Difference between revisions of "2016 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
| + | Plotting points B and C on the graph shows that they are at (-x,x^2) and (x,x^2), which is isosceles. By setting up the triangle angle formula you get: 64=1/2*2x*x^2 = 64=x^3 Making x 4, and the length of BC is 2x, so the answer is 8. | ||
==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 12:34, 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 B and C on the graph shows that they are at (-x,x^2) and (x,x^2), which is isosceles. By setting up the triangle angle formula you get: 64=1/2*2x*x^2 = 64=x^3 Making x 4, and the length of BC is 2x, so the answer is 8.
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. 
