Question

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 and $y=z$

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

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

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

$x+y+z=1$

Answer: D

Concepts:algebraic manipulation Diophantine Equation

Difficulty rating: 1370

Solution:

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 Problem 12

Answer: D

Difficulty rating: 1370

Solution:

Problem 12 in Other Years