2016 AMC 12B Problems/Problem 10

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

Solution 2

Solution by e_power_pi_times_i

By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{\textbf{(A)}\ 4}$ .

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

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2016 AMC 12B Problems/Problem 10

Contents

Problem

Solution

Solution 2

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