[/quote]
,
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 
and 
,
,
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 
,
be the projections of 
meets the sidelines at points 
and 
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 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a set of 
elements. Let 
be subsets of 
of them has more than 
elements. Show that there's a member of 
different subsets 
.
Y by Davi-8191, Smita, microsoft_office_word, OlympusHero, stayhomedomath, Jc426, centslordm, Adventure10, jhu08, mathlearner2357, megarnie, HWenslawski, bjump, lian_the_noob12, Mango247, ItsBesi, Sedro, Tastymooncake2, Rounak_iitr, ehuseyinyigit, DEKT, AlexCenteno2007, cubres, and 7 other users
Let 
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 
.
Prove that the six points, , , , and are concyclic.
Author: Andrey Gavrilyuk, Russia
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 
.Prove that the six points
, , , , and are concyclic.Author: Andrey Gavrilyuk, Russia
This post has been edited 4 times. Last edited by orl, Jul 20, 2008, 8:14 AM
the midpoints of 
and 
.
,
and 
are also perpendicular, and 
lies on the radical axis of 
and 
.
,
and 
all lie on a circle, whose center is clearly 
,
.
and 
also lie on the circle centered at 
and 
,
is perpendicular to the line connecting the centers of these circles, i.e. it is perpendicular to 
,
so 
lie on one circle. From the same reasoning 
and 
lie on a one circle as well. Suppose that this three circles doesn't coincide. Then we obtain a contradiction since radical axis of these circles - the sides of the triangle 
is cyclic consists in transforming the relation 
into 
(where 
and 
are the respective midpoints of ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
and ![$ [AC]$](http://latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
)
on both sides and applying the generalized version of Pythagoras' theorem (the one with cosines, I ignore the correct English name), we get a condition which is obviously true (simplifying cosines in right-angled triangles)...
is the midpoint of 
therefore 
are cyclic on circle 
lie on 
a
lie on circle 
.
,
be two isogonal points with respect to a given triangle 
,
,
be the orthogonal projections of 
and 
are concyclic. Let 
and 
.
.
circumcenters of the 
and 
respectively and 
midpoints of the segments 
.
.
.
.
.
and 
.
.
we have to prove that 
is symmetric in 
,
is the midpoint of 
is the midpoint 
.
.
is the radical axis of circle 
and 
.
cylcic, by similar argument on 
,
is independent from 
.
.
)
,
of the sides ![$ [CA]$](http://latex.artofproblemsolving.com/b/8/3/b8309224981a62a2c0df087562b92830bd7ad7b0.png)
, ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
respectively. Prove easily or is well-known that
. Thus, 
(remark that 
a.s.o.).
, 
, 
). In conclusion, 
.
obtain easily that 
.
.
,
,
and 
are concyclic then rest of the cases follow similarly.
and 
to be 
and 
respectively.
and 
respectively.
and 
and 
,
Also since line 
,
Hence, 
and 
intersect on their radical axis.
and 
,
,
and 
,
,
.![\[
\begin{cases}
A_1HA_2 &: A_1C \cdot CA_2 = CI \cdot CH \\
B_1HB_2 &: B_1C \cdot CB_2 = CI \cdot CH.
\end{cases}
\]](http://latex.artofproblemsolving.com/f/f/d/ffd30b16d16de24106a803e6299aeedd89e50b38.png)
Equating yields 
which, by Power of Point, implies 
is cyclic. Similarly, 
and 
are both cyclic, thus 
,
,
were concyclic, then their center would have to be the circumcenter of 
,
,
is the same as the perpendicular bisector of 
,
be the midpoints of sides 
and the same with the other points are equal. Let 
be the radius of the circumcircle of 
,
,
.
be the point on 
inside 
is tangent to 
be the point on 
inside 
is tangent to 
and 
,
.
because 
.
,
.
and the vertices be 
s.t. 
and the orthocenter is 
which is symmetric in 
and cyclic variants are all cyclic.
be the intersection of 
,
,
is just the 
and 
and 
lies on 
is the radical axis of 
,
.
are concyclic. Similarly 
and 
are separately concyclic, which follows from symmetry. Hence 
,
be the midpoints of the sides opposite 
be the circles centered at 
and 
and 
,
collinear.
,
are concyclic. Similarly, we can find that 
are concyclic, as are 
they are the same circle and we are done.
and 
by similar triangles. Thus, 
.
and 
;
is the radical axis of the two circles. Therefore, by power of a point, 
.
is 
by power of a point on circle 
.
is 
,
be the midpoint of 
and 
similarly. Note that we label the vertices as 
and 
are perpendicular to 
so it is also perpendicular to 
is cyclic. Similar conclusions follow for 
,
,
are distinct. This provides a contradiction as their radical axes will intersect at 
but they are not collinear so they must be the same circles.
be the midpoints opposite 
We begin by showing that 
are concyclic. Let 
but since 
by the Midpoint Theorem, it follows that 
so 
Therefore, 
and by Power of a Point our claim is proven.
are concyclic, and 
the circumcenter of 
so we are done. QED
.
are concyclic. Similarly, 
are concyclic, which finishes.
is parallel to 
are also cyclic. Since the perpendicular bisectors of 
is the same, which we call 
,
are concyclic.
to be collinear where 
is their second intersection. But 
(because perpendicular to line joining centers) and 
(orthocenter) are both perpendicular to 
be 
respectively.
must be perpendicular to 
and it also runs through 
Thus, 
are concurrent, we have that there must exist a circle running through 
and another one going through 
However, in all circles, the center is clearly 
be the midpoints of 
,
,
(and some other point). Now these conditions are enough to deduce that 
.
or 
.
,
are also the radical axes of the other two pairs, now by radical center they must be concurrent which means that 
.
