2021 AMC 10A Spring Problem 9

Attempt Problem 9 of the 2021 AMC 10A Spring below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Spring solutions, or check the answer key.

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

9.

What is the least possible value of (xy1)2+(x+y)2(xy-1)^2+(x+y)^2 for real numbers xx and y?y?

00

14\dfrac{1}{4}

12\dfrac{1}{2}

11

22

Answer: D
Concepts:optimizationalgebraic manipulation
Difficulty rating: 770
Video solution:
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Written solution:

Expanding, we get x2y22xy+1+x2+2xy+y2=x2y2+x2+y2+1. \begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gathered} Note that every square must be non-negative. Therefore, the minimum value is when all the terms except 11 are 0,0, making the sum 1.1.

This is attainable when x=y=0.x = y = 0.

Thus, D is the correct answer.

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