be a triangle with incentre 
and circumcircle 
.
and 
be the second intersection points of 
and 
,
meets 
,
.
be the intersection point of the line through 
.
and 
meet at a point 
.
and 
are either parallel or concurrent.
,
and 
be positive integers such that:
is divisible by 
is not divisible by 
.
Prove that
Let 
Prove that
Prove that
satisfy![\[
(\lvert x^2-12x+20 \rvert^{\log x^2})^{-1+\log x}=\lvert x^2-12x+20 \rvert^{1+\log(1/x)}?
\]](http://latex.artofproblemsolving.com/3/9/2/39233a1044197f10403587b51443c39b8568f4a9.png)
and 
n
.
![\[f(n) = \sum_{k=0}^{n} \binom{n}{k} a_k, \quad n = 1, 2, 3, \ldots\]](http://latex.artofproblemsolving.com/9/3/7/9377324cb6585daaf2cbec7fed6fce2209835e22.png)
where![\[a_k =
\begin{cases}
3^k, & \text{if } k \text{ is even}, \\
0, & \text{if } k \text{ is odd}.
\end{cases}
\]](http://latex.artofproblemsolving.com/8/b/e/8be520a30bdfd9272163470ee69120c2fd3e9b59.png)
Find the remainder when 
is divided by 
and 
Prove that
be positive real numbers such that 
.
is divided by 
is 
What is the remainder when 
is divided by 
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.
