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 positive reals such that 
. Prove that the inequality![\[a \sqrt[3]{1+b-c} + b\sqrt[3]{1+c-a} + c\sqrt[3]{1+a-b} \leq 1\]](http://latex.artofproblemsolving.com/f/2/d/f2db8f577625a01bcfaceaaae3fd6651b370ae80.png)
. Prove that
Let 
. Prove that
Let 
. Prove that
Let 
. Prove that
![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
such that:
with 
is an 
th prime and 
is an arbitrary positive integer.
that verfy the following equation :
be a positive integer. We say that a positive integer 
is -good if 
is divisible by 
for all positive integers 
with 
. Suppose is a positive integer such that is -good, but 
is not -good. Prove that 
is prime.
let 
and 
be the circumcenter and the orthocenter. The point 
is the reflection of 
with respect to 
. Assume that is not on the same side of 
as . Points 
lie on 
respectively such that 
. Let 
be the intersection point of 
. Prove that 
a grid of 9*9, we choose a little square in of this grid three times, we can choose three times the same.
be the incenter of triangle 
. Let 
, 
, and 
be the orthocenters of triangles 
, 
, and 
, respectively. Prove that the lines through , , and , parallel to 
, 
, and 
, respectively, are concurrent.
of acute triangle is 
. The ray drawn from through the circumcenter intersects at 
. Let the midpoint of 
be 
. Define 
, 
, 
, 
, 
, 
similarly. Prove that 
, 
, 
are concurrent.
be a triangle, and let 
be on 
so that 
. Let 
intersect the circumcircle of again at 
. Let 
be the midpoint of 
and let the perpendicular bisector of 
intersect 
at 
. If is the circumcenter of 
, prove that 
.
be a triangle, and let be on so that . Let intersect the circumcircle of again at . Let be the midpoint of and let the perpendicular bisector of intersect at . If is the circumcenter of , prove that .
and 
is perpendicular to 
and hence we just need to prove that lies on their radical axis. Power of to second circle is just 
and see that 
and hence 
and so power of to first circle is 
and obviously 
, done.
.
.
,
.
is the inverse of function 
and thus, they might intersect that 
axis (if they do, we are done). Thus: 
is the only solution.
.
Let 
.
,![\[\angle ZAD = \angle ZAM = \angle AMZ = \angle AMC = \angle ABC = \angle ABD\implies AZ \text{ is tangent to } \odot(ABD).\]](http://latex.artofproblemsolving.com/2/6/1/2617bd6fc6402446013b1b624b242db161020119.png)
Also![\[\angle MBD = \angle MBC = \angle MAC = \angle ABD \implies BM \text{ is tangent to } \odot(ABD).\]](http://latex.artofproblemsolving.com/a/c/b/acb407e74dc3f6c5b7b204b11193c667fef865f3.png)
Now 
is tangent to 
and 
lies on the radical axis of 
.
is tangent to 
lies on the radical axis of 
.
and 
are tangent to 
.
,
.
and 
.
is the radical axis of 
.
be the reflection of 
perpendicular to 
at 
at 
,
.
,
.
,
lie on circle 
.
by 
.
.
This means that 
,
must be the radical axis of 
is the line joining the centers of these two circles, which is perpendicular to the radical axis 
respectively, so 
.
.
tangent to 
be a circle centred at 
.
is tangent to 
.
is tangent to 
is the radical axis. Done