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Noke Lieu
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sun-tan-gent
« on: May 30th, 2010, 11:47pm » |
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Quite around here at the moment (this forum). I blame the northern hemisphere comign into summer, and folk leaving the safety of inside...
Given two different circles, separated by a distance greater than R1+R2; construct the common cross tangent.
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| « Last Edit: May 31st, 2010, 7:07pm by Noke Lieu » |
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towr
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See attachment.
Although I've assumed the centers of the circles were given. But they're constructed easily enough.
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| « Last Edit: May 31st, 2010, 2:19am by towr » |
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TenaliRaman
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Re: sun-tan-gent
« Reply #2 on: May 31st, 2010, 6:45am » |
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on May 31st, 2010, 2:06am, towr wrote:See attachment.
Although I've assumed the centers of the circles were given. But they're constructed easily enough. |
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Is it a well known property that the cross tangents are concurrent with Line CD ??
-- AI
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towr
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Re: sun-tan-gent
« Reply #3 on: May 31st, 2010, 7:58am » |
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on May 31st, 2010, 6:45am, TenaliRaman wrote:
Is it a well known property that the cross tangents are concurrent with Line CD ?? |
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Say CD and AB intersect at X; triangles ACX and BDX are similar triangles. Say Y is the intersection between tangent 1 and circle 2 (and E is the intersection with AB), then triangle AGE is similar to BYE.
So we have
|AE|/|AG| = |BE|/|BY|
|AX|/|AC| = |BX|/|BD|
from similarity of triangles, but also,
|AG| = |AC|
|BY| = |BD|
since they are the radius of their respective circle. So therefore |AE|/|BE| = |AX|/|BX|, so X=E.
It works as long as angle BAC = angle ABD (by the same reasoning); right angles were just the easiest choice for me.
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| « Last Edit: May 31st, 2010, 8:07am by towr » |
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Noke Lieu
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Re: sun-tan-gent
« Reply #4 on: May 31st, 2010, 7:07pm » |
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It's funny how coming up with the riddles takes more time than the solving! I'll keep scratching around to see what I can see...
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Noke Lieu
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Re: sun-tan-gent
« Reply #6 on: Jun 1st, 2010, 5:04am » |
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I must commend you on your new toy- they're beautiful pictures.
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towr
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Re: sun-tan-gent
« Reply #7 on: Jun 1st, 2010, 6:04am » |
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It's a combination of two toys, actually (and neither very new); Cinderella (1.4) for making constructions, and Inkscape to finish it.
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| « Last Edit: Jun 1st, 2010, 6:05am by towr » |
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TenaliRaman
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Re: sun-tan-gent
« Reply #8 on: Jun 2nd, 2010, 7:16am » |
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Thanks for the proof.
My question was more along the lines of, whether you were already aware of this property. Because, it looks like the key to the construction you gave.
@Noke,
I don't think anybody could solve problems as fast as towr does  . Atleast, I can assure you that I was still racking my brain when towr posted his solution.
on May 31st, 2010, 7:58am, towr wrote:
Say CD and AB intersect at X; triangles ACX and BDX are similar triangles. Say Y is the intersection between tangent 1 and circle 2 (and E is the intersection with AB), then triangle AGE is similar to BYE.
So we have
|AE|/|AG| = |BE|/|BY|
|AX|/|AC| = |BX|/|BD|
from similarity of triangles, but also,
|AG| = |AC|
|BY| = |BD|
since they are the radius of their respective circle. So therefore |AE|/|BE| = |AX|/|BX|, so X=E.
It works as long as angle BAC = angle ABD (by the same reasoning); right angles were just the easiest choice for me.
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towr
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Re: sun-tan-gent
« Reply #9 on: Jun 2nd, 2010, 7:48am » |
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on Jun 2nd, 2010, 7:16am, TenaliRaman wrote:| My question was more along the lines of, whether you were already aware of this property. Because, it looks like the key to the construction you gave. |
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I wasn't really aware of it before, it just came to me; I figured the ratio AE to EB should be the same in both cases. And I hadn't really thought it trough until you asked before.
The thing that gave me more trouble was figuring out how to construct the two tangents using that point.
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| « Last Edit: Jun 2nd, 2010, 7:52am by towr » |
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Noke Lieu
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Re: sun-tan-gent
« Reply #10 on: Jun 2nd, 2010, 7:53pm » |
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on Jun 2nd, 2010, 7:16am, TenaliRaman wrote:
@Noke,
I don't think anybody could solve problems as fast as towr does . Atleast, I can assure you that I was still racking my brain when towr posted his solution.
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 Too true!
That said, it's why I'm glad to come here. The number of really gifted folk that have drifted by these dark pages has left me inspired.
It used to be easier to get questions when the boards were busier, but even empires wax and wane. I don't doubt that one day we'll return to hive of activity, but I suspect that we may need to have more people posting wrongish answers so newbies aren't so intimidated.
At least, that's my excuse!
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TenaliRaman
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Re: sun-tan-gent
« Reply #11 on: Jun 3rd, 2010, 11:32am » |
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on Jun 2nd, 2010, 7:48am, towr wrote:
The thing that gave me more trouble was figuring out how to construct the two tangents using that point. |
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LOL, I would have never thought that would have been the case
http://www.mathopenref.com/consttangents.html
-- AI
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