Question

下列哪个条件足以保证整数 $x$ 、 $y$ 、 $z$ 满足方程？ $x(x-y)+y(y-z)+z(z-x)$ $= 1?$

Which of the following conditions is sufficient to guarantee that integers $x,$ $y,$ and $z$ satisfy the equation $x(x-y)+y(y-z)+z(z-x)$ $= 1?$

x > y 且 $y=z$

x > 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：代数变形丢番图方程

难度评级：1370

解答：

展开并重写： $E=x(x-y)+y(y-z)$ $+z(z-x)$ 。 $\begin{aligned} 2E={}&(x-y)^2+(y-z)^2\\ &+(z-x)^2. \end{aligned}$

要使值为 $1$ ，三个非负平方项的和必须为 $2$ 。因为 $x,y,z$ 是整数，这三个平方项只能是 $1,1,0$ 。

因此两个变量必须相等，第三个变量与它们相差 $1$ 。条件 $x=z$ 且 $y-1=x$ 正好保证这一点。

所以正确答案是 D。

Let $E=x(x-y)+y(y-z)$ $+z(z-x).$ Expanding gives $\begin{aligned} 2E={}&(x-y)^2+(y-z)^2\\ &+(z-x)^2. \end{aligned}$

For the value to be $1,$ the three nonnegative square terms must sum to $2.$ Since $x,y,z$ are integers, this means the squared differences are $1,1,0.$

Thus two of the variables must be equal, and the third must differ from them by $1.$ The condition $x=z$ and $y-1=x$ guarantees exactly that.

Thus, the answer is D .

2021 AMC 10B Fall 第 12 题

答案：D

难度评级：1370

解答：

其他年份的第 12 题