[/quote]
,
,
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.
be a scalene oblique triangle, and 
a point on the orthocentroidal circle of 
)
-
,
,
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)
such that 
is the square of an integer.
.
be the centroid of triangle 
be the foot of the perpendicular from 
to side 
.
intersects the circumcircle 
of triangle 
be the point where 
intersects the circle 
,
and 
such that there exist positive integers 
and 
with 
satifying the condition that,
is an integer.
.
is a nonzero polynomial with integer coefficients. Prove that there exists infinitely many prime numbers 
such that for some natural number 
.
Y by anantmudgal09, Mathuzb, Number_Ninjas_2003, Adventure10, Mango247
Given an acute-angled triangle 
with orthocenter 
. Reflection of nine-point circle about 
intersects circumcircle at points 
and 
. Prove that is the external bisector of 
.
Proposed by Mohammad Javad Shabani
with orthocenter 
. Reflection of nine-point circle about 
intersects circumcircle at points 
and 
. Prove that is the external bisector of 
.Proposed by Mohammad Javad Shabani
This post has been edited 1 time. Last edited by Dadgarnia, Apr 15, 2019, 7:42 PM
and 
meet 
at 
and 
.
where 
and let 
.
be the feet of perpendicular from 
on 
are concyclic, so 
.
and 
be the reflections of 
lie on the circumcircle, ![\[HA\cdot 2HD=HA\cdot HD'= HX\cdot HH' = HX\cdot 2HY'\implies HX\cdot 2HZ=HA\cdot HD = HX\cdot HY'\]](http://latex.artofproblemsolving.com/3/e/7/3e7587f811bf87686730d198944adbca2ca85e26.png)
giving 
.
and 
be the reflection of 
,
finishing the problem.
as the nine-point circle WRT 
,
as its reflection over 
as the reflections of 
over 
are concyclic points. Also reflection of 
lies on 
and 
.
be the midpoint of 
Also, 
bisects 
,
is the reflection of 
be the circumcircle and 
be the 
-
,
be their reflection in 
to 
in 
.
lies on 
with 
,
the circumcenter and 
.
so 
yields that 
bisects angle 
;
.
meet 
.
.
.
,
.
,
denote foot of altitude from 
to 
,
and similarly 
,
.
are coaxal with radical axis 
.
lie on the radical axis. First, note that 
so the first two circles are done. To see the third circle, recall that the projection 
of 
lie on both 
.
,
be the reflection of 
.
is the reflection of 
.
.
,
so 
are concyclic. Hence we are done.
at 
.
,
is the apollonial circle preserving 
,
we have that 
.
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the nine-point circle, and 
their respective reflections across 
.
denote negative inversion at 
.
and 
is an isosceles trapezoid and 
be a line through the exsimilicenter 
.
of 
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across 
.
inversion, 
is an isoceles trapezoid so 
the reflection of the nine-point circle respect 
the center of this circle.
describe a circle. 
all satisfy the equation. So this circle is 
.
and is such that 
be the projection of 
is the midpoint of 
because it's the projection also of the center of nine-point circle which is midpoint 
.
and 
is perpendicular to 
