perimeter

vipinz · #1 $Old$ November 1st, 2009, 06:57 AM

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?

Prove It · #2 $Old$ November 1st, 2009, 06:58 AM

Quote:

Originally Posted by vipinz $View Post$

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...

ukorov · #3 $Old$ November 1st, 2009, 07:12 AM

.....like this?

vipinz · #4 $Old$ November 1st, 2009, 07:16 AM

Very good diagram. But what is the answer.

ukorov · #5 $Old$ November 1st, 2009, 08:54 AM

Quote:

Originally Posted by vipinz $View Post$

Very good diagram. But what is the answer.

any other info on relationship between the two radii...?

vipinz · #6 $Old$ November 1st, 2009, 09:00 AM

There is no other information. The only information available is that the two radii are not equal.

nikhil · #7 $Old$ November 1st, 2009, 09:25 AM

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

ukorov · #8 $Old$ November 1st, 2009, 10:10 AM

Quote:

Originally Posted by nikhil $View Post$

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

what do you do with those angles at the centres?

nikhil · #9 $Old$ November 2nd, 2009, 06:25 AM

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.

aidan · #10 $Old$ November 3rd, 2009, 09:44 AM

Quote:

Originally Posted by vipinz $View Post$

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

Theta in radians.

vipinz · #11 $Old$ November 3rd, 2009, 10:43 AM

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

aidan · #12 $Old$ November 3rd, 2009, 04:16 PM

Quote:

Originally Posted by vipinz $View Post$

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?