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.
be a triangle with 
. Let 
be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of 
with the external bisector of angle 
. Prove that the midpoint of the segment 
lies on the circumcircle of triangle 
.
![$a,b,c\in[0,1]$](http://latex.artofproblemsolving.com/8/b/d/8bd8f441b63556fdf318ad140bbd3e8e358e9e1d.png)
, assume 
, compute the range of![\[f(a,b,c)=(1-a^2)^2+(1-b^2)^2+(1-c^2)^2.\]](http://latex.artofproblemsolving.com/7/c/1/7c1611d9b80f5f694afed1894acd57a93aa5b31c.png)
with orthocenter , circumcenter and incenter 
, circum-radius 
, in-radius 
.
.
satisfying
for all 
.
with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
, such that,
for all 
.
are primes, which satisfy:
is an integer
.
be circumscribed about circle 
, and let 
be the orthocenter of . The circle touches line 
at . The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
[hide=P#]a1b2c3[/hide] [hide=S#]a1b2c3[/hide]
be real numbers . Prove that
be reals such that 
Prove that
:o
so 
and since 
is just 
therefore 
so 
specifies the greatest common divisor of 
and 
.
meters
meters
therefore the length of the sides of the trac
square meters