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.
+1 w
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and 
such that 
Note that the indices are taken modulo 
Determine the greatest possible value of such that, regardless of numbering, there always exists a good vertex.
+1 w
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be a scalene triangle, 
be the median through 
, and 
be the incircle. Let touch 
at point 
and segment 
meet for the second time at point 
. Let 
be the triangle formed by lines and and the tangent to at . Prove that the incircle of triangle is tangent to the circumcircle of triangle .
+1 w
be real numbers such that 
Prove that
Where 
Where 
are are the lengths of the sides of a 
gon such that
then![$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$](http://latex.artofproblemsolving.com/9/3/e/93e5529518807b92cd9e0382f5fb07a293128b04.png)
be an acute triangle. Tangents on the circumscribed circle of triangle at points 
and 
intersect at point . Let 
and 
be a foot of the altitudes from onto 
and 
and let 
be the midpoint of . Prove: is the orthocenter of the triangle 
.
cuts 
in half.
be a positive integer and let 
be an irrational number. Show that for every integer 
, there exists a positive integer 
such that 
.
and be positive integers with 
. In a group of people, each one or always, what is the parity of people who speak to is a subset of size 
of the set of people. The answer can only
” or “
”. and is it possible to determine which people speak the truth and
be the Fermat point of a 
. Prove that the Euler line of the triangles 
, 
, 
are concurrent and the point of concurrence is 
, the centroid of .
until 
such that 
is equilateral now, taking as homothetic center the midpoint of (3:1) sending X to circumcenter of , P to centroid of and A to centroid of that's enough
, 
and 
are the bisectors of 
, 
cutting 
, at , respectively. If 
and 
, find the angles of 
, and are the bisectors of , cutting , at , respectively. If and , find the angles of
, be the midpoint of . Prove that
.
and 
are points on the circumcircle of such that 
is parallel to . Prove that 
is perpendicular to the Simson line of .
be the Fermat point of a . Prove that the Euler line of the triangles , , are concurrent and the point of concurrence is , the centroid of . then the problem is trivial by Liang-Zelich , consider the midpoint of , 
, as 
,let the centroid of be 
then we know 
so 
is parallel to 
which is also parallel to 
. now meets the point outside of , as 
such that 
is an equilateral triangle, and now by 
, where 
is the midpoint of we have 
is parallel to so 
lies on therefore it's done
be an acute triangle whose incircle touches sides and at and 
, respectively. Let the angle bisectors of 
and 
meet 
at and 
, respectively, and let the midpoint of be 
. Show that 
is equilateral if and only if 
.
are drawn perpendicular to the sides 
respectively of then prove that
Not been able to solve this even though I know I can do this.
is parallel to , show that tan tan 
.
but nvm
be the orthocenter of and let 
be the circumcenter
from and be and 
then just use 
, is the circumcentre and is the orthocentre. Then, prove that 
.
be the antipode and then 
is a rhombus
is parallel to , show that tan tan . but nvm be the orthocenter of and let be the circumcenter from and be and then just use 
and 
.
now they are the circumcenters of the equilateral triangle formed outside of 
,
,
is concurrent at 
.
,