AIME 2000 I · 第 9 题

Solution

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

$$\begin{aligned} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{aligned}$$ Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT, the above equations become (*)

$$\begin{aligned}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1 \end{aligned}$$ Small note from different author: $-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.$

From here, multiplying the three equations gives

$$\begin{aligned}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{aligned}$$ Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$. (Note from different author if you are confused on this step: if $\pm$ is positive then $\log y = \log 2 + 1 = \log 2 + \log 10 = \log 20,$ so $y=20.$ if $\pm$ is negative then $\log y = 1 - \log 2 = \log 10 - \log 2 = \log 5,$ so $y=5.$) This gives $y_1 = 20, y_2 = 5$, and the answer is $y_1 + y_2 = \boxed{025}$.

Solution 2

Subtracting the second equation from the first equation yields

$$\begin{aligned} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{aligned}$$ If $1-\log y=0$ then $y=10$. Substituting into the first equation yields $\log20000=4$ which is not possible.

If $\log\frac{x}{z}=0$ then $\frac{x}{z}=1\Longrightarrow x=z$. Substituting into the third equation gets

$$\begin{aligned} \log x^2-(\log x)(\log x) &= 0 \\ \log x^2-\log x^x &= 0 \\ \log x^{2-x} &= 0 \\ x^{2-x} &= 1 \\ \end{aligned}$$ Thus either $x=1$ or $2-x=0\Longrightarrow x=2$. (Note that here $x\neq-1$ since logarithm isn't defined for negative number.)

Substituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$, respectively. Thus $y_1+y_2=\boxed{025}$.

~ Nafer

Solution 3

Let $a = \log x$, $b = \log y$ and $c = \log z$. Then the given equations become:

$$\begin{aligned} \log 2 + a + b - ab = 1 \\ \log 2 + b + c - bc = 1 \\ a+c = ac \\ \end{aligned}$$ Equating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \implies{a = c}$. Plugging this result into the third equation, we get $c = 0$ or $2$. Substituting each of these values of $c$ into the second equation, we get $b = 1 - \log 2$ and $b = 1 + \log 2$. Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$, respectively, so our answer is $\boxed{025}$.

~ anellipticcurveoverq

Solution 4

All logs are base 10 by convention. Rearrange the given statements:

$$\begin{aligned} \log 2000 + \log x + \log y - \log x \log y = 4, \quad \text{which becomes} \quad \log x + \log y = \log x \log y + \log 5. \\ \log 2 + \log y + \log z - \log y \log z = 1, \quad \text{which becomes} \quad \log y + \log z = \log y \log z + \log 5. \\ \log z + \log x - \log z \log x = 0, \quad \text{which becomes} \quad \log x + \log z = \log z \log x. \\ \end{aligned}$$ Subtract the first two equations to obtain $(\log x - \log z) = \log y (\log x - \log z).$

This must mean $\log x = \log z$ because otherwise $\log y = 1$ turns the first equation into $\log y = \log 5$ which is self-contradictory.

With $\log x = \log z$ we know each value satisfies $2a = a^2$, so they are both $0$ or both $2$. Finally we arrive at our two solutions, where 0 gives us $0 + \log y = 0 + \log 5$, and $y = 5$, and 2 gives us $2 + \log y = 2 \log y + \log 5$, and $\log y = 2 - \log 5 = \log 20$, so $y = 20$. Similar to above we arrive at $\boxed{025}$.

~ GrindOlympiads

Video solution

https://www.youtube.com/watch?v=sOyLnGJjVvc&t