be non-negative real numbers such that 
Prove that
and 
Find the minimum 
be a positive integer and 
be real numbers such that ![\[a_1+a_2+\dots+a_n=0.\]](http://latex.artofproblemsolving.com/0/f/d/0fde8d75cebc35103ae9acf8ebe393653d336d60.png)
Define the set 
by![\[A=\left\{(i, j)\,|\,1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\}\]](http://latex.artofproblemsolving.com/6/9/7/6974ebf20c44bcb4d52d85adc0805e28aafa5d6e.png)
Prove that, if ![\[\sum_{(i, j) \in A} a_{i} a_{j}<0.\]](http://latex.artofproblemsolving.com/b/e/3/be3e65a633dc2ac092a264968f1cf8514d3d6124.png)
Y by Purple_Planet, Adventure10
Two circles are internally tangent at 
. The chords 
and 
of the larger circle are tangent to the smaller circle at 
and 
respectively. 
and 
are midpoint of arcs 
and respectively. Circumcircles of triangles 
and 
are intersect at 
. Show that 
is a parallelogram.
. The chords 
and 
of the larger circle are tangent to the smaller circle at 
and 
respectively. 
and 
are midpoint of arcs 
and respectively. Circumcircles of triangles 
and 
are intersect at 
. Show that 
is a parallelogram.
and 
are straight lines. 
so 
is a straight line. 
thus 
.
and we are done.
The chords 
Show that 
NKQ collinear, NMP collinear.
to 
.
are collinear. By similar reasoning, 
are collinear.
,
,
are cyclic,![\[ 180-\angle{BLM}+180-\angle{BLK}=\angle{BPM}+\angle{BQK}=180\]](http://latex.artofproblemsolving.com/9/4/a/94adab4035ef7cea4002220f2b8fa7fcb4af3a76.png)
![\[\implies \angle{BLM}+\angle{BLK}=180,\]](http://latex.artofproblemsolving.com/4/6/3/4631e332272ce180365553ec94b5e7812e15da9d.png)
are collinear. We then have 
,
.
,
.
will denote the measure of arc 
of the larger circle.
and 
are sets of collinear points.
Similarly, 
and thus 
are also collinear points. Indeed, 
as desired. Hence 
We may now conclude, since both pairs of opposite angles of 
and 
are collinear. Then since 
,
and 
,
,
.
and 
follows similarly
.
.
since 
,
we get that 
are sent to the midpoints of the arcs 
,
are sent to 
.![\[
\frac{KN}{NM} = \frac{QN}{NP} = \frac{BQ}{BP} = \frac{BQ}{QL},
\]](http://latex.artofproblemsolving.com/2/c/d/2cdbb543de5647e58acc4f35a6d28169ac20d19e.png)
and so the similarity holds and we have 
so 
.
and we finish. 
cyclic quadrilateral
