Difference between revisions of "2013 AMC 12B Problems/Problem 21"
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== Solution 2 == | == Solution 2 == | ||
| − | Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if | + | Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if their directrixes are parallel and the focus lies on the same side of both directrixes, and intersect twice otherwise. Thereby, as each parabola will intersect <math>30-3 = 27</math> other parabolas twice, we see that the answer is <cmath>2 \times \frac{30 \times 27}{2} = \boxed{810}.</cmath> |
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2013|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:51, 3 November 2024
Contents
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 
and 
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?
Solution 1
Being on two parabolas 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.
Solution 2
Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if their directrixes are parallel and the focus lies on the same side of both directrixes, and intersect twice otherwise. Thereby, as each parabola will intersect 
other parabolas twice, we see that the answer is ![\[2 \times \frac{30 \times 27}{2} = \boxed{810}.\]](http://latex.artofproblemsolving.com/9/8/3/983734dcc709e43cdca410e44b0e82bd882e91c6.png)
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 | |
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