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.
then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
and 
. Prove that
Equality holds when 
or 
and 
, Ivica and Marica play the following game: Given two piles of cookies, on each turn a player takes 
cookies from one of the piles, of which he eats 
and puts of is arbitrary in every move. Players take turns alternatively, with Ivica going
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
. Show that the tangent of 
at 
,the tangent of 
at 
, the tangent of 
at 
are concurrent.
and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
, 
and 
.
, where 
are real numbers such that 
is an integral multiple of 
. Using (a) or otherwise show that if 
, then 
for all positive integers .
be the orthocenter of an acute-angled triangle 
. The circle 
centered at the midpoint of 
and passing through intersects the sideline at points 
and 
. Similarly, define the points 
, 
, 
and 
., , , , and are concyclic.
with real coefficient. It is known that 
and 
. Prove that 
can't have 
distinct real roots.
is 
. Line 
intersects the circumcircle of 
at a point 
, in addition to the vertex . Prove that the line 
is perpendicular to the line 
.
and 
is 
. Perpendicular bisector of 
is 
is the intersection of these 
lines, so we get
Let the circumcenter of be 
.
, and we can find that the perpendicular bisector of 
is given by 
is the intersection of these lines, so setting them equal and solving, we get
is the point of intersection, other than , of the circle centered at and the line . We see the circle centered at is given by the equation
Line is given by the equation 
. Substituting this value of 
into the equation of the circle, and simplifying, we get
Solving for 
in the above equation, we get 
, which are the coordinates of point , and 
. Thus, 
. 
