Question

考虑如下定义的 $30$ 条抛物线：所有抛物线的焦点都是 $(0, 0)$ ，准线均形如 $y = ax + b$ ，其中 $a$ 和 $b$ 是整数，且 $a \in \{-2, -1, 0, 1, 2\}$ 、 $b \in \{-3, -2, -1, 1, 2, 3\}$ 。这些抛物线中没有三条有公共点。平面上有多少个点恰好位于其中两条抛物线上？

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?

$720$

$760$

$810$

$840$

$870$

答案：C

知识点：抛物线补集计数组合

难度评级：2360

小提示：

两条有共同焦点的抛物线通常交于 $2$ 点；例外是准线平行且焦点不在两准线之间。

Two parabolas sharing a focus meet in $2$ points unless their directrices are parallel with the focus not between them

大提示：

从所有 $\binom{30}{2}$ 对中减去不相交的对：斜率相同且 $y$ -截距同号。

From all $\binom{30}{2}$ pairs, subtract the non-intersecting ones: equal slope and same-sign $y$ -intercepts

解答：

两条共同焦点为 $O$ 的抛物线恰有 $2$ 个交点，除非它们的准线平行且 $O$ 在两准线之间的带状区域外，此时不相交。不相交的对具有相同斜率，且 $y$ -截距同号。共有 $5$ 个斜率；对每个斜率，同号截距对有 $2\binom{3}{2} = 6$ 对。每个相交对贡献 $2$ 个点，且没有三条共点，所以总数为 $2\left(\binom{30}{2} - 5\cdot 6\right)$ $= 2(435 - 30)$ $= 810$ 。所以正确答案是 C。

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

2013 AMC 12B 第 21 题

答案：C

难度评级：2360

小提示：

大提示：

解答：

其他年份的第 21 题