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

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

2016 AMC 12B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 ==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>
+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

Page

Toolbox

Search

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

Revision as of 16:27, 21 February 2016

Problem

Solution

See Also