[/quote]
,
in primes 
.
and 
are the feet of the altitude from 
and 
respectively. If 
intersect the line through 
.
,
be positive real numbers, such that 
.![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
and circumcircle 
.
,
intersect the circumcircle of triangle 
again at 
at 
and 
intersect on 
numbers on circle, and no one number is divided by other. In same time for all numbers we make next operation:
are two neighbors 
number 
and erase 
we can receive ?
numbers 
then after operation we receive circle with numbers 
.
.![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
?
,
,
,
.
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For 
represents 
.
are non-negative real numbers and 
Prove that
the equality hold for 
or 
or any
be positive real numbers. Prove 
Prove that
,
.
.
such that for all positive real numbers 
and 
:
Y by buratinogigle, Mualpha7, Amir Hossein, iarnab_kundu, Davi-8191, math1181, AlastorMoody, A-Thought-Of-God, jhu08, megarnie, HWenslawski, ImSh95, Adventure10, Tastymooncake2, ItsBesi, Rounak_iitr, and 1 other user
Given triangle 
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
Proposed by Evangelos Psychas, Greece
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)Proposed by Evangelos Psychas, Greece
and so 
is cyclic. But 
and so 
.
and hence 
.
,
,
are concyclic 
are concyclic 
.
be the midpoints of 
.
and 
so 
is cyclic. Thus 
is cyclic. Angle-chasing shows that this angle is actually 
,
and so 
is clearly the diameter of 
,
are concyclic.
and 
,
.
(since 
and 
.
,
.
,
.
,
.
.
.
which inplies 
we get 
[ASA] implying 
which means 
is cyclic. (1)
is cyclic. (2)
,
from (1)
.
is cyclic (3)
is cyclic (4)
to be cyclic.
.
.
.
,
(from 
and 
)
is the perpendicular bisector of 
,
,
is cyclic and similarly for 
.
,
bisects 
and 
bisects 
,
.
.
trivially, we have 
are cyclic, which concludes that 
.
and 
.
and easily get 
,
is cyclic because 
.
,
,
.
.
,
and similarly 
is the midpoint of 
,
is the medial triangle of 
and 
and 
are parallel to the internal bisectors of 
,
,
.
from tangency, and 
from the orthocenter, 
.
and 
with any of the methods above, or by considering the excenters 
,
is antiparallel to 
is the medial triangle of 
and we are done.
and 
,
.
and 
implies that 
is cyclic. Similarly, 
is cyclic. Thus with 
cyclic as well, 
.
and 
.
,
.
and 
This implies that 
too.
is a right triangle and therefore 
and 
and 
are congruent too.
and 
where 
is the semiperimeter of ABC.
and we are done.
.
is cyclic.
and
is cyclic. Since 
and 
.
are concylic.
is a parallelogram.
.
bisects 
.
excircle.
![\[\angle LFJ = \angle XBJ - \angle CXL = 90 - \frac B2 - \frac C2 = \frac A2 = \angle LAJ,\]](http://latex.artofproblemsolving.com/c/e/2/ce2b5041cd68f9f6c62d9fd65bbadc8fb3bbff3f.png)
implies 
is also cyclic. Instead exploiting symmetry like a normal person, we can instead notice that the converse of Pascal's on hexagon 
implies these six points lie on a conic, which must be a circle as we know five of the points define a circle.
,
and 
,![\[XS = AK = AL = XT. \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/d/6/cd617e38f059ff16229999f531460be70a3331d9.png)
![[asy]
size(330);
pair A, B, C, J, K, L, F, G, S, T, X;
A = dir(120);
B = dir(210);
C = dir(330);
J = extension(A, incenter(A, B, C), B, rotate(90, B)*incenter(A, B, C));
K = foot(J, A, B);
L = foot(J, A, C);
X = foot(J, B, C);
F = extension(B, J, X, L);
G = extension(C, J, X, K);
S = extension(A, F, B, C);
T = extension(A, G, B, C);
draw(A--K--J--L--A--J--X^^K--G--J--F--L^^A--S--T--cycle);
draw(circumcircle(A, K, L), dashed);
draw(circumcircle(X, K, L), dotted);
draw(A--K^^S--X, red+linewidth(1.5));
draw(A--L^^T--X, blue+linewidth(1.5));
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$K$", K, W);
dot("$L$", L, E);
dot("$J$", J, SE);
dot("$X$", X, N);
dot("$F$", F, NW);
dot("$G$", G, NE);
dot("$S$", S, W);
dot("$T$", T, E);
[/asy]](http://latex.artofproblemsolving.com/f/a/4/fa4ee7384a25b233f79911623a08bb2aedc050b4.png)
and 
and 
.
and 
.
,
,
,
and 
,
so 
,
.
is a paralelogram. We want to prove that 
but 
so we´re done. 
so 
and 
and 
so 
and 
,
and 
,
and 
.
so 
,
.
and 
,
,
,
,
.
,![\[\angle JF'L=\angle JML=90^{\circ}-\angle K,\]](http://latex.artofproblemsolving.com/c/6/9/c69aee0c018419aa54650797ae20dc15d69b51ba.png)
so since 
lies on 
,
Therefore 
Thus 
is cyclic. Similarly 
is cyclic and it follows that pentagon 
is cyclic.
Also, 
so 
But 
so by homothety 
Therefore 
where 
too, and we are done. QED
.
for some 
where![\[
\begin{vmatrix}
-a & t & c\\
0 & s-b & s-c\\
b-s & 0 & s
\end{vmatrix}=0.
\]](http://latex.artofproblemsolving.com/8/c/5/8c5952e88f4a12e4fd563abfed377801983d675f.png)
It follows that 
so 
.![\[
S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)
\]](http://latex.artofproblemsolving.com/a/6/a/a6a5c581322fb97e76df6b7ff1191b4a6faa6e9b.png)
and similarly![\[
T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).
\]](http://latex.artofproblemsolving.com/1/7/3/173aa9fd78970606124602ff9b0f362e8a965d5c.png)
Thus 
,
,
are concyclic.
are concyclic
.
is a parallelogram
.
,
by Claim III. The same is true of 
, therefore, 
.![[asy]
import olympiad;
import cse5;
size(11cm);
pair A = dir(75);
pair B = dir(210);
pair C = dir(-30);
pair X= dir((B+C)/2);
pair I = incenter(A,B,C);
pair J = (X-I)*2+I;
pair L= foot(J,A,C);
pair K= foot(J,A,B);
pair M= foot(J,B,C);
pair F =IP(L(L,M,5,5),L(J,B,5,5));
pair G =IP(L(K,M,5,5),L(J,C,5,5));
pair S =IP(L(A,F,5,5),L(B,C,5,5));
pair T =IP(L(A,G,5,5),L(B,C,5,5));
draw(A -- C--B--cycle);
draw(unitcircle);
draw(circumcircle(K,L,M),blue);
draw(circumcircle(A,J,F),red + dotted);
draw(L(A,C,0,1.5));
draw(L(A,B,0,1.5));
draw(L--F);
draw(J--F);
draw(J--G);
draw(K--G);
draw(S--T);
draw(A--T);
draw(S--A);
dot("$A$", A, dir(A));
dot("$B$", B,dir(145)*2);
dot("$C$", C, dir(33)*2);
dot("$I$",I,dir(45));
dot("$J$",J,dir(270));
dot("$K$", K, dir(180));
dot("$L$",L,dir(45));
dot("$M$",M,N);
dot("$F$",F,dir(135));
dot("$G$", G, dir(45));
dot("$S$",S,dir(180));
dot("$T$",T,E);
[/asy]](http://latex.artofproblemsolving.com/a/9/b/a9ba4d5f28b0b5aaa91b58928a4459ab536fd076.png)
.
is the reflection of 
is the polar of 
.
is the reflection of 
,
