such that:
with all 
is real number
be the set of integers. Determine all functions 
such that, for all integers 
and 
,
Proposed by Liam Baker, South Africa
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
Y by Adventure10
Let 
be a diameter of a circle 
and let 
be any point on the tangent drawn at 
to 
Define 
and let 
be the point diametrically opposite to 
If 
meets second time in 
then,
(i) Prove that
concur at 
(ii) If
is the radius of 
find such that the area of 
is maximum, and calculate the area in terms of 
be a diameter of a circle 
and let 
be any point on the tangent drawn at 
to 
Define 
and let 
be the point diametrically opposite to 
If 
meets second time in 
then,(i) Prove that
concur at 
(ii) If
is the radius of 
find such that the area of 
is maximum, and calculate the area in terms of 
to meet 
at 
.
and 
meet at 
.
becomes a median of 
(
becomes parallel to 
.
.
.
is a diameter
.
.
,
.
.
is cyclic.So 
and 
and the points 
we get ![\[\frac{PA}{CA}.\frac{CM}{MB}.\frac{BE'}{E'P}=1\]](http://latex.artofproblemsolving.com/a/e/2/ae22903ff31d1718c0d5925167d3a45528461d1a.png)
![\[\Rightarrow \left(\frac{PE'}{E'B}+1\right).\frac{BE'}{E'P}=\frac{MB}{CM}\]](http://latex.artofproblemsolving.com/5/a/8/5a82e796a3c5150ba2476226834e88617a76dc4b.png)
![\[\Rightarrow \frac{MB}{CM}=\frac{BP}{E'P}\]](http://latex.artofproblemsolving.com/4/f/1/4f1401cc7906cb7cf617727b27fd3809c40f6821.png)
![\[\Rightarrow \frac{MC}{BM}+1=\frac{BP+E'P}{BP}\]](http://latex.artofproblemsolving.com/d/2/0/d20bb262248010bf7f17b902b6e3ef4e4c658d38.png)
![\[\frac{\bigtriangleup AMC}{\bigtriangleup AMB}=\frac{MC}{BM}\]](http://latex.artofproblemsolving.com/0/9/b/09be557ce2060470862c3ff37b02a13cd4d59f79.png)
![\[{\Rightarrow \frac{\bigtriangleup ABC}{\bigtriangleup AMB}=\frac{MC}{BM}+1}\]](http://latex.artofproblemsolving.com/a/6/8/a6877bf0fa4cfe9025c903a633ac2ba89c112d63.png)
![\[\Rightarrow \bigtriangleup AMB=\frac{\bigtriangleup ABC.BP}{BE'+2E'P}\]](http://latex.artofproblemsolving.com/7/f/d/7fdaf1f7b210326b46618a340ee54488045a2d76.png)
![\[\bigtriangleup AMB=\frac{R.BE'.BP}{BE'+2E'P}\le \frac{R.BE'.BP}{2\sqrt{2BE'.E'P}}=\frac{R.BE'.BP}{2\sqrt{2}CE'}\]](http://latex.artofproblemsolving.com/d/0/b/d0b291ca3ef91004fdd09a276fdf536cf7db21cf.png)
and 
.![\[\bigtriangleup AMB\le \frac{R}{2\sqrt{2}}.\frac{BP}{\tan\angle CAB}=\frac{R}{2\sqrt{2}}.2R=\frac{R^2}{\sqrt{2}}\]](http://latex.artofproblemsolving.com/6/c/c/6ccd1b6d93a8e17bb98a1f3a8f54b41e0677cf3e.png)
.
implying that 
.
![\[\Rightarrow E'P^2=\frac{2R^2}{9}\]](http://latex.artofproblemsolving.com/9/d/6/9d6e3fa2c27ebe88477e0258fd9b9a7ac656bd3d.png)
![\[\Rightarrow BE'+E'P=BP=\frac{2\sqrt{2}R}{3}+\frac{\sqrt{2}R}{3}=\sqrt{2}R\]](http://latex.artofproblemsolving.com/f/e/5/fe59d2cb12f7f024b7fd8d725d7ee807ffd3fbca.png)
.So our required 
units away from 
![$[AMB]=\dfrac{1}{2}\cdot MB\cdot AB\cdot \sin \angle CBA.$](http://latex.artofproblemsolving.com/c/8/1/c81555dfcf4f7a8d664b2f8b94cc977ea3606631.png)
Let 
and 
we have 
Then 
Note that 
where 
![\begin{align*}[AMB]&=\dfrac{1}{2}\cdot 2R\cdot \left(\dfrac{BC\cdot PA}{PC+PA}\right)\sin \angle B\\&=2R^2\sin \angle A\cdot \cos \angle A\cdot \left(\dfrac{PA}{PA+PC}\right)\\&=2R^2\cdot\dfrac{k}{k^2+1}\cdot\dfrac{k^2+1}{k^2+2}\\&=R^2\left(\dfrac{2k}{k^2+2}\right).\end{align*}](http://latex.artofproblemsolving.com/3/3/3/333551162d36574ec99a77eceb05a860f53ec48a.png)
we get 
so 
at which point we have ![$[AMB]=\dfrac{R^2\sqrt{2}}{2}.$](http://latex.artofproblemsolving.com/4/b/d/4bdb1f9355093b143f651066fbdb676da7089476.png)
= 
. 
= 
. 
,
,
= 
= 
= 
= 
is midpoint of 
,
is area of 
= 
= 
= 
= 
= 
