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December 26th, 2011, 03:38 AM  #1 
Newbie Joined: Dec 2011 Posts: 6 Thanks: 0  The minimum radius
I have the problem presented in the attached image. Does someone know how can this be solved? Thank you. 
December 26th, 2011, 04:47 AM  #2 
Newbie Joined: Dec 2011 Posts: 6 Thanks: 0  Re: The minimum radius
I posted a image with the wrong Q angle. The correct image is attached here. 
December 26th, 2011, 05:16 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024  Re: The minimum radius
Please show WHOLE diagram. Also hard to tell what's "parallel"; AB is parallel to CD, right? So how can they be parallel to different lines? By "fitting", do you mean corners B and C will be on larger circle's circumference? 
December 26th, 2011, 06:45 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,632 Thanks: 2079 
Assuming that it's CB that's parallel to Y, not CD, try using coordinate geometry and trigonometry; r2 equals the longer of OC and OB.

December 26th, 2011, 07:39 AM  #5 
Newbie Joined: Dec 2011 Posts: 6 Thanks: 0  Re: The minimum radius
Yes, the text is wrong. Should be "CB is parallel to Y" not "CD is parallel to Y". By fit I mean C touches the larger circle and AD is tangent to the smaller circle. 
December 26th, 2011, 07:51 AM  #6  
Newbie Joined: Dec 2011 Posts: 6 Thanks: 0  Re: Quote:
 
December 26th, 2011, 08:42 AM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Re: The minimum radius 
December 26th, 2011, 08:47 AM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024  Re: The minimum radius Quote:
 
December 26th, 2011, 09:18 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024  Re: The minimum radius
You could have simply stated: rectangle ABCD has horizontal side AB = x and vertical side BC = y. That's what I'm doing(!) in this solution; also using r instead of your r1: Make E the point where the radius line meets AD; so angle AOE = Q Let a = AE: you can easily calculate length of a, since you have side r and angle Q (right triangle). Extend OE to OF (F on BC). You now have right triangle OFC: OF = r + x, CF = y  a So length of OC (the longer radius) = SQRT[(r + x)^2 + (y  a)^2] 
December 26th, 2011, 09:40 PM  #10  
Newbie Joined: Dec 2011 Posts: 6 Thanks: 0  Re: The minimum radius Quote:
Thank you  

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