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 scalene triangle with incenter 
and incircle 
. Let the tangency points of to 
be 
respectively. Let the line 
intersect the circumcircle of at the points 
. Assume that 
lies between the points 
and 
. Let 
be a circle that passes through and 
and that is tangent to at the point 
which lies on different semi-planes with 
with respect to the line . Let intersect 
at points 
and 
and let the second intersection point of the circumcircle of and the circumcircle of 
be 
. Prove that the intersection point of 
and 
lies on the circumcircle of if and only if the intersection point of 
and 
lies on .
people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
+1 w
be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers 
represent the number of unit squares that have exactly 
edges on the boundary of respectively. Find the largest real number 
such that the inequality 
holds for each polygon constructed with these conditions.
such that, for all 
,
+1 w, the function 
is defined as, 
and for all 
, if the prime factorization of is 
then 
. We have an arithmetic sequence 
. If for a positive integer 
, the sequence 
is also an arithmetic sequence, show that the sequence 
has to be constant.
, 
are coprime positive integers. Find all 
such that:
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
vertices, each edge has one of the colors 
, 
, or 
. For each 
, if the vertices can be divided into 
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call 
to be a grandson of if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
+1 w, for which 
for all functions 
, who satisfy the equation 
for all non-negative real numbers and 
.
and 
be sequences of real numbers for which 
and
for all positive integers . Prove that is eventually increasing (that is, there exists a positive integer for which 
for all 
). and 
be two concentric circles. Let and 
be any two equilateral triangles inscribed in and respectively. If and 
are any two points on and respectively, show that ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
, but your problem clearly stated 
.
in term of a function 
such that 
)
![\[ f(f(x)^2 + f(y)) = x f(x) + y \]](http://latex.artofproblemsolving.com/d/0/2/d022cecec0757daff00434c6a686225ccf384eef.png)
.
such that
.
to get 
for 
.
for 
and 
.
to yield
We then have
If 
and 
for nonzero 
This makes 
or 
,
to find 
.
is not constant, we have 
and 
.
denote the assertion, and observe that there are no constant solutions. By 
,
,
,
.
,
.
,
,
,
,
,
,
for some real 
is the only solution.
![\[ f(0) = 1 \]](http://latex.artofproblemsolving.com/8/7/8/8784a89d399b9480a41862bb77e7645094282249.png)
![\[ f^k(x) = f(x\sqrt{k}) \quad \forall k \in \mathbb{N}, \forall x \geq 0. \]](http://latex.artofproblemsolving.com/7/3/1/731d31076d43a297e6a9b249cfffd88bf5871f18.png)
,
and 
if 
gives injective and varying 
.
is a constant value, and since 
and therefore 
.
.
gives that 
,
implies that 
,
.
for some constant 
,
.
from 
to reals where 
and $f^k(x)=f(x\sqrt k).
and 
and 
Since 
This makes 
.
.
with 
in the given functional equation, we get![\[
g(x)^k = g(kx).
\]](http://latex.artofproblemsolving.com/e/c/2/ec229f3427eadce49a2b42957cd6cf6ee4210516.png)
also we can prove that it's positive always by contradiction, assume 
a 
,
> 0 take a sequence 
as n tends to 
from this we can contradict continuity therefore 
![\[
h(x) = \log g(x).
\]](http://latex.artofproblemsolving.com/7/e/f/7ef39ee05c2bf39ad71ee6a16459195daa8cf2b6.png)
and
,
,
that forms an Arithmetic Progression
Now, set 
Subtracting the two equations, we get that:-
for some 
value and 
be the set of real numbers. Determine all functions ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
such that 
and![\[
\Bigl(f(ab)\Bigr)^2 = \Bigl(af(b)+f(a)\Bigr)\Bigl(bf(a)+f(b)\Bigr) \quad \forall a,b \in (0,1).
\]](http://latex.artofproblemsolving.com/0/f/2/0f2d0a333bdd73ff468072cfd716f94865385710.png)
of pretty well known FEs 
we see 
.
and since 
for arbitrary 
we see 
.
and since 
is the identity.
works for all real 
.
twice to get 
which is equivalent to 
gives 
for nonnegative real 
gives 
(Cauchy FE) for nonnegative real 
so it still works for all nonnegative real 
for all nonnegative rational 
so 
be a positive rational
so 
which is equivalent to 
so 
.