AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a positive integer greater than 
. Find all functions 
satisfying:
for all integers 
has incentre 
and incircle touching 
at 
. Let 
be the antipode of 
in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
. Suppose that the sequence 
satisfies 
and 
. The sequence 
is the nondecreasing permutation of . Prove that![\[\sum_{i=1}^n \sum_{j=1}^n (j-i)^2a_ia_j \le \sum_{i=1}^n \sum_{j=1}^n (j-i)^2b_ib_j \]](http://latex.artofproblemsolving.com/e/5/9/e598b926b793c06cb1cc5529773e5c278279e59b.png)
be two integers such that their gcd has at least two prime factors. Let 
and call 
irreducible if it cannot be expressed as product of two or more elements of 
(not necessarily distinct). Show there exists 
such that any element of can be expressed as product of at most irreducible elements.
such that 
holds for all positive rational numbers 
.
, with help of 
and Moivre's formula.
![\[
\text{Find: }
10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z},
\]](http://latex.artofproblemsolving.com/f/3/3/f3360b85ce092804a9de263cac49aed025ea8cee.png)
![\[
11. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{(-1)^{n+m}}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/9/4/5/945aaa2340a4fb314e5eb218ba5286eb054c1b6f.png)
![\[12.
\int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos(2x) + 2} \tan^{-1}\left(\frac{\cos x \cot(2x)}{\sqrt{2}}\right) dx
= \frac{5\pi^2}{48\sqrt{2}} - \frac{\pi}{4\sqrt{2}} \cos^{-1}\left(\frac{1}{3}\right).
\]](http://latex.artofproblemsolving.com/b/1/3/b13a5b1c537926c8323170d3d533b0bc71bb06a1.png)
on the circle 
and write 
. Then 
. Also 
hence 
. It follows that we want 
for all 
hence 
.
we have 
and with 
we find 
. Hence all pairs with 
.
.
safisty: 
and 
.
![\[
I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du
\]](http://latex.artofproblemsolving.com/b/c/5/bc5d60ff6679870b47776c401658e1ba62af75aa.png)
defined on 
by![\[f(x, y) = x^4 + y^4 - 2(x - y)^2.\]](http://latex.artofproblemsolving.com/1/d/9/1d9e85722971f67e3ba2dee0e12d119ceabc671a.png)
(and determine them) such that![\[\forall (x, y) \in \mathbb{R}^2, f(x, y) \geq \alpha \| (x, y) \|^2 + \beta,\]](http://latex.artofproblemsolving.com/e/b/3/eb384916db4e019c7ba11f52fadf973f515891e2.png)
where 
denotes the Euclidean norm.
