[/quote]
,
,
.
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
intersect 
.
.
and 
.
.
be the point on 
.
by 
,
for every positive integer 
.
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
.
such that the n subsets 
are also disjoint.
s
satisfies 
.
,
.
such that for all irrational numbers 
and 
,
where 
Y by buratinogigle, AlastorMoody, valsidalv007, HWenslawski, son7, Adventure10, Mango247
Let 
be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides 
at 
and 
at 
. Let 
be the image of 
under 
. Prove that the circumcircle of the triangle 
is tangent to the circumcircle of triangle .
be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides 
at 
and 
at 
. Let 
be the image of 
under 
. Prove that the circumcircle of the triangle 
is tangent to the circumcircle of triangle .
cuts 
at 
=> 
cuts 
at 
. We prove that 
.
- 
cuts 
at 
=> 
// 
=> 
tangents to 
tangents to 
)))
at 
at 
and 
intersect 
at 
and 
resp. it's easy to notice that 
(antiparallel..). Let 
and 
. Since 
are collinear , applying Pascal's converse to the hexagon 
yields 
are tangent . in the other hand
therefore 
which ends the proof.
to be the tangency point of 
and 
meet again (other than 
.
is cyclic. This is because 
.
,
is tangent to circle 
cuts 
,
L
cut 
.
cut 
.
so 
,
is iscosseles trapezium.So R is the center of the homotety that takes circle with radiu
to 
at 
denote by 
WRT 
Line 
intersects 
at 
respectively. Easy angle-chasing we get 
and 
are collinear, this is because 
From Reim's Theorem we get 
is tangent to 
Clearly we have![\[\begin{aligned}
\measuredangle DQF&=\measuredangle DQB-\measuredangle FQB\\
&=\measuredangle DPB-\measuredangle QCB\\
&=\measuredangle PBC-\measuredangle QCB\\
&=\measuredangle QCY=\measuredangle QXY=\measuredangle QED
\end{aligned}\]](http://latex.artofproblemsolving.com/4/d/e/4de824b927a697e2ad34df20958b2950a91b022d.png)
i.e. 
as well. From 
is angle bisector of 
and 
,
are concyclic 
and 
are symmetric over 
or 
i.e. The reflection of 
as desired. 
![[asy]
size(9cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair A=D("A",dir(52),dir(52));
pair B=D("B",dir(-152),dir(180));
pair C=D("C",dir(-28),dir(0));
pair T=4*B/7+3*C/7;
pair O=D("O",origin,dir(45));
pair H=circumcenter(B,O,C);
pair P=D("P",IP(circumcircle(B,O,C),L(O,T,-0.01,6)),dir(-135));
pair D=D("D",extension(A,B,P,(H-P)*dir(90)+P),dir(-135));
pair E=D("E",extension(D,P,A,C),dir(-45));
pair Q=D("Q",OP(unitcircle,circumcircle(D,P,B)),dir(-90));
pair F=D("F",extension(E,D,Q,-Q*dir(90)+Q),dir(180));
pair X=D("X",OP(unitcircle,L(D,Q,0,5)),dir(-45));
pair Y=D("Y",IP(unitcircle,L(Q,E,5,-1)),dir(135));
D(unitcircle); D(circumcircle(B,O,C));
D(A--B--C--cycle);
D(F--E);
D(F--Q,blue);
D(D--B);
D(E--C);
D(D--X);
D(E--Y);
D(Y--P,dashed+red);
D(C--P,dashed+red);
D(X--Y,dashed);
D(B--Q,dashed);
D(C--Q,dashed);
D(O--P,dotted);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/1/8/f/18feaa679a10432ae338666c72ba35839c80a1bc.png)
then 
be the Miquel point of 
with respect to 
.
,
.
lying on both 
.
.
.
,
,
and 
,
and 
,
, we have that 
as well. Our objective is to show that 
is cyclic. This is true since
and 
intersect the circumcircle of 
,
,![\[\angle FGX = \angle FCX = \angle PCX = \angle PEX\]](http://latex.artofproblemsolving.com/4/b/c/4bc195e52a938a84283ef09476635856578fff5a.png)
so 
.
and 
are tangent at 
,
.
bisects 
.
is the reflection of 
over 
.
.
is on 
,
,
,
is tangent to 
.
,
(where 
.
we have that 
and 
intersect 
so 
which means that 
and 
are homothetic which implies that 
is the tangnet point of 
intersect 
and 
,
as cyclic.
is cyclic.
.
be tangent to 
.
