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.
, where 
, then the number of integers in the range of 
is __
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
. Prove that if the leading coefficient of the polynomial equals then the greatest common divisor of the other coefficients also equals .
be a triangle with circumcenter 
. Point 
is the intersection of the parallel line from to with the perpendicular line to from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
dwarves, whose weigths are 
, who want to cross a river. They have a small boat which can lift at most 
each time without sinking. For each journey of the boat a non-empty subset of the dwarves to be taken to the other side is chosen and one of these dwarves is chosen as the 
for that journey. Since return journeys are counter-current, no dwarf is able to do the rower for more than one return journey. Is it possible for all the dwarves to reach the other side of the river?
with circumcenter and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line such that 
. Let and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
be positive real numbers such that 
. Prove that![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
, we have 
.
. Note that 
is the - symmedian of triangle 
. It is enough to show that is the foot of the -symmedian of triangle 
. This is equivalent to proving that 
.
is harmonic since 
passes through the midpoint of 
. This can be easily seen by the fact that 
. Since 
and 
the conclusion follows.
is cyclic we have that 
and 
.
and 
as . It is enough to prove that 
, because that implies 
.
and 
via the side we get that 
and 
which along with 
and 
gives that 
.
to intersect line in point . We have that 
is cyclic quadrilateral. Quadrilateral is also cyclic. We have equal angles on the same arc, for example 
is also cyclic quadrilateral. We can use again same arc angles, so 
. And that is all. 
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
satisfying 
be a parallelogram and 
.
intersects line ![\[\angle AED=\angle PEB.\]](http://latex.artofproblemsolving.com/7/6/0/76003ac5897855eeecad24bcb8fd2bb04f96bf81.png)
.
hence 
and we're done.
,
.
and 
and 
We can also see that 
and 
and because 
,