2021 AMC 12A Problems/Problem 7: Difference between revisions

Latest revision as of 15:45, 8 November 2025

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

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

Remark: Honestly, this isn't even that overkill. I did it this way, and it was not very bashy. ~vaishnav

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

Yet another Video Solution

https://youtu.be/pUdeIBlp-y8

@@ Line 1: / Line 1: @@
+{{duplicate|[[2021 AMC 10A Problems/Problem 9|2021 AMC 10A #9]] and [[2021 AMC 12A Problems/Problem 7|2021 AMC 12A #7]]}}
 ==Problem==
 What is the least possible value of <math>(xy-1)^2+(x+y)^2</math> for real numbers <math>x</math> and <math>y</math>?
@@ Line 4: / Line 6: @@
 <math>\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2</math>
-==Solution==
+==Solution 1 (Expand)==
-Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014
+Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the Trivial Inequality (all squares are nonnegative) the minimum value for this is <math>\boxed{\textbf{(D)} ~1}</math>, which can be achieved at <math>x=y=0</math>.
+~aop2014
+==Solution 2 (Expand and then Factor)==
+We expand the original expression, then factor the result by grouping:
+<cmath>\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*}</cmath>
+Clearly, both factors are positive. By the Trivial Inequality, we have <cmath>\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{\textbf{(D)} ~1}.</cmath>
+Note that the least possible value of <math>(xy-1)^2+(x+y)^2</math> occurs at <math>x=y=0.</math>
+~MRENTHUSIASM
+==Solution 3 (Beyond Overkill)==
+Like solution 1, expand and simplify the original equation to <imath>x^2+y^2+x^2y^2+1</imath> and let <imath>f(x, y) = x^2+y^2+x^2y^2+1</imath>. To find local extrema, find where <imath>\nabla f(x, y) = \boldsymbol{0}</imath>. First, find the first partial derivative with respect to x and y and find where they are <imath>0</imath>:
+<cmath> \frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0</cmath>
+<cmath> \frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0</cmath>
+Thus, there is a local extremum at <imath>(0, 0)</imath>. 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 <imath>f(0, 0)</imath> is the minimum of <imath>f(x, y)</imath>. Plugging <imath>(0, 0)</imath> into <imath>f(x, y)</imath>, we find 1 <imath>\implies \boxed{\textbf{(D)} ~1}</imath>.
+~ DBlack2021
+Remark:
+Honestly, this isn't even that overkill. I did it this way, and it was not very bashy.
+~vaishnav
+==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
+==Yet another Video Solution==
+https://youtu.be/pUdeIBlp-y8
 ==See also==
+{{AMC10 box|year=2021|ab=A|num-b=8|num-a=10}}
 {{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

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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