AoPS Academy
be a triangle with 
, 
, 
, and let 
be a point in its interior. If 
, 
, 
are the centroids of 
, 
, 
, respectively, find the maximum possible area of 
.
red balls and blue balls. A draw consists of randomly selecting 
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.
and 
for each positive integer 
, what is the value of 
?
and for each positive integer , what is the value of ?
. Thus, 
. If we let 
, then we want to find 
. We know 
, so our expression is equal to 
. We actually wanted to multiply this by 
, so finally, 
.
then 
and 
is rather than 
.
should be rather than 
...
. Thus, . If we let , then we want to find . We know , so our expression is equal to . We actually wanted to multiply this by , so finally, .
is if 
for coprime 
, for 
for prime p, 
. This should be enough to find the sum.
and 
,
in degrees as 
ranges over all real numbers.
and let 
(ie. 
with 
removed). Determine the number of injective* functions 
such that 
for all 
.
for all distinct 
and 
in the domain of 
possible triangles instead of 
)
be the centroid of 
.
as the centroid of 
and define 
and 
similarly. Points 
and 
,
,
,![$\frac{[G_AG_BG_c]}{[ABC]}.$](http://latex.artofproblemsolving.com/d/3/8/d38c1ec527c34f1e1519bba772a0765849e6fe7d.png)
chessboard, 
for 
prime and 
positive then 
because 
puts one more factor of 
,
,
are 
with 
.
,
(jump!), 
,
(jump!). It appears that 
for all primes 
and we have a jump otherwise. Eventually, we conjecture that
To prove this, consider 
to 
,
increases by a factor of 
also increases by a factor of 
due to the primeness of 
to 
,
,
factors of 
out of the 
,
,
.
which computes to be 
;
which computes to be 
.
.
,
Now, notice that in order for 
we must have 
positive integers. For example, 
but 
because 
is prime and thus "introduces" a new prime factor.
,
for a prime 
More generally,
Similarly to the first case,
where 
where 
Now we just have to count the numbers that fit into each category.
primes less than 
,
to our total. The cubes of primes less than 
and 
,
to our running total. There is 
:
to our running total. There is 
-
to our total. So far, our sum of "special" numbers is 
so 
,