be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
.
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
Y by HoRI_DA_GRe8, Adventure10, Mango247
Let 
be an acute scalene triangle. Let 
and 
be points on the sides 
and 
, respectively, such that 
. Denote by 
and 
the circumcentres of the triangles 
and 
, respectively. Prove that the circumcircles of the triangles 
, and 
have a common point different from 
.
Proposed by Patrik Bak, Slovakia
be an acute scalene triangle. Let 
and 
be points on the sides 
and 
, respectively, such that 
. Denote by 
and 
the circumcentres of the triangles 
and 
, respectively. Prove that the circumcircles of the triangles 
, and 
have a common point different from 
.Proposed by Patrik Bak, Slovakia
This post has been edited 1 time. Last edited by Amir Hossein, Jul 2, 2018, 5:51 AM
of 
be 
.
due to 
and spiral similarity. Now we need to show that 
.
and then reflect about the perpendicular bisector of 
.
meets 
,
.
at 
,
and 
are collinear with 
meet on the internal bisector of 
be a point on the internal bisector of 
are 
respectively. Prove that 
passes through 
is 
is tangent to its circumcircle. Now since the lines 
determine an isosceles triangle with 
on the perpendicular bisector of 
meet the perpendicular bisectors of 
and 
.
.
But we also have, due to projections of 
on 
is the circumcenter of 
This means that 
,
the second intersection of 
and 
.
so 
also we have(by trivial angle chasing):
so all we need to prove is that 
is isosceles,which is immediate by angle chasing,so 
is isosceles trapezoid(because 
have the same perpendicular bisector and they are parallel) and cyclic,which completes the proof.
to 
and 
,
Claim: 
and 
![\[\frac{OO_1}{OO_2}=\frac{|DC||BO_1|}{|BE||CO_2|}.\]](http://latex.artofproblemsolving.com/9/1/3/91363bbf9c097e480e7b0cd1cad94a1f3d86ced3.png)
Which is equals to 
because of simple result of Sine's Law in triangles 
.
}
}
is isosceles trapezoid which is indeed cyclic. 
is perpendicular to 
(
)
is perpendicular to 
with center 
that send 
then 
but 
is on the mid arc of 
the midpoints of 
we know that 
so 
are cyclic Moreover let 
be the similarity with center 
that maps 
it sends also 
hence 
are cyclic so the circle 
with center 
is midcircle of 
)
thus 
and 
implies 
are collinear thus 
moreover 
thus 
is perpendicular bisector of 
is on 
then we can find with Spiral similarity 
the rest is easy.(Use 
and a little angle chasing.)
)
and
be the circumcentres of 
respectively. Let 
and 
denote the projections of 
on 
So, 
form a rhombus. Thus, 
is the perpendicular bisector of 
are collinear and hence their circumcircle are concurrent. 
is an isosceles trapezoid. Let 
be the centers of 
respectively.
is a rhombus.
because both are perpendicular to 
.
.
lies on the radical axis of 
where 
a
and 
.
such that 
.
is linear. Let 
.
which is equivalent to show ![\[ 0=f(F)=\frac{n}{a}f(B)+\frac{m}{a}f(C)\iff \frac{f(B)}{f(C)}=-\frac{m}{n}=-\frac{c}{b} \]](http://latex.artofproblemsolving.com/d/6/8/d683fe282dcb49c4da089c659522ee75feb5109b.png)
Now 
and we are done for the sub-claim. 
.
.
and 
therefore 
,
.
.
,
and 
.
,
,
