2000 AIME I Problems/Problem 9

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Problem

The system of equations

$\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\\log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\\end{eqnarray*}$

has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ .

Solution

Since $\log ab = \log a + \log b$ , we can reduce the equations to a more recognizable form:

$\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\-\log y \log z + \log y + \log z - 1 &=& - \log 2\\-\log x \log z + \log x + \log z - 1 &=& -1\\\end{eqnarray*}$

Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT, the above equations become (*)

$\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\(b-1)(c-1) &=& \log 2 \\(a-1)(c-1) &=& 1 \end{eqnarray*}$

From here, multiplying the three equations gives

$\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\(a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}$

Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$ . This gives $y_1 = 20, y_2 = 5$ , and the answer is $y_1 + y_2 = \boxed{025}$ .

Alternatively, at (*), notice that the RHS of the first two equations are the same, so $a=c$ . Substituting this into the third equation gives $a=c=0,2$ , which if we solve backwards for $y$ will give us the same answer.

2000 AIME I (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2000 AIME I Problems/Problem 9

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Problem

Solution

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