[/quote]
,
of positive integers, such that every sum
(with 
)
for all 
;
for all 
?
denotes the greatest common divisor of 
,
.
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
for all 
.
,
,
be positive real numbers such that 
.![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
steps and has to reach 
is valid but 
is not valid.
,
.
be the midpoints of 
and 
,
is acute, and let 
and 
lie on lines 
and 
,
and 
and 
and 
such for any 
.
satisfy,
Where 
}
be an integer. We are given a balance and 
weights of weight 
.
to satisfy that the equation 
has at least 1 roots 
of the equation![\[n^3=p^2-p-1\]](http://latex.artofproblemsolving.com/f/f/8/ff88093bb0aa93afcfdff3ae25149ec5fbacb61f.png)
is a prime number and 
?
Y by Amir Hossein, Adventure10, Mango247, and 1 other user
Fixed points 
and 
are on a fixed circle 
and point 
varies on this circle. We call the midpoint of arc 
(not containing ) 
and the orthocenter of the triangle 
, 
. Line 
intersects circle again in 
. Tangent in to circumcircle of triangle 
intersects line and circle again in 
and 
respectively. Prove that the value of 
is constant.
Proposed by Mehdi E'tesami Fard
and 
are on a fixed circle 
and point 
varies on this circle. We call the midpoint of arc 
(not containing ) 
and the orthocenter of the triangle 
, 
. Line 
intersects circle again in 
. Tangent in to circumcircle of triangle 
intersects line and circle again in 
and 
respectively. Prove that the value of 
is constant.Proposed by Mehdi E'tesami Fard
This post has been edited 1 time. Last edited by goodar2006, Sep 28, 2012, 1:24 PM
![\[
\angle AKH+\angle DAH=\angle AKD+\angle DAH
\]](http://latex.artofproblemsolving.com/7/e/f/7efb3d520faef0473945999c041d232aa1cc0b68.png)
is directed, and it is easy to angle chase and compute that this is 
.
lies on 
,
.
,![\[
\frac{AL}{AM}=\frac{1}{\frac{AM}{AL}}=\frac{1}{1-\frac{LM}{AL}}=\frac{1}{1-\frac{MD}{AH}}
\]](http://latex.artofproblemsolving.com/c/3/1/c31feeddfbef23834fc964537ee9a11cddf7fc76.png)
is fixed, so it suffices to show that 
is fixed, but this follows from the identity 
.
and 
.
.
.
,
.
and we have let 
.
.
.
.
so 
which is a constant.
cut the cirle 
.
is tangent to the circumcircle of 
,
,
collinear, 
.
,
cut 
,
is a parallelogram and 
= const.
thus 
.
.
.
.
is tangent to the circumcirle of 
.
which is independent of the position of 
,
.
