[/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 
such that 
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)
and in 
steps and has to reach 
is valid but 
is not valid.
,
lie on a line in this order and satisfy 
.
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 Some_one_, centslordm, HWenslawski, math_comb01, Adventure10, Rounak_iitr
In an acute angled triangle 
, let 
be the orthocenter, and let 
be the feet of altitudes from 
to the opposite sides, respectively. Let 
be the midpoints of the segments 
respectively. Let 
be the feet of altitudes from 
on to the line 
respectively. Prove that 
is perpendicular to 
.
, let 
be the orthocenter, and let 
be the feet of altitudes from 
to the opposite sides, respectively. Let 
be the midpoints of the segments 
respectively. Let 
be the feet of altitudes from 
on to the line 
respectively. Prove that 
is perpendicular to 
.
and 
and 
, hence 
are tangent to 
(EGMO lemma 
) So 
at 
Thus, 
is cyclic. So 
Similarly, 
and done. (here, 
since 
is a diameter of the nine-point circle.)
,
by some angle chasing (a nice diagram helped in that observation 
) hence done.
are tangent to 
.
are collinear
,
as EF is the radical axis of 
hence, 
.
.
,
since it is the diameter of the nine point circle. Now, LEFN cyclic and 
(
is angle bisector by angle bisector as 
and after two more, you get 
and 
so, 
.
,
be the feet of altitude from 
,
onto the line 
.
,
.
is collinear. Similarly 
is collinear.
,
are antipodes of 
.
is the Newton line of 
.
and 
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[/asy]](http://latex.artofproblemsolving.com/0/b/4/0b46d5829c587e8a858e2f13630a5abe0ea189d5.png)
be the incenter of 
and 
intersects 
at 
is a point on 
,
,
and 
are isogonal with respect to 
Then not that by angle chasing, 
and 
.
we obtain that 
and 
are isogonal with respect to 
.
(corresponding parts of similar triangles)
and 
are isogonal, in fact 
(they are tangents to 
)
is cyclic.
is cyclic.
and we are done.
and 
.
.
.
.
and 
are both cyclic quadrilaterals. This implies that 
and we are done.
and 
and because 
and 
are perpendicular to 
we can say that 
and 
so 
and 
are \textbf{isosceles} so we have 
Also,
So we have, 
and 
which allows us to delete 
and 
are midpoints of arc 
containing and not containing 
an
and 
gives us that 
and 
parallel to 
and let the line through 
cut 
and
so 
and 
are cyclic. Since 
and 
.
and 
,
,
