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.
of primes such that for all positive integers 
, there exists a positive integer 
and prime 
such that 
but 
.
chessboard contains a rook. One by one, rooks which attack an odd number of others are removed. Find the maximal number of rooks that can be removed. (A rook attacks another rook if they are on the same row or column and there are no other rooks between them.)
intersect at the point 
. On the segment 
, where 
is the altitude of the triangle, the point 
is marked. Points 
and 
are the feet of perpendiculars from point on straight lines 
and 
, respectively. The lines 
and 
intersect at 
. Prove that 
.
be real numbers such that 
. Find the value of 
be real numbers such that 
and 
. Find the value of 
, let 
be a point on arc 
of circle which doesn't contain 
. 
and intersect at 
. Let 
and 
be the reflection of about to 
and 
, respectively. 
intersects at 
, and 
intersects at . Prove that circumcenter of triangle 
lies on .
Prove that
Let 
Prove that
Let 
Prove that![$$ a+b^2+c\geq \frac{135 \sqrt[3]{90}-2}{2}$$](http://latex.artofproblemsolving.com/8/6/6/866623c65f54d6147a43b1b241c2cba96863b624.png)
such that 
for all 
such that 
.
be a positive integer such that the sum of its positive divisors is at least 
. Prove that has at least 
distinct prime factors.
is known as an interesting number if satisfies![\[{\ \{\frac{n}{10^k}} \} > \frac{n}{10^{10}} \]](http://latex.artofproblemsolving.com/0/c/a/0ca11fa76313b65fd76a3cceee90955960e44ef6.png)
.
is given. Find all functions 
such that for all 
we have
be an acute triangle, 
its orthocenter and the center of its nine point circle. Let be a point on the parallel through 
to such that 
and and are on different sides of and a point on the parallel through 
to such that 
and and are on different sides of . If 
and 
are the reflections of over and respectively, and 
are the intersections of 
and 
respectively with the circumcircle of , prove that the intersection of lines 
and 
lies on 
.
be an acute triangle and its orthocenter. Let be a point on the parallel through to such that 
. Define and as points on the parallels through to and through to similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
. The point on the continuation of is given such that 
. A point in the side of in which doesn’t lie is chosen such that 
. Prove that 
.
and let it intersect 
in 
and 
in . Also, let 
. Notice that we only need to have 
cyclic. Obviously is the orthocenter of 
, and thus 
. Thus 
is cyclic and as 
(because is the orthocenter) this implies 
and as 
, we have 
which implies 
parallel to the 
bisector and thus cyclic.
(because is the orthocenter) this implies . cyclic.
Prove that
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
(extended) and 
Let 
meets 
Let 
meets 
at 
is cyclic quadrilateral. So, 
(simson line). So, 
is cycic. Again 
is also cyclic cause 
be on the extension of 
such that 
.
.
at 
at 
.
cyclic.
.
as desired.