[/quote]
,
,
:
,
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 
(other than 
.
and 
be positive integers such that 
.
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)
\
.
is equal to
Y by Adventure10, PikaPika999
Suppose that 
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
,
-
and 
are collinear; therefore, the perpendicularity of 
and 
is equivalent with the fact that 
.
(the power of 
and so 
.
.
and 
(since the circumcircle of 
is the nine-point center of 
cuts the circumcircle 
of 
and 
is a circle with center 
cuts 
is cyclic and the angle 
is right. Since 
is a diameter of 
and 
goes through 
)
is perpendicular to 
,we are done.
.
and 
have 
as the radical axis and as 
are concurrent, then 
is cyclic. So 
is 
to 
as 
is the diameter of 
.
;
w.r.t. 
we get (already proven 
are collinear): 
,
is cyclic, i.e. 
.
,
,
are collinear, and we are done.
is the midpoint of the arc 
,
is the 
intersects 
is the incenter of 
and 
,
and ![\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]](http://latex.artofproblemsolving.com/6/7/7/677420e109fa1a2b2bed326bd115276ddb6cd335.png)
and 
,
and 
are collinea
,
be a point such that 
and 
,
are concyclic
and 
and 
are concyclic
, 
and 
are concyclic
and 
,
and 
are collinear
(
is a radical ceneter)
,
,
and 
,
.
.
be the line through 
perpendicular to 
.
.
are collinear on the external angle bisector of 
are collinear by the definition. Now consider the circles 
.
while 
since 
and 
since 
is a diameter, where 
is the 
.
is 
,
is 
,
is 
,
are collinear, showing 
.
,
,
has coordinates 
.
(as you don't believe check here 
be the 
excenter, By Brokard's Theorem on Quadrilateral 
and 
inversion, 
swap, so we're done because then we must have
which means that 
.
.
.
is cyclic.
,
.
is cyclic too. As 
Its consequence leads to the fact that 
.
)
the intersection of line 
with the circumcircle of 
.![\[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\]](http://latex.artofproblemsolving.com/9/6/4/96421bc720fe61b02e3919130ca77c79e47d729b.png)
Thus 
are concyclic. Now by the incenter/excenter lemma, we have that 
.![\[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\]](http://latex.artofproblemsolving.com/a/2/7/a27b81ccfacba67e9030325fdeee756a43c97e95.png)
Therefore 
are collinear, and 
,
.![\[BI\cdot BI_B = BB_1\cdot BB_2.\]](http://latex.artofproblemsolving.com/e/1/f/e1f805bf6fd4471a5ebaeb9de8300a2c7d69bce5.png)
Let 
be the intersections of line 
with 
.
,
.
and 
,
.![\[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\]](http://latex.artofproblemsolving.com/3/5/7/3579b2a835b31fdcea59a6196ca4462e6417e23f.png)
as desired.
is simply 
.![\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]](http://latex.artofproblemsolving.com/d/4/b/d4b74190fff0dbce1ca7c2ecc0aefacb690144b9.png)
and thus, 
.
,
also lies on 
.
.![\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]](http://latex.artofproblemsolving.com/d/4/0/d400d91248e047569cc5dd80d05fe0750e552886.png)
Thus, 
must also lie on 
and indeed, 
as required.
![[asy]
import olympiad; size(400);
draw(unitcircle);
pair B = dir(130), A = dir(200), C = dir(-20), M = midpoint(A--C), O = origin, B1 = intersectionpoint(O--(O+3*(O-M)),unitcircle);
pair I = incenter(A,B,C), L = intersectionpoint(B--(I+3*(I-B)),unitcircle), Ib = L + L-I;
pair exB = rotate(90,B) * I, B2 = extension(exB, B, A, C);
dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$I$",I); dot("$O$",O); dot("$M$",M); dot("$L$",L); dot("$I_B$",Ib); dot("$B_1$",B1); dot("$B_2$",B2);
draw(C--B2--B--C^^A--B--Ib^^Ib--B1--L^^B2--extension(B2,I,B1,Ib));
draw(B--B1);
// draw(incircle(A,B,C);
[/asy]](http://latex.artofproblemsolving.com/7/e/4/7e4b00a850d18aa54a1429fef14ca203eb05877b.png)
(since 
is diameter), all 
,
(from right triangle 
)
(from cyclic quad 
)
so from similar triangles 
.
,
so 
.
wrt 
and 
be 
are concyclic by incenter/excenter lemma we have that 
and because 
,
are concyclic.Then it succifies to show that 
are collinear which is trivial by the radical axis concurrence lemma on 
,
are concurrent at 
