2021 AMC 12A Spring Problem 7

Attempt Problem 7 of the 2021 AMC 12A Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

What is the least possible value of

(xy - 1)^2 + (x + y)^2

for real numbers

x

and

y?

0

\dfrac14

\dfrac12

1

2

Expand fully; the cross terms

-2xy

and

+2xy

cancel

The expression equals

(x^2 + 1)(y^2 + 1),

a product of two factors each at least

1

Expanding,

\begin{aligned} &(xy-1)^2 + (x+y)^2 \\ &= x^2y^2 - 2xy + 1 + x^2 \\ &\quad {}+ 2xy + y^2 \\ &= x^2y^2 + x^2 + y^2 + 1. \end{aligned}

This factors as

(x^2 + 1)(y^2 + 1).

Each factor is at least

1,

so the product is at least

1,

with equality when

x = y = 0.

Thus, the correct answer is D.

Problem 7 in other years