[/quote]
,
for which 
is the square of an integer
of natural numbers with 
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.
and 
be on segment 
of an acute triangle 
such that 
and 
.
be the points on 
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 
and let 
be the midpoint of 
.
be the foot of the altitude from 
and let 
be the centroid of the triangle 
be a circle through 
.
and 
are collinear.
![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
Y by Adventure10, Mango247
Parallelogram 
with obtuse angle 
is given. After rotation of the triangle 
around the vertex 
, we get a triangle 
, such that points 
and 
are collinear. Extensions of median of triangle that passes through intersects the straight line 
at point 
. Prove that 
is the bisector of the angle 
.
with obtuse angle 
is given. After rotation of the triangle 
around the vertex 
, we get a triangle 
, such that points 
and 
are collinear. Extensions of median of triangle that passes through intersects the straight line 
at point 
. Prove that 
is the bisector of the angle 
.This post has been edited 1 time. Last edited by parmenides51, Dec 16, 2022, 9:31 AM
.
is cyclic. 
-
=
.
