[/quote]
,
and 
intersect at points 
and 
.
,
intersects 
,
intersects 
.
is the circumcenter of 
,
.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
be a triangle with incenter 
,
.
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
,
past 
past 
be an even prime number and a sequence 
satisfy 
and 
Prove that always exists positive integer 
satisfying for all positive integers 
,
.
is "not congruence"
be pairwise coprime positive integers with 
being prime and 
.![$I = [0, a_1 a_2 \cdots a_n ]$](http://latex.artofproblemsolving.com/a/2/b/a2bce96b048c9b8fc94926db80aba37fd5037b4a.png)
of the real line, mark all integers that are divisible by at least one of the numbers 
. These points split 
units horizontally (left or right); or
and 
,
to 
using at most 
moves.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
.
such that:
such that:
is given, 
connected components respectively. Prove that 
has at least 
connected components.
and a subset of the numbers written on the board and adds 
be real numbers . Prove that : There exist positive integer 
such that 
Where 
,
,
are pairwise distinct real numbers satisfying 
Prove that
Let 
Prove that
and 
be positive integers.
and 
.
.
Y by Adventure10, Mango247
I have a very good solution of this but I want to see others.
Let the midpoint
of the side
of an inscribed quardiletar, 
.Let
the point of intersection of 
with 
. Let the parallel from the point 
to the
which intersects the at
. If 
angle=
angle= 
angle, prove that 
.
Let the midpoint
of the side
of an inscribed quardiletar, 
.Let
the point of intersection of 
with 
. Let the parallel from the point 
to the
which intersects the at
. If 
angle=
angle= 
angle, prove that 
.This post has been edited 1 time. Last edited by silouan, May 21, 2021, 3:22 PM
the angle 
is exterior, so angle 
.
, the angle 
from the hypothesis. So 
.
.
.
because 
is midpoint. Now at the triangle 
which is medium 
.
. But from hypothesis we have that AP// CF(F is the point where the CS intersects the AD). So angle 
.
.
.
is isosceles because 
so angle 
.
because it is going to the same arc of the circle with angle DAC. So angle 
.
.
,
is the ray of the circle.Now we make a trick. We extend the 
until CS intersects the circle in 
and 
.
because it is going to the middle of the circle. The 
has all of his angles=90 so it is an right-angled parallelogram.So 
.
and 
are equals because there are right-angled and they have: 
and angle DNS=angle PAB=x(angle DNS=x decause it is going to hte same arc of the circle with angle DAC). Finally because 
.
