Question

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

$x \gt y$ 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 casework

Difficulty rating: 1400

Solution:

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 Problem 7

Answer: D

Difficulty rating: 1400

Solution:

Problem 7 in Other Years