Question

El sistema de ecuaciones $\begin{aligned} &\log_{10}(2000xy) \\ &\quad {}- (\log_{10} x)(\log_{10} y) = 4 \end{aligned}$ $\log_{10}(2yz) - (\log_{10} y)(\log_{10} z) = 1$ $\log_{10}(zx) - (\log_{10} z)(\log_{10} x) = 0$ tiene dos soluciones $(x_1, y_1, z_1)$ y $(x_2, y_2, z_2).$ Halla $y_1 + y_2.$

The system of equations $\begin{aligned} &\log_{10}(2000xy) \\ &\quad {}- (\log_{10} x)(\log_{10} y) = 4 \end{aligned}$ $\log_{10}(2yz) - (\log_{10} y)(\log_{10} z) = 1$ $\log_{10}(zx) - (\log_{10} z)(\log_{10} x) = 0$ has two solutions $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2).$ Find $y_1 + y_2.$

Respuesta

Solución:

Sea $a = \log_{10} x,$ $b = \log_{10} y,$ $c = \log_{10} z.$ Usando $\log_{10} 2000 = 4 - \log_{10} 5,$ la primera ecuación se convierte en $a + b - ab = \log_{10} 5,$ que se factoriza como $(1 - a)(1 - b)$ $= 1 - \log_{10} 5$ $= \log_{10} 2.$ De forma similar, la segunda ecuación da $(1 - b)(1 - c)$ $= 1 - (1 - \log_{10} 2)$ $= \log_{10} 2,$ y la tercera da $(1 - c)(1 - a) = 1.$

Dividiendo las dos primeras (nota que $1 - b \ne 0$ ) se obtiene $1 - a = 1 - c,$ y entonces la tercera ecuación da $(1 - a)^2 = 1,$ así que $1 - a = \pm 1.$ Si $1 - a = 1,$ entonces $1 - b = \log_{10} 2,$ así que $b = 1 - \log_{10} 2 = \log_{10} 5$ e $y = 5$ (en efecto $(x, y, z) = (1, 5, 1)$ funciona). Si $1 - a = -1,$ entonces $1 - b = -\log_{10} 2,$ así que $b = \log_{10} 20$ e $y = 20$ (de $(x, y, z) = (100, 20, 100)$ ).

Por lo tanto, $y_1 + y_2 = 5 + 20 = 25.$

Let $a = \log_{10} x,$ $b = \log_{10} y,$ $c = \log_{10} z.$ Using $\log_{10} 2000 = 4 - \log_{10} 5,$ the first equation becomes $a + b - ab = \log_{10} 5,$ which factors as $(1 - a)(1 - b)$ $= 1 - \log_{10} 5$ $= \log_{10} 2.$ Similarly the second equation gives $(1 - b)(1 - c)$ $= 1 - (1 - \log_{10} 2)$ $= \log_{10} 2,$ and the third gives $(1 - c)(1 - a) = 1.$

Dividing the first two (note $1 - b \ne 0$ ) yields $1 - a = 1 - c,$ and then the third equation gives $(1 - a)^2 = 1,$ so $1 - a = \pm 1.$ If $1 - a = 1,$ then $1 - b = \log_{10} 2,$ so $b = 1 - \log_{10} 2 = \log_{10} 5$ and $y = 5$ (indeed $(x, y, z) = (1, 5, 1)$ works). If $1 - a = -1,$ then $1 - b = -\log_{10} 2,$ so $b = \log_{10} 20$ and $y = 20$ (from $(x, y, z) = (100, 20, 100)$ ).

Therefore $y_1 + y_2 = 5 + 20 = 25.$

2000 AIME I Problema 9

Solución:

El Problema 9 en otros años