AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, let 
be the foot of the perpendicular from 
to line 
. Point 
lies inside triangle such that 
and 
. The line through perpendicular to like 
meets the circle with diameter at points 
. Prove that 
such that there exist integers 
so that the polynomial 
has 4 distinct roots in the interval 
?
?
integers where 
. Show that there exists at least three positions 
such that ![$gcd(p[i], p[i+1])=1$](http://latex.artofproblemsolving.com/f/b/b/fbb304fb8ebdb8987edc830234698de2dc7a28a2.png)
(here ![$p[i]$](http://latex.artofproblemsolving.com/0/b/6/0b6cc7e1108ab6b64f3a644a3b28891f826b3bea.png)
refers to the -th position integer in the permutation)
be not cyclic and haves edges not parallel to each other.
as the intersection of the tangent line at 
with respect to circle 
and the 
-symmedian with respect to triangle 
and 
as the intersection of lines 
and 
, where all indexes taken cyclically.
, 
, 
, and 
are collinear.
(i know 
well). i want to study so if you guys have some easy or normal problems please send me
and 
. At each step, it is allowed to add to one of the polynomials and to multiply another one by the polynomial 
. Can the polynomials become equal at some point?
of degree . Sasha wants to guess it (knowing ). During a turn, Sasha can name a certain segment ![$[a;b]$](http://latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
and Maxim will give in response the maximum value of on the segment . Will Sasha be able to guess in a finite number of steps?
and 
Prove that
Let 
and 
Prove that
not isosceles, the incircle 
. 
is the intersection of with 
. 
is the projection of onto 
. 
cut at the second point . 
is the projection of onto 
. 
cut at the second point 
respectively. Prove 
are collinear
in 
.
to 
, so the same problem with orthocenter
, construct its circumcenter then we can draw the antipode of by first constructing 
then 
and 
, we can draw midpoint of by 
, we can also construct the intersection of symmedian with circumcircle by 
then draw s.t. 
by 
. 
to , so the same problem with orthocenter
, construct its circumcenter then we can draw the antipode of by first constructing then and , we can draw midpoint of by , we can also construct the intersection of symmedian with circumcircle by then draw s.t. by . 
.
as the second intersection between 
and . Construct 
, then 
. Since 
, 
. Construct 
as the second intersection between 
and . Since 
, 
is a diameter of so it passes through the circumcenter 
.
, and 
so we are done.
meet again at the midpoint of arc , and construct 
, 
similarly, then 
is the midpoint of (as is the perpendicular bisector of ). Construct 
, 
similarly. Then 
is the centroid of 
, so 
is the midpoint of and allows us to construct the perpendicular bisector 
of . Construct the other perpendicular bisectors similarly, then they meet at the circumcenter.
is an integer. Prove that 
are integers.
in natural numbers m and n.
![\[
v_p(n!) \leq \frac{n}{p - 1}
\]](http://latex.artofproblemsolving.com/5/7/c/57ca4d0a6532fb4b97687a97b9c1fb35ad02b9f2.png)
are drawn on the plane. Construct the circumcenter of 