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Algebra Difficulty 5.2 AIME, harder Find the answer

Let aa and bb be real numbers, and let r,sr, s, and tt be the roots of f(x)=x3+ax2+bx−1f(x)=x^{3}+a x^{2}+b x-1. Also, g(x)=x3+mx2+nx+pg(x)=x^{3}+m x^{2}+n x+p has roots r2,s2r^{2}, s^{2}, and t2t^{2}. If g(−1)=−5g(-1)=-5, find the maximum possible value of bb.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By Vieta's Formulae, m=−(r2+s2+t2)=−a2+2b,n=r2s2+s2t2+t2r2=b2+2am=-\left(r^{2}+s^{2}+t^{2}\right)=-a^{2}+2 b, n=r^{2} s^{2}+s^{2} t^{2}+t^{2} r^{2}= b^{2}+2 a, and p=−1p=-1. Therefore, g(−1)=−1−a2+2b−b2−2a−1=−5⇔(a+1)2+(b−1)2=5g(-1)=-1-a^{2}+2 b-b^{2}-2 a-1=-5 \Leftrightarrow(a+1)^{2}+(b-1)^{2}=5. This is an equation of a circle, so bb reaches its maximum when a+1=0⇒a=−1a+1=0 \Rightarrow a=-1. When a=−1a=-1, b=1±5b=1 \pm \sqrt{5}, so the maximum is 1+51+\sqrt{5}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.