[/quote]
,
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
and write the numbers 
down on a board. A move consists in erasing any two numbers 
and 
,
and 
on the board, and then removing repetitions (e.g., if the board contained the numbers 
,
and 
,
)
with real coefficients of degree 
)![\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]](http://latex.artofproblemsolving.com/6/5/e/65e5e4eefcadd5d0019b4929b3472cdf203c13aa.png)
has no real roots.
and 
,
,
the pair 
if for any elements 
we have 
.
be the greatest possible number of elements of a set that escapes the pair 
.
with incircle 
,
be the incenter and 
be the point where 
.
be the point on 
with 
and 
be the orthocenter of 
,
on 
also satisfies 
.
,
,
satisfying 
the following inequality holds:
of the triangle 
.
the orthogonal projection of 
on 
.
and 
are such that 
is a paralellogram, find the locus of point 
when 
.
be the circle tangent to 
passing through 
.
be the circle tangent to 
be the circle tangent to 
passing through 
.
its circumcenter and 
its circumcircle. Let 
and 
be points on 
,
.
,
.
.
and the tangent to 
are concurrent and that 
.
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Hello guys,
I think we should start a factoring marathon. Post your solutions like this , and your problems like this . Please make your own problems, and I'll start off simple:
I think we should start a factoring marathon. Post your solutions like this , and your problems like this . Please make your own problems, and I'll start off simple:
This post has been edited 1 time. Last edited by pican, Nov 15, 2015, 12:40 AM
.
or 
.
.
.![$\dfrac{1}{2}\left[((a-b)(a+b))^2+((b-c)(b+c))^2+((c-a)(c+a))^2\right]$](http://latex.artofproblemsolving.com/1/3/4/13492404f04cf780c7485c2ccfd4ce209a93a14e.png)
not as 
)
.
and 
would be factored such that they are independent. Furthermore, the expression is symmetric, meaning fi you replaced all 
with 
,
.
,
,
because every coefficient is positive. Plugging in 
,
,
so 
We can then simplify these two quadratics into linears, getting 
.
to get 
.
,
. Then we combine like terms to get 
.
t
or smth
.
