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2016 AMC 10B Problems/Problem 9

Revision as of 16:04, 24 March 2023 by Trex226 (talk | contribs) (→‎Solution)

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 $\frac{(2r)(r^2)}{2} = r^3$, so $r^3=64\implies r=4$, giving a total distance across the top of $8$, which is answer $\textbf{(C)}$.

Solution 2 (Guess and Check)

Let the point where the height of the triangle intersects with the base be $D$. Now we can guess what $x$ is and find $y$. If $x$ is $3$, then $y$ is $9$. The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$, respectively. The distance between $B$ and $C$ is $6$, meaning the area would be $\frac{6 \cdot 9}{2}=27$, not $64$. Now we let $x=4$. $y$ would be $16$. The cords of $B$ and $C$ would be $(-4,16)$ and $(4,16)$, respectively. $BC$ would be $8$, and the height would be $16$. The area would then be $\frac{8 \cdot 16}{2}$ which is $64$, so $BC$ is $\boxed{\textbf{(C)}\ 8}$.

Video Solution

https://youtu.be/1pi0eiD3jHc

~savannahsolver

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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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