AMC12 2013 B

Answer (C): If the directrices of two parabolas with the same focus intersect, then the corresponding parabolas intersect in exactly two points. The same conclusion holds if the directrices are parallel and the focus is between the two lines. Moreover, if the directrices are parallel and the focus is not between the two lines, then the corresponding parabolas do not intersect. Indeed, a point $C$ belongs to the intersection of the parabolas with focus $O$ and directrices $\ell_1$ and $\ell_2$, if and only if, $d(C,\ell_1)=OC=d(C,\ell_2)$. That is, the circle with center $C$ and radius $OC$ is tangent to both $\ell_1$ and $\ell_2$. If $\ell_1$ and $\ell_2$ are parallel and $O$ is not between them, then clearly such circle does not exist. If $\ell_1$ and $\ell_2$ intersect and $O$ is not on them, then there are are exactly two circles tangent to both $\ell_1$ and $\ell_2$ that go through $O$. The same is true if $\ell_1$ and $\ell_2$ are parallel and $O$ is between them. Thus there are $\binom{30}{2}$ pairs of parabolas and the pairs that do not intersect are exactly those whose directrices have the same slope and whose $y$-intercepts have the same sign. There are 5 different slopes and $2\cdot\binom{3}{2}=6$ pairs of $y$-intercepts with the same sign taken from $\{-3,-2,-1,1,2,3\}$. Because the pairs of parabolas that intersect do so at exactly two points and no point is in three parabolas, it follows that the total number of intersection points is \[ 2\left(\binom{30}{2}-5\cdot 6\right)=810. \] Note: It is possible to construct the two circles through $O$ and tangent to the lines $\ell_1$ and $\ell_2$ as follows: Let $\ell'$ be the bisector of the angle determined by the angular sector spanned by $\ell_1$ and $\ell_2$ that contains $O$ (or the midline of $\ell_1$ and $\ell_2$ if these lines are parallel and $O$ is between them). Let $Q$ be the symmetric point of $O$ with respect to $\ell'$ and let $P$ be the intersection of $\ell_1$ and the line $OQ$ (if $O=Q$ then let $P$ be the intersection of $\ell_1$ and a perpendicular line to $\ell'$ by $O$). If $C$ is one of the desired circles, then $C$ passes through $O$ and $Q$ and is tangent to $\ell_1$. Let $T$ be the point of tangency of $C$ and $\ell_1$. By the Power of a Point Theorem, $PT^2=PO\cdot PQ$. The circle with center $P$ and radius $\sqrt{PO\cdot PQ}$ intersects $\ell_1$ in two points $T_1$ and $T_2$. The circumcircles of $OQT_1$ and $OQT_2$ are the desired circles.