and 
then prove 
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
and 
.
are real number satisfying
show that the equation 
has at least one real root.
be real numbers. Find the ranges of 
converges.
converge?
,where 
,
it is positiv integer number.
to find the third side 
He uses logarithms as follows. He finds 
and doubles it; adds to that the double of 
subtracts the sum of the logarithms of 
and 
divides the result by 
and takes the anti-logarithm. Although his method may be open to suspicion his computation is accurate. What are the necessary and sufficient conditions on the triangle that this method should yield the correct result?
![\[ f(x) = \sum_{i=0}^{i=n} a_i x^{n - i}\]](http://latex.artofproblemsolving.com/0/0/5/0050577825cfc0208f72081efbfa8df9db902378.png)
be a polynomial of degree 
and 
are odd, prove that 
has no rational roots.
is positive, then ![\[ \log_e (1 + 1/x) > 1 / (1 + x).\]](http://latex.artofproblemsolving.com/f/3/0/f30f05414aace0d4c5ce8549afb430bfff34ccfc.png)
does the sequence 
defined by the initial condition 
and the recursion 
have 
for all 
Y by Adventure10, Mango247
Given an acute triangle 
where 
. The excircle 
and 
opposite 
and 
of 
centered at 
and 
respectively. Let 
be the midpoint of 
, 
be the intersection of the lines 
and 
and 
be the intersection of the lines 
and 
. Suppose 
is the midpoint of 
, prove that 
, the angle bisector of 
and the circumcircle of 
are concurrent.
where 
. The excircle 
and 
opposite 
and 
of 
centered at 
and 
respectively. Let 
be the midpoint of 
, 
be the intersection of the lines 
and 
and 
be the intersection of the lines 
and 
. Suppose 
is the midpoint of 
, prove that 
, the angle bisector of 
and the circumcircle of 
are concurrent.
be the intersection of the angle bisector of 
from circle angle properties. Letting 
and 
,
We also notice that
so this is equivalent to proving
Not too sure how to continue, but there's probably some way you could find a point on the circumcircle of 
such that you relate 
to some other angle, and use some angle chasing from there to get the answer (possibly extending line 
past 
to a point 
on the circle).
cut 
implies that 
.
is an external bisector in 
is a internal bisector.
.
because 
is cyclic.
is cyclic.
which implies that 
.
and angle bisector of 
concur.
.
bisects 
,
and therefore 
is cyclic, which gives us 
where 
is the antipode of 
collinear, done.
to be the reference triangle. It is well known that 
such that 
bisects 
.
are collinear. Hence, we can rephrase 
to be 
instead. Now, we know that if 
,![\[ -1 = (L,A;B_1,C_1) \overset{A'}{=} (L,F';C,B) \]](http://latex.artofproblemsolving.com/b/c/8/bc84a329e710f57b421a9c5e981556d703bac075.png)
Therefore, 
,
are concyclic, as desired.
