Kaleidoscope Of Circles

calculate the first radius circle we can find looking to a sector and connect the center of big circle to the first circle and a perpendicular of the center of first circle to the the line to sector and find a right triangle with the following relation

given $s=\sin { \scriptstyle({\pi}/n) }$ , we have $s={ R }_{ 1 }/({ r-R }_{ 1 })$ of the right triangle. so we can isolate ${ R }_{ 1 }$ and find ${ R }_{ 1 }=r(s/1+s)$ To find the other radius, you can analyse the k+1 circle and k circle.

anagously to the first circle we can find the following identity:

definying: ${ S }_{ k }={ R }_{ 1 }+{ R }_{ 2 }+...+{ R }_{ k }$

${ R }_{ k }=(r-2{ S }_{ k-1 })s/(1+s)\Longrightarrow { S }_{ k-1 }=(r-{ R }_{ k }(1+s)/s)/2$

So

${ R }_{ k+1 }(1+s)/s=r-{ 2S }_{ k }=r-2{ R }_{ k }-2{ S }_{ k-1 }=$ $=r-2{ R }_{ k }-(r-{ R }_{ k }(1+s)/s)={ R }_{ k }(1-s)/s\Longrightarrow { R }_{ k+1 }(1+s)={ R }_{ k }(1-s)$ , and we clearly find a geometric progression with ratio (1-s)/(1+s) and first therm r1=r(s/s+1).

so for a sector we have

$Area\quad of\quad sector={ \pi }({ R }_{ 1 }^{ 2 }+...+{ R }_{ k }^{ 2 })$ , we have a infinite PG with ratio less than 1 so its convergent and we can use that:

$Area of sector={ \pi }.{ R }_{ 1 }^{ 2 }{ /[1-(1-s)/(1+s)] }={ \pi }.r^{ 2 }.(s/(1+s))^{ 2 }.[(s+1)^{ 2 }/4s]={ \pi }.r^{ 2 }.s/4$

calculating An now:

using the fundamental trigonometrica limit:

$s=\sin { \scriptstyle({\pi}/n) }$ , so $\lim _{ n\xrightarrow { } \infty }{ \sin { ({ \pi }/n) } } /{ (\pi }/n)=1= \lim _{ n\xrightarrow { } \infty }{ n.s/{ \pi } }$ ,

$\lim _{ n\xrightarrow { } \infty }{ { A }_{ n } } =\lim _{ n\xrightarrow { } \infty }{ \quad n.s.{ \pi }.r^{ 2 }/4 } =\lim _{ n\xrightarrow { } \infty }{ \quad (n.s/{ \pi }).{ { \pi } }^{ 2 }.{ r }^{ 2 } } /4=({ { \pi }r/2) }^{ 2 }$

So the first expression is:

${ \pi }{ r }^{ 2 }-\lim _{ n\xrightarrow { } \infty }{ { A }_{ n } } ={ \pi }{ r }^{ 2 }-{ ({ \pi }r/2) }^{ 2 }={ \pi }{ r }^{ 2 }(1-{ \pi }/4)$

when we put $r=10u$ , we have that the expression is equal to aproximately $Expression=67,419u^{ 2 }$

Same solution as @Lucas Maia . And an epic problem by the way.

The radius of the largest incircles are given by: $r_{1}=10\frac{sin\frac{\pi}{n}}{1+sin\frac{\pi}{n}}$

The smaller ones can then be calculated by: $r_{k}=\frac{sin\frac{\pi}{n}}{1+sin\frac{\pi}{n}}(10-2\sum_{i<k}r_{i})$

The recursion above was realized with a subroutine receiving parameters n (count of segments) and i (count of iterations) and returning the sum of the squares of the radii:

Function kreise(n,i);

f=sin(pi/n)/(1+sin(pi/n));

r=10*f;

a=n*r^2;

for j=1 to i do

r1=f (10-2 r);

a=a+n*r1^2;

r=r+r1;

end

return(a)