HMMT-Feb 2013 alg, Problem 5

Let $a$ and $b$ be real numbers, and let $r, s$, and $t$ be the roots of $f(x)=x^{3}+a x^{2}+b x-1$. Also, $g(x)=x^{3}+m x^{2}+n x+p$ has roots $r^{2}, s^{2}$, and $t^{2}$. If $g(-1)=-5$, find the maximum possible value of $b$.

Solution: By Vieta's Formulae, $m=-\left(r^{2}+s^{2}+t^{2}\right)=-a^{2}+2 b, n=r^{2} s^{2}+s^{2} t^{2}+t^{2} r^{2}=$ $b^{2}+2 a$, and $p=-1$. Therefore, $g(-1)=-1-a^{2}+2 b-b^{2}-2 a-1=-5 \Leftrightarrow(a+1)^{2}+(b-1)^{2}=5$. This is an equation of a circle, so $b$ reaches its maximum when $a+1=0 \Rightarrow a=-1$. When $a=-1$, $b=1 \pm \sqrt{5}$, so the maximum is $1+\sqrt{5}$. $\fbox{1+\sqrt{5}}$.

Answer: 1+\sqrt{5}