Difference between revisions of "2016 AMC 10B Problems/Problem 9"

(Solution)
(Solution)
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==Solution==
 
==Solution==
Let the point in the first quadrant be <math>(a, a^2)</math>.
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<asy>import graph;size(7cm,IgnoreAspect);
Then, the area of the triangle is <math>\frac{2a\cdot a^2}{2}=a^3</math>.
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real f(real x) {return x*x;}
Solving the equation <math>a^3=64</math> for <math>a</math>, we get <math>a=4</math>, so <math>BC=2a=8</math>.
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draw((0,0)--(4,16)--(-4,16)--cycle,blue);
So the answer is <math>\boxed{\textbf{(C)}\ 8}</math>.
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draw(graph(f,-5,5,operator ..),gray);
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xaxis("$x$");yaxis("$y$",-1);
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label("$y=x^2$",(4.5,20.25),E);
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draw((4.2,0)--(4.2,16),Arrows);
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label("$r^2$",(4.2,0)--(4.2,16),E);
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draw((0,17)--(4,17),Arrows);
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label("$r$",(0,17)--(4,17),N);
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</asy>
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The area of the triangle is <math>r^3</math>, so <math>r^3=64\implies r=4</math>, giving a total distance across the top of <math>8</math>, which is answer <math>\textbf{(C)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:05, 21 February 2016

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

[asy]import graph;size(7cm,IgnoreAspect); real f(real x) {return x*x;} draw((0,0)--(4,16)--(-4,16)--cycle,blue); draw(graph(f,-5,5,operator ..),gray); xaxis("$x$");yaxis("$y$",-1); label("$y=x^2$",(4.5,20.25),E); draw((4.2,0)--(4.2,16),Arrows); label("$r^2$",(4.2,0)--(4.2,16),E); draw((0,17)--(4,17),Arrows); label("$r$",(0,17)--(4,17),N); [/asy] The area of the triangle is $r^3$, so $r^3=64\implies r=4$, giving a total distance across the top of $8$, which is answer $\textbf{(C)}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 10 Problems and Solutions

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