2021 AMC 12B Fall Problem 7

Below is the professionally curated solution for Problem 7 of the 2021 AMC 12B Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

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Concepts:algebraic manipulation casework

Difficulty rating: 1400

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

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

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

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

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

$x + y + z = 1$

Solution:

The expression satisfies $2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] = (x-y)^2 + (y-z)^2 + (z-x)^2.$ 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.

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