[/quote]
,
on the diagonal 
of a given rectangle 
,
.
is the intersection point between 
and the parallel line to 
that contains 
.
is equilateral if and only if 
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
checkered board is cut into several squares with integer side length. The size of a square is it's side length.
such that 
is divisible by 
,
is divisible by 
,
is divisible by 
.
?
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
board whose unit squares are colored in black and white in such way that starting with the top left square, every third diagonal is colored in black and the rest of the board is in white. In one move, one can take a 
square and change the color of all its squares in such way that white squares become orange, orange ones become black and black ones become white. Find all 
flips (if 
)
![$Q\in \{-1,1,0\}[x]$](http://latex.artofproblemsolving.com/b/4/0/b404c8d6415aaf3d4442cc5381f582c1f8f8b996.png)
with 
.![$\exists P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/2/e/a2e67dd74df9971ca90d1a5d81c1c8e1f3982276.png)
palindrom with 
?
to 
and city 
has a bidirectional road of lengths 
if and only if 
and 
is a prime number.
has incentre 
and incircle touching 
at 
.
be the antipode of 
in the circumcircle of 
is chosen on the internal angle bisector of 
such that 
.
,
be the midpoint of 
.
for which:
for all 
.
particles on a circle situated at the vertices of a regular 
be the smallest number of collisions after which all particles return to their original positions. Find 
Y by adityaguharoy
There is a board with the shape of an equilateral triangle with side 
divided into triangular cells with the shape of equilateral triangles with side 
(the figure below shows the board when 
). Each and every triangular cell is colored either red or blue. What is the least number of cells that can be colored blue without two red cells sharing one side?
divided into triangular cells with the shape of equilateral triangles with side 
(the figure below shows the board when 
). Each and every triangular cell is colored either red or blue. What is the least number of cells that can be colored blue without two red cells sharing one side?This post has been edited 1 time. Last edited by parmenides51, Nov 14, 2021, 8:33 AM
cells. So there should be at least 
a. And for the minimum case there is constraction like this: In any row write b,a,b,a,...,a,b. So the answer is 
where 
and 
.
just like the solution above. So, in each row of lenght 
there must necesseraly be 
red triangles and 
blue triangles. Now since each row is a consecutive odd number, there are exactly 
triangles in a big triangle of side lenght 
red triangles and 
blue triangles in each configuartion. More precisely you can divide the proof into the cases when 
for each integer 
and integer 
