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.
. 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 
.
, where 
, then the number of integers in the range of 
is __
about the perpendicular at onto (i.e. that line is the locus of 
's circumcenter); I am not able to provide more info as yet!
through !
) is actually the reflection of point about the perpendicular to at the point given by 
, or 
. be the center of the circumcircle of 
(of radius ) and the foot of the perpendicular from to . Then the given expression is equivalent to 
or (using the power of 
and around circle ) 
, or 
. That is, is a fixed, given point on . at . This perpendicular bisects the common chords 
of all circumcircles , and thus is the reflection of about this perpendicular.
is on so that 
, then 
is one limiting case of with 
and 
. Indeed, in this case, 
and 
. On the other hand, in the limiting case of 
, we get 
. To make sure that this is possible, we note that the previous quadratic in has a solution if 
, which is allowed by the given condition that 
.
with arbitrary radius sends 
to line 
and 
to points on lines 
such that 
Setting 
, we have 
for some constants 
(using 
, etc). passes through a fixed point different from iff line 
goes through a fixed point. Take an affine transformation sending to perpendicular axes 
and 
respectively. Then it is straightforward to check that line 
(the image of 
after the transformation) goes through the fixed point 
, as desired.
is fixed, not necessarily equal to 1.
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)
be reals such that 
P
satisfying 
.
,
.
let 
be on arc 
such that 
.
but 
so 
and now 
so 
passes through fixed point 
which is equivalent to 
or (if 
)
or 
,
and 
.
,
meets 
,
)
,
.
,
,
same solution as leader