.
Let 
be real numbers such that 
.
be a positive integer. Show that there exist a polynomial 
with integer coefficient that satisfy the following
for each 
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
Y by donotoven, HWenslawski, Miku_, TFIRSTMGMEDALIST, maolus, tiendung2006, Adventure10, Rounak_iitr
Let the points 
and 
be the circumcenter and orthocenter of an acute angled triangle 
Let 
be the midpoint of 
Let 
be the point on the angle bisector of 
such that 
Let 
be the point such that 
is a rectangle. Prove that 
are collinear.
and 
be the circumcenter and orthocenter of an acute angled triangle 
Let 
be the midpoint of 
Let 
be the point on the angle bisector of 
such that 
Let 
be the point such that 
is a rectangle. Prove that 
are collinear.This post has been edited 1 time. Last edited by ayan.nmath, Jul 17, 2019, 1:03 PM
be the reflection of 
.
be mid-point of arc 
of circle 
repectively and 
feet of 
concurrent
.
.
are cyclic.Now 
hence 
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be the midpoint of 
Then it is enough to show that 
are collinear. Suppose 
intersects the angle bisector of 
We know that 
Hence 
Thus 
Hence 
This completes our task.
be foot from 
.
is exterior bisector WRT 
and 
of 
passing through 
.
is Parallelogram, but 
where 
is the foot of perpendicular of the 
and 
which is the 
H
.
be the intersection of the diagonals 
.
be the feet of altitudes from 
respectively and let 
.
as 
.
and 
we get that 
.
.
be 
respectively.
are concyclic with diameter 
.
which implies that 
.
is harmonic.
are collinear.
be the Base of altitude from 
Respectively. Now Just observe 
is cyclic With Circumcenter 
So 
And also as 
Angle bisector of 
and also
Hence 
are tangents from 
Are collinear. And as 
So 
Are collinear 
,
and 
be 
,
is cyclic since 
(let this circle be 
)
,
and that 
,
.
,
is cyclic, with 
,
:
be the feets of the perpendicular from 
, Prove that 
be the midpoint of 
be the altitudes to 
and also its well known that 
are tangent to this circle. So, 
intersects the circle at the midpoints of the minor and major arcs 
,
is angle bisector, it passes through the midpoint of the minor arc and we easily see that 
cuts 
is well-known, it suffices to show 
.
,
But 
are isogonal, so 
as desired. 
are collinear, it suffices to show 
.
implying 
,
lies on the circle with diameter 
so 
lies on the perpendicular bisector of 
is also perpendicular to 
.
are collinear.
.
lies on 
(The nine point circle) where 
by some easy angle chasing (Just assuming variables and writing out all the angles will get the job done 
all of which I am ofc too lazy to type 
) which finishes the problem 
be the midpoint of AH. Diagram attached for reference
by a basic property of rectangles. So this is equivalent to proving 
.
.
.
since 
and as diagonals bisect each other 
is the mid-point of 
respectively.
with 
and 
and 
are tangents of 
is collinear implying 
collinear.
AND 
.
and 
are tangent to 
.
,
.
is harmonic and the result immediately follows. 
.
.
to meet 
,
be the reflection of 
.
are collinear.
.
is cyclic and 
,
,
.
and we are done 
be the feet of the feet of altitudes from 
.
,
.
.
and on 
at the same time, so 
.
.
linear with projective, why anyone didn't try to this?
be be circumcircles of 
,
.
,
are collinear. Since 
is kite, 
and projecting this to 
.
and 
are tangents to 
are collinear.
.
,
are collinear.
and 
be the feet of perpendiculars of 
are concyclic, so 
.
be the perpendicular bisector of 
.
.
.
,
is isosceles triangle. So, 
.
and 
,
,
, the midpoint of 
nine-point center
are collinear
be the orthic triangle of 
and 
so you get 
and because 
we get 
)
so 
and as 
concur...
is the reflection of 
by 
,
collinear. Let 
be the feet of the altitudes from 
is cyclic with center 
.
,
.
is cyclic with center 
.
.
since 
, so 
as 
as 
,
which means 
.
clearly 
, so it is well known that 
, so 
but also since 
.
be the unit circle. Let 
be the feet of the altitudes from 
and 
.
.
which means 
.
and 
.
.
Also,![\[\overline{\left(\frac{d-e}{f-e}\right)} = \overline{\left(\frac{yz}{y^2+z^2}\right)}\]](http://latex.artofproblemsolving.com/1/d/c/1dc48b1af1d87f429c32f04e3cfd4daf44c36189.png)
Now, note that since 
(which is known to be the complex number denoting the arc midpoint of 
and 
all lie on the unit circle,
Thus, 
which implies that, 
and thus, 
be Pedal triangle of 
respectively.
give us 
is angle bisector of 
.
.
(by considering 
.
and 
is cyclic with diameter 
is rectangle. Since, 
and it is also parallel to 
,
and 
are tangents to 
is a diameter of 
,
.
is the midpoint of 
be the foot of perpendiculars on 
and by the three tangents lemma we also have 
and 
tangent to this circle.
.
are diameters of 
so 
.
