Question

下列哪一个条件足以保证整数 $x$ 、 $y$ 、 $z$ 满足方程 $\begin{aligned} &x(x - y) + y(y - z) \\ &\quad {}+ z(z - x) = 1? \end{aligned}$

Which of the following conditions is sufficient to guarantee that integers $x,$ $y,$ and $z$ satisfy the equation $\begin{aligned} &x(x - y) + y(y - z) \\ &\quad {}+ z(z - x) = 1? \end{aligned}$

$x \gt y$ 且 $y = z$

$x \gt y$ and $y = z$

$x = y - 1$ 且 $y = z - 1$

$x = y - 1$ and $y = z - 1$

$x = z + 1$ 且 $y = x + 1$

$x = z + 1$ and $y = x + 1$

$x = z$ 且 $y - 1 = x$

$x = z$ and $y - 1 = x$

$x + y + z = 1$

答案：D

知识点：代数变形分类讨论

难度评级：1400

解答：

有恒等式所以原方程成立恰好当这个平方和等于 $2$ 。 $\small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{aligned}$

由于三个差的和为 $0$ ，这要求其中两个为 $\pm 1$ ，一个为 $0$ 。

选项 D 给出 $z - x = 0$ 、 $x - y = -1$ 、 $y - z = 1$ ，平方和为 $1 + 1 + 0 = 2$ 。因此对所有满足该条件的整数都成立。

所以正确答案是 D。

The expression satisfies $\small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{aligned}$ So the equation holds exactly when this sum of squares equals $2.$

Since the three differences sum to $0,$ this requires two of them to be $\pm 1$ and one to be $0.$

Option D gives $z - x = 0,$ $x - y = -1,$ and $y - z = 1,$ so the squares are $1 + 1 + 0 = 2.$ This works for all such integers.

Thus, the correct answer is D.

2021 AMC 12B Fall 第 7 题

答案：D

难度评级：1400

解答：

其他年份的第 7 题