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 
.
and 
Prove that
Let 
and 
Prove that
+1 w
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 .
is given. Prove that integers less than 
, it is possible to divide them into two non-empty sets such that the sum of the numbers in the first set will be congruent modulo p to the product of the numbers in the second set.
is given. Prove that integers less than , it is possible to divide them into two non-empty sets such that the sum of the numbers in the first set will be congruent modulo p to the product of the numbers in the second set.
and 
. Obviously 
or else divides 
.
and 
contain the other elements. Then the sum of the elements in is![\[ 1 + 2 + \cdots + p-1 - x - y = \frac{p(p-1)}{2} - x - y \pmod{p} . \]](http://latex.artofproblemsolving.com/9/a/f/9afdf7e0c9b138018311e729c6624f63b450279d.png)
On the other hand the product of the elements of just 
. Thus we simply need to check that 
:![\[ xy + x + y = (x+1)(y+1) -1 \equiv 2 \cdot 2^{-1} -1 \equiv 0\pmod{p}. \]](http://latex.artofproblemsolving.com/b/d/6/bd62c42ccbbb9db53ffa52cf362f8ad5452f6bb0.png)
Done.
and . Obviously or else divides . and contain the other elements. Then the sum of the elements in is![\[ 1 + 2 + \cdots + p - x - y = \frac{p(p+1)}{2} \equiv - x - y \pmod{p} . \]](http://latex.artofproblemsolving.com/a/7/b/a7b23d49c7e2e07fd678520fd3a46fc9d3fe0895.png)
On the other hand the product of the elements of just . Thus we simply need to check that :Done.
.. Is this correct or it is Prove that integers less than or equal to ? In the solution i could see terms used is till "P"
and 
(here both sets are non-empty).The product of the elements in first set is congruent mod to the sum of the elements in the second set.
are given such that 
.
in natural numbers m and n.
