Question

Supón que $x,$ $y,$ y $z$ son números complejos tales que $xy = -80 - 320i,$ $yz = 60,$ y $zx = -96 + 24i,$ donde $i = \sqrt{-1}.$ Entonces existen números reales $a$ y $b$ tales que $x + y + z = a + bi.$ Halla $a^2 + b^2.$

Suppose that $x,$ $y,$ and $z$ are complex numbers such that $xy = -80 - 320i,$ $yz = 60,$ and $zx = -96 + 24i,$ where $i = \sqrt{-1}.$ Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi.$ Find $a^2 + b^2.$

Respuesta

Solución:

Multiplicando las tres ecuaciones se obtiene $\begin{aligned} (xyz)^2 &= (-80 - 320i)(60) \\ &\quad {}\cdot (-96 + 24i) \\ &= 80 \cdot 60 \cdot 24 \\ &\quad {}\cdot \,(-1 - 4i)(-4 + i) \\ &= 240^2 (16 + 30i). \end{aligned}$ Como $16 + 30i = (5 + 3i)^2,$ obtenemos $xyz = \pm 240(5 + 3i).$

Dividiendo $xyz$ entre cada producto dado se obtiene $\begin{aligned} x &= \frac{xyz}{yz} = \pm(20 + 12i), \\ y &= \frac{xyz}{zx} = \pm(-10 - 10i), \\ z &= \frac{xyz}{xy} = \pm(-3 + 3i), \end{aligned}$ con signos coincidentes. Por lo tanto $x + y + z = \pm(7 + 5i),$ así que $(a, b) = \pm(7, 5)$ y $a^2 + b^2 = 49 + 25 = 74.$

Multiplying the three equations gives $\begin{aligned} (xyz)^2 &= (-80 - 320i)(60) \\ &\quad {}\cdot (-96 + 24i) \\ &= 80 \cdot 60 \cdot 24 \\ &\quad {}\cdot \,(-1 - 4i)(-4 + i) \\ &= 240^2 (16 + 30i). \end{aligned}$ Since $16 + 30i = (5 + 3i)^2,$ we get $xyz = \pm 240(5 + 3i).$

Dividing $xyz$ by each given product yields $\begin{aligned} x &= \frac{xyz}{yz} = \pm(20 + 12i), \\ y &= \frac{xyz}{zx} = \pm(-10 - 10i), \\ z &= \frac{xyz}{xy} = \pm(-3 + 3i), \end{aligned}$ with matching signs. Hence $x + y + z = \pm(7 + 5i),$ so $(a, b) = \pm(7, 5)$ and $a^2 + b^2 = 49 + 25 = 74.$

2018 AIME II Problema 5

Solución:

El Problema 5 en otros años