[/quote]
,
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
,
,
,
.
and 
are marked, draw the line connecting 
and 
are drawn, mark the intersection of 
is drawn, draw a line parallel to 
.
can be written as the sum of two square numbers
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
,
to be the set of all triplets 
of natural numbers, with 
odd and 
,
and 
.
for which 
be positive real numbers . Prove that
and 
in the plane, draws the perpendicular bisector of 
.
.
be the power of 
be the power of 
.
satisfy the equality
Find the maximum possible value of 
.
board. Julia
,
be its 
be the projection of 
be the intersection of 
and 
be the points on 
,
are on a line perpendicular to 
intersect 
,
intersects the circumcircle of 
.
Y by
Let 
be an acute-angled triangle with orthocenter 
. A line passing through intersects 
at points 
respectively so that 
. The bisector of 
intersects the circumcircle of the triangle 
at point 
. Show that 
bisects segment 
.
be an acute-angled triangle with orthocenter 
. A line passing through intersects 
at points 
respectively so that 
. The bisector of 
intersects the circumcircle of the triangle 
at point 
. Show that 
bisects segment 
.This post has been edited 1 time. Last edited by parmenides51, May 23, 2020, 7:37 AM
Reason: typo: replaced ''perimeter'' with ''the circumcircle of the triangle''
Reason: typo: replaced ''perimeter'' with ''the circumcircle of the triangle''
is the intersection of the 
bisector and 
.
is diameter of 
,
and 
.
and 
from the parallels we have that 
and 
and 
parallelogram. Now see that 
and since 
which means 
and so 
.
on 
then 
is a parallelogram and the conclusion follows.
