[/quote]
,
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 
and 
such that 
and 
lie on the opposite side of 
w.r.t 
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?
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 
.
Y by rkm0959, anantmudgal09, baopbc, ThE-dArK-lOrD, guangzhou-2015, buratinogigle, nguyendangkhoa17112003, Ya_pank, Adventure10, Rounak_iitr
Let 
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be the isogonal conjugate of 
WRT 
respectively. Clearly, 
and 
so one of the intersection 
of 
and 
lies on 
Let 
be the second intersection of these two circles, then note that 
and 
are concyclic we get 
hence 
Let 
be the midpoint of 
Since 
are concyclic (the 9-point circle of 
)
be the orthocenter of 
.
concur.
.
is cyclic on circle 
.
so 
.
.
and 
,
,
,
are collinear. But since 
is the 
altitude of 
,
as desired, and the lemma is proved. 
be the midpoints of 
.
,
are collinear. It remains to prove that this line passes through the other intersection of 
.
be the unique right circumhyperbola of 
of 
as well. Thus, 
w.r.t 
,
so 
are collinear. Thus, 
bisects 
be a non-cyclic quadrilateral and 
be the orthocenter of triangle 
.
be the pedal circle of 
be the pedal circle of 
.
passes through the mid-point of 
.
be the projections of 
respectively. Let 
meet the sides 
respectively.
be the midpoint of 
.
.
is the common chord of 
.
be the isogonal conjugate of 
;
be the isogonal conjugate of 
are symmetric in 
.
lies on the pedal circle 
and 
hence 
all lies on the circle with diameter 
.
.
be the midpoint of 
and 
meet at 
,
and 
meet at 
.
,
and 
hence 
.
and 
are both tangent to 
and 
meet at 
.
by Brokard's theorem.
,
.
,
,
.
is cyclic, and similarly, 
is cyclic. Let 
,
are tangent to circle 
w.r.t. 
,
and 
.
be the orthocenter of 
.
and 
are concurrent one of the intersections of 
.
,
.
,
.
and 
are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since 
and 
are cyclic, by Miquel's theorem it follows that 
it follows by homothety that 
passes through 
.
be the center of this hyperbola. Then by the fundamental theorem, this must be the other intersection point of 
with radical axis of circles 
,
.
the circle passing through 
.
,
lies on line 
is one of the intersections of circles 
lies on the radical axis of 
.
,
with respect to 
,
