[/quote]
,
and 
such that 
.
,
.
has a solution
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to 
is such that 
.
of triangle 
intersects 
be the common point of 
meets 
is tangent to 
such that
table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
,
is the circumcenter of 
,
are points on segments 
and 
respectively with 
.
be a point such that 
and 
.
intersect 
and 
intersect 
,
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
be altitudes in an acute triangle 
which meet at 
.
meets the circumcircle of 
and 
such that 
and 
.
and 
meet at 
.
and 
and the line 
concur.
aand 
Prove that 
Y by loveGeometrytothemoon
Let 
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed point
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed pointThis post has been edited 1 time. Last edited by 799786, Mar 4, 2022, 10:31 AM
Reason: typo
Reason: typo
intersects 
at 
and 
,
and 
.
.
and 
are corresponding points. So 
and therefore 
.
be the midpoints of 
,
.
and 
have the same radius, so 
are orthocenter of 
are collinear.
.
and 
.
we have 
.
and 
.
are collinear and 
are collinear. We also have 
.
.
or 
.
are midpoints of 
.
is parallelogram. So 
then 
and 
.
.
which is a fixed point.
