AoPS Academy
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
. Prove that
Equality holds when 
or 
and 
, Ivica and Marica play the following game: Given two piles of cookies, on each turn a player takes 
cookies from one of the piles, of which he eats 
and puts of is arbitrary in every move. Players take turns alternatively, with Ivica going
for which Marica has a winning strategy.
be a point inside triangle 
and let lines 
, 
and 
intersect sides 
, 
and 
in points 
, 
and 
, respectively. Prove that 
iff is incenter of 
![\[\frac{BX \cdot CX}{AX^2}=\frac{\sin \alpha_1 \sin \alpha_2}{\sin \beta\sin \gamma}\]](http://latex.artofproblemsolving.com/c/4/5/c454982d8ce3ebb18fcb324e435327d485cbef19.png)
where 
are the angles at 
of the line . But 
. So this gets maximal if 
i.e. is the angular bisector. any point not equal to the incenter is always less than the LHS for the incenter. So it suffices to check that the identity holds for 
. But here we can bash: Use that 
to find that 
. So we have![\[2LHS=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}-1\]](http://latex.artofproblemsolving.com/4/2/0/4209ba391b3e7810c7cb02ea18256b441a08e5a8.png)
On the other hand, we have 
and 
where 
is half of the perimeter. So![\[\frac{R}{r}=\frac{abcs}{4A^2}=\frac{abcs}{s(s-a)(s-b)(s-c)}=\frac{2abc}{(a+b-c)(a+c-b)(b+c-a)}\]](http://latex.artofproblemsolving.com/8/9/1/8918b7605cfb19da3b492a31b4384e09e7d9d9c6.png)
We are then left to check the identity![\[\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}+1=\frac{4abc}{(a+b-c)(a+c-b)(b+c-a)}\]](http://latex.artofproblemsolving.com/c/5/6/c56835ad1e645b33692b597d6ce9a574b1730b82.png)
which can be easily done by expanding.
