Difference between revisions of "2016 AMC 10B Problems/Problem 9"
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<math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math> | <math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math> | ||
| + | |||
| + | ==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 <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== | ==Video Solution== | ||
Latest revision as of 11:45, 2 July 2023
Contents
Problem
All three vertices of 
lie on the parabola defined by 
, with 
at the origin and 
parallel to the 
-axis. The area of the triangle is 
. What is the length of 
?
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]](http://latex.artofproblemsolving.com/f/f/1/ff1ff1c3607f4680df88baa2e03b81634391f450.png)
The area of the triangle is 
, so 
, giving a total distance across the top of 
, which is answer 
.
Solution 2 (Guess and Check)
Let the point where the height of the triangle intersects with the base be 
. Now we can guess what is and find 
. If is 
, then is 
. The cords of 
and 
would be 
and 
, respectively. The distance between and is 
, meaning the area would be 
, not . Now we let 
. would be 
. The cords of and would be 
and 
, respectively. would be , and the height would be . The area would then be 
which is , so is 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
