[/quote]
,
and 
points on a plane is 
,
such that: in every set of 
points of a plane, no three of which are collinear, consisting of 
points of every color there exists a 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
,
be the set of all positive integers not exceeding ![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
be reals such that 
and 
.
integers are such that 
is prime and 
.
such that 
.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a 
,
,
is __
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
satisfying 
is inscribed in circle 
.
denote its incenter and 
its 
-
denote the midpoint of arc 
.
meets 
.
at 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
meets 
a second time at 
.
passes through the antipodal to 
.
Y by ChineseHacker, thin21
Point 
lies inside the triangle 
. The centers of the circumcircles of triangles 
are 
, respectively. Denote by 
, the center of the circumcircle of the triangle 
. Prove that the point P satisfies the condition 
if and only if is the orthocenter 
.
(Bilokopitov Eugene)
lies inside the triangle 
. The centers of the circumcircles of triangles 
are 
, respectively. Denote by 
, the center of the circumcircle of the triangle 
. Prove that the point P satisfies the condition 
if and only if is the orthocenter 
.(Bilokopitov Eugene)
This post has been edited 2 times. Last edited by parmenides51, Jul 31, 2021, 12:48 PM
Reason: latex
Reason: latex
is the orthocenter of 
then 
.
.
.
.
we have that 
.
,so 
and the proof is complete.
