Question

Sean $a, b, c,$ y $d$ números reales que satisfacen el sistema de ecuaciones $\begin{aligned} a + b &= -3, \\ ab + bc + ca &= -4, \end{aligned}$ $\begin{aligned} abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{aligned}$ Existen enteros positivos primos entre sí $m$ y $n$ tales que $a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.$ Halle $m + n$ .

Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations $\begin{aligned} a + b &= -3, \\ ab + bc + ca &= -4, \end{aligned}$ $\begin{aligned} abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{aligned}$ There exist relatively prime positive integers $m$ and $n$ such that $a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.$ Find $m + n.$

Respuesta

Solución:

Como $a + b = -3$ , la segunda ecuación se lee $ab + c(a + b) = ab - 3c = -4$ , así que $ab = 3c - 4$ . Agrupar la tercera ecuación como $ab(c + d) + cd(a + b) = 14$ da $(3c - 4)(c + d) - 3cd = 14$ , que se simplifica a $3c^2 - 4c - 4d = 14$ , así que $d = \frac{3c^2 - 4c - 14}{4}$ . La cuarta ecuación se convierte en $(3c - 4)\,cd = 30$ .

Sustituyendo $d$ se obtiene $c(3c - 4)(3c^2 - 4c - 14) = 120$ , es decir $\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}$ El factor cuadrático tiene discriminante negativo, así que $c = -2$ o $c = \frac{10}{3}$ . Si $c = \frac{10}{3}$ , entonces $ab = 6$ con $a + b = -3$ , imposible para $a, b$ reales, ya que $9 - 24 \lt 0$ . Por lo tanto $c = -2$ , lo que da $ab = -10$ y $d = \frac{12 + 8 - 14}{4} = \frac{3}{2}$ .

Entonces $a^2 + b^2$ $= (a + b)^2 - 2ab$ $= 9 + 20 = 29$ y $c^2 + d^2 = 4 + \frac{9}{4} = \frac{25}{4}$ , así que $a^2 + b^2 + c^2 + d^2 = \frac{141}{4}$ y $m + n = 141 + 4 = 145$ .

Since $a + b = -3,$ the second equation reads $ab + c(a + b) = ab - 3c = -4,$ so $ab = 3c - 4.$ Grouping the third equation as $ab(c + d) + cd(a + b) = 14$ gives $(3c - 4)(c + d) - 3cd = 14,$ which simplifies to $3c^2 - 4c - 4d = 14,$ so $d = \frac{3c^2 - 4c - 14}{4}.$ The fourth equation becomes $(3c - 4)\,cd = 30.$

Substituting for $d$ yields $c(3c - 4)(3c^2 - 4c - 14) = 120,$ i.e. $\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}$ The quadratic factor has negative discriminant, so $c = -2$ or $c = \frac{10}{3}.$ If $c = \frac{10}{3},$ then $ab = 6$ with $a + b = -3,$ impossible for real $a, b$ since $9 - 24 \lt 0.$ So $c = -2,$ giving $ab = -10$ and $d = \frac{12 + 8 - 14}{4} = \frac{3}{2}.$

Then $a^2 + b^2$ $= (a + b)^2 - 2ab$ $= 9 + 20 = 29$ and $c^2 + d^2 = 4 + \frac{9}{4} = \frac{25}{4},$ so $a^2 + b^2 + c^2 + d^2 = \frac{141}{4}$ and $m + n = 141 + 4 = 145.$

2021 AIME II Problema 7

Solución:

El Problema 7 en otros años