and 
be the points in which the median and the angle bisector, respectively, at 
meet the side 
.
and 
meets 
and 
,
produced.
is perpendicular to 
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that there exist positive integers 
for which the following holds 
for all 
.
and 
Prove that
be points on a plane such that 
,
be the set of all points 
,
is a real number. The path that 
and 
are defined in the following manner:
then find the value of 
then find the value of 
and ![$e\in [-2,2]$](http://latex.artofproblemsolving.com/b/b/7/bb78c9dedb18af5e6a904f5492917921e41a7ebc.png)
such that 
Find the value of 
Find the maximum positive integer 
such that for every subset 
with n elements, there always exist two elements a, b in T such that:
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 
.
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 
.
and 
the two intersections of line 
and 
are the solutions of the equation 
,
and 
.
is its own projection on chord 
.
is the projection of 
.
and 
,
,
