2016 AMC 10B Problems/Problem 9: Difference between revisions

Latest revision as of 16:57, 18 October 2025

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 (CREATIVE THINKING)

https://youtu.be/pSJkO6kQGOs

~Education, the Study of Everything

Video Solution

https://youtu.be/1pi0eiD3jHc

~savannahsolver

@@ Line 6: / Line 6: @@
 ==Solution==
-Let the point in the first quadrant be <math>(a, a^2)</math>.
+<asy>import graph;size(7cm,IgnoreAspect);
-Then, the area of the triangle is <math>\frac{2a\cdot a^2}{2}=a^3</math>.
+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>.
+draw((0,0)--(4,16)--(-4,16)--cycle,blue);
-So the answer  is <math>\boxed{(C) 8}</math>.
+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 <math>\frac{(2r)(r^2)}{2} = 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>.
+==Solution 2 (Guess and Check)==
+Let the point where the height of the triangle intersects with the base be <math>D</math>. Now we can guess what <math>x</math> is and find <math>y</math>. If <math>x</math> is <math>3</math>, then <math>y</math> is <math>9</math>. The cords of <math>B</math> and <math>C</math> would be <math>(-3,9)</math> and <math>(3,9)</math>, respectively. The distance between <math>B</math> and <math>C</math> is <math>6</math>, meaning the area would be <math>\frac{6 \cdot 9}{2}=27</math>, not <math>64</math>. Now we let <math>x=4</math>. <math>y</math> would be <math>16</math>. The cords of <math>B</math> and <math>C</math> would be <math>(-4,16)</math> and  <math>(4,16)</math>, respectively. <math>BC</math> would be <math>8</math>, and the height would be <math>16</math>. The area would then be <math>\frac{8 \cdot 16}{2}</math> which is <math>64</math>, so <math>BC</math> is <math>\boxed{\textbf{(C)}\ 8}</math>.
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/pSJkO6kQGOs
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/1pi0eiD3jHc
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}
+[[Category: Introductory Geometry Problems]]

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