[/quote]
,
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be a scalene triangle with orthocenter 
and circumcenter 
.
,
the midpoints of 
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
and incenter 
and let 
are selected on sides 
,
such that 
,
,
.
intersects 
other than 
.
and 
meet on 
,
,
,
,
,
initially contains one coin. The following operations are allowed
,
,
;
,
,
and 
.
coins.
Prove that
its circumradius. Let 
,
across the line 
,
across the line 
.
.
be a real number and let 
be arbitrary real numbers in 
.
and 
are then defined recursively by
.
for all sufficiently large 
.
is a prime number
.
its 
.
meets 
at 
and 
meets 
a second time at 
meets 
a second time at 
.
passes through the antipodal to 
intersects the lines 
,
at points 
,
,
of the segments 
,
,
.
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
,
.
Y by buratinogigle, AlastorMoody, Adventure10, Mango247, lian_the_noob12, ItsBesi, Captainscrubz
Let 
be an acute-angled triangle, and let 
be the foot of the altitude from 
The angle bisector of 
intersects 
at 
and meets the circumcircle 
of triangle 
again at 
If
, show that 
is tangent to 
be an acute-angled triangle, and let 
be the foot of the altitude from 
The angle bisector of 
intersects 
at 
and meets the circumcircle 
of triangle 
again at 
If
, show that 
is tangent to 
,
,
be the incenter of 
.
we get 
is the external bisector of 
,
is the B-excenter of 
are concyclic ,
. ... 
,
is the midpoint 
of 
and 
,
is tangent to 
)
are concyclic we can finish the proof as following :
we get 
is the angle bisector of 
so 
.
so we are finished.
and 
Since angle bisectors of 
and 
is the excenter of the triangle 
.
(
.
.
.
.
thus 
.
are concyclic.
intersects 
.
and 
,
,
is a 45-45-90 triangle. Angle chasing yields 
,
.
,
be the angle bisector of 
with 
.
and 
.
then 
which means 
Then 
implying 
.
.
so we can rotate 
.
.
we easily get quadrilateral 
is cyclic. This gives 
.
quadrilateral 
is cyclic as well. This gives 
.
and so quadrilateral 
is cyclic. This gives 
because 
.
which gives 
as desired.
and center 
.
,![\[\angle BXA = 180^\circ - 2(\angle EBA + \angle EAB) = 180^\circ - 2 \cdot 45^\circ = 90^\circ.\]](http://latex.artofproblemsolving.com/1/f/e/1fe9c0b732c8926c22eafec67487627d06eaff7a.png)
So 
is cyclic, so 
.
,
is cyclic. So we have 
,
is tangent to 
![[asy]
import olympiad;
import cse5;
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(0);
pair B = dir(180);
pair F = dir(65);
pair D = extension(A, B, F, dir(205));
pair K = F*F;
pair C = extension(D, D+dir(90), B, K);
pair E = extension(B, F, C, D);
filldraw(unitcircle, palecyan+opacity(0.2), heavyblue);
filldraw(circumcircle(A, D, E), palered+opacity(0.2), red+dotted);
filldraw(CP(E,D), pink+opacity(0.2), magenta);
draw(arc(F,abs(E-F),170,310), orange+dashed);
draw(K--A--B--C--D, deepcyan);
draw(C--A, deepcyan+dashed);
draw(B--F--A, deepgreen);
draw(F--D, deepgreen);
draw(C--F, mediumred);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$F$", F, dir(F));
dot("$D$", D, dir(225));
dot("$K$", K, dir(K));
dot("$C$", C, dir(C));
dot("$E$", E, dir(165));
[/asy]](http://latex.artofproblemsolving.com/9/1/f/91fbfdb0ecf6c3503c33fd288ba36b11407ae460.png)
.
.
,
is cyclic and thus 
as needed. One can also realize 
is cyclic by noting 
.
.
.
,
,
we have 
,
,![\[ F = (a(a+c) : -b^2 : c(a+c))
= (a(a+c) : a^2-c^2 : c(a+c))
= (a : a-c : c). \]](http://latex.artofproblemsolving.com/2/6/c/26c2027d6452c514d2983670ccb717b5145fcc19.png)
But I guess I'll spare this problem...
as 
,
since 
bisects 
.
is a harmonic bundle and from there it follows (provable with LoS) that 
is right.
to meet 
at 
and 
opposite to side 
,
is tangent to 
and 
.
,
=
.
.
,
and 
,
=
,
.
=
.
=
,
=
=
.
=
=
,
.
and 
which is obvious.
as well.
.
Also, 
To finish, we have
,
It follows that 
so 
is tangent to circle 
as desired.
,
(since 
is diameter in 
is isosceles.
.
.
gives 
are collinear, i.e. 
,
and 
,
and 
are congruent. This implies that 
shows that 
is an isosceles right triangle. By symmetry, so is triangle 
.
and 
by symmetry, we can also get a quick angle chase to show that those two angles are also equal with 
.
.
and 
,
,
.
,
,
isosceles with 
.
is isosceles, with 
,
,
is the angle bisector of 
.
,
,
.
.
.
.
and 
.
and 
,
.
,
,
-
,
,
.
,
.
.
.
.
.
and thus, 
,
and we are done.
,
.
.
,
so 
.
is cyclic. Now, as 
,
so 
.
so that 
.
,
,
.
by the alternate segment theorem.
is tangent to 
.
.
.
.
which implies the result and concludes the proof. 
Hence, 
and 
We see 
.
is the external bisector of 
excenter of 
.
and 
Note that 
.
.
Also, 
,
,
.
we are done.
and 
with 
we have:
we have:
now from the 
we get:
Combining with 
we get: 
.
,
,
.
and 
,
,
and we win. 
in 
we have incenter of 
.
we have 
which gives 
.
is cyclic , hence we have 
.
.![\[\angle CFB = 180^{\circ} - \frac{\angle B}{2} - \frac{180 - \angle C}{2} = 45^{\circ},\]](http://latex.artofproblemsolving.com/d/1/2/d12267f852283cfd38890d82893ac00b6bad4f15.png)
and since 
too, the problem follows.
