Concentric Circles Equally Split This Chord

We can use Stewart´s Theorem to find ${ r }_{ 9 }, { r }_{ 8 }, ...., { r }_{ 1 }$ in terms of ${ r }_{ 10 }$ . Then:

${ { r }_{ 9 } }^{ 2 }={ { r }_{ 10 } }^{ 2 }+2\cdot { 106 }^{ 2 }$

${ { r }_{ 8 } }^{ 2 }={ { r }_{ 10 } }^{ 2 }+6\cdot { 106 }^{ 2 }$

...

${ { r }_{ 1 } }^{ 2 }={ { r }_{ 10 } }^{ 2 }+90\cdot { 106 }^{ 2 }$

Then

$[{ O }_{ 1 }]+...+[{ O }_{ 10 }]=\pi \cdot \left( { { r }_{ 1 } }^{ 2 }+...+{ { r }_{ 10 } }^{ 2 } \right) =\pi \cdot \left( { { 10\cdot r }_{ 10 } }^{ 2 }+330\cdot { 106 }^{ 2 } \right)$

and solving

$\pi \cdot \left( { { 10\cdot r }_{ 10 } }^{ 2 }+330\cdot { 106 }^{ 2 } \right) <20140000$

for ${r}_{10}$ , it gives that ${ r }_{ 10 }<519,89$ . So, the largest INTEGER possible for ${r}_{10}$ is $\boxed{519}$ .

For this problem, knowledge of Holditch's Theorem helps.

The length of each equally divided segment is $\dfrac{2014}{19}=106$ .

Let the radius of $O_{10}$ be $r$ .

We have that $[O_9]=[O_{10}]+\text{area of annulus between the two circles}$ .

By Holditch's Theorem, the area of the annulus is $(106\cdot 1)\cdot (106\cdot 2)\cdot \pi$ .

Similarly, $[O_8]=[O_{10}]+\text{area of annulus between the two circles}$ .

By Holditch's, the area of this annulus is $(106\cdot 2)\cdot (106\cdot 3)\cdot \pi$ .

We continue on in this manner, until we find that $[O_1]+[O_2]+\cdots +[O_{10}]=10[O_{10}]+106^2\pi(1\cdot 2+2\cdot 3+\cdots +9\cdot 10)$ .

Since $[O_{10}]=\pi r^2$ , we have that $10\pi r^2+3707880\pi < 20140000$ .

Solving for $r$ gives that $r<519.892$ , so our answer is $\boxed{519}$ .

let, $r_{10}=r$ and here, chord length of the smallest circle is $x=\frac{2014}{19}=106$ now, $r_{9}=\sqrt{r^{2}+2x^{2}}$ $r_{8}=\sqrt{r^{2}+6x^{2}}$ $r_{7}=\sqrt{r^{2}+12x^{2}}$ $r_{6}=\sqrt{r^{2}+20x^{2}}$ ................... $$ According to the question, $[O_{10}]+[O_{9}]+[O_{8}]+[O_{7}]+......+[O_{1}]<20140000$ $=> \pi r^{2}+\pi \left ( r^{2}+2x^{2} \right )+\pi \left ( r^{2}+6x^{2} \right )+... <20140000$ $=> r^{2}+\left ( r^{2}+2x^{2} \right )+\left ( r^{2}+6x^{2} \right )+... <\frac{20140000}{\pi }$ $=> 10r^{2}+x^{2}\left \{ 2+6+12+20+...+n(n+1) \right \}<6410761.1077$ $=> 10r^{2}+x^{2} \sum_{n=1}^{9}n(n+1)<6410761.1077$ $=> 10r^{2}+x^{2} \left (\frac{n^{3}}{3}+n^{2}+\frac{2n}{3}\right )<6410761.1077 \; [n=9]$ $=> 10r^{2}+106^{2}\times 330<6410761.1077$ now, solving and considering integer value we get 519

Length of secant segment inside Circle i,

$d(i) = d(1 - (2( i-1) / 19 )) = 2014 ( 1 - 2 (i -1)/19 ) = 106 (21 - 2 i )$

Since the normal to the secant from the center is equal for all circles,

$r(i)^2 - (d(i)/2)^2 = constant, for i = 1, 2, ..., 10$

Therefore

$r(i)^2 = constant + (d(i)/2)^2, for i = 1, 2, ...., 10$

$\pi \sum_{i=1}^{10} ( r(i)^2 ) = 10 \pi (constant) + \pi sum_{i=1}^{10} ( (d(i)/2)^2 )$

$\pi \sum_{i=1}^{10} ( r(i)^2 ) = 10 \pi (constant) + \pi (53)^2 ( 441 (10) - 84 (10)(11)/2 + 4/6 (10)(11)(21) ) < 20140000$

$constant < 267479.11077415441$

$r(10) = \sqrt { constant + (d(10)/2)^2 } < \sqrt { 267479.11077415441 + (53)^2 } = 519.89$