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Topic: Wrapped Circles (Read 755 times) |
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Sir Col
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Wrapped Circles
« on: Apr 14th, 2004, 10:29am » |
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An inelastic band (of negligible thickness) fits perfectly around two touching circles with radii 1 cm and 3 cm. Find the exact length of the band.
What would be the length of a band surrounding three touching circles with radii 1 cm, 2 cm, and 3 cm?
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Noke Lieu
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Re: Wrapped Circles
« Reply #1 on: Apr 14th, 2004, 4:47pm » |
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Aw, its much easier when all three are the same size 
But I love these.
The centre of the two circles (A,B) are 4cm apart.
Draw common tangents of A and B. Gives the angles (if you draw the trap, remove the 4x1 rectangle, leaves a right triangle sides 2, 12.5,4) Hey hey, 30,60,90.
so the band's points of contact/leaving of the circles are 120o apart relative to their centres.
This means that the band has 2x12.5+(2pi/3)+(12pi/3) (that is there are two lengths from the gap between the circles, 1/3 of the circumference of the small circle and 2/3 the circumference of the big one.
Simplifies down to (14pi/3)+2x12.5
Will figure out how to get the sqrt sign shortly. Gives someone a chance to get the second part.
[e} okay, that's easy... will wait a while before answering the next bit- I have a time zone advantage here... have left the [pi] [sqrt] simple- they hide better[/e]
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| « Last Edit: Apr 14th, 2004, 4:53pm by Noke Lieu » |
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Sir Col
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Re: Wrapped Circles
« Reply #2 on: Apr 14th, 2004, 5:41pm » |
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That's true about hiding [sqrt] and [pi]; when you highlight the region to read the hidden text they kind of invert making them difficult to read. I tend to write sqrt() in hidden text.
Very nice, Noke Lieu!
You could write 2sqrt(12)+14pi/3=4sqrt(3)+14pi/3. I suppose we could say that the first part is all wrapped up! 
Actually, as an extension to the first part...
Given that R=3r, find the perimeter, P, in terms of r.
Can you find any other relationships between r and R, such that P can be given in terms of [pi]?
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| « Last Edit: Apr 15th, 2004, 2:31am by Sir Col » |
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Barukh
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For those who prefer vizualization - the drawing for the second question is attached...
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Sir Col
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Re: Wrapped Circles
« Reply #4 on: Apr 15th, 2004, 4:28am » |
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Thanks for that, Barukh; I'm sure it will be helpful.
Out of interest, what did you use to create your drawing?
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Barukh
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Re: Wrapped Circles
« Reply #5 on: Apr 15th, 2004, 4:47am » |
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on Apr 15th, 2004, 4:28am, Sir Col wrote:| Out of interest, what did you use to create your drawing? |
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I use an old demo version of Geometry Sketchpad - a dynamic geometry tool. You may want to visit the following page to learn more about it.
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Barukh
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Re: Wrapped Circles
« Reply #6 on: Apr 15th, 2004, 9:01am » |
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on Apr 14th, 2004, 5:41pm, Sir Col wrote:Given that R=3r, find the perimeter, P, in terms of r.
Can you find any other relationships between r and R, such that P can be given in terms of [pi]? |
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As stated, the question is too general. Probably, you had in mind to restrict the relations somehow (e.g. R/r is rational, and/or [pi] has a rational coefficient in P)?
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Sameer
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Re: Wrapped Circles
« Reply #7 on: Apr 20th, 2004, 7:23am » |
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Using basic geometry for two touching circles, I came to this formula
For derivation r != R, but it also holds for r=R, we have
P=4*sqrt(r*R) + 2*pi*R - 2*(R-r)*tan-1(2*sqrt(r*R)/(R-r))
Using original problem of r=1,R=3, we get the same answer as Noke had.
Note: I think the term inside tan-1 is of the form 2xy/(x2-y2) which can be converted to tangent formulas and get rid of tan-1. Maybe I will leave it to someone else.
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| « Last Edit: Apr 20th, 2004, 7:24am by Sameer » |
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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