[/quote]
,
such that 
.
and 
.
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that 
?
and 
is a subset of 
then 
Find the smallest of 
that satisfy the following property: if 
,
,
,
are all the positive divisors of 
with 
,
divides 
for every 
.
,
:
,
and 
Prove that
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be non-negative real numbers such that 
Prove that
be a scalene triangle with 
,
be the foot of the altitude from 
.
be a point in the interior of the segment 
be the point on the segment 
such that 
.
be the point on the segment 
such that 
.
intersects segment 
at a second point 
.
and 
.
for 
.
are integers, then 
is divisible by an odd prime.
![\[ \frac{\displaystyle\sum_{k=0}^r \binom nk \binom{n-2k}{r-k}}{\displaystyle\sum_{k=r}^n \binom nk \binom{2k}{2r}\left(\frac{3}{4}\right)^{n-k}\left(\frac{1}{2}\right)^{2k-2r}}\]](http://latex.artofproblemsolving.com/9/d/e/9dedd490eaf9c6d22aaf93bbd74d82375adcfa85.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.
be the second intersection point of the circles 
.
,
.
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 
)
to 
,
and also 
.
.
passes through 
is a cyclic quadrilateral. Using the fact that 
,
by spiral similarity 
is a cyclic quadrilateral, and 
passes through 
,
.
.
,
,
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 
,
,
and 
,
which is the desired point. Suppose not:
with center 
to 
is 
,
.
,
.
,
.
.
concur at some point 
suffices. 
are coaxial. Indeed, if 
,
,
lies on 
.
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, 
