Two unequal circles (radius x and y, x > y) are touching each other. A rubber band is passed around both of them. What would be the length of the rubber band?

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- Nov 1st 2009, 04:57 AMvipinzperimeter
Two unequal circles (radius x and y, x > y) are touching each other. A rubber band is passed around both of them. What would be the length of the rubber band?

- Nov 1st 2009, 04:58 AMProve It
- Nov 1st 2009, 05:12 AMukorov
.....like this?

- Nov 1st 2009, 05:16 AMvipinzAnswer
Very good diagram. But what is the answer.

- Nov 1st 2009, 06:54 AMukorov
- Nov 1st 2009, 07:00 AMvipinz
There is no other information. The only information available is that the two radii are not equal.

- Nov 1st 2009, 07:25 AMnikhilno other information is needed
try pythagoras theorem

use co-axial circles (it will only help in easy visualization)

answer is

pi(x+y)+4[xy]^(1/2) - Nov 1st 2009, 08:10 AMukorov
- Nov 2nd 2009, 04:25 AMnikhil
in the give figure move the line that joins the centres of the circles downwards(or upwards).

now you will get a right angle triangle whose one side is x+y, other side is x-y and the third side can be found (pythagoras theorem)which is the length of segment of rubber band.by symmetry other other segment of band will also have same length. - Nov 3rd 2009, 07:44 AMaidan
nikhil's formula will work in rough cut situations.

However, if you need an exact fit, try this:

Let R be the radius of the larger circle.

Let r be the radius of the smaller circle.

Extend the two tangent lines until they intersect.

Let theta be the angle at the intersection.

$\displaystyle \theta = 2 sin^{-1}\left( \dfrac{R-r}{R+r}\right)$

LengthOfRubberBand= $\displaystyle R(\pi+\theta) + r(\pi -\theta) + 2\sqrt{4Rr}$

Theta in radians. - Nov 3rd 2009, 08:43 AMvipinz
Nikhil must explain how he got the sum of the lengths of the two arcs as

pi (x + y). The explanation about the part of the rubber band between the two circles is OK.

Aidan, I think you are introducing a new variable "theta". Your answer must be only in terms of x and y.

Vipin - Nov 3rd 2009, 02:16 PMaidan
Sorry, I never saw that condition until now.

However, it is easily solved:

Let x be the radius of the larger circle.

Let y be the radius of the smaller circle.

Thus:

LengthOfRubberBand = x[pi + (1/1)(x^1/1) + (1/2)(x^3/3) + (1/2)(3/4) x^5/5) + (1/2)(3/4)(5/6)(x^7/7) + (1/2)(3/4)(5/6)(7/8)(x^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(x^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(x^13/13)] + y(pi - [ (1/1)(((x-y)(x+y))^1/1) + (1/2)(((x-y)(x+y))^3/3) + (1/2)(3/4)(((x-y)(x+y))^5/5) + (1/2)(3/4)(5/6)(((x-y)(x+y))^7/7) + (1/2)(3/4)(5/6)(7/8)(((x-y)(x+y))^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(((x-y)(x+y))^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(((x-y)(x+y))^13/13)] ) + 4 sqrt(xy)

There all in terms of x and y.

ok? - Nov 28th 2009, 01:35 AMnikhiloopssss
actually i did a mistake so let me give the correct solution

it will be

pi(x+y)+2arcsin[(x-y)/(x+y)](x-y)+4(xy)^(1/2)

you might have got the solution but you may ask me how i got it. its easy

hint: (fit the two circles in a cone then solve)