My Math Forum The minimum radius

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December 26th, 2011, 03:38 AM   #1
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I have the problem presented in the attached image.
Does someone know how can this be solved?

Thank you.
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 geom.png (21.5 KB, 162 views)

December 26th, 2011, 04:47 AM   #2
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I posted a image with the wrong Q angle.
The correct image is attached here.
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 geom.png (21.2 KB, 157 views)

 December 26th, 2011, 05:16 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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: 20,965 Thanks: 2214 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
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Re:

Quote:
 Originally Posted by skipjack 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.
Yes, but I have no ideea how to calculate OC.

 December 26th, 2011, 08:42 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: The minimum radius $OC\,=\,\sqrt{$$r_{\small{1}}\,+\,x_{\small{1}}$$^2 \,+\,$$y_{\small{1}}\,-\,\tan(Q)r_{\small{1}}$$^2}$
December 26th, 2011, 08:47 AM   #8
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Quote:
 Originally Posted by Denis Please show WHOLE diagram.
Sorry, didn't notice the whole diagram was available by "dragging" ......

 December 26th, 2011, 09:18 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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
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Quote:
 Originally Posted by Denis 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]

Thank you

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