[/quote]
,
with circumcircle 
,
that is tangent to segment 
-
of 
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 Adventure10, Mango247
On the circumcircle of the acute triangle 
, 
is the midpoint of 
. Let 
be a point on 
, 
the midpoint of 
, and let 
lie on 
. The lines 
and 
have intersection 
, and the lines 
and 
have intersection 
. If 
, prove that the incenter of the triangle is on the line 
, 
is the midpoint of 
. Let 
be a point on 
, 
the midpoint of 
, and let 
lie on 
. The lines 
and 
have intersection 
, and the lines 
and 
have intersection 
. If 
, prove that the incenter of the triangle is on the line 
This post has been edited 5 times. Last edited by zhaobin, Jan 19, 2011, 1:40 AM
cut 
at point 
= 
,
and 
,
similiar to 
.
:![\[
DX^2 = \frac{RD^2*SE^2}{AE^2}=\frac{RD^2*SE*SE}{SE*TE}=\frac{RD^2*SE}{TE}=\frac{RD^2*SD}{RD} = SD*RD = DC^2
\]](http://latex.artofproblemsolving.com/7/3/7/7378d94868c499ff60342015c54a7e521592ed6f.png)
is tangent to ciccumcircle of 
and 
is tangent to circumcircle of 
.
the intersection of 
is cyclic because 
Let 
.
and of course 
which gives that 
.
and 
is the bisector of 
.
and 
are homothetic and thus 
is tangent to 
at 
.
is tangent to 
and similarly 
