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.![$f \colon [0,1] \to \mathbb{R} $](http://latex.artofproblemsolving.com/2/1/8/21859d37ed035abd07d0be112f49d52598fa477e.png)
be a differentiable function such that its derivative is an integrable function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
, and 
. Prove that ![\[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]](http://latex.artofproblemsolving.com/2/3/e/23e33eed39fc1e20486bbb86261d3566e8fd2017.png)
be a positive integer. Prove that there exists a positive integer 
such that exactly 
of the integers 
are squarefree.
which verify relation![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
for which there is a positive real number such that![\[ f'\left(\frac ax\right)=\frac x{f(x)} \]](http://latex.artofproblemsolving.com/b/1/4/b1428d1e639f7c515404d6f9e37e027a6ec82abe.png)
cells of an 
square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell 
, one can reach any other black cell 
through a sequence 
of black cells where each consecutive pair 
are adjacent (sharing a common side) for every 
. Let 
denote the maximum possible cost over all initial colorings with exactly black cells. Determine a constant 
such that![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](http://latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
+1 w
be the orthocenter of triangle 
. Let 
be the circumradius of . Prove that triangle is equilateral iff
be a circle, 
be a diameter of the circle and for each point 
let 
be the foot of the perpendicular from on 
and 
the base of the perpendicular from 
on 
. Let 
be the geometric locus of all points if passes through all points on the circle . Prove that actually has no axes of symmetry other than , as shown in the picture.
and be positive integers. In an 
grid, two cells are considered neighboring if they share a common edge. Kritesh performs the following actions:
in any cell of the grid.
. and .
, s.t :
of real numbers such that![\[ a_{m^2 + m + n} = a_{m}^2 + a_m + a_n\]](http://latex.artofproblemsolving.com/b/b/9/bb97053e86e7733fe5c26f7859c87566eb444b52.png)
for all positive integers and .
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
, and let 
for 
. Determine whether![\[ \sum_{n=1}^{\infty}a_n^2 \]](http://latex.artofproblemsolving.com/5/4/0/540a4f3740d9b561c40dcc36ea9350d94421dd86.png)
converges.
be a function defined on 
with values in such that 
. Then, from the hypothesis, we have 
for every real 
. Since 
and 
are continuous, then 
is also continuous. is not constant. Then, since is continuous, there exists an interval 
such that either
for every 
. We will consider the first case (the other being similar). Obviously, if 
for everythen
for every .
for every , then 
is strictly increasing and injective and 
is strictly convex, so the equationhas at most 3 roots in , which is a contradiction. Therefore, must be constant and equal to zero for every real . This yields
Then, 
and 
, where and 
are real constants.
which implies
and consider another function, 
, defined by
and 
, where p and q are polynomials of degree at most 2 and equal leading coefficient from ![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
for every , which means that 
for every (using the same arguments we used above for h(x)). Then, 
for every real and 
, so
This means that
and we are done.
,![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
