Odd Circles

1. Study the visual proof above, hence prove that three mutually tangent circle sharing a common tangent, their radii are related by $\frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} = \frac{1}{\sqrt{r_3}}$

2. By using the formula, proof that the radius of white circle is $\frac{1}{4}$ , the radii of gray circles are $1 , \frac{1}{3^2} , \frac{1}{5^2} , \frac{1}{7^2} , \frac{1}{9^2} , ...$ 2b. (Optional) Use mathematical induction to prove the radius of k-th gray circle is $\frac{1}{(2k-1)^2}$ .

3. By using the definition of Riemann zeta function, show that $1 + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \frac{1}{9^4} + ... = (1-\frac{1}{2^4})\zeta(4)$

4. Given $\zeta(4) = \frac{\pi^4}{90}$ , show that the total area of gray circles is $2 \cdot \pi \cdot \frac{15}{16} \cdot \zeta(4) = \frac{1}{48} \pi^5$

$\therefore A=1, \ B=48, \ C=5, \ 10000A+100B+C=14805$

We can use Descarte's Theorem to find the radii of four mutually tangent circles.

$k_4=k_1+k_2+k_3-2\sqrt{k_1k_2+k_2k_3+k_3k_1}$

where $k_1,k_2,k_3,k_4$ = $\frac{1}{r_1},\frac{1}{r_2},\frac{1}{r_3},\frac{1}{r_4}$

If $k_1, k_2,k_3=0,1,1$ , then $r_4=\frac{1}{4}$ , which is the radius of the white circle.

Then if $k_1,k_2,k_3=0,4,k_i$ , then $k_{i+1}=(2+\sqrt{k_i})^2$

This means that the radii of the gray circles are $1, \frac{1}{3^2}, \frac{1}{5^2}, \frac{1}{7^2}, ...$ , so that the total area is (with help from the hint)

$\displaystyle \sum _{ r=1 }^{ \infty }{ \frac { 2\pi }{ { \left( 2r-1 \right) }^{ 4 } } } =\dfrac { { \pi }^{ 5 } }{ 48 }$

so that the answer is $14805$

Note:
$\displaystyle \sum _{ r=1 }^{ \infty }{ \frac {1}{ { \left( 2r-1 \right) }^{ 4 } } } = (1-\frac{1}{2^4})\zeta(4)=\dfrac { { \pi }^{ 4 } }{ 96 }$