[/quote]
,
satisfying
Prove
and 
.
intersects circle 
at 
,
intersects circle 
at 
.
and 
are isogonal conjugates in triangle 
,
.
satisfying
Prove
be a real or complex inner product space. Suppose that 
and that 
Then
,
such that 
,![\[
f(x^2+f(y))=f(xf(x))+y
\]](http://latex.artofproblemsolving.com/d/e/d/dedd2338779fcc6723077ce586713fb4a68b9931.png)
be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at 
and 
the midpoints of 
and 
.
and 
meet 
and 
at 
,
,
bisects both angles 
,
,
.
Prove that
and 
Prove that
be positive real numbers such that 
Find the minimum value of 
lies on the bisector of an acute angle with vertex 
meets the sidelines at points 
such that 
.
and the circle with center 
are tangent.
are nonzero real numbers such that![\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]](http://latex.artofproblemsolving.com/6/d/9/6d9aabd8130b99d7128de9a3f063eb4c7124febf.png)
let 
be the maximum value of the expression![\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]](http://latex.artofproblemsolving.com/a/b/2/ab28769126365574d4c0108500c370826082de6b.png)
Determine the sum of the numerator and denominator of the simplified fraction representing 
Prove that
Where 
Y by Akacool, IMUKAT, ohhh
Let 
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
,
,
belongs to 
(which I don't even use in the solution, but it's what gives us the initial path to simplify the problem).
,
and 
,
coincide, but the centers of the first two don't seem too bad to work with,
and the center of 
.
!
,
is perpendicular to the radical axis 
unpretentiously, we see that if 
!
.
and the projections that I take in lemma 
.
on the line 
and 
on the line 
)
with the circle 
,
of 
is in 
.
and 
almost disjoint lemmas, that is, solve 
Lemma 
