[/quote]
,
with circumcircle 
,
that is tangent to segment 
at 
.
of 
is tangent to 
at 
.
is the midpoint of the arc 
,
.
have the same radius, then tangents to 
intersect on 
.
such that 
is a perfect square for all integers 
and 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
be a convex hexagon, such that the triangles 
are equilateral and the diagonals 
and 
are concurrent. Prove that 
or 
to the right;
squares north of the initial square after some finite number of moves if it is initially facing:
satisfying 
is a perfect square
,![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
be the set of the first 
perfect squares of nonzero integers. Suppose that 
for some 
.
stands for the set 
.
holds for every 
.
such that 
,
satisfy that: 
be a polynomial with integer coefficients. Suppose there exist infinitely many integer pairs 
such that 
.
![\[\sqrt x - \frac 1y - 2w + 3z = 1,\]](http://latex.artofproblemsolving.com/d/0/f/d0ff4819db851e6c12f738dcdb6b81bbd2ed0ca7.png)
![\[x + \frac{1}{y^2} - 4w^2 - 9z^2 = 3,\]](http://latex.artofproblemsolving.com/6/5/3/653c1547620f85ff761817a5cc333f16bce9c4c6.png)
![\[x \sqrt x - \frac{1}{y^3} - 8w^3 + 27z^3 = -5,\]](http://latex.artofproblemsolving.com/1/b/9/1b9c2ed383acc931aa3607ba82031e182e18f3d1.png)
![\[x^2 + \frac{1}{y^4} - 16w^4 - 81z^4 = 15.\]](http://latex.artofproblemsolving.com/e/0/3/e032ce109cd74bbae47dc6bca04dcab2d8f9ef29.png)
Y by A_Math_Lover, Adventure10
Let 
be a circle and 
are two points inside . 
are two circles passing though and tangent internally to at 
, respectively. 
cut again at 
, respectively. 
cuts 
at 
. Prove that reflection of 
though 
lies on line 
.
See more IMO Shortlist 2011, G4.
be a circle and 
are two points inside . 
are two circles passing though and tangent internally to at 
, respectively. 
cut again at 
, respectively. 
cuts 
at 
. Prove that reflection of 
though 
lies on line 
.See more IMO Shortlist 2011, G4.
it follows that 
If 
cut 
then 
are concyclic 
and the tangent of 
of 
and 
If 
denotes the reflection of 
then 
and 
is tangent to 
is internally tangent to 
of 
is also insimilicenter of their circumcircles 
In this inversion, 
and 
of 
Since 
then by conformity 
is internally tangent to 
coincide 
be the intersection of 
and 
, respectively.
be the intersection of 
and 
and 
be the midpoint of 
.
be the intersection of 
.
is the exsimilicenter of 
, respectively,
.
) we get 
are collinear.
is an isosceles trapezoid,
.
and 
) we get 
are collinear.
are collinear.
and 
,
. ie. 
with respect to 
