[/quote]
,
and 
. Prove that
. Prove that 
and 
. Prove that
Where 
![$ a,b\in [0,1] $](http://latex.artofproblemsolving.com/9/7/5/975edfc0cf5316d78283b313c6b5cfbdd7b92b6e.png)
. Prove that
Where 
Where 
where 
are lengths of parallel sides and 
are other two sides.
be real numbers satisfying 
and ![\[a_{i+1} - 2a_i + a_{i-1} = a^2_i\]](http://latex.artofproblemsolving.com/c/d/8/cd878e4ac230853182da75c54dcbf7be05b305bc.png)
for 
Prove that 
for 
,
letters 
can be placed in each cell of a table with 
columns, and has the following property: For any two distinct rows in the table, there exists at most one column, such that the entries of these two rows in such a column are the same letter.
be an continuous function defined in ![$\text{[0,2015]},f(0)=f(2015)$](http://latex.artofproblemsolving.com/c/9/9/c99f63e726b504aea8c1a65ba2926155a873effa.png)
pairs of 
such that 
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
.
and 
,
be the point on line 
different from 
such that the line through 
is tangent to 
be the point on line 
such that the line through 
is tangent to 
intersect the circumcircle of triangle 
.
and 
be the midpoints of 
.
Y by wiseman, Adventure10
Given a nonequilateral triangle 
, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles 
(the vertices listed counterclockwise) for which the triples of points 
and 
are collinear.
Proposed by Poland.
, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles 
(the vertices listed counterclockwise) for which the triples of points 
and 
are collinear.Proposed by Poland.
of 
meet at the Miquel point of 
i.e. the 1st Fermat point 
of 
Let 
be the midpoint of 
running on midcircle 
of 
and let 
with 
midpoint of its arc 
Since 
bisects 
it follows that 
is the external bisector of 
and 
be the centroids of 
and 
Since 
then 
But 
are in the same ratio 
coaxal with 
but from 
we deduce that 
Therefore, locus of 
and 
concur at the Fermat point 
then 
cut 
of its arcs 
these are none other than the reflections of 
on 
Thus 
is the inner equilateral Napoleon triangle of 
whose center is the centroid 
since 
and 
The conclusion follows.
.
such that 
and 
passes through 
respectively, is a circle centered at the centroid 
passing through 
are the circumcircles of 
.
Now let 
traverses a distance equal to 
on the perimeter of 
and makes the point 
)
and 
with 
and 
as 
respectively, we obviously have 
If 
be the centroid of 
then 
is the rotatated vector of 
with angle 
(
and 
are also traversing clockwise). So the length of 
.
be the midpoint of 
,
be the anti-pedal triangle of 
WRT 
.
be the centroid of 
is the midpoint of 
,
are similar,
pass through a fixed point 
.
pass through a fixed point 
,
are concurrent at 
,
are concyclic .
passing through 
.
is the diameter of 
,
is the projection of 
.
is the projection of 
,
is the diameter of 
is the center of 
orbit on a circle with center 
that passes through 
My proof is exactly the same what I did in my 1st post.
cuts 
at 
and cuts 
again at 
we have 
Powers of 
are in constant ratio 
is the intersection 
that verifies 
Now, powers of 
because 
and passing through 
be first Fermat point of 
)
)
)
,
,
)
,
,
of 
)
)
+ 
)
)
)
denote the midpoints of arcs 
of circles 
respectively. Then, we claim that the locus 
of all the possible centroids of 
is all the points on 
except point 
meet at 
.
be a point on 
and let it intersect 
respectively. We will prove that 
are collinear, from which it follows that 
and 
.
which implies 
are collinear. Therefore, 
