.
and 
the midpoints of 
and 
.
and 
meet 
and 
at 
,
bisects both angles 
,
,
.
and 
Prove that
be positive real numbers such that 
Find the minimum value of 
lies on the bisector of an acute angle with vertex 
,
be the projections of 
meets the sidelines at points 
and 
such that 
.
and the circle with center 
are tangent.
are nonzero real numbers such that![\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]](http://latex.artofproblemsolving.com/6/d/9/6d9aabd8130b99d7128de9a3f063eb4c7124febf.png)
let 
be the maximum value of the expression![\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]](http://latex.artofproblemsolving.com/a/b/2/ab28769126365574d4c0108500c370826082de6b.png)
Determine the sum of the numerator and denominator of the simplified fraction representing 
Prove that
Where 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
elements. Let 
be subsets of 
of them has more than 
elements. Show that there's a member of 
different subsets 
.
of the set 
,
such that:
,
,
,
.
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 
and 
are equal.
direction.
