Difference between revisions of "2016 AMC 12B Problems/Problem 10"

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==Solution==
 
==Solution==
By distance formula we have <math>(a-b)^2+(b-a)^2*2*(a+b)^2 = 256</math>. SImplifying we get <math>(a-b)(a+b) = 8</math>. Thus <math>a+b</math> has to be a factor of 8 and the only answer that's a factor of 8 is <math>\boxed{\textbf{(A)}\ 4}</math>
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By distance formula we have <math>(a-b)^2+(b-a)^2*2*(a+b)^2 = 256</math>. SImplifying we get <math>(a-b)(a+b) = 8</math>. Thus <math>a+b</math> and <math>a-b</math> have to be a factor of 8. The only way for them to be factors of <math>8</math> and remain integers is if <math>a+b = 4</math> and <math>a-b = 2</math>. So the answer is <math>\boxed{\textbf{(A)}\ 4}</math>
  
 
Solution by I_Dont_Do_Math
 
Solution by I_Dont_Do_Math

Revision as of 16:27, 21 February 2016

Problem

A quadrilateral has vertices $P(a,b)$, $Q(b,a)$, $R(-a, -b)$, and $S(-b, -a)$, where $a$ and $b$ are integers with $a>b>0$. The area of $PQRS$ is $16$. What is $a+b$?

$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ 13$

Solution

By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$. SImplifying we get $(a-b)(a+b) = 8$. Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$. So the answer is $\boxed{\textbf{(A)}\ 4}$

Solution by I_Dont_Do_Math

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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