[/quote]
,
.
and 
are drawn internally tangent to 
and 
while 
and 
.
which is simultaneously tangent to both 
be a prime number. Prove that there exists a natural number 
such that![\[
p \mid m^n - n.
\]](http://latex.artofproblemsolving.com/3/a/3/3a38b299f0328d176286d65e02b92e9918e8109d.png)
is almost squarefree if there exists a prime number 
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
the ratio 
is an integer.
for some nonnegative 
and 
.
such that 
and 
.
over 
,
be integers. Suppose that for each 
there exists an integer 
such that 
is divisible by 
.
for some integer 
.
and 
be positive real numbers such that 
.
and 
.
and 
share circumcircle 
so that points 
and 
occur in this order along 
be the triangle formed by lines 
and 
and define triangles 
similarly. Furthermore, let 
and 
similarly.
and the two common external tangents to 
,
and 
similarly. Prove that points 
,
Prove that
be an acute-angled triangle with 
.
be the orthocenter of triangle 
be the midpoint of the side 
be a point on the side 
and the points 
is perpendicular to the common chord of the circumscribed circles of triangle 
.
Y by
Given a cyclic quadrilateral 
with 
and 
. Lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
with 
and 
. Lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
direction but not pretty sure with the 
direction, can anyone help me?
is a parallelogram and 
.
iff 
.
and 
.
and 
.![\[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]](http://latex.artofproblemsolving.com/e/0/c/e0c170e9bf04bfbd87a4e72b3b7f2f1c93858b3b.png)
Again, by the similarity this means ![\[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]](http://latex.artofproblemsolving.com/c/9/0/c90aaa3c7f32f7fdd231d270e38b43fc934e2ce2.png)
which essentially means the areas of 
are equal.
direction.
