[/quote]
,
black and 
red cards. You pull out 
cards, one after another, and check their colour. If both cards are the same colour, then a black card is added to the deck. However, if the cards are of different colours, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards keeps reducing. What is the likelihood the last card left in the deck is black?
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
such that 
and 
.
be a given positive real and 
be the set of all positive reals. Find all functions 
such that ![\[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]](http://latex.artofproblemsolving.com/b/0/6/b069e7b7ec1e277f4a4ce85a99434fdb54eb02f3.png)
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and 
such that 
Note that the indices are taken modulo 
Determine the greatest possible value of 
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 the median through 
,
be the incircle. Let 
at point 
and segment 
meet 
.
be the triangle formed by lines 
be real numbers such that 
Prove that
Where 
Where 
are are the lengths of the sides of a 
g
then![$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$](http://latex.artofproblemsolving.com/9/3/e/93e5529518807b92cd9e0382f5fb07a293128b04.png)
Y by NO_SQUARES, sami1618
In triangle 
, interior point 
is chosen such that triangle 
is equilateral. Let 
be the isogonal conjugate of point with respect to triangle . Define point 
on the ray 
such that 
. Similarly, define point 
on the ray 
such that 
. Prove that line 
bisects segment 
.
Proposed by Áron Bán-Szabó, Budapest
, interior point 
is chosen such that triangle 
is equilateral. Let 
be the isogonal conjugate of point with respect to triangle . Define point 
on the ray 
such that 
. Similarly, define point 
on the ray 
such that 
. Prove that line 
bisects segment 
.Proposed by Áron Bán-Szabó, Budapest
be the reflection of 
be the point such that 
is a parallelogram. Since 
,
and 
inside 
such that 
.
and 
are also equilateral triangles.![[asy]
import geometry;
size(12cm);
pair A=dir(175);
pair B=dir(330);
pair C=dir(390);
pair D=.5(B+C)+rotate(90)*sqrt(3)*.5*(C-B);
pair F=.5(B+C)+rotate(90)*sqrt(3)*.5*(B-C);
triangle t=triangle(A,B,C);
pair E=isogonalconjugate(t,D);
pair X=B+rotate(300)*(E-B);
pair Y=C+rotate(60)*(E-C);
pair Q=A+Y-E;
pair P=A+X-E;
pair Ep=B+C-E;
fill(A--P--Q--cycle,lightred);
fill(C--Y--F--cycle, palered);
fill(C--E--B--Ep--cycle, palered);
fill(B--X--F--cycle, palered);
fill(E--X--Y--cycle, lightred);
draw(F--C--A--B--C--D--B--F--Y--C--Ep--B--X--F--B--E--C);
draw(A--Ep,dashed);
draw(P--Q);
draw(X--Y);
draw(Y--E--X);
draw(E--F);
dot(.5(X+Y));
dot(.5(P+Q));
dot("A",A,dir(A));
dot("B",B,dir(B));
dot("C",C,dir(C));
dot("D",D,dir(100));
dot("E",E,dir(180));
dot("F",F,dir(0));
dot("X",X,dir(300));
dot("Y",Y,dir(60));
dot("P",P,dir(240));
dot("Q",Q,dir(120));
dot("E'",Ep,dir(-10));
[/asy]](http://latex.artofproblemsolving.com/d/4/d/d4d8405f66d5a447e3fc6474a4c2ba262312ff9d.png)
and 
.
and 
,
and 
are shifted copies of each other. Therefore, the 
is a parallelogram. Thus line 
is the 
.
,
,![\[
\frac{[ACE]}{[ABE]} = \frac{AC \cdot CE \cdot \sin(120^\circ)}{AB \cdot BE \cdot \sin(120^\circ)} = \frac{AC \cdot AQ}{AB \cdot AP}.
\]](http://latex.artofproblemsolving.com/9/b/5/9b570228a98e508d21b458c78546c15625be8ba0.png)
Hence,![\[
\frac{[AQE]}{[APE]} = \frac{[ACE]}{[ABE]} \cdot \frac{AQ/AC}{AP/AB}= \frac{AQ^2}{AP^2},
\]](http://latex.artofproblemsolving.com/d/c/0/dc067ec77761ba374519f9a57f0c28ce36038a5d.png)
implying that 
is the 
-
.
is the 
