Difference between revisions of "2013 AMC 12B Problems/Problem 21"
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So out of <math>2\dbinom{30}{2}</math> possible intersection points, only <math>2*5*2*\dbinom{3}{2}</math> fail to exist. This leaves <math>870-60=810=\boxed{\textbf{(C)} 810}</math> solutions. | So out of <math>2\dbinom{30}{2}</math> possible intersection points, only <math>2*5*2*\dbinom{3}{2}</math> fail to exist. This leaves <math>870-60=810=\boxed{\textbf{(C)} 810}</math> solutions. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=B|num-b=20|num-a=22}} | ||
Revision as of 17:14, 22 February 2013
Problem
Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form 
with a and b integers such that 
and 
. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D}}\ 840\qquad\textbf{(E)}\ 870$ (Error compiling LaTeX. Unknown error_msg)
Solution
Being on two parabolae means having the same distance from the common focus and both directrices. In particular, you have to be on an angle bisector of the directrices, and clearly on the same "side" of the directrices as the focus. So it's easy to see there are at most two solutions per pair of parabolae. Convexity and continuity imply exactly two solutions unless the directrices are parallel and on the same side of the focus.
So out of 
possible intersection points, only 
fail to exist. This leaves 
solutions.
See also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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 | |
