2021 AMC 12B Fall Problem 7

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

Multiply by

2

: the left side equals

(x - y)^2 + (y - z)^2 + (z - x)^2

For integers, that sum equals

2

only when two differences are

\pm 1

and one is

0

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.

Problem 7 in other years