Difference between revisions of "2021 AMC 12A Problems/Problem 7"

Revision as of 21:05, 28 October 2022

The following problem is from both the 2021 AMC 10A #9 and 2021 AMC 12A #7, so both problems redirect to this page.

1 Problem
2 Solution 1 (Expand)
3 Solution 2 (Expand and then Factor)
4 Solution 3 (Beyond Overkill)
5 Video Solution (Simple & Quick)
6 Video Solution by Aaron He (Trivial Inequality)
7 Video Solution by North America Math Contest Go Go Go (Trivial Inequality, Simon's Favourite Packing Theorem)
8 Video Solution by Hawk Math
9 Video Solution by OmegaLearn(Trivial Inequality, Simon's Favorite Factoring)
10 Video Solution 6
11 Video Solution by TheBeautyofMath
12 Video Solution by The Learning Royal
13 See also

Problem

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

$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution 1 (Expand)

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{\textbf{(D)} ~1}$ , which can be achieved at $x=y=0$ .

~aop2014

Solution 2 (Expand and then Factor)

We expand the original expression, then factor the result by grouping: $\begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right). \end{align*}$ Clearly, both factors are positive. By the Trivial Inequality, we have $\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{\textbf{(D)} ~1}.$ Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$

~MRENTHUSIASM

Solution 3 (Beyond Overkill)

Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ : $\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0$ $\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0$ Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{\textbf{(D)} ~1}$ .

~ DBlack2021

Video Solution (Simple & Quick)

https://youtu.be/2CZ1u4J9yk4

~ Education, the Study of Everything

Video Solution by Aaron He (Trivial Inequality)

https://www.youtube.com/watch?v=xTGDKBthWsw&t=6m58s

Video Solution by North America Math Contest Go Go Go (Trivial Inequality, Simon's Favourite Packing Theorem)

https://www.youtube.com/watch?v=PbJK4KKfQjY&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=8

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn(Trivial Inequality, Simon's Favorite Factoring)

https://youtu.be/DP0ppuQzFPE

~ pi_is_3.14

Video Solution 6

https://youtu.be/hmOGYmVmY1c

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=640 (for AMC 10A)

https://youtu.be/cckGBU2x1zg?t=95 (for AMC 12A)

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

@@ Line 46: / Line 46: @@
 https://www.youtube.com/watch?v=P5al76DxyHY
-== Video Solution (Trivial Inequality, Simon's Favorite Factoring) ==
+== Video Solution by OmegaLearn(Trivial Inequality, Simon's Favorite Factoring) ==
 https://youtu.be/DP0ppuQzFPE

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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