[/quote]
,
Show that there exists a constant 
such that for all positive integer 
![\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]](http://latex.artofproblemsolving.com/f/e/9/fe933e886e28c2ecf1fa880d0e650ac7a2998aaa.png)
Proposed by Cheng Jiang
is written on the board. A move consists of adding either 
or 
to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is 
,
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be a circle passing through 
and 
at 
.
be the intersection of 
and 
be the intersections of 
and 
with the tangent to 
at 
.
and 
and 
be a polynomial with integer coefficients such that 
,
be an integer. Define 
and 
for all integers 
.
.
such that 
coprime, 
and 
.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
and 
such that 
.
and 
,
.
,
,
be non-negative real numbers, no two of which are zero. Prove that :
Y by Adventure10
Let 
be a triangle inscribed in circle 
with center 
. Let 
be the orthocenter of triangle and let 
be the midpoint of 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Prove that 
and 
are perpendicular.
by Michael Sarantis, Greece
be a triangle inscribed in circle 
with center 
. Let 
be the orthocenter of triangle and let 
be the midpoint of 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Tangent of at 
intersects the perpendicular bisector of 
at 
. Prove that 
and 
are perpendicular.by Michael Sarantis, Greece
This post has been edited 1 time. Last edited by parmenides51, May 5, 2019, 6:32 PM
Reason: author added
Reason: author added
be the feet of the altitudes from 
respectively and denote by 
the intersection of the tangents to 
be the center of 
.
and 
.
collinear. Next, note that 
is the radical center of 
,
imply that 
is the radical axis of 
and 
.
.
and 
and 
.
.
.
.
be a point such that 
is a parallelogram. Let 
be points on 
,
.
are collinear. Also, 
as 
form a parallelogram. So, 
.
are the polars of 
respectively. Since 
,
.
.
Two equations give 
.
Finally
Clearly 
![[asy]
size(250);
pair P=dir(105),Q=dir(-120),R=dir(-60),O=incenter(P,Q,R),A=foot(O,Q,R),B=foot(O,R,P),C=foot(O,P,Q),K=circumcenter( 1/2*(B+C),1/2*(C+A),1/2*(A+B)),L=extension(P,B,Q,O),M=extension(P,C,R,O),IP=2*dir(-90)-O,T=circumcenter(P,Q,R),IR=extension(Q,IP,R,O),IQ=extension(R,IP,P,IR);
draw(circumcircle(A,B,C),blue);
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$K$",K,dir(230));
dot("$M$",M,dir(180));
dot("$L$",L,dir(90));
dot("$I_P$",IP,dir(IP));
draw(P--Q--R--P,red);
draw(M--L,brown);
draw(IP--T^^A--K,green);
draw(Q--L^^R--M,purple);
[/asy]](http://latex.artofproblemsolving.com/e/f/0/ef0e400710fdb6d62066c0aef8ef3537b5c9398a.png)
be the tangential triangle of 
are angle bisectors. Let 
be the excenters of 
.
and nine-point center of 
is perpendicular to 
.
are homothetic, thus 
.
are at 
,
and 
.
at 
.
and 
.
we have 
.
and rearrange to get 
Moreover, since 
we have
Combining 
and 
we get that
Due to symmetry we also obtain
Now we have to prove that 
,
.
.
Now 
.
Therefore 
and thus 
,
be the reflection of 
the reflection of 
and 
.
.
and 
.
is a parallelogram. So 
.
,
but this is obvious because 
is a parallelogram so we are done.
