AoPS Academy
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points 
and 
are taken so that 
and 
. Prove that 
.
and points and on sides 
and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
with
such that the following statement is true: real polynomials, 
such that the sum of any two of them have no real roots but the sum of any three does.
such that the following statement is true: real polynomials, such that the sum of any two of them have no real roots but the sum of any three does.
, with the construction being 
, which clearly work. Now we show that 
fails.
, define 
. Out of 
, we may WLOG assume that 
has the largest absolute value. Then 
and 
all have the same sign, again WLOG let it be positive.
were positive, say 
. Since each of 
are positive, and 
have no real roots, we conclude that ![\[P_1(x)+P_2(x) > 0, P_2(x) +P_3(x) >0,P_3(x) + P_1(x) > 0\]](http://latex.artofproblemsolving.com/3/8/1/381fdbc7081eb911f43f3ffbead4eb1ef7cf554e.png)
for all 
. Summing and dividing by 2, we find that 
for all , hence it has no real roots, contradiction.
are all negative, so we can use a similar argument to show that 
has no real roots, contradiction and we are done
