[/quote]
,
and 
Prove that
Let 
Prove that
.
passing through 
and 
passing through 
are tangent at 
of 
and 
,
of 
and 
.
and 
meet at 
.
be a point such that 
is tangent to 
is tangent to 
are collinear.
and 
are on segments 
.
and 
intersect minor arcs 
and 
and 
are either parallel or they are the same line.
and 
.![$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$](http://latex.artofproblemsolving.com/9/6/3/9633b5849794285e44519c745c1f616eceb77711.png)
of them there were either 
who were all friends among each other or 
and 
,
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
,
be the circumcircle of a scalene triangle 
be a circle internally tangent to 
.
and 
and 
concur on the bisector of 
.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
and 
are the midpoints of sides 
and 
.
.
,
,
of a hexagon 
inscribed in a circle 
intersect at a point 
,
.
be the radius of the circle tangent to segments 
and 
and internally tangent to 
,
,
,
,
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
be the circumcentre, 
satisfy the following condition:
is constant.
.
are both perfect squares.
such that ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
,
and 
to be the midpoint of the smaller arcs.
Y by
For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle 
, define:
Points 
, 
and 
are collinear, that is is the Feuerbach point.
Points 
, 
and 
are collinear
, define:
as its circumcentre, as its orthocentre and as its nine-point centre.
, 
and 
to be the midpoint of the smaller arcs.- as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
- as the ninja point of
- as the ninja point of the contact triangle
Points 
, 
and 
are collinear, that is is the Feuerbach point.
Points 
, 
and 
are collinear
as the feuerbach hyperbola, then redefine 
,
which therefore shows that (notice 
is by homothety euler line of 
where incircle of 
at 
respectively) in fact 
orthocenters of 
respectively, then they also lie on 
and 
,
are colinear but also recall that the poles of 
are the 
-
are colinear and lie on the polar of 
we have to notice from La'Hire twice that 
lies on the polar of 
w.r.t. 
,
meet at a point 
meet at a point 
lie on 
and the 
,
then 
but we will prove 
which is enough by the radax that gives an harmonic showing it lies on line 
however by the angle condition mentioned we force 
that since 
we have that by invariant of pedal circles through the center of 
is cyclic but also reflecting 
let them be 
then you also reflect this recently found circle to see that 
are cyclic and thus note 
which shows that 
are colinear so we have that 
as desired.
is trivial by considering homothety at 
to 
thus we are done 
.
