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 the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
, and 
for all integers 
. Prove that 
.
+1 w
and 
Prove that
Let 
and 
Prove that
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
and 
intersect at 
. The common tangent line nearer to 
of the two circles touches 
at 
respectively. Let 
be the symmetric points of with respect to respectively. The circumcircle of triangle 
intersects circles and at points 
respectively which are distinct from . Prove that the circumradii of the triangles 
and 
are equal.
, 
. 
intersects 
at 
. A line 
cuts segment 
at 
respectively. If 
and 
, prove that the diameter of the circumcircle of 
equals 
.
are points on a circle 
, satisfying 
, 
. meets 
at 
. 
is on 
such that 
. Suppose 
intersects again at 
. 
is the reflection of 
wrt 
. Prove that 
lies on the same circle.
in its exterior
we draw the triangles 
so that 
, 
, 
.
and
and be on segment of an acute triangle 
such that 
and 
. Let and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
has a right angle at 
. Let be the point on line 
such that 
and 
lies between and 
. Point 
is chosen so that 
and is the bisector of 
. Point 
is chosen so that 
and 
is the bisector of 
. Let be the midpoint of . Let 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and a point such that 
and 
are tangent to the circumcircle of . Let 
be the reflection of over and 
be the reflection of over 
. If 
is the circumcenter of 
, help Oscar prove that 
is perpendicular to .
. It is easy to see that for any prime 
, we must have 
. Hence we get 
So we are done by discrete continuity if 
and 
for some 
. So we have two cases left:
for all .
. Let 
be the smallest non-negative integer such that 
Hence we have 
. Call this common value 
. Note that 
.
for 
. Also by our assumption 
. So 
form a set of at least 
consecutive integers, and hence every prime not exceeding divides at least one of those terms. But 
for all 
is divisible by every prime not exceeding 
, which is absurd because 
and 
. 
for all sufficiently large .
for all . Call a prime 
new if 
but 
for some 
. Since 
is unbounded and 
is bounded, the sequence 
must be unbounded. Therefore there must be infinitely many new primes. Choose any new prime 
, and let 
such that 
but 
. Contradiction! 
are given such that 
.
in natural numbers m and n.
,
denote the product of distinct prime factors of 
.
,
for 
with 
.
,
?