2013 AMC 12B Problem 21

Attempt Problem 21 of the 2013 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Consider the set of 3030 parabolas defined as follows: all parabolas have as focus the point (0,0)(0, 0) and the directrix lines have the form y=ax+by = ax + b with aa and bb integers such that a{2,1,0,1,2}a \in \{-2, -1, 0, 1, 2\} and b{3,2,1,1,2,3}.b \in \{-3, -2, -1, 1, 2, 3\}. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?

720720

760760

810810

840840

870870

Answer: C
Concepts:parabolacomplementary countingcombinations
Difficulty rating: 2360
Solution:

Two parabolas with common focus OO meet in exactly 22 points, except when their directrices are parallel and OO lies outside the strip between them, in which case they do not meet. The non-intersecting pairs have directrices of equal slope and yy-intercepts of the same sign. There are 55 slopes, and for each, 2(32)=62\binom{3}{2} = 6 same-sign intercept pairs. Since every intersecting pair meets in 22 points and no point lies on three parabolas, the total is 2((302)56)2\left(\binom{30}{2} - 5\cdot 6\right) =2(43530)= 2(435 - 30) =810.= 810. Thus, the correct answer is C.

← Problem 20#20
Full Exam

Problem 21 in Other Years