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        riddles >> medium >> Wrapped Circles
        
(Message started by: Sir Col on Apr 14^th, 2004, 10:29am)

Title: Wrapped Circles
Post by Sir Col on Apr 14^th, 2004, 10:29am

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?

Title: Re: Wrapped Circles
Post by Noke Lieu on Apr 14^th, 2004, 4:47pm

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.[hide]
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.
[/hide]
so the band's points of contact/leaving of the circles are [hide] 120^o apart relative to their centres.[/hide]

This means that the band has [hide] 2x12^.5+(2pi/3)+(12pi/3)[/hide] (that is there are two lengths from the gap between the circles, [hide] 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
[/hide]
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]

Title: Re: Wrapped Circles
Post by Sir Col on Apr 14^th, 2004, 5:41pm

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 [hide]2sqrt(12)+14pi/3=4sqrt(3)+14pi/3[/hide]. I suppose we could say that the first part is all wrapped up! ;D

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]?

Title: Re: Wrapped Circles
Post by Barukh on Apr 15^th, 2004, 3:28am

For those who prefer vizualization - the drawing for the second question is attached...

Title: Re: Wrapped Circles
Post by Sir Col on Apr 15^th, 2004, 4:28am

Thanks for that, Barukh; I'm sure it will be helpful.

Out of interest, what did you use to create your drawing?

Title: Re: Wrapped Circles
Post by Barukh on Apr 15^th, 2004, 4:47am

on 04/15/04 at 04:28:04, Sir Col wrote:

Out of interest, what did you use to create your drawing?

I use an old demo version of Geometry Sketchpad - a dynamic geometry tool. You may want to visit the following page (http://www.keypress.com/sketchpad/) to learn more about it.

Title: Re: Wrapped Circles
Post by Barukh on Apr 15^th, 2004, 9:01am

on 04/14/04 at 17:41:04, 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]?

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)?

Title: Re: Wrapped Circles
Post by Sameer on Apr 20^th, 2004, 7:23am

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.