AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
were selected 
pairwise distinct points, abscissas of which are integer numbers from the segment ![$[0; 100000]$](http://latex.artofproblemsolving.com/0/5/e/05e1d2855ef6a6dfaf9b24f2713455ab1017a30d.png)
. Prove that it is possible to choose six different selected points 
, 
, 
, 
, 
, 
such that areas of triangles 
and 
are equals.
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-regular graph that contains no triangles or quadrilaterals, there exists a coloring using 
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
+1 w
, a single parity swapping is to choose 
terms in this sequence, say 
and 
, such that 
is odd, then switch their placement, while the other terms stay in place. This creates a new sequence.
to 
, using only single parity swapping.
be an odd prime and 
is a bijection and is an integer such that 
Show that 
is a multiple of .
be a positive integer. Show that there exist a polynomial 
with integer coefficient that satisfy the following is at most 
for each 
,such that the inequality 
holds for all positive reals 
and for every positive real 
, there exist positive reals 
, such that the equality holds.
be a polynomial where 
and 
are coprime integers in the set 
, and suppose it satisfies the following condition: there exists at most one prime 
such that for every positive integer 
, 
. Prove that for every prime 
there exists a positive integer for which 
be a triangle with 
. On the side 
we consider points 
such that 
and 
. Let 
be the circumcenter of triangle 
and
, 
be the points of intersection of the lines 
, 
and 
, 
respectively. Let 
be
, 
the circle with center and radius 
, and 
the circle and radius 
., , and pass through the same point and that this point of intersection lies on the line 
.
be positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
be a function satisfying the following conditions for all 
:![\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]](http://latex.artofproblemsolving.com/0/4/7/047e28c9d899d7de443ae40ac2c756fda851fafb.png)
Find the value of 
.
is the foot of the altitude from of triangle . On the lines 
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.
board with a number written in each cell. Every two neighbour rows sum up to at least 
, and every two neighbour columns sum up to at most 
. Find the sum of all numbers on the board.
where ![$a \in [10, 16]$](http://latex.artofproblemsolving.com/2/3/7/2379f0f94d6dcb54024b80fa1bdce0e7baaf7b53.png)
fits this condition which means the sum can be anywhere in ![$ [20, 32]$](http://latex.artofproblemsolving.com/7/4/3/743b32f68a688d26582878dbdf576921cfb3dcd5.png)
. There might be even loser restrictions too but I don't actually know how to solve this mathematically.
where fits this condition which means the sum can be anywhere in . There might be even loser restrictions too but I don't actually know how to solve this mathematically.
we get that total sum is at least 
. By taking pairs of columns without some odd column we get that sums in columns are 
. So the total sum is .
.
.![\[\sum ab +\sum ab^2 \le 6\]](http://latex.artofproblemsolving.com/9/f/6/9f632ab1e6df5029bcd4416eb06623123ad2cd8c.png)
board, not a 
board.