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.
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
such that
.
denote the number of positive integers less than 
that are relatively prime to . Prove that there exists a positive integer 
for which the equation 
has at least 
solutions in .
is a rational number.
and other permutations.
with real coefficients for which there exists a polynomial 
with real coefficients such that for all real 
, 
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points in such that 
.
, there exists a balanced set consisting of points. for which there exists a balanced centre-free set consisting of points.
, such that every positive integer appears exactly once as a member of the sequence and 
divides 
if and only if 
?
be a sequence of real numbers, and let be a fixed positive integer less than . We say an index 
with 
is good if there exists some 
with 
such that 
, where the indices are taken modulo . Let 
be the set of all good indices. Prove that 
.
such that for any positive real numbers and 
,
, with 
.Initially, he folds the sheet along a straight line 
, where 
is a point on the side 
, so that vertex 
is located on side 
, as shown in the figure. Then folds the sheet again along a straight line 
, where 
is a point on side , so that vertex lies on the line ; and finally folds the sheet along the line 
. Matías observed that the vertices and were located on the same point of segment after making the folds. Calculate the measure of the angle 
.
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
. Prove that 
. (
is area of triangle ).
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that . ( is area of triangle ).
be any point on the Steiner circumellipse of 
and let 
be the cevian triangle of WRT 
Then ![$[DEF]=2 \cdot [ABC].$](http://latex.artofproblemsolving.com/3/3/e/33ec9a123e2b3b63c128a4fe7fe4aa12551261a1.png)
be the antimedial triangle of (
against 
). The Steiner circumellipse of is obviously the Steiner inellipse of tangent to its sides at 
are collinear on the polar of WRT and similarly 
and 
are respectively collinear. Thus by Pappus theorem for the hexagon 
it follows that 
is at infinity, i.e. 
and similarly we get 
![$[DEF]=[DYZ]=[CYZ]=2\cdot [ABC]. \ \blacksquare$](http://latex.artofproblemsolving.com/b/4/0/b4028b012dc90ae29c0bc707cbf447409f6ab058.png)
be the perspector of the parabola. Since the center of any inconic is the isotomcomplement of its perspector (in the parabola case at infinity), then it follows then that the isotomic conjugate of WRT must be on the line at infinity is on the Steiner circumellipse of and the result follows from the lemma.
and 
Prove that
![$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$](http://latex.artofproblemsolving.com/5/5/b/55be0378562556d1aa4400bef11d1510ff722854.png)
Where 
.
