Question

$A,B,C$ 和 $D$ 的值要从 $\{1, 2, 3, 4, 5, 6\}$ 中不重复地选出。共有多少种选择，使得两条曲线 $y=Ax^2+B$ 和 $y=Cx^2+D$ 相交？

两条曲线列出的顺序不重要；例如选择 $A=3$ $B=2$ $C=4$ $D=1$ 与选择 $A=4$ $B=1$ $C=3$ $D=2$ 视为相同。

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e., no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect?

(The order in which the curves are listed does not matter; for example, the choices $A=3,$ $B=2,$ $C=4,$ $D=1$ is considered the same as the choices $A=4,$ $B=1,$ $C=3,$ $D=2.$ )

$30$

$60$

$90$

$180$

$360$

答案：C

知识点：抛物线组合

难度评级：1540

视频讲解：

Click to load, then click again to play

文字解答：

令两式相等，得到，因为平方数非负。 $Ax^2 + B = Cx^2 + D$ $x^2(A - C) = D - B$ $x^2 = \dfrac{D - B}{A - C} \geq 0$

因此 $D - B$ 与 $A - C$ 必须同号。

若先为 $(A, C)$ 和 $(B, D)$ 各选两个不同值，则有 $2$ 种安排使两个差同号。

不过还要除以 $2$ ，因为两条曲线列出的顺序不区分。

所以所求选择数为 $\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.$

所以正确答案是 C。

Setting the equations equal to each other, we get $Ax^2 + B = Cx^2 + D$ $x^2(A - C) = D - B$ $x^2 = \dfrac{D - B}{A - C} \geq 0$ since squares are non-negative.

This means $D - B$ and $A - C$ must both have the same sign.

If we choose two distinct values for $(A, C)$ and $(B, D),$ there are $2$ ways to arrange them such that the numerator and denominator both have the same sign.

We have to divide by $2,$ however, since the two curves are not considered distinct.

Therefore, the total number of tuples is $\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.$

Thus, C is the correct answer.

2021 AMC 10A Spring 第 15 题

答案：C

难度评级：1540

视频讲解：

文字解答：

其他年份的第 15 题