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.
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 
.
are concyclic and that leads to an easy solution.
, while more angle chasing yields 
. Hence 
is cyclic, and if we denote the foot of the perpendicular from to as , then 
. Now, let 
. It is easy to find 
, hence 
. Therefore, 
.
be a triangle with as midpoint of minor arc of circumcircle of be incentre of triangle than we prove that is centre of circumcircle of 
. excentre of triangle lie on circumcircle of 
and thus 
are collinear.
where 
is excentre.
is circumcentre of .
we get that 
are concyclic.
we have 
we thus get 
since 
.
are concyclic.
in points 
. Now, 
.
. Using that quadrilateral 
is cyclic we obtain:
and 
so 
lie on 
is isosceles with 
We use complex numbers. Put on the unit circle and note that 
so 
Now note that 
so 
Thus 
so 
Thus 
as desired.
be the midpoint of minor arc of 
and be its center. We have that 
is equilateral and hence 
. Also evidently 
, hence 
is a parallelogram. But since 
it's a rhombus, whence 
. Now from the excenter/incenter lemma we have that 
is cyclic. We have 
, as desired.
and since 
, then 
. Now, it is well-known that 
so as a result we have 
so is cyclic. Now, reflecting over and letting that point be 
we know is on the circumcircle of , so 
so 
.
are concyclic. I claim that lies on this circle. Let 
denote the circumcircle of and 
denote 
. Then, arc 
has measure 
. Then, since 
, looking at , 
not containing has measure as well. Hence, and are reflections of each other across . Since the reflection of across lies on , must lie on .![\[3\angle ABC=6\angle IBC=6(180-\angle IHC)=1080-6(180-\angle AHI+\angle ABC)=6\angle AHI-6\angle ABC,\]](http://latex.artofproblemsolving.com/6/d/7/6d76b96a388821e2e23200dded1a196c99f90eef.png)
which implies the desired result
Will lie on same circle..
with orthocenter , 
.
. Also, 
and 
since and are isogonal conjugates. Thus, 
.
is cyclic.
and multiplying both sides by 
gives the desired 
. 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/2/9/a/29a6230ae84bd5663eb157b123a2ce40a7b7563f.png)
are concylic
and 
so we get 
hence points are concylic 
are concyclic, as 
from some simple angle chasing.
, and we are done. 
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 reals such that 
P
and 
.
.
,
is centered at 
.
is obvious.
since 
.
.
.
as desired.
,
and NOT 
.
and 
are
(which we will call 
and 
respectively) are reflections over 
and 
where 
is the refllection of 
and 
as needed.
bisects 
is isosceles with 
,
.
and 
.
since 
.
.
,
is the perpendicular bisector of 
.
(or 
is the center of the nine-point circle of 
,
is cyclic since 
.
where 
,
,
.
and check that, this is the result we wanted.
and 
.
.
is cyclic. This is because 
and 
Let 
to 
so 
hence done.