Difference between revisions of "2013 AMC 12B Problems/Problem 6"
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+ | {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #6]] and [[2013 AMC 10B Problems|2013 AMC 10B #11]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>. What is <math>x + y</math>? | + | Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2+y^2=10x-6y-34</math>. What is <math>x+y</math>? |
+ | |||
+ | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | If we move every term dependent on <math>x</math> or <math>y</math> to the LHS, we get <math>x^2 - 10x + y^2 + 6y = -34</math>. Adding <math>34</math> to both sides, we have <math>x^2 - 10x + y^2 + 6y + 34 = 0</math>. We can split the <math>34</math> into <math>25</math> and <math>9</math> to get <math>(x - 5)^2 + (y + 3)^2 = 0</math>. Notice this is a circle with radius <math>0</math>, which only contains one point. So, the only point is <math>(5, -3)</math>, so the sum is <math>5 + (-3) = 2 \implies \boxed{\textbf{(B)}}</math>. ~ asdf334 | ||
+ | |||
+ | ==Solution 2== | ||
+ | If we move every term including <math>x</math> or <math>y</math> to the LHS, we get <cmath>x^2 - 10x + y^2 + 6y = -34.</cmath> We can complete the square to find that this equation becomes <cmath>(x - 5)^2 + (y + 3)^2 = 0.</cmath> Since the square of any real number is nonnegative, we know that the sum is greater than or equal to <math>0</math>. Equality holds when the value inside the parhentheses is equal to <math>0</math>. We find that <cmath>(x,y) = (5,-3)</cmath> and the sum we are looking for is <cmath>5+(-3)=2 \implies \boxed{\textbf{(B)}}.</cmath> - Honestly | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ba6w1OhXqOQ?t=1810 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/a-3CAo4CoWc | ||
+ | |||
+ | –no one | ||
+ | |||
+ | ==See Also== | ||
− | + | {{AMC12 box|year=2013|ab=B|num-b=5|num-a=7}} | |
+ | {{AMC10 box|year=2013|ab=B|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:04, 14 July 2021
- The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.
Problem
Real numbers and
satisfy the equation
. What is
?
Solution 1
If we move every term dependent on or
to the LHS, we get
. Adding
to both sides, we have
. We can split the
into
and
to get
. Notice this is a circle with radius
, which only contains one point. So, the only point is
, so the sum is
. ~ asdf334
Solution 2
If we move every term including or
to the LHS, we get
We can complete the square to find that this equation becomes
Since the square of any real number is nonnegative, we know that the sum is greater than or equal to
. Equality holds when the value inside the parhentheses is equal to
. We find that
and the sum we are looking for is
- Honestly
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1810
~ pi_is_3.14
Video Solution
–no one
See Also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.