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.![$P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/0/5/a05997bd7ba17001c62bc17ae35c76dd78e24789.png)
be a nonconstant polynomial such that 
and 
are both odd numbers. Prove that 
has no factor of the form 
where 
is an arbitrary rational number.
+1 w
, define 
Assume that, when fully expanded, we have 
Prove that 
.
be the circumcircle and 
the incircle of an acute scalene triangle 
with incenter 
. Let the reflection of 
over hit 
at 
respectively. Let 
touch at 
respectively and let 
hit again at 
. Suppose that 
hit again at 
respectively and 
are points on 
with 
and 
.
contains the intersection of lines 
and 
.
be non-negative real variables with 
Prove that
Equality holds iff 
or 
where 
so that, for all 
,![\[f(x+y, z+t)\leq f(z,x)+f(t,y)\]](http://latex.artofproblemsolving.com/8/7/1/87131a7d558961fa5da5369fe3a8177ef3ed3f72.png)
and![\[f\left(\frac{x}{y}, \frac{y}{x}\right)(x+y-f(y,x))=f(x,y)\]](http://latex.artofproblemsolving.com/9/2/3/92340f0294313cda0c52ea0e84e364b97d98bc3a.png)
(
),and such 
. Show that:there exist positive constant 
, such for any positive integer 
have
.
are non-negative real numbers. Prove that![\[\frac{a(b+9c)}{(b+c)^2}+\frac{b(c+9a)}{(c+a)^2}+\frac{c(a+9b)}{(a+b)^2} \geqslant 6.\]](http://latex.artofproblemsolving.com/0/6/8/0683bd20c51ff0ababe1bc11c9961ad72fa6fca3.png)
The equality holds for 
(or cyclic permutation). be a circle and a point outside it; distinct points 
and 
are selected on so that 
and 
are tangent to . Let 
be the reflection of across the point , and denote by 
the circumcircle of triangle 
. Suppose and meet at 
and line 
intersects at 
. Prove that line 
is tangent to the circle .
are non-negative real numbers such that 
and 
. Find the maximum value of 
and determine all equality cases.
. I note the middlepoint 
and the point 
which belongs to the crease. Thus,
, i.e. 
in the point 
, then 
. If I note the intersections 
, 
, then 
, 
) a.s.o.
is so folded that falls on the mid point of of . Prove that the crease will divide in the ration 
.
, B
, C
, D
. Then, CD's middle point is M
,
. BM's middle point is N
.
.
, which divides BC in the ratio 
.
is isosceles 
Therefore, 
, where 
lies on ,
to be 
~ 
~ 
. I note the middlepoint and the point which belongs to the crease. Thus,, i.e. in the point , then . If I note the intersections , , then , ) a.s.o.
but it must be equal to 
also M is the reflection of B in line NP hence NP is the perpendicular bisector of 
and 
in which C is on the origin A is at(0,a) ...
other than 
such that 
and 
are both perfect squares.
such that for all 
:
we have![\[\frac{a(b+k^2c)}{(b+c)^2}+\frac{b(c+k^2a)}{(c+a)^2}+\frac{c(a+k^2b)}{(a+b)^2} \geqslant 2k.\]](http://latex.artofproblemsolving.com/f/1/c/f1c7edec977b1eaae03a4e2153f05f23f8aa3c02.png)
The equality holds for 
are real numbers. Prove that![\[4(2+xyz-x-y-z)+(x^2+y^2+z^2)(x^2+y^2+z^2-2) \geqslant 0.\]](http://latex.artofproblemsolving.com/0/8/8/088d3bc97f61f42fcc08bfe4c9dafc41de7121c8.png)
Prove that
be the crease, where 
is the point on 
be the midpoint of 
and 
.
.
,
be the square. Let its sides be = 
units.
is the midpoint of 
.
is the crease formed.
and 
.
.
.
.
,
.
PMC
from (i)
(ii)
HENCE PROVED