Problem:
Triangle ABC is inscribed in a circle of radius 2 with ∠ABC≥90∘, and x is a real number satisfying the equation x4+ax3+bx2+cx+1=0, where a=BC,b=CA,c=AB. Find all possible values of x.
Solution:
The given equation can be rewritten as
(x2+2ax​)2+(b−4a2+c2​)x2+(2cx​+1)2=0
Noting that we must have xî€ =0, the equation holds if and only if
b=4a2+c2​ and x=−2a​=−c2​
The assumption ∠ABC≥90∘ and the fact that the circle's diameter is 4 imply a2+c2≤b2≤4b; but since we saw that b=(a2+c2)/4, both of these inequalities are equalities. We conclude that ∠ABC=90∘,b=4,a2+c2=16, and ac=4. These last two equations imply (a+c)2=16+2⋅4=24 and (a−c)2=16−2⋅4=8. Since a,c>0, we have a+c=26​ and a−c=±22​. Hence the only possible values of x=−a/2 are −21​(6​+2​) or −21​(6​−2​). Conversely, these are indeed possible, by having a right triangle with sides a=6​+2​,b=4,c=6​−2​ or a=6​−2​,b=4,c=6​+2​, respectively.
Remark. One can also show that the acute angles of the triangle are 15 degrees and 75 degrees.