[/quote]
,
at 
and 
meet at 
.
meets sides 
and 
again at 
and 
respectively. Let 
be the circumcentre of 
is perpendicular to 
.
for which 
is the square of an integer
of 
if the product of the elements of 
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
be on segment 
of an acute triangle 
such that 
and 
.
and 
,
and 
.
and 
is on the circumference of triangle 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
.
)
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
,
:
,
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
.
be the midpoint of 
be the midpoint of 
and let 
be the centroid of the triangle 
be a circle through 
.
and 
Y by ACCCGS8, Amir Hossein, andria, Adventure10
Let 
be a point in the plane and let 
be a circle which does not contain . Two distinct variable lines 
and 
through meet the circle at points 
and 
, and 
and 
, respectively. Let 
and 
be the antipodes of in the circles 
and 
, respectively. Prove that the line 
passes through a fixed point.
Mihai Chis
be a point in the plane and let 
be a circle which does not contain . Two distinct variable lines 
and 
through meet the circle at points 
and 
, and 
and 
, respectively. Let 
and 
be the antipodes of in the circles 
and 
, respectively. Prove that the line 
passes through a fixed point.Mihai Chis
be the centers of 
and 
respectively. 
and 
are the perpendicular bisectors of 
and 
Since 
and 
are antiparallel WRT 
it follows that 
Similarly, 
Hence 
and 
is a parallelogram 
passes through the midpoint of 
Then, the image 
goes through 
and 
lie in that order.
be the midpoint of 
,
be the foot of the perpendicular from 
,
and 
and 
.
.
as 
is a rectangle and Midpoint Theorem. Similarly, if 
is the midpoint of 
and 
is the point on 
,
.
so the perpendicular bisectors of 
.
.
are conlinear.
.
.
are conlinear.
be the polar of 
WRT 
.
be the projection of 
be the center of 
be the projection of 
,
we get 
pass through 
we get 
passes through 
and 
and 
are right angles, 
must be a rectangle. Similarly, by intersecting lines 
and 
with 
and 
reveals another rectangle 
.
is symmetric with respect to 
