,
holds for all positive reals 
and for every positive real 
,
,
be a polynomial where 
and 
are coprime integers in the set 
,
such that for every positive integer 
,
.
there exists a positive integer 
.
.![\[\sum ab +\sum ab^2 \le 6\]](http://latex.artofproblemsolving.com/9/f/6/9f632ab1e6df5029bcd4416eb06623123ad2cd8c.png)
.
we consider points 
such that 
and 
.
be the circumcenter of triangle 
and
,
be the points of intersection of the lines 
,
and 
,
respectively. Let 
be
,
the circle with center 
,
the circle
.
.
be positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
be a function satisfying the following conditions for all 
:![\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]](http://latex.artofproblemsolving.com/0/4/7/047e28c9d899d7de443ae40ac2c756fda851fafb.png)
Find the value of 
.
of triangle 
and 
points 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles 
.
.
be a triangle with side‐lengths 
,
,
the area of 
be the areas of triangles 
,
,
,![\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]](http://latex.artofproblemsolving.com/1/3/9/139a3e15f08afb78ca0b8301f24d30db8f202207.png)
Y by
Let 
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
.![$[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$](http://latex.artofproblemsolving.com/3/a/0/3a0d813ceb846745ae927892e0a6bbddb6eec0bd.png)
is the incenter of 
.![$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$](http://latex.artofproblemsolving.com/4/5/6/4567bb6a0790e2821b3a2b4ba1592e30649baf97.png)
and 
belong to the 
mixtilinear incircle of 
.
is the midpoint of arc 
and 
belongs to the 
be the feet of perpendicular from 
meet the line 
.
centered at 
.
and 
.
.
and similarly 
and we have proved the lemma. 
.
at 
.
,![\[CL'^2 = CD\cdot CA = CX^2 = CL^2.\]](http://latex.artofproblemsolving.com/d/c/9/dc9e1ea07a744020713f6dc1137a2caeb0135af7.png)
Also 
lie on same side of 
lies on 
or 
.
.
is the internal angle bisector of 
.
gives 
and 
m
is the 
and 
at 
and 
respectively. We know (well-known) that 
are collinear and 
.
and we are given 
are collinear. This gives 
and 
which further gives that 
is the 
by 
.
are tangent to 
is tangent to 
.
is parallel to 
,
.
and 
.![\[\measuredangle XRR' = \measuredangle XPB = \measuredangle R'YX,\]](http://latex.artofproblemsolving.com/e/d/0/ed0b04ceb8f66d55c4dd0f96740056f508bbb1a4.png)
so 
is cyclic. Similarly, 
is cyclic as well.
,
,
.
and 
.
is tangent to 
.
and 
are tangents to 
and 
are also tangents to 
form the 
.
,
by 
.
,
and 
are tangent to 
is harmonic. This means that the tangents to 
,
and 
.
,
.
,
lies on 
(as 
and 
)
are collinear (as 
)
,
(as 
and 
)
be the tangent to 
Thus, 
,
.
such that 
so 
is tangent to 
and also 
is tangent to 
and hence by PoP we have 
and so 
.
and see that 
and so 
and hence it is the 
is 
