AMC12 2012 B

AMC12 2012 B · Q13

AMC12 2012 B · Q13. It mainly tests Quadratic equations, Probability (basic).

Two parabolas have equations $y= x^2 + ax +b$ and $y= x^2 + cx +d$, where $a, b, c,$ and $d$ are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have at least one point in common?

两条抛物线的方程为 $y= x^2 + ax +b$ 和 $y= x^2 + cx +d$，其中 $a, b, c,$ 和 $d$ 为整数，且各自独立地通过掷一枚公平的六面骰子确定。两条抛物线至少有一个公共点的概率是多少？

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{25}{36} \frac{25}{36}

(D) \frac{31}{36} \frac{31}{36}

(E) 1 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

Set the two equations equal to each other: $x^2 + ax + b = x^2 + cx + d$. Now remove the x squared and get $x$'s on one side: $ax-cx=d-b$. Now factor $x$: $x(a-c)=d-b$. If $a$ cannot equal $c$, then there is always a solution, but if $a=c$, a $1$ in $6$ chance, leaving a $1080$ out $1296$, always having at least one point in common. And if $a=c$, then the only way for that to work, is if $d=b$, a $1$ in $36$ chance, however, this can occur $6$ ways, so a $1$ in $6$ chance of this happening. So adding one thirty sixth to $\frac{1080}{1296}$, we get the simplified fraction of $\frac{31}{36}$; answer $\boxed{(D)}$.

令两式相等：$x^2 + ax + b = x^2 + cx + d$。消去 $x^2$ 并把含 $x$ 的项放在一边：$ax-cx=d-b$。提取 $x$：$x(a-c)=d-b$。若 $a\ne c$，则总有解；但若 $a=c$（概率为 $\frac{1}{6}$），则前面共有 $1296$ 种情况中有 $1080$ 种必有至少一个公共点。若 $a=c$，要使其成立还需 $d=b$（概率为 $\frac{1}{36}$），但这可发生 $6$ 种方式，因此该事件发生的概率为 $\frac{1}{6}$。将 $\frac{1}{36}$ 加到 $\frac{1080}{1296}$ 上，得到化简后的分数 $\frac{31}{36}$；答案为 $\boxed{(D)}$。

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