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.
+2 w
such that the equality 
holds for all pairs of real numbers 
.
such that 
and 
.
be a given positive real and 
be the set of all positive reals. Find all functions 
such that ![\[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]](http://latex.artofproblemsolving.com/b/0/6/b069e7b7ec1e277f4a4ce85a99434fdb54eb02f3.png)
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.
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 .
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 triangle with 
. 
is bisector of 
. A line, which passes through 
intersect side 
at 
. Then 
is bisector of 
.
.
in triangle 
for bisector 
, line 
and obtain that 
is bisector of angle 
. By analogy 
is bisector of angle 
.
.
, 
. Let the bisectors of angles 
, 
and 
, 
and respectively. Prove that 
. 
, 
and 
. Observe that the ray 
is the bisector of 
and 
is harmonically. Is well-known that the division 
is harmonically. Thus, 
, i.e. the ray 
. Analogously prove that the ray 
is the bisector of 
. In conclusion, 
. See and here .
.
, then 
is harmonical pencil, i.e. 
is harmonical division 
.If denote the second intersection 
of 
with the circumcircle of , then 
is equilateral.
(Pompeiu's theorem). But 
( 
). In conclusion, 
. equilateral triangles a.s.o.
bisects the exterior angle 
, must be the -excenter of 
. By symmetry, 
must be the -excenter of 
. Thus, 
bisects 
, while 
bisects 
, from which we can find that 
. The desired conclusion follows immediately.
is an obvious solution. Now, let 
.
or 
.
.
.
a
meet the opposite sides at 
an
passes through 
and 
.
be incenter. Then quadrilateral 
is cyclic (because 
)
and 
,
,
so 
Hence, 
and similarly 
Thus, 
