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.
interesting if 
is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
is irreducible for every positive integer n.
, 
and 
with real coefficients and leading coefficient 
, such that each of the polynomials![\[
P(Q(x)), \quad Q(R(x)), \quad R(P(x))
\]](http://latex.artofproblemsolving.com/5/5/4/55478b1313c73a45ed64be4e9f4075ac61475545.png)
has at least one real root, while each of the polynomials![\[
Q(P(x)), \quad R(Q(x)), \quad P(R(x))
\]](http://latex.artofproblemsolving.com/f/f/b/ffb016543e621fec0a5e8a668a9702318ec1f1d7.png)
has no real roots.
, with help of 
and Moivre's formula.
of the triangle 
consider a mobile point 
. Let 
the orthogonal projection of 
on 
. If the mobile points 
and 
are such that 
is a paralellogram, find the locus of point 
when goes through 
.
real numbers, not all zero, such that their sum is zero. Prove that one can label these numbers 
in such a manner that![\[a_1a_2 + a_2a_3 +\cdots + a_{n-1}a_n + a_na_1 < 0\]](http://latex.artofproblemsolving.com/6/9/4/6949824017301ebae714606e15eb1af825cbac33.png)
valid? I feel that this is valid for independent events 
and 
but may not be for dependent events. In this case would 
? If that is the case the problem becomes trivialized.
and 
. Prove that
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let and 
. Prove that
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
is a triangle with circumcenter 
and 
is a circle tangent to and internally to 
The tangent
to intersects again at 
If 
are the contact points of with 
respectively, prove that 
, 
and 
, a triangle can be formed. Can it happen that from segments of lengths
an obtuse triangle can be formed?
. Easy to guess that the answer is , but I haven't found the reason yet. :(
be a function that is at least twice differentiable on an open interval containing ![\( [0, 2\pi] \)](http://latex.artofproblemsolving.com/9/7/7/977adaa4f83dc8164975a3e78539f27cbb354adf.png)
. Given that![\[
f(0) = f(2\pi) = f'(0) = f'(2\pi) = 0
\]](http://latex.artofproblemsolving.com/d/9/9/d998bdd53ffec1fa4c88688d7402ad25efa31112.png)
and![\[
f(x) + f''(x) \geq 0, \quad \forall x \in [0,2\pi].
\]](http://latex.artofproblemsolving.com/a/1/c/a1c24260aa02cd989630295d9de8741af659e958.png)
Prove that 
for all ![\( x \in [0,2\pi] \)](http://latex.artofproblemsolving.com/d/c/5/dc59f105fb3b9f6be175d2c035ba2ead4b9d984a.png)
.
such that
for all 
