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 .
doesn't work. here is a proof for all the others:
such that 
is not expressible as the sum of 2 nonnegative squares but expressible as the sum of 3 nonnegative squares we will be done. The aim is to get 
as this meets both the above requirements. If 
we can pick as 
modulo 
. If 
, take 
; if 
, take 
.
. So, either 
is the condition, or 
must be there (I suspect the former). Anyways, here is a solution for `fixed' version.
. Take 
, consider the following sets in modulo 
:![\[
A=\left\{x^2:0\le x\le\frac{p-1}{2}\right\}\quad\text{and}\quad B=\left\{-y^2-1:0\le y\le \frac{p-1}{2}\right\}.
\]](http://latex.artofproblemsolving.com/7/b/5/7b53f0c9ea78002a634f2efc2b858da7e4d75bc0.png)
I claim 
. Assume the contrary. Then, 
. Notice, however, that 
, absurd. Hence, there exists 
so that 
. Check that 
as . Thus, 
is divisible by . Using 
, it is easy to verify 
, as requested. and 
. Let 
and 
. Note that there exists an 
such that 
and 
as 
. Furthermore, we can assume 
. Doing the math, it is not hard see that for , as requested.
is allowed in that problem. This one, however, restricts to 
, for which you need more work. I suspect they took Korea verbatim, added positive integer condition without realizing that then would not work.
with 
not all 
such that the above condition holds. Using Legendre's three-square theorem, if 
we may take 
and otherwise 
.
square theorem;
is not of the form 
for nonnegative integers 
and 
. As said above, always works.
are given such that 
.
in natural numbers m and n.
![\[x^2+y^2+z^2=tp.\]](http://latex.artofproblemsolving.com/f/b/4/fb48ac67780ee637cae32f3f5bb03eb577075ccc.png)
would always work. So 