Three circles are drawn inside another circle such that each of the four circles is tangent to the other three circles. Two circles with radii a and b, where a>b, have their centers on the diameter of the largest circle. (a) What is the radius of the largest circle in terms of a and b? (b) For what natural numbers a and b will the radius of the smallest circle be closest to one? I'm trying to use Descartes' Theorem to solve it. http://en.wikipedia.org/wiki/Descartes'_theorem
(a) is a+b
do u have a diagram?
No, this one does not have a diagram, but they are referred to as "kissing circles"
well twice the radius of a circle is it's diameter and since both circles touch the bigger one and their centers lie on the diameter of the bigger one. 2a +2b = diameter of larger circle, divided by 2 will give the radius... which will end up with a+b so you're right on the first one... i'm not taking a look at the second part of the question
that looks complex for my level, I'll see if someone else can help u on that one
Yeah, I figured that (a) is a+b. The second part is utterly confusion, but thank you very much! You have been extremely helpful.
ok, just brainstorming here. but would this be a visual of what they're trying to say...?
|dw:1330496529420:dw|
yeah something like that :)
I'm assuming that's what they're talking about.
according to the question, what natural numbers will make c close to 1. If a increases, b reduces and yet still c will reduce. but there should be a relationship between a b and c as well as the diameter of the largest circle such that we can tell the size of the circles relative to their radius and hence find c or b or a. this is what I'm thinking, i have no knowledge of this though..... educated guess. |dw:1330496767233:dw|
ummm i meant what natural numbers of a and b ***
The Descartes Circle Theorem provides a relationship between all four radii. In these formulas (on the Wikipedia page), k=1/radius. http://en.wikipedia.org/wiki/Descartes'_theorem
Any progress? What are your thoughts. This problem has me completely stumped.
Good luck with that.
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