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 ![\[A=\left\{(i, j)\,|\,1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\}\]](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
Suppose that two circles 
with centers 
, respectively , intersect orthogonally at 
,
. Let 
be a diameter of 
that is exterior to 
. Let 
be points on such that 
are tangent to , with 
on one side of 
and 
on the other side of . Let 
be the intersection of 
and 
be the intersection of 
. Prove that 
is parallel to and is tangent to
with centers 
, respectively , intersect orthogonally at 
,
. Let 
be a diameter of 
that is exterior to 
. Let 
be points on such that 
are tangent to , with 
on one side of 
and 
on the other side of . Let 
be the intersection of 
and 
be the intersection of 
. Prove that 
is parallel to and is tangent to 
![[asy]
import graph; size(11.3cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.29, xmax = 5.24, ymin = -1.65, ymax = 2.88; /* image dimensions */
pen uququq = rgb(0.25,0.25,0.25); pen xdxdff = rgb(0.49,0.49,1);
pair P = (0,0), A = (0.76,0.66), B = (0.88,-0.48), Q = (1.21,0.13), C = (1.26,0.82), D = (1.16,-0.56), F = (0.43,-0.9), T = (0.98,-0.29), G = (-1,0.08), H = (0.83,-0.06), I = (1,-0.08), J = (-0.81,2.46);
/* draw figures */
draw(circle(P, 1));
draw(circle(Q, 0.69));
draw(D--F);
draw(C--(0.15,0.99));
draw((1.03,0.34)--T);
draw(D--(0.15,0.99));
draw(C--F);
draw(A--Q);
draw(B--Q);
draw(A--P);
draw(P--B);
draw(C--D);
draw(D--G);
draw(G--C);
draw(A--B);
draw(D--A);
draw(C--B);
draw(G--I);
draw(J--A);
draw(J--G);
draw((0.15,0.99)--J);
/* dots and labels */
dot(P,uququq);
label("$P$", (0.02,0.03), NE * labelscalefactor,uququq);
dot(A,xdxdff);
label("$A$", (0.77,0.68), NE * labelscalefactor,xdxdff);
dot(B,xdxdff);
label("$B$", (0.9,-0.44), NE * labelscalefactor,xdxdff);
dot(Q,uququq);
label("$Q$", (1.23,0.16), NE * labelscalefactor,uququq);
dot(C,xdxdff);
label("$C$", (1.28,0.85), NE * labelscalefactor,xdxdff);
dot(D,xdxdff);
label("$D$", (1.17,-0.53), NE * labelscalefactor,xdxdff);
dot((0.15,0.99),uququq);
label("$E$", (0.17,1.02), NE * labelscalefactor,uququq);
dot(F,uququq);
label("$F$", (0.45,-0.87), NE * labelscalefactor,uququq);
dot((1.03,0.34),uququq);
label("$S$", (1.05,0.37), NE * labelscalefactor,uququq);
dot(T,uququq);
label("$T$", (0.96,-0.42), NE * labelscalefactor,uququq);
dot(G,uququq);
label("$G$", (-0.98,0.11), NE * labelscalefactor,uququq);
dot(H,uququq);
label("$H$", (0.72,-0.13), NE * labelscalefactor,uququq);
dot(I,uququq);
label("$I$", (1.06,-0.08), NE * labelscalefactor,uququq);
dot(J,uququq);
label("$J$", (-0.79,2.49), NE * labelscalefactor,uququq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/0/2/f028b37fc09aa35b39aeddbcab801a4a8668bda0.png)
and 
so we get that 
is the orthocenter of 
so it also lies on the circle 
at 
.
are collinear so 
.
is perpendicular to both 
and 
.
and 
theorem but i wasn't able to complete that idea, unfortunately.
be a cyclic quadrilateral and let 
where by 
we mean the tangent to the circle at point 
are collinear.
and 
we get 
and 
are collinear so 
are collinear.
the polar line of 
with respect to 
because the midpoint of 
is on the radical axis of 
(because 
)
.
.
lies on 
.
is perpendicular to 
.
so because 
,
is tangent to 
and the magazine you are refering is from 2021.
.
but we have 
and 
but 
is cyclic so we are done.
collinear.
is right angled and 
.
.
then 
and thus 
.
collinear.
cyclic.
,
cyclic. Also, 
and 
collinear. Thus, 
cyclic. Thus, the claim follows.
and similarly 
.
for the parallel condition.
.
.
.
is diameter of 
.
.
(
)
and 
and we are done.
