[/quote]
,
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.
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 
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.
and incircle touching 
be the antipode 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 TheBarioBario, ImSh95
Let 
be a right-angled triangle with 
. Let the height from 
cut its side
at 
. Let 
be the incenters of triangles 
respectively. Let also
be the excenters of with respect to vertices 
and 
respectively. If 
is the
point of intersection of the circumcircles of
and 
, show that 
passes through
the midpoint
of side .
be a right-angled triangle with 
. Let the height from 
cut its side
at 
. Let 
be the incenters of triangles 
respectively. Let also
be the excenters of with respect to vertices 
and 
respectively. If 
is thepoint of intersection of the circumcircles of
and 
, show that 
passes throughthe midpoint
of side .
,
and 
being the circumcenter of 
,
which is symmetric with respect to 
,
and hence 
is cyclic. As 
,
,
(with 
)
.
to the computation of nicer segments, note that by the Cosine Law in triangle 
we have (with 
and 
)
.
)
Now from 
we have 
and from 
with 
we get 
and so the above expression becomes
(from the angle bisector formula), 
,
,
,
the main expression becomes
To argue that the latter is symmetric with respect to 
and multiply with 
and so work instead with
.
define:
which is a linear function. We then have 
and also, using 
we get:
Then we have, by linearity:
We thus have:
which is symmetric under 
so 
which is what we wanted.
and 
be centers of 
and 
and Let 
and 
be their radius.
and 
are cyclic.
Note that 
.
.
so 
or 
and Note that 
and 
and 
then just expand both sides and we're Done.
be the excenters of 
and 
,
.
at 
and 
.
to 
.
As 
,
.
.
are concyclic (they lie of circle with diameter 
)
are concyclic. Now ignore all points in the figure and only look at points
Invert at 
be points on sides 
,
concur. If 
.
and 
respectively. we want to prove that 
lies on the radical axis of the two circles, we do the following. for a point 
we define 
we know that 
is linear. this means 
= 
= 
= 
similarly we get 
= 
we wish to prove 
or 
we know 
is cyclic. we also know 
.
let 
= 
by similarity we know 
meaning 
cyclic this means 
where 
is the intersection of 
and 
which we get by simple Angle chasing. this indicates 
meaning 
cyclic. since 
is a trapezoid. we get 
= 
which we know equals 
by antiparallel lines 
and 
