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.
. Denote by 
the number of changes of signs in the
Then prove that
.
be an acute triangle with orthocenter 
, and let 
be a point on the side 
, lying strictly between 
and 
. The points 
and 
are the feet of the altitudes from and , respectively. Denote by 
is the circumcircle of 
, and let 
be the point on such that 
is a diameter of . Analogously, denote by 
the circumcircle of triangle 
, and let 
be the point such that 
is a diameter of . Prove that 
and are collinear.
greater than 
, such that, if 
, then 
such that 
is a perfect square but none of the integers 
are perfect squares.
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
and are externally tangent at a point 
. Let 
be a line tangent to at 
and at 
. Let 
and 
be diameters in and respectively. Suppose points 
and 
lies on such that 
and 
are tangent to , and points 
and 
lies on such that 
and 
are tangent to ., , , lie on a circle 
., , , determine a quadrilateral with an incircle 
, and its radius is 
times the radius of . be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through and which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from ; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
. lies on 
. The Simson line of intersects Line 
at . and are the reflections of and over . Prove that the circles 
and 
intersects at a point on line .
+1 w![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
, 
, 
, 
be non-negative real numbers such that![\[2a+2b+2c+2d+ab+bc+cd+da+3=abcd.\]](http://latex.artofproblemsolving.com/d/e/8/de8ccd816cfe032a97d940982192b1abea21dc18.png)
prove that : ![\[\sqrt[4]{abc}+\sqrt[4]{bcd}+\sqrt[4]{cda}+\sqrt[4]{dab}\le\sqrt[4]{27(1+a)(1+b)(1+c)(1+d)}.\]](http://latex.artofproblemsolving.com/a/f/c/afc556cd8f5c7f4dedf153ffb92fb1f978c56823.png)
, but your problem clearly stated 
.
and 
Prove that
and 
Prove 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 
,
,
,
.
,
is nonconstant, we must have 
.
,
,
,
,
,
,
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 
.