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

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==Solution==
 
==Solution==
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Plotting points B and C on the graph shows that they are at (-x,x^2) and (x,x^2), which is isosceles. By setting up the triangle angle formula you get: 64=1/2*2x*x^2 = 64=x^3 Making x 4, and the length of BC is 2x, so the answer is 8.
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By: Albert471
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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 17:47, 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 (Problems • Answer Key • Resources)
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

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