# Difference between revisions of "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 $a+b$ has to be a factor of 8 and the only answer that's a factor of 8 is $\boxed{\textbf{(A)}\ 4}$

Solution by I_Dont_Do_Math

 2016 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 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