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

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{{duplicate|[[2021 AMC 10A Problems#Problem 9|2021 AMC 10A #9]] and [[2021 AMC 12A Problems#Problem 7|2021 AMC 12A #7]]}}
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{{duplicate|[[2021 AMC 10A Problems/Problem 9|2021 AMC 10A #9]] and [[2021 AMC 12A Problems/Problem 7|2021 AMC 12A #7]]}}
  
 
==Problem==
 
==Problem==
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<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>
 
<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>
  
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==Solution 1==
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==Solution 1 (Expand)==
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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
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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>.  
  
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==Solution 2 (Beyond Overkill)==
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~aop2014
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Like solution 1, expand and simplify the original equation to <math>x^2+y^2+x^2y^2+1</math> and let <math>f(x, y) = x^2+y^2+x^2y^2+1</math>. To find local extrema, find where <math>\nabla f(x, y) = \boldsymbol{0}</math>. First, find the first partial derivative with respect to x and y and find where they are <math>0</math>:
 
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<cmath> \frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0</cmath>
 
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<cmath> \frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0</cmath>
 
  
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Thus, there is a local extreme at <math>(0, 0)</math>. Because this is the only extreme, 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 <math>f(0, 0)</math> is the minimum of <math>f(x, y)</math>. Plugging <math>(0, 0)</math> into <math>f(x, y)</math>, we find 1 <math>\implies \boxed{\textbf{(D)} ~1}</math>
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==Solution 2 (Expand and then Factor)==
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~ DBlack2021
 
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==Solution 3 (Expand then Factor)==
 
 
We expand the original expression, then factor the result by grouping:
 
We expand the original expression, then factor the result by grouping:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
==Solution 3 (Beyond Overkill)==
 +
Like solution 1, expand and simplify the original equation to <math>x^2+y^2+x^2y^2+1</math> and let <math>f(x, y) = x^2+y^2+x^2y^2+1</math>. To find local extrema, find where <math>\nabla f(x, y) = \boldsymbol{0}</math>. First, find the first partial derivative with respect to x and y and find where they are <math>0</math>:
 +
<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 <math>(0, 0)</math>. 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 <math>f(0, 0)</math> is the minimum of <math>f(x, y)</math>. Plugging <math>(0, 0)</math> into <math>f(x, y)</math>, we find 1 <math>\implies \boxed{\textbf{(D)} ~1}</math>.
 +
 +
~ DBlack2021
  
 
==Video Solution (Simple & Quick)==
 
==Video Solution (Simple & Quick)==
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https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
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== Video Solution (Trivial Inequality, Simon's Favorite Factoring) ==
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== Video Solution by OmegaLearn (Trivial Inequality, Simon's Favorite Factoring) ==
 
https://youtu.be/DP0ppuQzFPE
 
https://youtu.be/DP0ppuQzFPE
  
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~IceMatrix
 
~IceMatrix
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 +
==Video Solution by The Learning Royal==
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https://youtu.be/AWjOeBFyeb4
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==Yet another Video Solution==
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https://youtu.be/pUdeIBlp-y8
  
 
==See also==
 
==See also==

Latest revision as of 11:12, 25 December 2023

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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