[/quote]
,
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
,
,
be nonzero real numbers such that 
and 
.
.
kawaii if it satisfies 
is the number of integers not more than 
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers 
and 
.
and 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
,
,
is a multiple of 
.
a point in the space and 
the centroid of a tetrahedron 
.
be a 
on it, line 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from 
with exactly 9 elements that is subset of 
.
and 
that satisfy the following:
is an empty subset.
is 
and 
to 
is 
and 
.
and 
is 
is 
is tangent to the 
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 
again at a point D and consider the cyclic quadrilateral ADBC with the diagonal intersection F. Let the perpendicular to OF at F meet AC at P, DB at P', the circumcircle arc DC opposite to the vertex B at X, and the circumcircle arc DC opposite to the vertex A at X'. Since 
,
.
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 
and cuts the circumcirle 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, 
.
be the circle with center 
at 
of 
.
at F
is cyclic quadrilateral of 
.
.
meet 
at 
.
by SAS congruency.
.
.
be a point on 
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 
.
the two intersections of line 
,
and 
.
is its own projection on chord 
.
is the projection of 
.
and 
,
,
