,
be the sum of the digits of 
such that there are infinite positive integers 
.
,
be fixed positive integers. There are 
ducks sitting in a
be the circumcenter of triangle 
intersects the altitude from 
at point 
.
be the midpoints of 
,
respectively. If 
intersects 
at 
,
meets 
at 
,
is cyclic.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
such that ![$\lfloor \sqrt[3]{n}\rfloor\mid n$](http://latex.artofproblemsolving.com/5/6/2/562cedff73965e1dce784a360514b931e36158b1.png)
.
red balls and 
of the remaining balls from bag without replacement, and then setting them aside. A draw is called a match if the two balls have the same color. Compute the expected number of draws until either a match occurs or the bag is empty.
where 
.
but none of my attempts worked and the fractions looked increasingly complex. I don't think finding an explicit form is possible (I would love to be disproven), so I decided to look at the relation asymptotically.
.
tends to 
,
is a fixed point. This is also obvious when looking at the original problem, since the probability of drawing a match gets closer to 
.
.
to any function.
?
,
is the greatest integer less than or equal to 
is prime?
Prove that
Let 
Prove that
be complex numbers. For a complex number 
where 
,
to be the distance of 
from the origin, or 
.
be the minimum value and 
be the maximum value of![\[
\frac{ |a + b| + |b + c| + |c + a| }{ |a| + |b| + |c| }
\]](http://latex.artofproblemsolving.com/1/7/5/1752b55a5bca8490d3dbb398fb2c4e4f9f71f47c.png)
for all complex numbers 
.
.
.
and 
are added, their sum 
is also a vector. Geometrically, this is represented by placing the tail of 
.
be the same distance from the origin as 
Substituting this in the given equation, we get
.
.
![\[
x^2 + 2|x| - 6x + 15 = 0.
\]](http://latex.artofproblemsolving.com/1/1/d/11dbf7a605e8c49bb730eb0bd3737019b277fa41.png)
(Note: the modulus of a complex number 
is 
.
and 
.
and 
.
.
![\[
\frac{ab + bc + ca}{a^2 + b^2 + c^2} + \frac{1}{6} \left( \frac{(a - b)^2}{a^2 + b^2} + \frac{(b - c)^2}{b^2 + c^2} + \frac{(c - a)^2}{c^2 + a^2} \right) \leq 1.
\]](http://latex.artofproblemsolving.com/d/8/b/d8b3869addff6715d86b73083c77395d64ab1995.png)
are the 256 roots (not necessarily distinct) of the equation 
,
.
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 
be the 2nd intersection of 
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 
be centroid 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, 
