2015 AMC 12A 第 24 题

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24.

从区间 [0,2)[0, 2) 内所有可以写成分数 nd\dfrac{n}{d} 的有理数中随机选择有理数 aabb,其中 nndd 是整数且 1d51 \le d \le 5。表达式 是实数的概率是多少? (cos(aπ)+isin(bπ))4(\cos(a\pi) + i\sin(b\pi))^4

Rational numbers aa and bb are chosen at random among all rational numbers in the interval [0,2)[0, 2) that can be written as fractions nd\dfrac{n}{d} where nn and dd are integers with 1d5.1 \le d \le 5. What is the probability that (cos(aπ)+isin(bπ))4(\cos(a\pi) + i\sin(b\pi))^4 is a real number?

350\dfrac{3}{50}

425\dfrac{4}{25}

41200\dfrac{41}{200}

625\dfrac{6}{25}

1350\dfrac{13}{50}

答案:D
知识点:复数棣莫弗定理基本计数
难度评级:2520
解答:

各有 2020 个可能的 aab.b. 写成最简分数时,分母 1,2,3,4,51,2,3,4,5 分别贡献 2,2,4,4,82,2,4,4,8 个位于 [0,2),[0,2), 中的值,总计 20.20.

x=cos(aπ)x = \cos(a\pi)y=sin(bπ),y = \sin(b\pi),四次方 (x+iy)4(x + iy)^4 为实数,当且仅当 x=0,x = 0, y=0,y = 0,x=±y.x = \pm y.

情形 x=0x = 0 意味着 a{12,32},a \in \left\{\dfrac12, \dfrac32\right\},给出 220=402\cdot 20 = 40 对;情形 y=0y = 0 意味着 b{0,1},b \in \{0, 1\},再给出 4040 对,其中 44 对已经计数。

对于剩余条件 cos(aπ)=±sin(bπ),\cos(a\pi) = \pm\sin(b\pi),且两边都非零,允许的 bb 值为 14,\dfrac14, 12,\dfrac12, 34,\dfrac34, 54,\dfrac54, 32,\dfrac32,74.\dfrac74. 对应的 aa 值个数依次为 4,2,4,4,2,4.4,2,4,4,2,4. 例如,当 b=14,b = \dfrac14, 时,各值为 a=14,a = \dfrac14, 34,\dfrac34, 54,\dfrac54,74;\dfrac74; 其他各行同样由四分之一周的恒等式得到。因此此条件再贡献 4+2+4+4+2+4=204+2+4+4+2+4=20 对。

总共有 40+404+20=9640 + 40 - 4 + 20 = 96 对有效组合,全部组合为 400,400, 对,所以概率为 96400=625.\dfrac{96}{400} = \dfrac{6}{25}.

因此,正确答案是 D

There are 2020 possible values for each of aa and b.b. In reduced form, denominators 1,2,3,4,51,2,3,4,5 contribute respectively 2,2,4,4,82,2,4,4,8 values in [0,2),[0,2), for a total of 20.20.

Writing x=cos(aπ)x = \cos(a\pi) and y=sin(bπ),y = \sin(b\pi), the fourth power (x+iy)4(x + iy)^4 is real if and only if x=0,x = 0, y=0,y = 0, or x=±y.x = \pm y.

The case x=0x = 0 means a{12,32},a \in \left\{\dfrac12, \dfrac32\right\}, giving 220=402\cdot 20 = 40 pairs; the case y=0y = 0 means b{0,1},b \in \{0, 1\}, giving another 4040 pairs, of which 44 were already counted.

For the remaining condition cos(aπ)=±sin(bπ),\cos(a\pi) = \pm\sin(b\pi), with neither side zero, the allowed values of bb are 14,\dfrac14, 12,\dfrac12, 34,\dfrac34, 54,\dfrac54, 32,\dfrac32, and 74.\dfrac74. The corresponding numbers of allowed values of aa are respectively 4,2,4,4,2,4.4,2,4,4,2,4. For example, when b=14,b = \dfrac14, the values are a=14,a = \dfrac14, 34,\dfrac34, 54,\dfrac54, and 74;\dfrac74; the other rows follow from the same quarter-turn identities. Hence this condition contributes 4+2+4+4+2+4=204+2+4+4+2+4=20 more pairs.

In all there are 40+404+20=9640 + 40 - 4 + 20 = 96 valid pairs out of 400,400, so the probability is 96400=625.\dfrac{96}{400} = \dfrac{6}{25}.

Thus, the correct answer is D.

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