[/quote]
,
be reals such that 
and 
.
and a set 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
be positive real numbers satisfying 
.![\[
3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.
\]](http://latex.artofproblemsolving.com/f/3/c/f3cd114f909f388eacee142d3f90d07f6b149131.png)
be real numbers such that 
.
Let 
.
and 
,
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be a convex quadrilateral. Suppose that the circles with diameters 
intersect at points 
and 
.
and 
.
,
,
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
.
?
and 
points on a plane is 
,
points of a plane, no three of which are collinear, consisting of 
,
be the set of all positive integers not exceeding ![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
Y by Adventure10
Let 
be a triangle. Let 
be a midpoint of 
and 
be a point on the segment 
. 
and 
lies on the segment 
between 
and 
. Let 
be a midpoint of 
. 
cuts circumcircle of 
in 
and 
. Prove that circumcircle of 
is tangent to .
be a triangle. Let 
be a midpoint of 
and 
be a point on the segment 
. 
and 
lies on the segment 
between 
and 
. Let 
be a midpoint of 
. 
cuts circumcircle of 
in 
and 
. Prove that circumcircle of 
is tangent to .This post has been edited 1 time. Last edited by Pinionrzek, Mar 3, 2015, 7:15 PM
.
is cyclic, hence 
,
.
.
so the claim is proved.
and 
be a point on 
such that 
.
,
are concyclic ,
is tangent to 
at 
.
.
.
.
is tangent to 
and the points 
and 
and 
respectively.Miquel theorem assure that 
a
and the points 
and 
and 
respectively.Miquel theorem assure 
a
then 
and 
are coaxal thus the ratio of the powers wrt 
and 
are the same for 
which leads to the power of 
wrt 
is 
ie 
is a parallelogram. Then 
tangent to 
.
,
.
so 
so 
and 
are similar. Let 
meet 
at 
.
so 
is tangent to 
Since 
is cyclic. Since 
it is the center of 
Hence, 
and 
