Author |
Topic: Rectangle inside rectangle (Read 924 times) |
|
Barukh
Uberpuzzler
Gender: 
Posts: 2276
|
 |
Rectangle inside rectangle
« on: Apr 28th, 2008, 11:28am » |
 Quote  Modify
|
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.
|
|
 IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Rectangle inside rectangle
« Reply #1 on: Apr 28th, 2008, 11:55pm » |
Quote Modify
|
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.
There is some x in [0,pi/2] such that
A >= acos(x) + bsin(x)
and
B >= asin(x) + bcos(x)
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: Rectangle inside rectangle
« Reply #2 on: Apr 29th, 2008, 2:01am » |
Quote Modify
|
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])
|
| « Last Edit: Apr 30th, 2008, 7:55am by Hippo » |
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Rectangle inside rectangle
« Reply #3 on: Apr 29th, 2008, 3:48am » |
Quote Modify
|
on Apr 28th, 2008, 11:55pm, 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. 
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Rectangle inside rectangle
« Reply #4 on: Apr 30th, 2008, 12:49pm » |
Quote Modify
|
OK. I think we can give a "simpler" condition:
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!)
|
| « Last Edit: Apr 30th, 2008, 12:49pm by Aryabhatta » |
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Rectangle inside rectangle
« Reply #5 on: Apr 30th, 2008, 1:30pm » |
Quote Modify
|
Well, I seem to arrive at the following result:
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
|
|
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Rectangle inside rectangle
« Reply #6 on: May 1st, 2008, 12:19am » |
Quote Modify
|
Excellent, Grimbal! 
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Rectangle inside rectangle
« Reply #7 on: May 1st, 2008, 10:51am » |
Quote Modify
|
Yes, that is a good looking formula!
btw, I think my working is all wrong, so please ignore it.
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: Rectangle inside rectangle
« Reply #8 on: May 3rd, 2008, 7:12am » |
Quote Modify
|
on Apr 30th, 2008, 1:30pm, Grimbal wrote:Well, I seem to arrive at the following result:
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
|
|
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.
|
|
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
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  as a auxiliary variable?
In fact, Aryabhatta's post contains similar ideas.
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: Rectangle inside rectangle
« Reply #10 on: May 5th, 2008, 5:07am » |
Quote Modify
|
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.
|
| « Last Edit: May 5th, 2008, 5:18am by Hippo » |
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Rectangle inside rectangle
« Reply #11 on: May 5th, 2008, 5:15pm » |
Quote Modify
|
Nice problem. But the answer is only interesting when a/b > 1+ 2
|
| « Last Edit: May 5th, 2008, 5:15pm by Eigenray » |
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: Rectangle inside rectangle
« Reply #12 on: May 11th, 2008, 9:44am » |
Quote Modify
|
A >=sqrt(a^2 + b^2), B >=sqrt(a^2 + b^2)
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Rectangle inside rectangle
« Reply #13 on: May 12th, 2008, 3:23am » |
Quote Modify
|
on May 11th, 2008, 9:44am, 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.
|
| « Last Edit: May 12th, 2008, 3:24am by Barukh » |
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: Rectangle inside rectangle
« Reply #14 on: May 13th, 2008, 7:09pm » |
Quote Modify
|
Then counterprove my theory, or prove your theory.
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: Rectangle inside rectangle
« Reply #15 on: May 13th, 2008, 7:32pm » |
Quote Modify
|
on May 13th, 2008, 7:09pm, 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!
|
| « Last Edit: May 14th, 2008, 7:27am by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
JiNbOtAk
Uberpuzzler
Hana Hana No Mi
Gender:
Posts: 1187
|
|
Re: Rectangle inside rectangle
« Reply #16 on: May 14th, 2008, 4:33am » |
Quote Modify
|
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. 
|
|
IP Logged |
Quis custodiet ipsos custodes?
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: Rectangle inside rectangle
« Reply #17 on: May 14th, 2008, 5:38am » |
Quote Modify
|
on May 14th, 2008, 4:33am, 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. |
|
I am kewl. (That's what people tell me anyway.) 
And I am not riled up - just having a bit of sport. He may as well serve some useful purpose while he is around.
|
|
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print