Question

Los números reales $r$ y $s$ son raíces de $p(x) = x^3 + ax + b,$ y $r + 4$ y $s - 3$ son raíces de $q(x) = x^3 + ax + b + 240.$ Halla la suma de todos los valores posibles de $|b|.$

Real numbers $r$ and $s$ are roots of $p(x) = x^3 + ax + b,$ and $r + 4$ and $s - 3$ are roots of $q(x) = x^3 + ax + b + 240.$ Find the sum of all possible values of $|b|.$

Respuesta

Solución:

Ambas cúbicas tienen coeficiente $x^2$ igual a cero, así que sus raíces suman $0:$ la tercera raíz de $p$ es $t = -r - s,$ y la tercera raíz de $q$ es $-(r+4) - (s-3) = t - 1.$ El coeficiente de $x$ es $a$ en ambas, así que $\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4), \end{aligned}$ lo cual se simplifica a $t = 4r - 3s + 13.$

Los términos constantes dan $b = -rst$ y $b + 240 =$ $-(r+4)(s-3)(t-1),$ así que $240 =$ $rst - (r+4)(s-3)(t-1),$ es decir $rs - 4st + 3tr - 3r$ $+ 4s + 12t - 252 = 0.$ Sustituir $t = 4r - 3s + 13$ reduce esto a $12\left[(r-s)^2 + 7(r-s) - 8\right] = 0,$ así que $r - s = 1$ o $r - s = -8.$

Si $r - s = 1,$ entonces $t = 4r - 3s + 13 = r + 16$ y $t = -r - s = -2r + 1,$ así que $r = -5:$ las raíces son $-5,$ $-6,$ $11,$ y $b = -rst = -330.$ Si $r - s = -8,$ entonces $t = r - 11 = -2r - 8,$ así que $r = 1:$ las raíces son $1,$ $9,$ $-10,$ y $b = 90.$ La suma pedida es $330 + 90 = 420.$

Both cubics have zero $x^2$ coefficient, so their roots sum to $0:$ the third root of $p$ is $t = -r - s,$ and the third root of $q$ is $-(r+4) - (s-3) = t - 1.$ The coefficient of $x$ is $a$ in both, so $\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4), \end{aligned}$ which simplifies to $t = 4r - 3s + 13.$

The constant terms give $b = -rst$ and $b + 240 =$ $-(r+4)(s-3)(t-1),$ so $240 =$ $rst - (r+4)(s-3)(t-1),$ i.e. $rs - 4st + 3tr - 3r$ $+ 4s + 12t - 252 = 0.$ Substituting $t = 4r - 3s + 13$ reduces this to $12\left[(r-s)^2 + 7(r-s) - 8\right] = 0,$ so $r - s = 1$ or $r - s = -8.$

If $r - s = 1,$ then $t = 4r - 3s + 13 = r + 16$ and $t = -r - s = -2r + 1,$ so $r = -5:$ the roots are $-5,$ $-6,$ $11,$ and $b = -rst = -330.$ If $r - s = -8,$ then $t = r - 11 = -2r - 8,$ so $r = 1:$ the roots are $1,$ $9,$ $-10,$ and $b = 90.$ The requested sum is $330 + 90 = 420.$

2014 AIME II Problema 5

Solución:

El Problema 5 en otros años