# Difference between revisions of "2013 AMC 12B Problems/Problem 21"

## 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 $y=ax+b$ with a and b integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?

$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D}}\ 840\qquad\textbf{(E)}\ 870$ (Error compiling LaTeX. ! Extra }, or forgotten \$.)

## 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 $2\dbinom{30}{2}$ possible intersection points, only $2*5*2*\dbinom{3}{2}$ fail to exist. This leaves $870-60=810=\boxed{\textbf{(C)} \ 810}$ solutions.