[/quote]
,
and its prime fatorization 
,
Prove that there exist infinitely many naturals 
such that 
.
that satisfy![\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]](http://latex.artofproblemsolving.com/4/7/f/47f094ce57fe35d339ec3b93152ec3459a22ec48.png)
for every 
,
for every 
and 
,
.
be a set of 
points in a plane that lie within a disk of radius 
billion. Define a 
as picking a point 
and reflecting it across 
moves after which all points of 
Prove that
such that for every real number 
,
is a perfect square if and only if 
is a perfect square. Prove that 
is the square of a polynomial with real coefficients.
be positive integers such that 
and 
Prove that if 
and 
have the same prime divisors, then 
is a power of 2.
be non-empty subsets of the positive integers, and let ![\[\sum_{i=1}^m \sum_{(a,b)\in H_i^2}\gcd(a,b)\ge\frac1m \sum_{(a,b)\in S^2}\gcd(a,b).\]](http://latex.artofproblemsolving.com/1/c/d/1cdf561d3af92f524628e0724e0c5ee24d8d6a09.png)
Proposed by Dávid Matolcsi, Berkeley
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
,
as 
,
as 
.
and 
Find the value of 
,
Prove that
be an infinite sequence of real numbers, for which there exists a real number 
for all 
,![\[\left\lvert a_i-a_j \right\rvert\geq \frac{1}{i+j} \quad \text{for all }i,\ j \text{ with } i \neq j. \]](http://latex.artofproblemsolving.com/a/4/d/a4d87553590a450365b6525ca62a5d0416fe41cc.png)
Prove that 
.
Prove that
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 
,
