[/quote]
,
such that there exist positive integers 
and 
with 
satifying the condition that,
is an integer.
.
is a nonzero polynomial with integer coefficients. Prove that there exists infinitely many prime numbers 
such that for some natural number 
.
with circumcenter 
.
and 
intersect the altitude from 
to 
and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
is equal to the largest prime divisor of 
.
and 
.
Let 
be positive reals such that 
.
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
Prove that
Let 
Prove that
Prove that
![\[
\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.
\]](http://latex.artofproblemsolving.com/5/5/e/55ee0fba8392ee81b7b37da45d99755ea6d6754a.png)
.
.
.
Y by Davi-8191, Math-Ninja, Adventure10, megarnie, HWenslawski, Mango247, Mango247, Mango247, ItsBesi, ehuseyinyigit, and 3 other users
Let 
be a quadrilateral, and let 
and 
be points on sides 
and 
, respectively, such that 
. Ray 
meets rays 
and 
at 
and 
, respectively. Prove that the circumcircles of triangles 
, 
, 
, and 
pass through a common point.
be a quadrilateral, and let 
and 
be points on sides 
and 
, respectively, such that 
. Ray 
meets rays 
and 
at 
and 
, respectively. Prove that the circumcircles of triangles 
, 
, 
, and 
pass through a common point.
.
,
.
are directly similar, and hence the figures 
are also similar. It follows from here that 
are similar, meaning that 
,
.
.
be the point of intersection of the lines AD and BC. Let the circumcircles of triangles 
and 
intersect at a point 
apart from 
). Then, 
,
.
.
and 
are directly similar. The points F and E are corresponding points in these similar triangles, since they lie on the respective sides BC and AD and divide them in the same ratio 
.
.
.
.
.
.
.
.
.
. Actually, it's an application of Miquel points. The basic fact is that four lines in the plane enclose four triangles, and that the circumcircles of these triangles have a common point; this point is called the Miquel point of the four lines. Now, the whole problem comes down to proving that the Miquel point of the lines AD, BC, AB, EF coincides with the Miquel point of the lines AD, BC, CD, EF; in other words, we have to prove that the circumcircles of triangles 
,
and 
,
isn’t parallel to 
,
have a common point
have a common point
have a common point
,
is circumcircle of triangle 
)
.
passes through 
is a cyclic quadrilateral. Using the fact that 
,
by spiral similarity 
is a cyclic quadrilateral, and 
passes through 
,
.
.
,
,
to 
.
to 
and 
to 
,
and 
,
is 
,
is 
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/f/3/3/f33e1da81676e0f6af5078ccf9bd8ba02daaf085.png)
.
and 
.
.
and 
concur at 
to 
,
so 
share a Miquel point and thus the circles concur.
,
.
,
.![\[EA \stackrel{M'}{\longmapsto} FB \text{ and } ED \stackrel{M''}{\longmapsto} FC.\]](http://latex.artofproblemsolving.com/c/e/6/ce66f32ee1b302125004a08f579c32897dfafab6.png)
Now observe that the maps that send 
to 
and 
to 
are unique, so that 
,
to 
,
and 
,
which is the desired point. Suppose not:
with center 
to 
is 
,
.
,
.
,
.
.
concur at some point 
suffices. 
are coaxial. Indeed, if 
,
,
.
we have 
,![\[\frac{EA}{FB} = \frac{ME}{MF} = \frac{ED}{FC}. \]](http://latex.artofproblemsolving.com/b/2/6/b26575a7d01bb70badd4b6350af0154c40b579bf.png)
Thus 
so that 
and as the Miquel point for any quadrilateral as well as any spiral similarity is unique, 
Clearly, 
