2000 AMC 12 第 20 题

先试着解答 2000 AMC 12 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2000 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

若正数 xx、yy 和 zz 满足 x+1y=4,x + \frac{1}{y} = 4\text{,} y+1z=1,y + \frac{1}{z} = 1\text{,} 以及 z+1x=73,z + \frac{1}{x} = \frac{7}{3}\text{,} 则 xyzxyz 等于多少?
If x,x, y,y, and zz are positive numbers satisfying x+1y=4,x + \frac{1}{y} = 4, y+1z=1,y + \frac{1}{z} = 1, and z+1x=73,z + \frac{1}{x} = \frac{7}{3}, then what is xyz?xyz?
23\dfrac{2}{3}
11
43\dfrac{4}{3}
22
73\dfrac{7}{3}
答案:B
知识点:方程组代数变形对称性(代数)
小提示:
将三个方程相加,并另外将三个方程相乘。
Add the three equations, and separately multiply all three together
大提示:
展开乘积会得到 xyz+1xyzxyz + \dfrac{1}{xyz},再加上你已经算出的和。
Expanding the product produces xyz+1xyzxyz + \dfrac{1}{xyz} plus the sum you already computed
解答:
三个方程相加得 (x+1y)+(y+1z)+(z+1x)=4+1+73=223。 \begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}\text{。} \end{gathered}
三个方程相乘得 4⋅1⋅73=283。 4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}\text{。}
展开乘积,得到 (x+1y)⋅(y+1z)⋅(z+1x)=xyz+(x+y+z+1x+1y+1z)+1xyz。 \begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}\text{。} \end{aligned} 中间括号中的一组之和就是 223\tfrac{22}{3},所以 xyz+1xyz=283−223=2xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2。
因此 (xyz−1)2=0(xyz - 1)^2 = 0,所以 xyz=1xyz = 1。
因此,正确答案是 B。
Adding the three equations gives (x+1y)+(y+1z)+(z+1x)=4+1+73=223. \begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered}
Multiplying them gives 4⋅1⋅73=283. 4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.
Expanding the product, (x+1y)⋅(y+1z)⋅(z+1x)=xyz+(x+y+z+1x+1y+1z)+1xyz. \begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned} The middle group is the sum 223,\tfrac{22}{3}, so xyz+1xyz=283−223=2.xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.
Hence (xyz−1)2=0,(xyz - 1)^2 = 0, so xyz=1.xyz = 1.
Thus, the correct answer is B.
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