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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 7"
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<math>\textbf{(E)} \: x+y+z=1</math>
 
<math>\textbf{(E)} \: x+y+z=1</math>
  
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==Solution 1==
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== Solution 1 (Completing the Square)==
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It is obvious <math>x</math>, <math>y</math>, and <math>z</math> are symmetrical. We are going to solve the problem by Completing the Square.
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<math>x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1</math>
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<math>2x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2</math>
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<math>(x-y)^2 + (y-z)^2 + (z-x)^2 = 2</math>
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Because <math>x, y, z</math> are integers, <math>(x-y)^2</math>, <math>(y-z)^2</math>, and <math>(z-x)^2</math> can only equal <math>0, 1, 1</math>. So one variable must equal another, and the third variable is <math>1</math> different from those <math>2</math> equal variables. So the answer is <math>\boxed{D}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 2==
 
Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.  
 
Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.  
  
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~kingofpineapplz
 
~kingofpineapplz
  
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==Solution 2 (Bash) ==
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==Solution 3 (Bash) ==
  
 
Just plug in all these options one by one, and one sees that all but <math>D</math> fails to satisfy the equation.
 
Just plug in all these options one by one, and one sees that all but <math>D</math> fails to satisfy the equation.
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~Wilhelm Z
 
~Wilhelm Z
  
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==Solution 3 (Strategy)==
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==Solution 4 (Strategy)==
  
 
Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that <math>x=z</math> and <math>y-1=x</math> means the equation becomes <math>x(x-(x+1)) + (x+1)(x+1 - x)  = 1 \implies -x + x + 1 = 1 \implies 1 =  1</math>, which is always true, so the answer is <math>\boxed{D}</math>
 
Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that <math>x=z</math> and <math>y-1=x</math> means the equation becomes <math>x(x-(x+1)) + (x+1)(x+1 - x)  = 1 \implies -x + x + 1 = 1 \implies 1 =  1</math>, which is always true, so the answer is <math>\boxed{D}</math>
  
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~KingRavi
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==Video Solution (Just 2 min!)==
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https://youtu.be/HeHu_ZlXi8E
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<i> ~Education, the Study of Everything </i>
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==Video Solution by Interstigation==
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https://www.youtube.com/watch?v=lJ-RHZXPV_E
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==Video Solution by WhyMath==
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https://youtu.be/gp0xqp5BsfY
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/R7TwXgAGYuw?t=236
 +
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For AMC 12: https://www.youtube.com/watch?v=4qgYrCYG-qw&t=392
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~IceMatrix
  
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==See Also==
 +
{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:53, 10 July 2023

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

Problem

Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation \[x(x-y)+y(y-z)+z(z-x) = 1?\]

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

Solution 1 (Completing the Square)

It is obvious $x$, $y$, and $z$ are symmetrical. We are going to solve the problem by Completing the Square.

$x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1$

$2x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2$

$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2$

Because $x, y, z$ are integers, $(x-y)^2$, $(y-z)^2$, and $(z-x)^2$ can only equal $0, 1, 1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. So the answer is $\boxed{D}$.

~isabelchen

Solution 2

Plugging in every choice, we see that choice $\textbf{(D)}$ works.


We have $y=x+1, z=x$, so \[x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.\] Our answer is $\textbf{(D)}$.

~kingofpineapplz

Solution 3 (Bash)

Just plug in all these options one by one, and one sees that all but $D$ fails to satisfy the equation.

For $D$, substitute $z=x$ and $y=x+1$:

$LHS=x(x-(x+1))+(x+1)(x+1-x)+x(x-x)=(-x)+(x+1)=1=RHS$

Hence the answer is $\boxed{\textbf{(D)}}.$

~Wilhelm Z

Solution 4 (Strategy)

Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that $x=z$ and $y-1=x$ means the equation becomes $x(x-(x+1)) + (x+1)(x+1 - x) = 1 \implies -x + x + 1 = 1 \implies 1 = 1$, which is always true, so the answer is $\boxed{D}$

~KingRavi


Video Solution (Just 2 min!)

https://youtu.be/HeHu_ZlXi8E

~Education, the Study of Everything

Video Solution by Interstigation

https://www.youtube.com/watch?v=lJ-RHZXPV_E

Video Solution by WhyMath

https://youtu.be/gp0xqp5BsfY

~savannahsolver

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/R7TwXgAGYuw?t=236

For AMC 12: https://www.youtube.com/watch?v=4qgYrCYG-qw&t=392

~IceMatrix

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 Fall AMC 12B (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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