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

(Created page with "==Problem== A quadrilateral has vertices <math>P(a,b)</math>, <math>Q(b,a)</math>, <math>R(-a, -b)</math>, and <math>S(-b, -a)</math>, where <math>a</math> and <math>b</math>...")
 
m (Video Solution by TheBeautyofMath)
 
(7 intermediate revisions by 4 users not shown)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
 +
Note that the slope of <math>PQ</math> is <math>\frac{a-b}{b-a}=-1</math> and the slope of <math>PS</math> is <math>\frac{b+a}{a+b}=1</math>.  Hence, <math>PQ\perp PS</math> and we can similarly prove that the other angles are right angles.  This means that <math>PQRS</math> is a rectangle. 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 2==
 +
Solution by e_power_pi_times_i
 +
 +
 +
By the Shoelace Theorem, the area of the quadrilateral is <math>2a^2 - 2b^2</math>, so <math>a^2 - b^2 = 8</math>. Since <math>a</math> and <math>b</math> are integers, <math>a = 3</math> and <math>b = 1</math>, so <math>a + b = \boxed{\textbf{(A)}\ 4}</math>.
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://www.youtube.com/watch?v=Eq2A2TTahqU
 +
with a second example of Shoelace Theorem done after this problem
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:08, 19 February 2021

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

Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$. Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. 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 $a + b = \boxed{\textbf{(A)}\ 4}$.

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=Eq2A2TTahqU with a second example of Shoelace Theorem done after this problem

~IceMatrix

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

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

Invalid username
Login to AoPS