Infinite Number of Circles

Let $O_n$ be the centre of $\gamma_n$ and $P_n$ the point of contact of $\gamma_n$ with $l$ . Let $S$ be the centre of $C_n$ and $Q$ the point of contact of $C_n$ with $l$ .

Let $T$ be the point on $O_nP_n$ such that $TP_nP_{n+1}O_{n+1}$ is a rectangle. Applying Pythagoras’s theorem to triangle $O_nTO_{n+1}$ gives $TO_{n+1}$ =1/2^(0.5(2n-1)) $=P_nP_{n+1}$

Let $U$ be the point on $O_nP_n$ such that $UP_nQS$ is a rectangle. Applying Pythagoras’s theorem to triangle $O_nQS$ gives $US=\sqrt r_n/2^(0.5(n-2))=P_nQ$

Then $P_nQ+QP_{n+1}=P_nP_{n+1}$ . Solving , we get $\sqrt r_n$ $(3+\sqrt 8)=1/2^n$ . \displaystyle \sum {n=1}^{\infty} r n\(=1/(3+\sqrt 8\=3-\sqrt 8. Therefore we have got a as 3 and b as 8. Hence our answer is 11.

In Descartes theorem

Setting one of the terms to 0.as the curvature of a circle is the reciprocal of its radius and as the radius of a straight line can be thought of to be infinity.

Thus the formula reduces to

a=b+c +/-(plus or minus) $\sqrt{b}{c}$ where a,b.c stand for the curvatures of the 3 circles

Which is just the reciprocals of their radii

Thus the radius of the circle that is tangent to the 2 circles and a line is given by

R= \frac{uv}{u+v +/- \(\sqrt{b}{c} }).

Where R is the radius of the circle which is tangent to the 2 circles and the line.u is the radius of one of the circles and v is the radius of the other

It can be shown that this is in a geometric progression. By plugging isn't he radii of the nth n+1 th and n-1 th circles taking ratios and showing that the ratio is a constant(after some simplification)

Then apply the infinite sum of a gp formula and the answer turns out to be 3-(\sqrt{8})

Hence 3+ 8= 11

P.s sorry for the lack of simplification and latex edits.have an exam tomorrow and as said in my email to u.

I couldn't scan my copy as there was some problem with Please do forgive me sir.

To make this problem easier to see. Let your first big circle be r=2. Let your second big circle be r=1. Draw the inbetween circle. If you draw the radii to make a trapezoid. It will have two sides of length 1 and 2, as well as a diagonal segment of length 3 (the two radii added together). The bottom side by pythagorean theorem must be 2sqrt(2).

If you draw the radii between the large and inbetween circles, You can make two new right triangles. The first has sides (2+r),(2-r), (a) The second has sides (1+r),(1-r),(b) (a) = 2sqrt(r) (b) = 2 sqrt2 * sqrt(r) (a) + (b) = 2 sqrt2

if you simplify from there, r = 6 - 4*sqrt(2)

Because our situation is just the original scaled up by a factor of 4... r1 = (6-4 sqrt2) / 4 With the infinite series, each r will be 1/2 of the previous... rtotal = r1 + (1/2) r1 + (1/4) r1 + ... = 2 r1 rtotal = (6-4 sqrt2) / 2 = 3 - 2 sqrt2 = 3 - sqrt(8) Answer = 3+8 = 11

In the infinite sequence of circles $\Gamma_n$ . Just consider any two circles $\Gamma_n$ & $\Gamma_{n+1}$ . we see that the circle $C_n$ is tangent to the two circles and a line $\ell$ . this can be solved by a special case of descartes theorem when one of the four kissing or mutually tangent circles is a line .

The curvatures of the line , $\Gamma_n$ & $\Gamma_{n+1}$ are 0 , $2^n$ , $2^{n+1}$ . By putting these values in the equation of the theorem gives that the curvature of $C_n$ $(k_1)$ is $k_1=2^n+2^{n+1} \pm2\sqrt{2^n\ast2^{n+1}}$ $\Rightarrow k_1=3(2^n)\pm2\sqrt{2^{2n}\ast2}$ $\Rightarrow k_1=(2^n)(3\pm2\sqrt{2})$ ignore the negative sign as it creates a solution of circle which isn't contained in the region between two circles . hence we get the radius of the circles of sequence $C_n$ as $r_i=\frac{1}{k_1}=\frac{1}{2^n(3+2\sqrt{2})}=\left(\frac{1}{2^n}\right)(3-\sqrt{8}$

Hence , $\sum_{i=1}^{\infty}r_i=\sum_{i=1}^{\infty}\frac{(3-\sqrt{8}}{2^i}$ $\sum_{i=1}^{\infty}r_i=(3-\sqrt{8})\sum_{i=1}^{\infty}\frac{1}{2^n}=3-\sqrt{8}$

we get our answer as $a+b = 3+8 = \boxed{11}$ .

Lemma: The radius of any smaller circle that is like in the configuration of the problem is