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 a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
, centroid 
and orthocenter 
. Prove that 
is constant.
such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
points in a circle and number them arbitrarily with numbers from 
to . The chords cannot share common endpoints, also, the endpoints of these chords should be among the points.
pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most 
. cannot be improved.
satisfies that for any positive integer 
and 
Suppose there is 
non-negative reals 
such that for 
Determine the least possible value of 
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
to it. It cannot add two times in a row. By several operations it transformed a number 
into a number 
(
are positive integers). Prove that 
.
be a triangle and the points 
and 
on 
, 
and 
on 
, and 
and 
on 
be such that 
and 
. The intersections of 
with 
and 
are 
and 
respectively, and the intersections of 
with 
and 
are 
and 
, respectively. Prove that the lines 
and 
are concurrent.
rows and 
columns. Alice and Eva take turns filling the empty cells of the first row of the table, Alice plays first. In each move, Alice chooses an empty cell and puts a cross in it; Eva in each move chooses an empty cell and puts a zero. When no empty cells remain in the first row, the players move on to the second row, and so on (in each new row Alice plays first).
so that:
with all real numbers 
satisfies 
. Prove that
Let 
. Prove that
the parallelogram internal angles, 
. Using the cosine theorem, for example, for the triangle 
together with 
, we get
, because their corresponding side lines are parallel: 
and 
. Hence,
into the last cosine theorem formula,
and substituting for 
,
together with 
, we get
be a parallelogram. Suppose a line passing through 
and lying outside the parallelogram meets and 
produced at and 
respectively. Show that ![\[ AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF . \]](http://latex.artofproblemsolving.com/2/9/c/29ce31b24e798276b0e04b2b069a1e07501aba76.png)
.
.
.
, becomes 
. Division by AC transforms this into AC + CG = AU + AV.
.
.
.
, so that AU = VG, and we are done.
for the cevian-line 
and 
results: 
and 
Prove that
and 
Prove that
