SAT1000 - P844

Let the position coordinates of $A, B, M, N$ be $(t^2,2t), (u^2,2u), (p^2,2p), (q^2,2q)$ respectively. Let the mid point of $\overline {AB}$ be $G(g,h)$ .

Then,

$\dfrac {2t}{t^2-1}=k$

$\implies t=\dfrac {1+\sqrt {k^2+1}}{k}$ ,

$u=\dfrac {1-\sqrt {k^2+1}}{k}$

$\implies g=\dfrac {k^2+2}{k^2}, h=\dfrac {2}{k}$

$\dfrac {2p-h}{p^2-g}=-\dfrac {1}{k}$

$\implies p=\dfrac {\sqrt {k^4+3k^2+2}-k^2}{k}$ ,

$q=-\dfrac {\sqrt {k^4+3k^2+2}+k^2}{k}$

So, $|\overline {MG}|=\dfrac {2(k^2+1)(\sqrt {k^2+2}-\sqrt {k^2+1})}{k}$

$|\overline {GN}|=\dfrac {2(k^2+1)(\sqrt {k^2+2}+\sqrt {k^2+1})}{k}$

$|\overline {AG}|=\dfrac {2(k^2+1)}{k^2}$

Since the points $A, M, B, N$ are concyclic, therefore

$|\overline {MG}|\times |\overline {GN}|=|\overline {AG}|^2$

$\implies \dfrac {4(k^2+1)^2}{k^4}=\dfrac {4(k^2+1)^2}{k^2}$

$\implies |k|=1$

Therefore the required answer is $\boxed {1000}$ .

Solution to this problem :

From the given value of $a_1$ and the recurrance relation, we get $a_{2n+1}=\dfrac {F_{n+1}}{F_{n+2}}$ ,

where $F_n$ is the $n'^{th}$ Fibonacci number.

From the relation $a_{2010}=a_{2012}$ we get $a_{2010}=\phi -1$ , where $\phi =\dfrac {\sqrt 5+1}{2}$ is the golden ratio .

Using this we see that $a_{2n}=\phi -1$ for all positive integers $n$ .

So, $a_{11}=\dfrac {F_6}{F_7}=\dfrac {8}{13}, a_{20}=\phi -1=\dfrac {\sqrt 5-1}{2}$ , and

$a_{20}+a_{11}\approx 1.2334186$

Hence the required answer is $\boxed {1233}$ .