[/quote]
,
be a positive integer larger than 
.
of subsets of 
having the following properties:
distinct sets of 
that satisfy the equation 
?
and 
Find the value of 
such that 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of 
is the reflection of 
over 
lie on 
are parallel to 
.
lie on the circumcircle of 
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
be an acute triangle with 
and let 
be a point on the perpendicular bisector of 
such that 
and 
are on different sides of 
,
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
.
and 
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
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 
.
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 
repectively and 
feet of 
concurrent
.
.
are cyclic.Now 
hence 
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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 
,
,
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 
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 
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 
.
,
.
.
and on 
at the same time, so 
.
.
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 
,
collinear. Let 
is cyclic with center 
.
,
.
is cyclic with center 
.
.
since 
, so 
as 
as 
,
which means 
.
clearly 
, 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 
.
