# perimeter

• Nov 1st 2009, 04:57 AM
vipinz
perimeter
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 AM
Prove It
Quote:

Originally Posted by vipinz
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?

It would help you to draw a diagram of this situation first...
• Nov 1st 2009, 05:12 AM
ukorov
.....like this?
• Nov 1st 2009, 05:16 AM
vipinz
Very good diagram. But what is the answer.
• Nov 1st 2009, 06:54 AM
ukorov
Quote:

Originally Posted by vipinz
Very good diagram. But what is the answer.

any other info on relationship between the two radii...?
• Nov 1st 2009, 07:00 AM
vipinz
There is no other information. The only information available is that the two radii are not equal.
• Nov 1st 2009, 07:25 AM
nikhil
no other information is needed
try pythagoras theorem
use co-axial circles (it will only help in easy visualization)
pi(x+y)+4[xy]^(1/2)
• Nov 1st 2009, 08:10 AM
ukorov
Quote:

Originally Posted by nikhil
try pythagoras theorem
use co-axial circles (it will only help in easy visualization)
pi(x+y)+4[xy]^(1/2)

what do you do with those angles at the centres?
• Nov 2nd 2009, 04:25 AM
nikhil
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 AM
aidan
Quote:

Originally Posted by vipinz
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?

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.

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

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

• Nov 3rd 2009, 08:43 AM
vipinz
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 PM
aidan
Quote:

Originally Posted by vipinz
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

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 AM
nikhil
oopssss
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)