HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

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 $b$ .

Your answer

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

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 $1+\sqrt{5}$ .

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