[/quote]
,
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
.
and 
.
?
. Prove that
be 
unit vectors on a plane, where 
for 
such that 
st for every two subsets 
of 
, there exists a polynomial 
with integer coefficients st either 
or 
where the equations are considered mod n.
is the set of reminders of 
where 
mod n.
be a positive integer such that 
.
.
be Rutthee sequence if 
be a positive integer and
for all 
and there are some positive integer 
such that 
and 
also find all possible value of 
Prove that
Let 
Prove that
and 
on 
and 
are on the same side of 
be midpoint of segment 
be intersection of 
with 
and 
be two positive integers, such that 
.
can take, where ![\[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]](http://latex.artofproblemsolving.com/c/b/2/cb27ccbcde6d90f98e10f8f6876c36ae8972af7f.png)
be positive real integer such that 
Prove that
such that 
and 
real number on the board such that each number is more than 
.
on the board and write 
.
of positive integers satisfying the equation 
Y by narutomath96, Adventure10, Mango247
Let 
be the circumcenter of an acute-angled triangle 
, let 
be the circumcenter of the triangle 
, and let 
be the midpoint of the segment 
. We take a point 
on the side 
and a point 
on the side 
that satisfy 
. Show that the straight lines 
and 
are perpendicular.
be the circumcenter of an acute-angled triangle 
, let 
be the circumcenter of the triangle 
, and let 
be the midpoint of the segment 
. We take a point 
on the side 
and a point 
on the side 
that satisfy 
. Show that the straight lines 
and 
are perpendicular.
intersects the sides 
and 
respectively. The foot of the perpendicular from 
.
in 
cuts 
cuts 
,
.
.
and 
be the midpts of 
and 
are obviously similar, so 
(#).
the point 
(##).
with center in centroid of 
,
.
and 
.
.
and 
,
and 
.
and
.
be a bisector of 
.
.
and 
because 
is isocles
.
and in the end 
.
, 
be the circumcircles of an acute 
and 
respectively. Let 
be the midpoint of ![$ [AC]$](http://latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
.![$ D\in [AB]$](http://latex.artofproblemsolving.com/4/9/5/495f45fe51200cd2ffec9a32db797abadb1c6a01.png)
and ![$ E\in [BC]$](http://latex.artofproblemsolving.com/a/c/c/accd6bac634f0b69b4099af39def05b80d1390f4.png)
which satisfy 
. Show that 
.
.
, i.e. the points 
, 
belong to the circle 
, 
of the segments ![$ [AA']$](http://latex.artofproblemsolving.com/0/6/9/069cac5ca343ab4b294172c52bf001038e5857a7.png)
, ![$ [CC']$](http://latex.artofproblemsolving.com/1/7/e/17e54c2e5ba3157e0d3c023e50337a9557bca056.png)
respectively. Observe that 
.
the quadrilateral 
is ciclically (the circumcenter is the midpoint of the segement ![$ [BT]$](http://latex.artofproblemsolving.com/c/a/7/ca719302eb5216efee157601decf2e94708da4ff.png)
).
obtain that the quadrilateral 
is cyclically.
from the point 
to the line 
passes through the center of the circumcircle of 
. In conclusion, 
.
and 
meet 
and 
at 
.
.
is cyclic, so 
since 
.
,
is cyclic. Now, notice that 
.
,
,
.
,
is cyclic, so 
.
,
.
,
![\[2d=2a+c-\frac{ab}{c}\]](http://latex.artofproblemsolving.com/6/8/b/68bfdb8e15f285609b564822c8a4a784c2847de6.png)
and![\[2e=a+2c-\frac{bc}{a}.\]](http://latex.artofproblemsolving.com/5/f/4/5f45125811c7bc5195485bf269de95f6a35dbafd.png)
and 
.
.
,
.
is cyclic. Since 
,
is also cyclic. Thus, we have 
,
as desired.
be midpoints of 
and the center of the 9-point circle of 
is the foot of perpendicular of 
on 
and symmetry WRT the internal bisector of 
pictures 
to 
then it pictures 
to 
so 
(
and 
then 
is an isosceles trapezoid so since 
then 
Likewise 
so 
is the circumcenter of 
.
.
lie on a circle. Let 
be the intersection of lines 
and 
be the intersection of lines 
and 
followed by reflection in the internal bisector of angle 
passes through the midpoint of 
we conclude that 
.
and we obtain 
.
Similarly, 
.
is the radical axis of 
.
denote the reflection of 
and 
are cyclic. So, it suffices to show 
and 
.
be the isogonal conjugate of 
are collinear or that 
goes through the circumcenter of 
,
be a quadrilateral with 
.
and 
.
denote midpoints of 
and let 
and 
.
and 
,
,
meet 
again at 
respectively; 
Observe that 
(analogously 
)
Hence by the Thales 
are midpoints of 
Next 
so 
are cyclic.
is perpendicular to 
as desired.
and 
be the feet of the altitudes from 
and 
respectively. Then it is clear that 
as the unit circle, setting 
,
.
and 
,
and 
.
.
is imaginary, which is clear since its conjugate is its negative.
