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.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to .
of positive integers such that for every central sequence 
there are infinitely many positive integers with 
.
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .
be a positive integer and let 
, 
be two sequences of positive real numbers. Suppose 
is a sequence of positive real numbers such that 
for all 
.
. Prove that ![\[
\left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2
\ge
\left( \frac{x_1+\dots+x_n}{n} \right)
\left( \frac{y_1+\dots+y_n}{n} \right). \]](http://latex.artofproblemsolving.com/8/c/4/8c4a72f57363eb1ef114825c8d36191bd5f42cb7.png)
be a triangle. On side 
take points 
and 
such that 
take points 
and such that 
and on side 
take points 
and 
such that 
Let 
and 
are concurrent.
that verify the following equation :
lie the following points: 
and 
on 
, 
on 
, and 
on 
. Let![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet at 
. Prove that the points 
, , and 
are collinear.
be inscribed in the circle 
. A line through point 
intersects 
and at points 
and 
, respectively. Let be the reflection of across the midpoint of , and be the reflection of across the midpoint of . Prove that:
of triangle across line 
lies on the circle .
and 
coincide.
be a triangle and 
and 
be two points on 
, 
such that 
and let 
. Let the parallel through 
to the interior angle bisector of 
intersect 
in 
. Prove that 
.
be an isosceles triangle with 
, and let 
be its incenter. Incircle touches sides 
at 
, respectively. Foot of altitudes from 
to are 
, respectively. Rays 
intersect 
at 
, respectively. Prove that 
touches incircle at 
.
be a triangle with incenter , incircle 
and circumcircle 
. Let 
be the midpoints of sides 
, 
, 
and let be the tangency points of with and , respectively. Let 
be the intersections of line 
with line 
and line 
, respectively, and let 
be the midpoint of arc 
of . lies on ray 
.
bisects 
.
, where both 
and 
are integers between 
and 
inclusive, such that 
.
can be factored as 
. Thus, we consider three cases: 
, 
, or both.
pairs. The second case yields 
pairs. The third case yields 
pair. This yields 
pairs.
is a solution then divide by 
, sub 
to get 
, so 
. Casework gives 
.
at the end
and 
can equal 
can equal 
so we have to subtract for overcount.
,assume some constant 
allows us to factor by grouping, thus the expression becomes 
or 
. In order to factor we want 
,we find 
and we can factor and solve as above
We get 
or 
so 
,assume some constant allows us to factor by grouping, thus the expression becomes or . In order to factor we want ,we find and we can factor and solve as above
, where both and are integers between and inclusive, such that .) this equation is equivalent to:
We now have two mutually exclusive cases:
. Since 
, we can constraint 
which rearranges to 
. Likewise, since 
, 
. All such 
with will produce a valid and unique 
satisfying the original equation, so it just remains to count the number of solutions for which is 
.
and 
and we proceed to bound . Since 
, we have 
, and similarly 
. All such 
with will produce a valid and unique solution for except for 
, which violates the constraint. Thus our answer for this case is (subtracting at the end so as not to count ) 
.
solutions.
, and solve over , we have![\[
x=\frac{-y\pm17y}{24}
\]](http://latex.artofproblemsolving.com/a/6/c/a6c330741d1e09ecbd859c435f9b3dbae664ebbc.png)
so either 
, or 
.
if , then we have 
, hence there are 
solutions. Similarly for , there are 
solutions.
as a solution, so the answer is 
.
,
or 
.
,
,
possible ordered pairs. If it is divisible by 
but not 
,
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
.
.
to 
.
.
,
The pairs of integers that satisfy this equation are 
and 
for 
For the first type of solution, we have 
so the number of solutions of that type are 
Then, for the second type, we get 
so the number of solutions of the second type is 
However, note that the solution 
So when we get that our first case is 3x+2y and 
that then we get 
but we over count (0,0) so we have 
then just find x and y and make sure not to double counted the zero 