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.
on the diagonal 
of a given rectangle 
, such that 
. The point 
is the intersection point between 
and the parallel line to 
that contains 
. Prove that the triangle 
is equilateral if and only if is a square.
![$P \in \mathbb R[x,y]$](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.
be a positive integer. We are given a 
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 for which, using a finite number of moves, we can make all the squares which were initially black white, and all squares which were initially white black.
-polygon can be constructed from any other triangulation by at most 
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 
?
cities numbered 
to . City 
and city 
has a bidirectional road of lengths 
if and only if 
and 
is a prime number. dollar to fix, and Laura wants to minimize the cost. Prove that Laura would have one unique way to do this.
has incentre 
and incircle touching 
at 
. Let 
be the antipode of 
in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
for which:
for all 
.
particles on a circle situated at the vertices of a regular -gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.
be the smallest number of collisions after which all particles return to their original positions. Find .
and its permutations. For construction, use 
rectangles repeteadly.
-cells. Call 
rotated trominos of any tromino 
of tromino.
rectangles.-cell. If it is not the tromino, the other trominos we add will make 
square without 
cells, but we have to occupy these too, then using more trominos, we can't make -cells 
since adding a tromino makes at least another -cell, hence we must have -cell from start, which is square, hence conclusion.
for which an 
rectangle can be tiled with (possibly rotated) L-shaped trominos.
