[/quote]
,
lies on side 
lies on side 
.
and 
be points on segments 
and 
,
is parallel to 
.
be a point on line 
,
.
be the point on line 
,
.
,
.
meet at point 
.
on line 
be 
.
meet 
respectively. Let 
lie on a circle.
such that 
satisfying 
where 
.
cannot be a perfect cube of a positive integer.
has an incenter 
Suppose 
.
.
meets 
again at point 
.
and 
.
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
.
.
and 
with another positive integer 
,
is a perfect square. For example, if the number on the blackboard is 
,
,
,
,
.
,
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
is perpendicular to 
.
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
.
and 
are perpendicular.
and 
are perpendicular.
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
and 
and 
)
Prove that if lines 
are similar.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses.
,
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 
.
Y by Davi-8191, Adventure10, Mango247, Funcshun840
Let 
be the circumcircle of the triangle 
. Two circles 
are tangent to each of the circle and the rays 
, with 
interior to , 
exterior to . The common tangent of 
and the common tangent of 
intersect at the point 
. Let 
be the midpoint of the arc 
(not containing the point 
) on the circle , and the segment 
be the diameter of . Prove that 
, and 
are collinear.
be the circumcircle of the triangle 
. Two circles 
are tangent to each of the circle and the rays 
, with 
interior to , 
exterior to . The common tangent of 
and the common tangent of 
intersect at the point 
. Let 
be the midpoint of the arc 
(not containing the point 
) on the circle , and the segment 
be the diameter of . Prove that 
, and 
are collinear.
be the tangency points of 
and 
with 
It's well-known that 
go through the exsimilicenter 
and insimilicenter 
of 
and the incircle 
(for this use Monge & d'Alembert theorem for 
and similarly for 
)
then the pencil 
is harmonic 
quadrilateral 
is harmonic 
be the Incenter, A-excenter of 
,
and 
be the projection of 
on 
.
be the tangent point of 
with 
be the tangent point of 
with 
,
are isogonal conjugate of 
(well-known) ,
,
is a harmonic quadrilateral 
and 
are collinear .
is radical axis of 
be incenter and A-excenter of 
.
be the points of tangency of 
.
be the inversion followed by reflection about 
such that 
.
,
.
,
,
.
are on the radical axis of 
,
are all on this radical axis.
in 
with radius 
composed with reflection 
be the touch point of incircle,excircle,the foot of A-altitude,the foot of 
than it's well-known that 
are collinear.Projecting 
on 
be points on 
respectively. Let 
be points on 
,
.
,
are concyclic.
be the midpoint of arc 
in 
and 
be the tangency points between 
and 
,
and 
are collinear, where 
are the incenter and 
,
is the point at infinity along 
is harmonic and we are done.
,
the points of tangency of the excirlce and incircle 
(homothety) and 
and 
are isogonals wrt angle 
reflecting in the 
,
.
we get 
,
,
.
be the 
.
is cyclic.
,
are collinear. From 
.
be the point at infinity on line 
and 
are internal and external angle bisectors of 
,
.
as the touchpoints of 
resp. with 
resp. as the incenter and 
excenter and 
the midpoint of 
Then 
.
is the symmedian of 
which is obvious. Darn
the intouch,extouch point of inMixtilinear and exMixtilinear with 
be the incenter A-excenter and let 
be on 
, 
, 
are the midpoints of 
are on the A-mixtilinear circle which is 
also on the exterior A-mixtilinear which is 
is the radical axis of 
are collinear so 
are colinear
and 
are collinear
![[asy]
size(10cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.35));
dotfactor *= 1.5;
pair A = dir(130), B = dir(200), C = dir(340), I = incenter(A,B,C), D = foot(I,B,C), E = extension(A,I,B,C), M = (B+C)/2, F = 2M-D, G = foot(A,B,C), X = foot(E,A,M);
draw(A--B--C--A);
draw(incircle(A,B,C), gray);
draw(circumcircle(A,D,X)^^arc(circumcenter(A,F,X),circumradius(A,F,X),150,315), heavygreen);
draw(A--G^^A--E^^A--M^^A--F, blue);
draw(circumcircle(A,G,E), purple+dashed);
dot("$A$", A, dir(135));
dot("$B$", B, dir(200));
dot("$C$", C, dir(340));
dot("$D$", D, dir(270));
dot("$E$", E, dir(270));
dot("$F$", F, dir(270));
dot("$M$", M, dir(270));
dot("$G$", G, dir(270));
dot("$X$", X, dir(90));
[/asy]](http://latex.artofproblemsolving.com/1/1/7/117ff36ba92e521c053242551bddf0e70316e785.png)
,
,
.
at 
is cyclic.
,
.
be a function of point 
such that ![\[ f(\bullet)=P(\bullet, O_1)-P(\bullet, O_2). \]](http://latex.artofproblemsolving.com/4/2/2/4222e494302b8bcce2d6d66edece9065305760ef.png)
It is known 
,
and 
,
be the tangency point of 
respectively. We want 
.![\[ 4f(M)=2\left(f(I)+f(I_A)\right)=f(P)+f(Q)+f(R)+f(S)=0. \]](http://latex.artofproblemsolving.com/a/2/d/a2dc0d0dc937b917998ae302dea75183a1bd489b.png)
Done.
-
and 
be the circles passing through 
.
,
is tangent to 
being tangent is similar. It suffices to show that 
.
.
and
as desired. 
