Difference between revisions of "2016 AMC 12B Problems/Problem 6"
m |
m (→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | 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 angle formula you get: <math>64=\frac{1}{2}*2x*x^2 = 64=x^3</math> Making <math>*4</math>, and the length of <math>BC</math> is <math>2x</math>, so the answer is <math>8</math>. | + | 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 angle formula you get: <math>64=\frac{1}{2}*2x*x^2 = 64=x^3</math> Making <math>*4</math>, and the length of <math>BC</math> is <math>2x</math>, so the answer is <math>\boxed{\textbf{(D)} 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}} | ||
Revision as of 13:22, 21 February 2016
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
Plotting points 
and 
on the graph shows that they are at 
and 
, which is isosceles. By setting up the triangle angle formula you get: 
Making 
, and the length of is 
, so the answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
