Finding radius of circleWhat is the formula for finding the radius of a circle?

[Saitama]

Homework Statement

(see attachment 1)

In this problem, we have a row of circles placed on a line. All points of tangency are distinct. The circle $C_n$ is uniquely determined.

Homework Equations

The Attempt at a Solution

Here's the sketch I drew for the problem.

Radius of $C_n$ = $r_n$
Radius of $\Gamma_n$ = $1/2^n$
Radius of $\Gamma_{n+1}$ = $1/2^{n+1}$.

Therefore AB, BC and AC can be easily calculated but how do I calculate $r_n$. I can't perform the summation until I don't find $r_n$.

Any help is appreciated. Thanks!

 

[Curious3141]

Homework Statement

(see attachment 1)

In this problem, we have a row of circles placed on a line. All points of tangency are distinct. The circle $C_n$ is uniquely determined.

Homework Equations

The Attempt at a Solution

Here's the sketch I drew for the problem.

Radius of $C_n$ = $r_n$
Radius of $\Gamma_n$ = $1/2^n$
Radius of $\Gamma_{n+1}$ = $1/2^{n+1}$.

Therefore AB, BC and AC can be easily calculated but how do I calculate $r_n$. I can't perform the summation until I don't find $r_n$.

Any help is appreciated. Thanks!

Try working in a coordinate system with the origin at the point of contact between the largest circle and the line.

Use Pythagoras theorem to solve for the x-coordinate of the center of the circle of intermediate size.

Now set up quadratic simult. equations using Pythagoras theorem to solve for the radius of the small circle. The other variable is the x-coordinate of the small circle, which you don't really need.

It should reduce to a nice expression. The rest becomes quite easy, two geometric sums, etc.

 

[Saitama]

Try working in a coordinate system with the origin at the point of contact between the largest circle and the line.

Use Pythagoras theorem to solve for the x-coordinate of the center of the circle of intermediate size.

With centre B? How will I find the x-coordinate, I don't have the distance of B from origin. :(

 

[ehild]

You can find some right-angle triangles...

ehild

 

[Curious3141]

With centre B? How will I find the x-coordinate, I don't have the distance of B from origin. :(

What ehild said. Red triangle for the first part. Blue and green for the second part (simultaneous quadratics).

The first equation is easy peasy. The simult. quadratics are a bit icky. But you get a nice expression for $r_n$ in the end.

 

[Saitama]

You can find some right-angle triangles...

ehild

I marked the points. (see attachment)

AH=\frac{1}{2^n}-\frac{1}{2^{n+1}}
AB=\frac{1}{2^n}+\frac{1}{2^{n+1}}
BH^2=\left(\frac{1}{2^n}+\frac{1}{2^{n+1}}\right)^2-\left(\frac{1}{2^n}-\frac{1}{2^{n+1}}\right)^2
BH=\frac{\sqrt{2}}{2^n}

Calculating EG:
AE=\frac{1}{2^n}-r_n
AG=\frac{1}{2^n}+r_n
Hence
EG^2=\left(\frac{1}{2^n}+r_n\right)^2-\left(\frac{1}{2^n}-r_n\right)^2
EG=2\sqrt{\frac{r_n}{2^n}}

Calculating FG:
BF=\frac{1}{2^{n+1}}-r_n
BG=\frac{1}{2^{n+1}}+r_n
FG^2=\left(\frac{1}{2^{n+1}}+r_n \right)^2-\left(\frac{1}{2^{n+1}}-r_n \right)^2
FG=\sqrt{2r_n}{2^n}

Since, EG+FG=BH
\sqrt{\frac{2r_n}{2^n}}+2\sqrt{\frac{r_n}{2^n}}= \frac { \sqrt {2} }{ 2^n }
Solving this for $r_n$
r_n=\left(\frac{1}{\sqrt{2}}\right)^{2n}(\sqrt{2}-1)^2

Thanks a lot both of you. If I perform the summation with this expression. I get the right answer.

The answer I get is $3-\sqrt{8}$.