no idea how to get the result.
(may be I didn't try enough ;) ).

I expect there will be some very nice picture hidden under reasoning ...

In all cases good work.

Title: Re: Rectangle inside rectangle
Post by Barukh on May 5^th, 2008, 3:10am

Hippo, sorry for not answering promptly.

I don't want to give away the whole thing, but here's some big hint.

Consider the case when the smaller rectangle is "enclosed" in the bigger one, like in the attached drawing.

Can you derive the relationship between four quantities A, B, a, b using the angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif as a auxiliary variable?

In fact, Aryabhatta's post contains similar ideas.

Title: Re: Rectangle inside rectangle
Post by Hippo on May 5^th, 2008, 5:07am

Barukh: Of course this was what I had done first, but I had ended with very complicated formulas. ... Ok I will make another try ...

OK ;) I have got it ;) so no nice picture :'( just an algabra.

Title: Re: Rectangle inside rectangle
Post by Eigenray on May 5^th, 2008, 5:15pm

Nice problem. But the answer is only interesting when a/b > [hide]1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2[/hide] ;)

Title: Re: Rectangle inside rectangle
Post by temporary on May 11^th, 2008, 9:44am

A >=sqrt(a^2 + b^2), B >=sqrt(a^2 + b^2)

Title: Re: Rectangle inside rectangle
Post by Barukh on May 12^th, 2008, 3:23am

on 05/11/08 at 09:44:17, temporary wrote:

A >=sqrt(a^2 + b^2), B >=sqrt(a^2 + b^2)

If this is correct, somebody needs to prove it.

But if it's not, a single counterexample will suffice. In the drawing attached to my previous post, A < a < sqrt(a^2 + b^2), and so your condition is not satisfied.

Title: Re: Rectangle inside rectangle
Post by temporary on May 13^th, 2008, 7:09pm

Then counterprove my theory, or prove your theory. [hide]

[/hide]

Title: Re: Rectangle inside rectangle
Post by ThudanBlunder on May 13^th, 2008, 7:32pm

on 05/13/08 at 19:09:25, temporary wrote:

Then counterprove my theory,

He just did! Jeez. ::)

All you do is wander from thread to thread degrading them by spouting nonsense and generally making a nuisance of yourself. Listen, why don't you do something useful: check out of here and return to that Rock, Paper, Scissors site, thus raising the average IQ at both sites!

Title: Re: Rectangle inside rectangle
Post by JiNbOtAk on May 14^th, 2008, 4:33am

Cool down T&B. It's not worth it getting riled up over this guy. Think of calm, pleasant things, like a field of sunflowers, or maybe an aquarium of colourful fishes. ;D

Title: Re: Rectangle inside rectangle
Post by ThudanBlunder on May 14^th, 2008, 5:38am

on 05/14/08 at 04:33:32, JiNbOtAk wrote:

Cool down T&B. It's not worth it getting riled up over this guy. Think of calm, pleasant things, like a field of sunflowers, or maybe an aquarium of colourful fishes. ;D

I am kewl. (That's what people tell me anyway.) :P
And I am not riled up - just having a bit of sport. He may as well serve some useful purpose while he is around.