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> | ||
− | ==Solution 1== | + | == Solution 1 (Completing the Square)== |
+ | |||
+ | 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. | ||
+ | |||
+ | <math>x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1</math> | ||
+ | |||
+ | <math>2x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2</math> | ||
+ | |||
+ | <math>(x-y)^2 + (y-z)^2 + (z-x)^2 = 2</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==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 | ||
− | ==Solution | + | ==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 | ||
− | ==Solution | + | ==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 | ~KingRavi | ||
− | |||
− | |||
− | ~ | + | ==Video Solution (Just 2 min!)== |
+ | https://youtu.be/HeHu_ZlXi8E | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | ==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== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | {{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 21:53, 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.
Contents
Problem
Which of the following conditions is sufficient to guarantee that integers ,
, and
satisfy the equation
and
and
and
and
Solution 1 (Completing the Square)
It is obvious ,
, and
are symmetrical. We are going to solve the problem by Completing the Square.
Because are integers,
,
, and
can only equal
. So one variable must equal another, and the third variable is
different from those
equal variables. So the answer is
.
Solution 2
Plugging in every choice, we see that choice works.
We have , so
Our answer is
.
~kingofpineapplz
Solution 3 (Bash)
Just plug in all these options one by one, and one sees that all but fails to satisfy the equation.
For , substitute
and
:
Hence the answer is
~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 and
means the equation becomes
, which is always true, so the answer is
~KingRavi
Video Solution (Just 2 min!)
~Education, the Study of Everything
Video Solution by Interstigation
https://www.youtube.com/watch?v=lJ-RHZXPV_E
Video Solution by WhyMath
~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.