[/quote]
,
such that for all naturals 
the function satisfies:![\[a + b \mid a^{f(a)} + b^{f(b)} \]](http://latex.artofproblemsolving.com/e/b/b/ebb44041cedd29de516192d983a334d46e7fe7d6.png)
Bob wants to find all possible functions Alice could have. Help Bob and find all functions that Alice could have.
be the medial triangle of an acute-angle triangle 
.
perpendicular to 
meet 
at 
.
analogously. Let 
.
and 
.
meet the 
,
and 
similarly. Let 
be the circumcenter of 
be the circle with diameter 
,
at 
.
and 
.
are collinear.
lie on a circle centered at 
are coaxial.
denote the set of all real numbers. Find all functions 
such that
for all real numbers 
,
.
![\[x^2-cx+1 = \dfrac{f(x)}{g(x)},\]](http://latex.artofproblemsolving.com/1/3/3/1339acdaf1999e06e1a4b7dec7d1217389e2afef.png)
where 
and 
are polynomials with nonnegative real coefficients. For each 
,
exist.
such that every collection of 
.
are in that order and 
and 
.
on line 
on line 
and 
are equidistant from the circumcenter of 
be six positive real numbers. Prove that![\[a^2+b^2+c^2+\frac{4(ax+by+cz)\sqrt{ab+bc+ca}}{x+y+z} \geqslant 2(ab+bc+ca).\]](http://latex.artofproblemsolving.com/f/8/0/f808f7857c2150960f2a430837286487666797d8.png)
be a cyclic quadrilateral with circumcentre 
.
be the intersection of tangents at 
and 
to 
be the circumcircle of triangle 
and let 
,
denote the second intersections of 
with 
and 
and 
be the circumcircle of triangle 
.
,
is also tangent to 
be the midpoint of 
be the intersection of the tangents from 
.
intersects 
.
and 
meet at 
and 
meet at 
.
be a positive integer. Alice and Bob play the following game, with Alice going first after which they alternate turns. They determine the numbers 
in the following way.
to be an integer such that:
for 
.
successfully.
in the circumcircle of 
the point where the 
meets 
and intersects 
or 
is tangent to 
and 
are tangent to the circumcircle of 
,
is perpendicular to 
denote the set of all positive integers. Find all real numbers 
for which there exists a function 
satisfying:
,
is an integer if and only if 
;
,
.
be a positive integer larger than 
,
be integers. It is known that the equation 
has 
and 
are relatively prime.
Y by nguyendangkhoa17112003, Adventure10, mathematicsy, Mango247, ehuseyinyigit
Let 
be the circumcenter and 
the orthocenter of an acute-angled triangle 
such that 
. Let 
be the foot of the altitude 
of triangle . The perpendicular to the line 
at the point intersects the line 
at 
. Prove that 
.
be the circumcenter and 
the orthocenter of an acute-angled triangle 
such that 
. Let 
be the foot of the altitude 
of triangle . The perpendicular to the line 
at the point intersects the line 
at 
. Prove that 
.
,
,
concur if and only if the lines 
,
,
concur. Hereby, for any point P and any line g, the notion 
means the perpendicular from the point P to the line g.
.
.
.
.
and AC. Since OF || C'Z, we can rewrite 
,
.
,
,
concur. In order to prove this, we denote by S the point of intersection of the lines 
.
,
,
,
,
,
.
,
.
from 
,
.
has right angle 
.
,
are midlines of this right angle triangle, i.e., 
.
are (oppositely) congruent and 
.
,
,
intersects the circumcircle of 
at 
.
are concyclic because 
and we are asked to show 
.
.
,
,
.
.
.
,
.
and 
.
.
.
.
and 
it follows that 
are isogonal conjugates with respect to 
Consequently, if 
denote the midpoints of 
then 
is the pedal triangle of 
Since 
is N-isosceles, then 
is cyclic, so 
but 
.
,
,
,
and 
,
,
,
.
.
so that 
,
cuts 
at 
.
is a well known property, 
.
then 
and 
,
and we are done.
and 
,
.
,
and 
,
at 
respectively, thus 
and 
.
,
,
are corresponding points concerning similar triangles 
and 
.
be the point that corresponds to 
and 
we now just need to prove 
.
and 
,
be the orthic triangle of 
and we already have that 
as isogonal wrt 
,
are isogonal conjugates wrt 
.
(since they are the feet of the pedals from 
so 
so 
as desired.
,
is perpendicular to the sides of the pedal triangles.
instead of 
.
![[asy]
size(8cm);
pair A=(0,0), B=(120,0), C=(34.5,140.9), F=(34.5,0), O=(60,60), P=(3.3,13.3), Q=(136,97.7), H=(34.5,20.9), X=(19.4,-14.5), Y=(34.4,-20.9);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NW);
label("$Q$", Q, NE);
label("$H$", H, NW);
label("$O$", O, NW);
label("$P$", P, NW);
label("$H_1$", X, SW);
label("$H_2$", Y, S);
draw(A--B--C--cycle, linewidth(0.5)+blue);
draw(circle(O, 60sqrt(2)), linewidth(0.4)+blue);
draw(C--F, linewidth(0.5)+blue);
draw(F--P, linewidth(0.5)+blue);
draw(X--O--F--cycle, linewidth(0.5)+red);
draw(F--Y--O, linewidth(0.5)+red);
draw(F--Q--C, linewidth(0.4)+dashed+blue);
draw(circle((17.2,2.8), 17.42), linewidth(0.4)+dashed+grey);
dot(A);
dot(B);
dot(C);
dot(F);
dot(O);
dot(P);
dot(Q);
dot(H);
dot(X);
dot(Y);
[/asy]](http://latex.artofproblemsolving.com/6/0/3/6034660f955b984745b16a09de43eb32c60af7ae.png)
and 
be the reflections of 
and 
respectively, and let 
.
too.
,
.
,
,
so 
,
and 
.
as well. This yields that
or that 
.
so 
is cyclic and
Yep...
.
.
.
and so, 
and so, 
.
gives power of 
power of 
is isosceles and thus, 
.
,
at 
,
of 
.
at F
is cyclic quadrilateral of 
.
.
meet 
at 
.
by SAS congruency.
.
.
such that 
.
,
.
.
and together with 
it follows that 
and that 
is the perpendicular bisector of 
.
and we are done.
it's clear that 
.
be the intersection of 
.
so 
is a parallelogram 
and we're done. 
and 
.
,
and 
.
is its own projection on chord 
.
is the projection of 
.
and 
,
,
