Difference between revisions of "2016 AMC 12B Problems/Problem 6"

(Solution)
m (Solution)
 
Line 8: Line 8:
 
By: Albert471
 
By: Albert471
  
Plotting points <math>B</math> and <math>C</math> on the graph shows that they are at <math>\left( -x,x^2\right)</math> and <math>\left( x,x^2\right)</math>, which is isosceles. By setting up the triangle area formula you get: <math>64=\frac{1}{2}*2x*x^2  =  64=x^3</math> Making x=4, and the length of <math>BC</math> is <math>2x</math>, so the answer is <math>\boxed{\textbf{(C)} 8}</math>.
+
Plotting points <math>B</math> and <math>C</math> on the graph shows that they are at <math>\left( -x,x^2\right)</math> and <math>\left( x,x^2\right)</math>, which is isosceles. By setting up the triangle area formula you get: <math>64=\frac{1}{2}*2x*x^2  =  64=x^3</math> Making x=4, and the length of <math>BC</math> is <math>2x</math>, so the answer is <math>\boxed{\textbf{(C)}\ 8}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:47, 4 August 2017

Problem

All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

Solution

By: Albert471

Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$, which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2  =   64=x^3$ Making x=4, and the length of $BC$ is $2x$, so the answer is $\boxed{\textbf{(C)}\ 8}$.

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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