AoPS Academy
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray 
at points 
and 
, respectively. Given that 
, prove that 
.
be an acute triangle, and let 
and 
be its circumcenter and orthocenter, respectively. For 
, points 
and 
lie on lines 
and 
(where 
), respectively, such that 
is a parallelogram. Prove that![\[\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.\]](http://latex.artofproblemsolving.com/b/a/5/ba563336236b6a04fc9ef661760fa1251bfb07d6.png)
in this order. Prove that![\[a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.\]](http://latex.artofproblemsolving.com/9/e/e/9ee73bbdc4b4f2cad2eb3fcfb3dbdf76b6200b4d.png)
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
and the distribution function is 
then
and find the expression of 
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
be a continuous function . Suppose![\[f(x)=\frac{1}{t} \int^t_0 (f(x+y)-f(y))\,dy\]](http://latex.artofproblemsolving.com/c/0/c/c0c48c4259d4e98562c21a23cf4a6841704b8e82.png)
and all 
. Then show that there exists a constant 
such that 
, 
, 
, if f ' (0)=p and f ' (5)=q, then prove f ' (-5) = q
is differentiable and 
for all 
. Show that for some 
, 
.
on the real line such that for all 
, ![\[ \left( f(x) \right)^2 = \displaystyle\int_0^x \left[ \left( f(t) \right)^2 + \left( f'(t) \right)^2 \right] \, \mathrm{d}t + 1990. \]](http://latex.artofproblemsolving.com/e/7/9/e79b1d8b17010c15a0f6397c8775e4e268a66a2e.png)
be the midpoints of the diagonals 
Since 
then 
Parallelogram 
is a rhombus 
bisects 
internally. Let 
Since 
and have corresponding parallel sides, it follows that their angle bisectors are parallel as well. By Menelaus' theorem for 
cut by 
keeping in mind that 
and 
we get
is parallel to the bisector of the angle 
, hence the triangles 
and 
are pseudo-similar and our problem is solved then.
be the center of spiral similarity that sends 
, 
. Therefore, it's also the center of spiral similarity that sends 
and 
. Hence, 
are cyclic quadrilaterals, also 
. Hence, there is a spiral similarity center at that sends 
. But 
.
.
be a convex quadrilateral such that 
.
and 
be the midpoints of 
and 
,
and 
and 
,
.
and 
,
and
Since, 
,
and 
,
and
Since, 
,
Because, 
and 
,
Since 
,
,
.
and 
.