be a parallelogram with 
.
,
,
intersects diagonal 
at points 
and 
with 
,
,
,
.
,
and 
are positive integers, and 
.
and 
intersect at points 
and 
.
is tangent to 
and 
,
closer to point 
passes through 
and intersecting 
.
,
are collinear, 
,
,
.
.
and 
intersect at 
,
intersect at 
,
and 
intersect at 
.
intersects line 
at 
intersects line 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
,
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
,
,
,
are collinear if and only if 
.
data values 
are all equal to 
.
plot 
distinct moves available; two from each point. Let 
count the number of ways to advance around the circle exactly twice, beginning and ending at 
for all 
.
Y by wu2481632, Davi-8191, samrocksnature, HWenslawski, math31415926535, jhu08, Adventure10, Mango247, ItsBesi, cubres, NicoN9
Find, with proof, the least integer 
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .This post has been edited 1 time. Last edited by NormanWho, Apr 20, 2016, 9:31 PM
.
basically match up pairs to make 
and gg.
.
,
,
is achievable. There are 
unordered pairs with sum 
of them. Those 
.
.![\[(1,6048), (2,6047), \ldots, (3024,3025),\]](http://latex.artofproblemsolving.com/3/c/6/3c605115e0b235fc3d163eeca788e427f14c7c3e.png)
.
(hopefully I typed that right, it's supposed to be 
)![\[2017+2018+\dots+4032=1008*6049,\]](http://latex.artofproblemsolving.com/2/f/5/2f54abf69dc0c24431ebcb670c5a4184c60904ab.png)
which we need when we remove the positive integers from 
to 
,
,
,
,
.
numbers and are looking for 
,
to 
.
.
from the set and this is the smallest sum we can get from the remaining elements. Now we prove this satisfies the condition. We note that there are 
.
such pairs left in the set, which allows us to sum 
,
.
as always possible 
N = 1008.6049.
.
.
.
.
.
.
,
which is 
and so we are done..
are removed than the least sum is 
Just like we wanted.
works. To prove this consider the pairs: 
Regardless of which 
are deleted, at least 
of these pairs have both elements remaining. Since each pair has sum 6049, we can take these pairs to be the desired numbers.
.
clearly follows from removing 
.
works, take the pairs 
.
remaining numbers summing to 
,
doesn't work. 
.
and 
that sum to 
By removing the first 
we see that 
Now, we prove validaty. If we don't remove any of 
we have that this will always work. Now, assume that we remove exactly 
is valid. For each removed 
from our sum, and replacing with a valid 
pair. Thus, we are done.
.
,
.
works. Consider the pairs 
.
