[/quote]
,
be a triangle with points 
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
,
,
,
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
.
for every 
.
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide 
,
.
,
,
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that 
coprime, 
and 
.
such that for every 
is satisfied? Explain the answer.
)
-
,
,
through 
.![\[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]](http://latex.artofproblemsolving.com/1/f/3/1f39a54e4cf378bfd4e66d97949350b9b37dc1ec.png)
Y by
Let 
be a triangle and the points 
and 
on 
, 
and 
on 
, and 
and 
on 
be such that 
and 
. The intersections of 
with 
and 
are 
and 
respectively, and the intersections of 
with 
and 
are 
and 
, respectively. Prove that the lines 
and 
are concurrent.
be a triangle and the points 
and 
on 
, 
and 
on 
, and 
and 
on 
be such that 
and 
. The intersections of 
with 
and 
are 
and 
respectively, and the intersections of 
with 
and 
are 
and 
, respectively. Prove that the lines 
and 
are concurrent.
,
,
,
.
,
,
multiplying everything gives 
So from Menelaus 
are collinear. Since 
,
,
are collinear 
are concurrent(From Desaurges Theorem). So 
are collinear. Similar for 
and 
.
, 
,
are collinear 
are concurrent(From Desaurges Theorem). We are done.
and let 
.
passing through the point 
,
,
,
.![[asy]
import geometry;
size(10cm);
pair A=dir(120);
pair B=dir(210);
pair C=dir(330);
pair L= .2A+.8B;
pair K=.8A+.2B;
pair N=.3B+.7C;
pair M=.7B+.3C;
pair P=.25A+.75C;
pair Q=.25C+.75A;
pair R=intersectionpoint(line(K,N),line(M,Q));
pair T=intersectionpoint(line(K,N),line(L,P));
pair D=intersectionpoint(line(P,N),line(M,L));
pair E=intersectionpoint(line(P,N),line(K,Q));
pair X=intersectionpoint(line(E,B),line(D,R));
pair F=intersectionpoint(line(K,Q),line(L,M));
pair S=intersectionpoint(line(M,Q),line(L,P));
draw(conic(K,L,M,P,Q),deepgreen);
draw(B--E,grey); draw(A--D,grey); draw(C--F,grey); draw(N--K); draw(Q--M); draw(L--P);
draw(A--B--C--cycle, red);
draw(D--E--F--cycle);
dot("A",A,dir(A));
dot("B",B,dir(B));
dot("C",C,dir(C));
dot("K",K,dir(120));
dot("L",L,dir(185));
dot("M",M,dir(250));
dot("N",N,dir(-50));
dot("P",P,dir(0));
dot("Q",Q,dir(30));
dot("R",R,dir(10));
dot("T",T,dir(60));
dot("D",D,dir(280));
dot("E",E,dir(20));
dot(X);
dot("F",F,dir(165));
dot("S",S,dir(120));
[/asy]](http://latex.artofproblemsolving.com/b/1/4/b148db34973f8fc27cf21c4755da3ff5bd2cf2dd.png)
