AoPS Academy
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and 
, 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
is a triangle, let 
is orthocenter of 
, let 
is midpoint of 
. Let 
is a line perpendicular with 
at point . Let meet 
at 
respectively. Prove that 
.
be, the orthogonal projections of 
respectively, on the line segment 
and we denote the points 
from 
and 
and then, the points 
and 
are the reflexions of respectively, with respect to 
with 
as their orthocenters respectively and so, their distances from their homologous vertices, are equal.
and the proof is completed.
is a chord of a circle 
and 
is a point on 
Two distintc chords 
and 
pass through 
and 
cut at 
respectively. Then we have![\[ \frac{1}{PA}-\frac{1}{PX}=\frac{1}{PB}-\frac{1}{PY} \]](http://latex.artofproblemsolving.com/0/f/6/0f6e88f2cc6b36865336290a1ea0bfab7c9359f6.png)
the diametrically oposite point of 
in the circumcircle of 
. 
are collinear since by angle chasing 
is a parallelogram. Also, 
and 
are both cyclic (opposite angles equal 
, sum 
. Since 
this implies is angle bisector of 
then 
.
the diametrically oposite point of in the circumcircle of . are collinear since by angle chasing is a parallelogram. Also, and are both cyclic (opposite angles equal 
, sum . Since 
this implies is angle bisector of then 
.
cut 
at 
and 
cut 
at 
.
so the circle from Butterfly Theorem is of center 
.
and 
.
perpendicular to (d), we have that 
.
and 
,
.
