[/quote]
,
be positive integers.Define injective function 
from 
to itself such that 
.
divides 
.
board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
such that 
.
that divides 
for the positive integer 
.
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of positive integers such that 
in all of this cases :
3-Every prime number less than 
can be represented in the form 
by taking an Appropriate pair 
and 
. The number 
is the least prime number that cannot be represented as a sum or an difference of a power of 
, 
, 
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
be real positive numbers such that: 
.
,
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
![\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]](http://latex.artofproblemsolving.com/9/5/9/9591f7c842d1b6afcba11b15b3ea3f9e01bc937a.png)
![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
of injective functions from the set of real numbers to itself satisfying
and 
.
is injective if 
for any distinct real numbers 
and 
.
,
such that any graph with 
Y by HWenslawski, PHSH, Rounak_iitr
Given a non-right triangle 
with 
. Two points 
on the plane satisfy that, for 
, if 
and 
intersect the circumcircle of the triangle at 
, and 
, respectively, then 
and 
. Let the line 
intersects the circumcircle of at 
and 
. The Simson lines of , with respect to intersect at 
.
Prove that lies on the nine-point circle of .
with 
. Two points 
on the plane satisfy that, for 
, if 
and 
intersect the circumcircle of the triangle at 
, and 
, respectively, then 
and 
. Let the line 
intersects the circumcircle of at 
and 
. The Simson lines of , with respect to intersect at 
.Prove that
lies on the nine-point circle of .This post has been edited 1 time. Last edited by JustPostChinaTST, Mar 28, 2022, 4:33 PM
be a point lie on 
,
be the Simson line of 
be the orthocenter of 
,
.
intersect 
,
at 
respectively. 
intersect 
at 
respectively.
,
is the midpoint of 
.
,
be a diameter of 
.
intersect 
at 
respectively.
is the diameter of the nine-point circle of 
.
is a diameter of 
are collinear.
intersect 
at 
,
intersect 
at 
.
,
are collinear. (Let this line be 
)
intersect 
at 
,
intersect 
at 
.
,
are collinear.
,
are collinear.
,
and that they satisfy 
which is not granted !
but only one works ...
is on the 
-
is on the circle 
of the points 
s.t. 
but 
which means that 
hence it orthogonal to 
-
are collinear and the result follows .
(Simson line's equation is well known; the Lemma_2 by little_dinosaur above is also quite well known)
and 
to 
be the circle with center 
Further, let 
be the apolloniam circle of 
Then, points 
and 
are the intersections of 
be the projections of 
respectively. It is well known that 
(I know it as Lemma 2.5 from Geometry of Conics, A. V. Akopyan, A. A. Zaslavsky; I hope this is well known; it is just angle chasing anyways ).![[asy]
import graph;
import geometry;
import olympiad;
size(200);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1,400);
p=Circle(K,distance(K,B),400);
q=Circle(L,distance(L,A),400);
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Da,Ea,Fa;
Da=foot(Px,B,C);
Ea=foot(Px,A,C);
Fa=foot(Px,A,B);
fill(A--Fa--Px--Ea--cycle, 0.9*white+ 0.1*cyan);
fill(B--Da--Px--Fa--cycle, 0.9*white+0.1*green);
fill(C--Da--Px--Ea--cycle,0.9*white+0.1*magenta);
draw(A--B--C--cycle);
draw(Ea--Da--Fa--cycle);
draw(A--Px--C);
draw(Px--B);
draw(Ea--Px--Fa);
draw(Px--Da);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$P_1$",Px,dir(140));
dot("$X_1$",Da,dir(Da));
dot("$Y_1$",Ea,dir(Ea));
dot("$Z_1$",Fa,dir(Fa));
[/asy]](http://latex.artofproblemsolving.com/f/8/a/f8a99bc692fefdc03aa4bf56bcd40364cdece50a.png)
and
We can easily manipulate this to get 
so it follows easily that 
Finally, we have that by law of sines at 
and 
so it follows that 
![[asy]
import graph;
import geometry;
import olympiad;
size(300);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1);
p=Circle(K,distance(K,B));
q=Circle(L,distance(L,A));
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Ota=2*B-Oba;
draw(u,red);
draw(p,green);
draw(q,blue);
draw(A--B--C--cycle);
draw(Px--Py--Qx--Qy--cycle,dotted+magenta);
draw(K--Ota,gray);
draw(K--Oca,gray);
dot("$A$",A,dir(90));
dot("$B$",B,dir(150));
dot("$C$",C,dir(30));
dot("$K$",K,dir(K));
dot("$O$",O,dir(90));
dot("$P_1$",Px,dir(150));
dot("$P_2$",Py,dir(Py));
dot("$Q_1$",Qx,dir(-90));
dot("$Q_2$",Qy,dir(Qy));
[/asy]](http://latex.artofproblemsolving.com/f/8/b/f8bd28fc9d3ce90545d0ef5eb139dea448a7f0e9.png)
we must have that 
the circumcenter of 
must lie on 
and 
be the reflection of 
By above, we must have that 
are diametrically oppositive and it follows easily from angle chasing that their Simson lines are perpendicular 
.
(respectively 
)
is parallel to the Simson line of 
repsectively; then it follows from conclusion 
But since points 
centered at 
